dynamics-13th-hibbeler-edition-solutions.pdf

S O L U T I O N M A N U A L CONTENTS

Chapter 12 General Principles

1

Chapter 13 Force Vectors

245

Chapter 14 Equilibrium of a Particle

378

Chapter 15 Force System Resultants

475

Review 1 Kinematics and Kinetics of a Particle

630

Chapter 16 Equilibrium of a Rigid Body

680

Chapter 17 Structural Analysis

833

Chapter 18 Internal Forces

953

Chapter 19 Friction

1023

Review 2 Planar Kinematics and Kinetics of a Rigid Body

1080

Chapter 20 Center of Gravity and Centroid

1131

Chapter 21 Moments of Inertia

1190

Chapter 22 Virtual Work

1270

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

12–1. A baseball is thrown downward from a 50-ft tower with an initial speed of 18 ft>s. Determine the speed at which it hits the ground and the time of travel.

SOLUTION v22 = v21 + 2ac(s2 - s1) v22 = (18)2 + 2(32.2)(50 - 0) v2 = 59.532 = 59.5 ft>s

Ans.

v2 = v1 + ac t 59.532 = 18 + 32.2(t) t = 1.29 s

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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





12–2. When a train is traveling along a straight track at 2 m/s, it begins to accelerate at a = 160 v-42 m>s2, where v is in m/s. Determine its velocity v and the position 3 s after the acceleration.

s

SOLUTION a =

dv dt

dt =

dv a v

3

dt =

L0

3 =

dv -4 L2 60v 1 (v5 - 32) 300

v = 3.925 m>s = 3.93 m>s

Ans.

ads = vdv ds = s

s =

1 60 L2

3.925

v5 dv

1 v6 3.925 a b` 60 6 2

= 9.98 m

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L0

ds =

1 5 vdv = v dv a 60






v

12–3. From approximately what floor of a building must a car be dropped from an at-rest position so that it reaches a speed of 80.7 ft>s 155 mi>h2 when it hits the ground? Each floor is 12 ft higher than the one below it. (Note: You may want to remember this when traveling 55 mi>h.)

SOLUTION ( + T)

v 2 = v20 + 2ac(s - s0) 80.72 = 0 + 2(32.2)(s - 0) s = 101.13 ft

# of floors =

101.13 = 8.43 12 Ans.

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The car must be dropped from the 9th floor.






*12–4. Traveling with an initial speed of 70 km>h, a car accelerates at 6000 km>h2 along a straight road. How long will it take to reach a speed of 120 km>h? Also, through what distance does the car travel during this time?

SOLUTION v = v1 + ac t 120 = 70 + 6000(t) t = 8.33(10 - 3) hr = 30 s

Ans.

v2 = v21 + 2 ac(s - s1) (120)2 = 702 + 2(6000)(s - 0) s = 0.792 km = 792 m

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12–5. A bus starts from rest with a constant acceleration of 1 m>s2. Determine the time required for it to attain a speed of 25 m>s and the distance traveled.

SOLUTION Kinematics: v0 = 0, v = 25 m>s, s0 = 0, and ac = 1 m>s2. + B A:

v = v0 + act 25 = 0 + (1)t t = 25 s

or

v2 = v02 + 2ac(s - s0) teaching

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laws

+ B A:

Ans.

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in

252 = 0 + 2(1)(s - 0)

Ans.

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s = 312.5 m






12–6. A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later.

SOLUTION + T s = s1 + v1 t + sA = 0 + 0 +

1 2 a t 2 c

1 (32.2)(2)2 2

sA = 64.4 ft sA = 0 + 0 +

1 (32.2)(1)2 2

sB = 16.1 ft ¢s = 64.4 - 16.1 = 48.3 ft

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12–7. A bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 km/h. Determine his acceleration if it is constant. Also, how long does it take to reach the speed of 30 km/h?

SOLUTION v2 = 30 km>h = 8.33 m>s v22 = v21 + 2 ac (s2 - s1) (8.33)2 = 0 + 2 ac (20 - 0) ac = 1.74 m>s2

Ans.

v2 = v1 + ac t 8.33 = 0 + 1.74(t) t = 4.80 s

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*■12–8. A particle moves along a straight line with an acceleration of a = 5>(3s1>3 + s5>2) m>s2, where s is in meters. Determine the particle’s velocity when s = 2 m, if it starts from rest when s = 1 m. Use Simpson’s rule to evaluate the integral.

SOLUTION 5

a =

A 3s3 + s2 B 1

5

a ds = v dv 2

v

5 ds

L1 A 3s + s 1 3

0.8351 =

5 2

B

=

L0

v dv

1 2 v 2

v = 1.29 m>s

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12–9. If it takes 3 s for a ball to strike the ground when it is released from rest, determine the height in meters of the building from which it was released. Also, what is the velocity of the ball when it strikes the ground?

SOLUTION Kinematics: v0 = 0, ac = g = 9.81 m>s2, t = 3 s, and s = h.

A+TB

v = v0 + act = 0 + (9.81)(3) = 29.4 m>s

or laws

A+TB

Ans.

1 s = s0 + v0t + act2 2 1 h = 0 + 0 + (9.81)(32) 2 Ans.

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= 44.1 m






12–10. The position of a particle along a straight line is given by s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in seconds. Determine the position of the particle when t = 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled.

SOLUTION Position: The position of the particle when t = 6 s is s|t = 6s = 1.5(63) - 13.5(62) + 22.5(6) = -27.0 ft

Ans.

Total DistanceTraveled: The velocity of the particle can be determined by applying Eq. 12–1. v =

ds = 4.50t2 - 27.0t + 22.5 dt

The times when the particle stops are

laws

or

4.50t2 - 27.0t + 22.5 = 0 t = 5s

and

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t = 1s

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The position of the particle at t = 0 s, 1 s and 5 s are

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s t = 0 s = 1.5(03) - 13.5(02) + 22.5(0) = 0

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s t = 1 s = 1.5(13) - 13.5(12) + 22.5(1) = 10.5 ft

the

(including

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s t = 5 s = 1.5(53) - 13.5(52) + 22.5(5) = - 37.5 ft

this

Ans.

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stot = 10.5 + 48.0 + 10.5 = 69.0 ft

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From the particle’s path, the total distance is






12–11. If a particle has an initial velocity of v0 = 12 ft>s to the right, at s0 = 0, determine its position when t = 10 s, if a = 2 ft>s2 to the left.

SOLUTION + B A:

s = s0 + v0t +

1 2 at 2 c

= 0 + 12(10) +

1 (- 2)(10)2 2 Ans.

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= 20 ft






*12–12. Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m>s2 and decelerate at 2 m>s2.

SOLUTION Using formulas of constant acceleration: v2 = 1.5 t1 1 (1.5)(t21) 2

x =

0 = v2 - 2 t2 1 (2)(t22) 2

1000 - x = v2t2 -

or

Combining equations:

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t1 = 1.33 t2; v2 = 2 t2

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x = 1.33 t22

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1000 - 1.33 t22 = 2 t22 - t22 the

t1 = 27.603 s use

United

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learning.

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of

t2 = 20.702 s;

and

t = t1 + t2 = 48.3 s

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Ans.






12–13. Tests reveal that a normal driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ft>s) and their cars can decelerate at 2 ft>s2, determine the shortest stopping distance d for each from the moment they see the pedestrians. Moral: If you must drink, please don’t drive!

v1

44 ft/s

d

SOLUTION Stopping Distance: For normal driver, the car moves a distance of d¿ = vt = 44(0.75) = 33.0 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 33.0 ft and v = 0. + B A:

v2 = v20 + 2ac (s - s0) 02 = 442 + 2( -2)(d - 33.0) d = 517 ft

Ans.

laws

or

For a drunk driver, the car moves a distance of d¿ = vt = 44(3) = 132 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 132 ft and v = 0. teaching

Web)

v2 = v20 + 2ac (s - s0) in

+ B A:

Dissemination

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02 = 442 + 2( -2)(d - 132)

permitted.

d = 616 ft

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Ans.






12–14. A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can accelerate at 0.6 ft>s2, decelerate at 0.3 ft>s2, and reach a maximum speed of 8 ft>s, determine the shortest time to make the lift, starting from rest and ending at rest.

SOLUTION +c

v2 = v20 + 2 ac (s - s0)

v2max = 0 + 2(0.6)(y - 0) 0 = v2max + 2( -0.3)(48 - y) 0 = 1.2 y - 0.6(48 - y) y = 16.0 ft, +c

vmax = 4.382 ft>s 6 8 ft>s

v = v0 + ac t

4.382 = 0 + 0.6 t1

Web)

laws

or

t1 = 7.303 s

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0 = 4.382 - 0.3 t2

Ans.

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t = t1 + t2 = 21.9 s






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Dissemination

t2 = 14.61 s

12–15. A train starts from rest at station A and accelerates at 0.5 m>s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m>s2 until it is brought to rest at station B. Determine the distance between the stations.

SOLUTION Kinematics: For stage (1) motion, v0 = 0, s0 = 0, t = 60 s, and ac = 0.5 m>s2. Thus, + B A:

s = s0 + v0t +

s1 = 0 + 0 + + B A:

1 2 at 2 c

1 (0.5)(602) = 900 m 2

v = v0 + act

laws

or

v1 = 0 + 0.5(60) = 30 m>s

in

teaching

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For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus, 1 2 at 2 c

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s = s0 + v0t +

copyright

+ B A:

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s2 = 900 + 30(900) + 0 = 27 900 m

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the

v = v0 + act

assessing

is

+ B A:

for

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the

by

and

For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and ac = - 1 m>s2. Thus,

provided

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this

0 = 30 + ( - 1)t

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of

s = s0 + v0t +

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t = 30 s

s3 = 27 900 + 30(30) + = 28 350 m = 28.4 km






1 ( - 1)(302) 2 Ans.

*12–16. A particle travels along a straight line such that in 2 s it moves from an initial position sA = +0.5 m to a position sB = - 1.5 m. Then in another 4 s it moves from sB to sC = +2.5 m. Determine the particle’s average velocity and average speed during the 6-s time interval.

SOLUTION ¢s = (sC - sA) = 2 m sT = (0.5 + 1.5 + 1.5 + 2.5) = 6 m t = (2 + 4) = 6 s vavg =

2 ¢s = = 0.333 m>s t 6 sT 6 = = 1 m>s t 6

Ans.

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(vsp)avg =

Ans.






12–17. The acceleration of a particle as it moves along a straight line is given by a = 12t - 12 m>s2, where t is in seconds. If s = 1 m and v = 2 m>s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period.

SOLUTION v

L2

t

dv =

L0

(2 t - 1) dt

v = t2 - t + 2 s

L1

t

ds = s =

L0

(t2 - t + 2) dt

1 3 1 t - t2 + 2 t + 1 3 2

When t = 6 s, Ans.

s = 67 m

Ans. teaching

Web)

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or

v = 32 m>s

Wide

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Since v Z 0 then

Ans.

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Dissemination

d = 67 - 1 = 66 m






12–18. A freight train travels at v = 6011 - e -t2 ft>s, where t is the elapsed time in seconds. Determine the distance traveled in three seconds, and the acceleration at this time.

v

s

SOLUTION v = 60(1 - e - t) s

L0

3

ds =

L

v dt =

L0

6011 - e - t2dt

s = 60(t + e - t)|30 s = 123 ft

Ans.

dv = 60(e - t) dt

a =

or

At t = 3 s laws

a = 60e - 3 = 2.99 ft>s2

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Ans.






12–19. A particle travels to the right along a straight line with a velocity v = [5>14 + s2] m>s, where s is in meters. Determine its position when t = 6 s if s = 5 m when t = 0.

SOLUTION 5 ds = dt 4 + s s

L5

(4 + s) ds =

t

L0

5 dt

4 s + 0.5 s2 - 32.5 = 5 t When t = 6 s, s2 + 8 s - 125 = 0

laws

or

Solving for the positive root s = 7.87 m

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Ans.






*12–20. The velocity of a particle traveling along a straight line is v = (3t2 - 6t) ft>s, where t is in seconds. If s = 4 ft when t = 0, determine the position of the particle when t = 4 s. What is the total distance traveled during the time interval t = 0 to t = 4 s? Also, what is the acceleration when t = 2 s?

SOLUTION Position: The position of the particle can be determined by integrating the kinematic equation ds = v dt using the initial condition s = 4 ft when t = 0 s. Thus, + B A:

ds = v dt s

L4 ft s2

t

ds =

s

L0

2 A 3t - 6t B dt

= (t 3 - 3t2) 2

4 ft

t

0

s = A t - 3t + 4 B ft

or

2

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3

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When t = 4 s, s|4 s = 43 - 3(42) + 4 = 20 ft

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The velocity of the particle changes direction at the instant when it is momentarily brought to rest. Thus,

(including

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work

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the

v = 3t2 - 6t = 0

assessing

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the

t(3t - 6) = 0

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t = 0 and t = 2 s the part any

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s|2 s = 23 - 3 A 22 B + 4 = 0

of

and

s|0 s = 0 - 3 A 02 B + 4 = 4 ft

is

This

The position of the particle at t = 0 and 2 s is

Using the above result, the path of the particle shown in Fig. a is plotted. From this figure, sTot = 4 + 20 = 24 ft

Ans.

Acceleration: + B A:

a =

dv d = (3t2 - 6t) dt dt

a = 16t - 62 ft>s2 When t = 2 s, a ƒ t = 2 s = 6122 - 6 = 6 ft>s2 :






Ans.

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Dissemination

Ans.

12–21. If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation a = 9.81[1 - v2(10-4)] m>s2, where v is in m>s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s, and (b) the body’s terminal or maximum attainable velocity (as t : q ).

SOLUTION Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2. (+ T)

dt = t

L0

dv a

v

dv 2 L0 9.81[1 - (0.01v) ]

dt = v

t =

v

1 dv dv c + d 9.81 L0 2(1 + 0.01v) 2(1 0.01v) L0 1 + 0.01v b 1 - 0.01v laws

or

9.81t = 50ln a

100(e0.1962t - 1)

teaching

Web)

(1) in

e0.1962t + 1

Wide

copyright

v =

instructors

States

World

permitted.

Dissemination

a) When t = 5 s, then, from Eq. (1) of

the

not

100[e0.1962(5) - 1]

Ans. and

use

United

on

learning.

is

= 45.5 m>s

e0.1962(5) + 1

work the

(including

: 1. Then, from Eq. (1)

of is

work

solely

e0.1962t + 1

for

e0.1962t - 1

protected

b) If t : q ,

student

the

by

v =

Ans.

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this

assessing

vmax = 100 m>s






12–22. The position of a particle on a straight line is given by s = 1t3 - 9t2 + 15t2 ft, where t is in seconds. Determine the position of the particle when t = 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled.

SOLUTION s = t3 - 9t2 + 15t ds = 3t2 - 18t + 15 dt

v =

v = 0 when t = 1 s and t = 5 s t = 0, s = 0 t = 1 s, s = 7 ft

Ans.

sT = 7 + 7 + 25 + (25 - 18) = 46 ft

Ans.

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the

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for

work

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and

use

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on

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of

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not

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World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

t = 6 s, s = - 18 ft

or

t = 5 s, s = - 25 ft






12–23. Two particles A and B start from rest at the origin s = 0 and move along a straight line such that a A = (6t - 3) ft>s2 and aB = (12t2 - 8) ft>s2, where t is in seconds. Determine the distance between them when t = 4 s and the total distance each has traveled in t = 4 s.

SOLUTION Velocity: The velocity of particles A and B can be determined using Eq. 12-2. dvA = aAdt vA

t

dvA =

L0

(6t - 3)dt

L0

vA = 3t2 - 3t dvB = aBdt vB

t

dvB =

L0

(12t2 - 8)dt

L0

vB = 4t3 - 8t The times when particle A stops are or

t = 0 s and = 1 s

laws

3t2 - 3t = 0

in

teaching

Web)

The times when particle B stops are

World

not

instructors

States

Position:The position of particles A and B can be determined using Eq. 12-1.

permitted.

Dissemination

Wide

copyright

t = 0 s and t = 22 s

4t3 - 8t = 0

United

on

learning.

is

of

the

dsA = vAdt

and

use the

by

(3t2 - 3t)dt

work (including

L0

student

L0

t

dsA =

for

sA

work

solely

of

protected

the

3 2 t 2

assessing

is

sA = t3 -

integrity

of

and

work

this

dsB = vBdt

provided

part

the

is

This

(4t3 - 8t)dt

any

L0

courses

L0

t

dsB =

and

sB

will

their

destroy

of

sB = t4 - 4t2 sale

The positions of particle A at t = 1 s and 4 s are sA |t = 1 s = 13 -

3 2 (1 ) = - 0.500 ft 2

sA |t = 4 s = 43 -

3 2 (4 ) = 40.0 ft 2

Particle A has traveled dA = 2(0.5) + 40.0 = 41.0 ft

Ans.

The positions of particle B at t = 22 s and 4 s are sB |t = 12 = (22)4 - 4(22)2 = -4 ft sB |t = 4 = (4)4 - 4(4)2 = 192 ft Particle B has traveled dB = 2(4) + 192 = 200 ft

Ans.

At t = 4 s the distance beween A and B is ¢sAB = 192 - 40 = 152 ft © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

*12–24. A particle is moving along a straight line such that its velocity is defined as v = ( - 4s2) m>s, where s is in meters. If s = 2 m when t = 0, determine the velocity and acceleration as functions of time.

SOLUTION v = - 4s2 ds = - 4s2 dt s

L2

s - 2 ds =

t

L0

-4 dt

Web)

2 8t + 1

laws

s =

teaching

1 -1 (s - 0.5) 4

in

t =

or

-s - 1| s2 = - 4t|t0

2 2 16 b = m>s 8t + 1 (8t + 1)2

Wide

copyright

permitted.

Dissemination

Ans. States

World

v = -4a

of

the

not

instructors

16(2)(8t + 1)(8) dv 256 = = m>s2 dt (8t + 1)4 (8t + 1)3

is

on

learning. United

and work the

(including

for

work

solely

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protected

this

assessing

is

integrity

of

and

work

provided

any

courses

part

the

is

This

destroy

of

and

will

their

sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

use student

the

by

a =

12–25. A sphere is fired downwards into a medium with an initial speed of 27 m>s. If it experiences a deceleration of a = ( -6t) m>s2, where t is in seconds, determine the distance traveled before it stops.

SOLUTION Velocity: v0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have

A+TB

dv = adt v

L27

t

dv =

L0

-6tdt

v = A 27 - 3t2 B m>s

(1)

At v = 0, from Eq. (1) 0 = 27 - 3t2

t = 3.00 s

teaching

Web)

laws

or

Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result v = 27 - 3t2 and applying Eq. 12–1, we have

A+TB

Dissemination

Wide

copyright States

World

permitted.

A 27 - 3t2 B dt

the

not

instructors

is

L0

learning.

L0

t

ds =

of

s

in

ds = vdt

(2)

student

the

by

and

use

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on

s = A 27t - t3 B m

work

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of

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this

assessing

is

s = 27(3.00) - 3.003 = 54.0 m

of

protected

the

(including

for

work

At t = 3.00 s, from Eq. (2)






Ans.

12–26. When two cars A and B are next to one another, they are traveling in the same direction with speeds vA and vB , respectively. If B maintains its constant speed, while A begins to decelerate at aA , determine the distance d between the cars at the instant A stops.

A

B

d

SOLUTION Motion of car A: v = v0 + act 0 = vA - aAt

t =

vA aA

v2 = v20 + 2ac(s - s0) 0 = v2A + 2( - aA)(sA - 0) v2A 2aA laws

or

sA =

teaching

Web)

Motion of car B: in

vA vAvB b = aA aA instructors

States

World

permitted.

Dissemination

Wide

copyright

sB = vBt = vB a

on

learning.

is

of

the

not

The distance between cars A and B is

and

use

United

v2A vAvB 2vAvB - v2A ` = ` ` aA 2aA 2aA

work

student

the

by

Ans.

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is

This

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this

assessing

is

work

solely

of

protected

the

(including

for

sBA = |sB - sA| = `






12–27. A particle is moving along a straight line such that when it is at the origin it has a velocity of 4 m>s. If it begins to decelerate at the rate of a = 1- 1.5v1>22 m>s2, where v is in m>s, determine the distance it travels before it stops.

SOLUTION 1 dv = - 1.5v2 dt

a = v

L4 1

t

1

v- 2 dv =

v

L0

- 1.5 dt

t

2v2 4 = - 1.5t 0 1

2av2 - 2 b = - 1.5t v = (2 - 0.75t)2 m>s t

L0

t

(2 - 0.75t)2 dt =

L0

(4 - 3t + 0.5625t2) dt

or

L0

ds =

laws

s

(1)

teaching

Web)

s = 4t - 1.5t2 + 0.1875t3

Wide

copyright

in

(2)

States

World

permitted.

Dissemination

From Eq. (1), the particle will stop when

United

on

learning.

is

of

the

not

instructors

0 = (2 - 0.75t)2

student

the

by

and

use

t = 2.667 s

work

s|t = 2.667 = 4(2.667) - 1.5(2.667)2 + 0.1875(2.667)3 = 3.56 m

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This

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the

(including

for

Ans.






*12–28. A particle travels to the right along a straight line with a velocity v = [5>14 + s2] m>s, where s is in meters. Determine its deceleration when s = 2 m.

SOLUTION 5 4+s

v =

v dv = a ds dv =

-5 ds (4 + s)2

5 - 5 ds b = a ds a (4 + s) (4 + s)2

laws

or

- 25 (4 + s)3

a =

in

teaching

Web)

When s = 2 m

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This

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is

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solely

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protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

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not

instructors

States

World

permitted.

Dissemination

Wide

Ans.

copyright

a = -0.116 m>s2






12–29. A particle moves along a straight line with an acceleration 1>2 a = 2v m>s2, where v is in m>s. If s = 0, v = 4 m>s when t = 0, determine the time for the particle to achieve a velocity of 20 m>s. Also, find the displacement of particle when t = 2 s.

SOLUTION Velocity: + B A:

dt =

dv a

t

L0

v

dt =

dv

L0 2v1>2

t

v

t ƒ 0 = v1>2 ƒ 4

t = v1>2 - 2

teaching

Web)

laws

or

v = (t + 222

copyright

in

When v = 20 m>s,

instructors

States

World

permitted.

Dissemination

Wide

20 = 1t + 222

Ans. and

use

United

on

learning.

is

of

the

not

t = 2.47 s

work

student

the

by

Position:

(including

for

+ B A:

assessing

is

work

solely

of

protected

the

ds = v dt

integrity

of

and

work

(t + 2)2 dt

provided

the part

L0

is

L0

this

t

ds =

This

s

any

courses

destroy

of

and

sale

will

0

t

1 (t + 2)3 2 3 0

their

s

s2 =

s =

1 [(t + 2)3 - 23] 3

=

1 2 t(t + 6t + 12) 3

When t = 2 s, s =

1 (2)[(2)2 + 6(2) + 12] 3

= 18.7 m






Ans.

12–30. As a train accelerates uniformly it passes successive kilometer marks while traveling at velocities of 2 m>s and then 10 m>s. Determine the train’s velocity when it passes the next kilometer mark and the time it takes to travel the 2-km distance.

SOLUTION Kinematics: For the first kilometer of the journey, v0 = 2 m>s, v = 10 m>s, s0 = 0, and s = 1000 m. Thus, + B A:

v2 = v0 2 + 2ac (s - s0) 102 = 22 + 2ac (1000 - 0) ac = 0.048 m>s2

For the second kilometer, ac = 0.048 m>s2. Thus,

s0 = 1000 m,

s = 2000 m,

and

v2 = v0 2 + 2ac (s - s0) or

+ B A:

v0 = 10 m>s,

teaching

Web)

laws

v2 = 102 + 2(0.048)(2000 - 1000) v = 14 m>s

Wide

copyright

in

Ans.

instructors

States

World

permitted.

Dissemination

For the whole journey, v0 = 2 m>s, v = 14 m>s, and ac = 0.048 m>s.2 Thus, the

not

v = v0 + act use

United

on

learning.

is

of

+ B A:

for

work

student

the

by

and

14 = 2 + 0.048t

sale

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is

This

provided

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this

assessing

is

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protected

the

(including

t = 250 s






Ans.

12–31. The acceleration of a particle along a straight line is defined by a = 12t - 92 m>s2, where t is in seconds. At t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity.

SOLUTION a = 2t - 9 v

L10

t

dv =

L0

12t - 92 dt

v - 10 = t2 - 9 t v = t2 - 9 t + 10 s

or

13 t - 4.5 t2 + 10 t 3 in

teaching

Web)

13 t - 4.5 t2 + 10 t + 1 3

Wide Dissemination

s =

1t2 - 9t + 102 dt

laws

s-1 =

L0

copyright

L1

t

ds =

of

the

not

instructors

States

World

permitted.

Note when v = t2 - 9 t + 10 = 0:

work

s = - 36.63 m

assessing

is

work

solely

of

protected

the

(including

When t = 7.701 s,

student

s = 7.13 m

for

When t = 1.298 s,

the

by

and

use

United

on

learning.

is

t = 1.298 s and t = 7.701 s

integrity

of

and

work

this

s = - 30.50 m

provided

When t = 9 s,

s = - 30.5 m

(b)

sTo t = (7.13 - 1) + 7.13 + 36.63 + (36.63 - 30.50)

Ans.

sTo t = 56.0 m (c)






v = 10 m>s

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courses

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is

This

(a)

Ans. Ans.

*12–32. The acceleration of a particle traveling along a straight line 1 1>2 is a = s m>s2, where s is in meters. If v = 0, s = 1 m 4 when t = 0, determine the particle’s velocity at s = 2 m.

SOLUTION Velocity: + B A:

v dv = a ds v

L0

s

v dv =

1 1>2 s ds L1 4

v

s v2 2 1 = s3>2 ` 2 0 6 1

23

1s3>2 - 121>2 m>s

or

1

laws

v =

When s = 2 m, v = 0.781 m>s.

sale

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of

and

any

courses

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is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

Ans.






12–33. At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

SOLUTION + c sA = (sA)0 + (vA)0 t + sA = 0 + 450 t +

1 2 a t 2 c

1 (- 9.81) t2 2

+ c sB = (sB)0 + (vB)0 t + sB = 0 + 600(t - 3) +

1 a t2 2 c 1 ( -9.81)(t - 3)2 2

laws

or

Require sA = sB

Ans. Ans.

sale

will

their

destroy

of

and

any

courses

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the

is

This

provided

integrity

of

and

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this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

h = sA = sB = 4.11 km






permitted.

Dissemination

copyright

t = 10.3 s

Wide

in

teaching

Web)

450 t - 4.905 t2 = 600 t - 1800 - 4.905 t2 + 29.43 t - 44.145

12–34. A boy throws a ball straight up from the top of a 12-m high tower. If the ball falls past him 0.75 s later, determine the velocity at which it was thrown, the velocity of the ball when it strikes the ground, and the time of flight.

SOLUTION Kinematics: When the ball passes the boy, the displacement of the ball in equal to zero. Thus, s = 0. Also, s0 = 0, v0 = v1, t = 0.75 s, and ac = - 9.81 m>s2.

A+cB

s = s0 + v0t +

1 2 at 2 c

0 = 0 + v110.752 +

1 1-9.81210.7522 2

v1 = 3.679 m>s = 3.68 m>s

or

Ans.

in

teaching

Web)

laws

When the ball strikes the ground, its displacement from the roof top is s = - 12 m. Also, v0 = v1 = 3.679 m>s, t = t2, v = v2, and ac = - 9.81 m>s2. 1 2 at 2 c

Wide

permitted.

Dissemination States

World

s = s0 + v0t +

copyright

A+cB

United

on

learning.

is

of

the

not

instructors

1 1- 9.812t22 2

and

use

- 12 = 0 + 3.679t2 +

the of

protected

3.679 ; 21 - 3.67922 - 414.90521 -122 work

solely

214.9052

and

any

courses

part

the

is

This

Choosing the positive root, we have

provided

integrity

of

and

work

this

assessing

is

t2 =

(including

for

work

student

the

by

4.905t22 - 3.679t2 - 12 = 0

Ans.

sale

will

their

destroy

of

t2 = 1.983 s = 1.98 s Using this result,

A+cB

v = v0 + act

v2 = 3.679 + 1 -9.81211.9832 = - 15.8 m>s = 15.8 m>s T






Ans.

12–35. When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf. If this variation of the acceleration can be expressed as a = 1g>v2f21v2f - v22, determine the time needed for the velocity to become v = vf>2 . Initially the particle falls from rest.

SOLUTION g dv = a = ¢ 2 ≤ A v2f - v2 B dt vf v

dy

L0 v2f - v2¿

=

t

g v2f L0

dt

vf + v y g 1 ln ¢ ≤` = 2t 2vf vf - v 0 vf

vf 2g

ln ¢

vf + v vf - v

≤

vf + vf> 2 vf - vf> 2

≤

or

2g

ln ¢

b

Ans.

sale

will

their

destroy

of

and

any

courses

part

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is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

g

Wide

vf

copyright

t = 0.549 a

in

teaching

Web)

t =

vf

laws

t =






*12–36. A particle is moving with a velocity of v0 when s = 0 and t = 0. If it is subjected to a deceleration of a = - kv3, where k is a constant, determine its velocity and position as functions of time.

SOLUTION a =

dv = - ky3 dt

v

Lv0

t

v-3 dv =

-k dt

L0

1 - 1v-2 - v0-22 = - kt 2 1

-2 1 v = ¢ 2kt + ¢ 2 ≤ ≤ v0

Ans.

or

ds = v dt laws

teaching

Web)

1

2 1 ≤≤ v20

Wide

¢ 2kt + ¢

in

L0

dt

permitted.

Dissemination

L0

t

ds =

copyright

s

the

not

instructors

States

is

of

and

use

United

on

learning.

4 by

2k

t

student

the

s =

World

1

2 1 2 ¢ 2kt + ¢ 2 ≤ ≤ v0

protected

1

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

s =

is

work

solely

2 1 1 1 C ¢ 2kt + ¢ 2 ≤ ≤ S v0 k v0

of

the

(including

for

work

0






Ans.

12–37. As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula a = -g0[R2>(R + y)2], where g0 is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If g0 = 9.81 m>s2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth. Hint: This requires that v = 0 as y : q.

SOLUTION v dv = a dy q

0

Ly

v dv = - g0R

2

dy

L0 (R + y)

2

g0 R2 q v2 2 0 2 = 2 y R + y 0

teaching

Web)

laws

or

v = 22g0 R

Wide

copyright

in

= 22(9.81)(6356)(10)3

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

= 11167 m>s = 11.2 km>s






12–38. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–37), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude y0 from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y0 = 500 km? Use the numerical data in Prob. 12–37.

SOLUTION From Prob. 12–37, (+ c )

a = -g0

R2 (R + y)2

Since a dy = v dv then v

=

L0

v dv

y 1 v2 d = R + y y0 2

g0 R2[

1 1 v2 ] = R + y R + y0 2

Web)

laws

or

g0 R2 c

teaching Dissemination

Wide

Ly0 (R + y)

2

in

dy

copyright

y

-g0 R2

States

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permitted.

Thus

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on

learning.

is

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the

not

instructors

2g0 (y0 - y) A (R + y)(R + y0)

and

use

v = -R

by

y = 0,

(including

for

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the

When y0 = 500 km,

the

2(9.81)(500)(103)

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protected

A 6356(6356 + 500)(106)

this

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v = - 6356(103)

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v = - 3016 m>s = 3.02 km>s T






Ans.

12–39. A freight train starts from rest and travels with a constant acceleration of 0.5 ft>s2. After a time t¿ it maintains a constant speed so that when t = 160 s it has traveled 2000 ft. Determine the time t¿ and draw the v–t graph for the motion.

SOLUTION Total Distance Traveled: The distance for part one of the motion can be related to time t = t¿ by applying Eq. 12–5 with s0 = 0 and v0 = 0. + B A:

1 ac t2 2

s = s0 + v0 t + s1 = 0 + 0 +

1 (0.5)(t¿)2 = 0.25(t¿)2 2

The velocity at time t can be obtained by applying Eq. 12–4 with v0 = 0. + B A:

v = v0 + act = 0 + 0.5t = 0.5t

(1)

laws

or

The time for the second stage of motion is t2 = 160 - t¿ and the train is traveling at a constant velocity of v = 0.5t¿ (Eq. (1)).Thus, the distance for this part of motion is teaching

Web)

s2 = vt2 = 0.5t¿(160 - t¿) = 80t¿ - 0.5(t¿)2 in

+ B A:

permitted.

Dissemination

Wide

copyright

If the total distance traveled is sTot = 2000, then

is

of

the

not

instructors

States

World

sTot = s1 + s2

by

and

use

United

on

learning.

2000 = 0.25(t¿)2 + 80t¿ - 0.5(t¿)2

the

(including

for

work

student

the

0.25(t¿)2 - 80t¿ + 2000 = 0

assessing

is

work

solely

of

protected

Choose a root that is less than 160 s, then

Ans.

provided

integrity

of

and

work

this

t¿ = 27.34 s = 27.3 s

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destroy

of

and

any

courses

part

the

is

This

v–t Graph: The equation for the velocity is given by Eq. (1).When t = t¿ = 27.34 s, v = 0.5(27.34) = 13.7 ft>s.






*12–40. A sports car travels along a straight road with an acceleration-deceleration described by the graph. If the car starts from rest, determine the distance s¿ the car travels until it stops. Construct the v- s graph for 0 … s … s¿ .

a(ft/s2)

6 1000

SOLUTION v

4

s Graph: For 0 … s 6 1000 ft, the initial condition is v = 0 at s = 0.

+ B A:

vdv = ads v

s

vdv =

L0

L0

6ds

v2 = 6s 2

A 212s1>2 B ft>s

v =

or

When s = 1000 ft,

in

teaching

Web)

laws

v = 212(1000)1>2 = 109.54 ft>s = 110 ft>s

Dissemination

World

permitted.

vdv = ads

not

instructors

States

+ B A:

Wide

copyright

For 1000 ft 6 s … s¿ , the initial condition is v = 109.54 ft>s at s = 1000 ft.

v

is

of

and

use

United

on

learning.

-4ds by

L1000 ft

the

L109.54 ft>s

the

s

vdv =

student

v

is

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protected

the

(including

for

work

s v2 2 = - 4s 1000 ft 2 109.54 ft>s

this

assessing

A 220 000 - 8s B ft>s

part

the

is and

any

courses

When v = 0,

This

provided

integrity

of

and

work

v =

s¿ = 2500 ft

The v–s graph is shown in Fig. a.






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0 = 220 000 - 8s¿

Ans.

s¿

s(ft)

12–41. A train starts from station A and for the first kilometer, it travels with a uniform acceleration. Then, for the next two kilometers, it travels with a uniform speed. Finally, the train decelerates uniformly for another kilometer before coming to rest at station B. If the time for the whole journey is six minutes, draw the v- t graph and determine the maximum speed of the train.

SOLUTION For stage (1) motion,

vmax = 0 + 1ac21t1 vmax = 1ac21t1

(1)

v12 = v02 + 21ac211s1 - s02

vmax 2 = 0 + 21ac2111000 - 02 1ac21 =

vmax 2 2000

(2)

or

+ B A:

v1 = v0 + 1ac21t

laws

+ B A:

copyright

in

teaching

Web)

Eliminating 1ac21 from Eqs. (1) and (2), we have

Wide

2000 vmax

World

permitted.

Dissemination

(3) not

instructors

States

t1 =

the

by

and

use

United

on

learning.

is

of

the

For stage (2) motion, the train travels with the constant velocity of vmax for t = 1t2 - t12. Thus, (including

for

work

student

1 1a 2 t2 2 c 2

the

s2 = s1 + v1t +

1000 + 2000 = 1000 + vmax1t2 - t12 + 0 integrity

of

and provided

(4)

part

the

is and

any

courses

2000 vmax

This

t2 - t1 =

work

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assessing

is

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solely

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protected

+ B A:

+ B A:

v3 = v2 + 1ac23t

sale

+ B A:

will

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destroy

of

For stage (3) motion, the train travels for t = 360 - t2. Thus,

0 = vmax - 1ac231360 - t22

vmax = 1ac231360 - t22

(5)

v32 = v22 + 21ac231s3 - s22

0 = vmax 2 + 23 - 1ac23414000 - 30002

1ac23 =

vmax 2 2000

(6)

Eliminating 1ac23 from Eqs. (5) and (6) yields 360 - t2 =

2000 vmax

(7)

Solving Eqs. (3), (4), and (7), we have t1 = 120 s

t2 = 240 s

vmax = 16.7 m>s Based on the above results, the v-t graph is shown in Fig. a. © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





v = vmax for 2 min < t < 4 min.

Ans.

12–42. A particle starts from s = 0 and travels along a straight line with a velocity v = (t2 - 4t + 3) m > s, where t is in seconds. Construct the v - t and a -t graphs for the time interval 0 … t … 4 s.

SOLUTION a–t Graph: a =

dv d 2 = 1t - 4t + 32 dt dt

a = (2t - 4) m>s2 Thus, a|t = 0 = 2(0) - 4 = - 4 m>s2 a|t = 2 = 0

teaching

Web)

laws

or

a|t = 4 s = 2(4) - 4 = 4 m>s2

permitted. not

instructors

States

World

dv = 0. Thus, dt

v– t Graph: The slope of the v -t graph is zero when a =

Dissemination

Wide

copyright

in

The a -t graph is shown in Fig. a.

the

t = 2s

and

use

United

on

learning.

is

of

a = 2t - 4 = 0

for

work

student

the

by

The velocity of the particle at t = 0 s, 2 s, and 4 s are

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solely

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protected

the

(including

v|t = 0 s = 02 - 4(0) + 3 = 3 m>s

any sale

will

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and

The v- t graph is shown in Fig. b.

courses

part

integrity the

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This

provided

v|t = 4 s = 42 - 4(4) + 3 = 3 m>s

of

and

work

this

assessing

is

v|t = 2 s = 22 - 4(2) + 3 = - 1 m>s






12–43. If the position of a particle is defined by s = [2 sin [(p>5)t] + 4] m, where t is in seconds, construct the s - t, v- t, and a - t graphs for 0 … t … 10 s.

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permitted.

Dissemination

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copyright

in

teaching

Web)

laws

or

SOLUTION






*12–44. An airplane starts from rest, travels 5000 ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi>h. It then climbs in a straight line with a uniform acceleration of 3 ft>s2 until it reaches a constant speed of 220 mi>h. Draw the s–t, v–t, and a–t graphs that describe the motion.

SOLUTION v1 = 0 v2 = 162

mi (1h) 5280 ft = 237.6 ft>s h (3600 s)(1 mi)

v22 = v21 + 2 ac(s2 - s1) (237.6)2 = 02 + 2(ac)(5000 - 0) ac = 5.64538 ft>s2 v2 = v1 + act

laws

or

237.6 = 0 + 5.64538 t

in

teaching

Web)

t = 42.09 = 42.1 s mi (1h) 5280 ft = 322.67 ft>s h (3600 s)(1 mi) instructors

States

World

permitted.

Dissemination

Wide

copyright

v3 = 220

United

on

learning.

is

of

the

not

v23 = v22 + 2ac(s3 - s2)

student

the

by

and

use

(322.67)2 = (237.6)2 + 2(3)(s - 5000)

protected

the

(including

for

work

s = 12 943.34 ft

this

assessing

is

work

solely

of

v3 = v2 + act

the

is

This

provided

integrity

of

and

work

322.67 = 237.6 + 3 t

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destroy

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and

any

courses

part

t = 28.4 s






12–45. The elevator starts from rest at the first floor of the building. It can accelerate at 5 ft>s2 and then decelerate at 2 ft>s2. Determine the shortest time it takes to reach a floor 40 ft above the ground. The elevator starts from rest and then stops. Draw the a–t, v–t, and s–t graphs for the motion.

40 ft

SOLUTION + c v2 = v1 + act1 vmax = 0 + 5 t1 + c v3 = v2 + ac t 0 = vmax - 2 t2 Thus t1 = 0.4 t2

or

1 2 a t1 2 c laws

+ c s2 = s1 + v1t1 +

in

teaching

Web)

1 (5)(t21) = 2.5 t21 2

Dissemination

Wide

copyright

h = 0 + 0 +

the

not

instructors

States

World

permitted.

1 (2) t22 2

learning.

is

of

+ c 40 - h = 0 + vmaxt2 -

the

by

and

use

United

on

+ c v2 = v21 + 2 ac(s - s1)

the

(including

for

work

student

v2max = 0 + 2(5)(h - 0)

assessing

is

work

solely

of

protected

v2max = 10h

part

the

is and

any

courses

v2max = 160 - 4h

This

provided

integrity

of

and

work

this

0 = v2max + 2(-2)(40 - h)

sale

will

their

destroy

of

Thus, 10 h = 160 - 4h h = 11.429 ft vmax = 10.69 ft>s t1 = 2.138 s t2 = 5.345 s t = t1 + t2 = 7.48 s When t = 2.145, v = vmax = 10.7 ft>s and h = 11.4 ft.






Ans.

12–46. The velocity of a car is plotted as shown. Determine the total distance the car moves until it stops 1t = 80 s2. Construct the a–t graph.

v (m/s)

10

SOLUTION

t (s) 40

Distance Traveled: The total distance traveled can be obtained by computing the area under the v - t graph. s = 10(40) +

1 (10)(80 - 40) = 600 m 2

Ans.

dv a – t Graph: The acceleration in terms of time t can be obtained by applying a = . dt For time interval 0 s … t 6 40 s, a =

or

v - 10 0 - 10 1 , v = a - t + 20 b m>s. = t - 40 80 - 40 4

teaching

Web)

laws

For time interval 40 s 6 t … 80 s,

dv = 0 dt

in

dv 1 = - = - 0.250 m s2 dt 4 instructors

States

World

permitted.

Dissemination

Wide

copyright

a =

by

and

use

United

on

learning.

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For 0 … t 6 40 s, a = 0.

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the

(including

for

work

student

the

For 40 s 6 t … 80, a = - 0.250 m s2 .






80

12–47. The v–s graph for a go-cart traveling on a straight road is shown. Determine the acceleration of the go-cart at s = 50 m and s = 150 m. Draw the a–s graph.

v (m/s)

8

SOLUTION For 0 … s 6 100 v = 0.08 s,

100

dv = 0.08 ds

a ds = (0.08 s)(0.08 ds) a = 6.4(10 - 3) s At s = 50 m,

a = 0.32 m>s2

Ans.

For 100 6 s 6 200

or

v = - 0.08 s + 16,

teaching

Web)

laws

dv = - 0.08 ds

Wide

copyright

in

a ds = (-0.08 s + 16)(- 0.08 ds)

not

instructors

States

World

permitted.

Dissemination

a = 0.08(0.08 s - 16)

Ans.

the

a = - 0.32 m>s2 by

and

use

United

on

learning.

is

of

At s = 150 m,

for

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the

Also,

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the

(including

v dv = a ds

integrity

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this

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dv ) ds

part

the

is

This

provided

a = v(

will

8 ) = 0.32 m>s2 100

sale

a = 4(

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destroy

of

and

any

courses

At s = 50 m,

Ans.

At s = 150 m, a = 4(

-8 ) = -0.32 m>s2 100

At s = 100 m, a changes from amax = 0.64 m>s2 to amin = -0.64 m>s2 .






Ans.

200

s (m)

*12–48. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure. The acceleration and deceleration that occur are constant and both have a magnitude of 4 m>s2. If the plates are spaced 200 mm apart, determine the maximum velocity vmax and the time t¿ for the particle to travel from one plate to the other. Also draw the s–t graph. When t = t¿>2 the particle is at s = 100 mm.

smax

v s

vmax

SOLUTION ac = 4 m/s2

t¿/ 2

s = 100 mm = 0.1 m 2 v2 = v20 + 2 ac(s - s0) v2max = 0 + 2(4)(0.1 - 0) vmax = 0.89442 m>s

= 0.894 m>s

Ans.

v = v0 + ac t¿

in

teaching

Web)

laws

or

t¿ 0.89442 = 0 + 4( ) 2

Ans. Dissemination

Wide

= 0.447 s

copyright

t¿ = 0.44721 s

the

not

instructors

States

World

permitted.

1 a t2 2 c United

on

learning.

is

of

s = s0 + v0 t +

work

student

the

by

and

use

1 (4)(t)2 2

(including

for

s = 0 + 0 +

assessing

is

work

solely

of

protected

the

s = 2 t2

provided

integrity

of

and

work

this

0.44721 = 0.2236 = 0.224 s, 2

part

the

is

This

When t =

of

and

any

courses

s = 0.1 m

L0.894

v = - 4 t +1.788 s

t

ds =

L0.1

1 -4t+ 1.7882 dt L0.2235

s = - 2 t2 + 1.788 t - 0.2 When t = 0.447 s, s = 0.2 m






will

4 dt L0.2235

sale

t

ds = -

their

destroy

v

t¿

t

12–49. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure, where t¿ = 0.2 s and vmax = 10 m>s. Draw the s–t and a–t graphs for the particle. When t = t¿>2 the particle is at s = 0.5 m.

smax

v s

vmax

SOLUTION For 0 6 t 6 0.1 s, t¿/ 2

v = 100 t dv = 100 dt

a =

ds = v dt s

L0

t

ds =

L0

100 t dt

or

s = 50 t 2

in

teaching

Web)

laws

When t = 0.1 s,

Dissemination

Wide

copyright

s = 0.5 m

of

the

not

instructors

States

World

permitted.

For 0.1 s 6 t 6 0.2 s,

by

and

use

United

on

learning.

is

v = -100 t + 20

work

student

the

dv = - 100 dt

of

protected

the

(including

for

a =

this

assessing

is

work

solely

ds = v dt t

part

the

is

destroy

of

and

sale

s = - 50 t 2 + 20 t - 1

will

their

s - 0.5 = ( -50 t 2 + 20 t - 1.5)

any

courses

1 -100t + 202dt L0.1

This

L0.5

ds =

provided

integrity

of

and

work

s

When t = 0.2 s, s = 1m When t = 0.1 s, s = 0.5 m and a changes from 100 m/s2 to -100 m/s2. When t = 0.2 s, s = 1 m.






t¿

t

12–50. The v -t graph of a car while traveling along a road is shown. Draw the s -t and a- t graphs for the motion.

v (m/s)

SOLUTION

20

0 … t … 5

20 ¢v = = 4 m>s2 ¢t 5

a =

5 … t … 20

a =

20 … t … 30

a =

5

¢v 20 - 20 = = 0 m>s2 ¢t 20 - 5 ¢v 0 - 20 = = -2 m>s2 ¢t 30 - 20

From the v–t graph at t1 = 5 s , t2 = 20 s , and t3 = 30 s,

or

1 (5)(20) = 50 m 2 laws

s1 = A1 =

in

teaching

Web)

s2 = A1 + A2 = 50 + 20(20 - 5) = 350 m

permitted.

Dissemination

Wide

copyright

1 (30 - 20)(20) = 450 m 2 of

the

not

instructors

States

World

s3 = A1 + A2 + A3 = 350 +

and

use

United

on

learning.

is

The equations defining the portions of the s–t graph are t

ds =

s = 2t2 work

student

4t dt;

L0

(including

the

L0

the

ds = v dt;

for

v = 4t;

by

s

0 … t … 5s

s = 20t - 50

work

solely

of

t

20 dt;

L5

integrity

of

and

this

L50

ds =

assessing

is

ds = v dt; work

v = 20;

protected

s

5 … t … 20 s

provided

s

ds = v dt;

part

the

is

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destroy

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For 0 … t 6 5 s, a = 4 m>s2. For 20 s 6 t … 30 s, a = - 2 m>s2. At t = 5 s, s = 50 m. At t = 20 s, s = 350 m. At t = 30 s, s = 450 m. © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





t

ds =

L350

courses

v = 2(30 - t);

This

20 … t … 30 s

L20

2(30 - t) dt;

s = - t2 + 60t - 450

20

30

t (s)

12–51. The a–t graph of the bullet train is shown. If the train starts from rest, determine the elapsed time t¿ before it again comes to rest. What is the total distance traveled during this time interval? Construct the v–t and s–t graphs.

a(m/s2)

a

0.1t

30

v t Graph: For the time interval 0 … t 6 30 s, the initial condition is v = 0 when t = 0 s. dv = adt v

L0

t

dv =

L0

0.1tdt

v = A 0.05t2 B m>s When t = 30 s, v t = 30 s = 0.05 A 302 B = 45 m>s laws

or

or the time interval 30 s 6 t … t¿ , the initial condition is v = 45 m>s at t = 30 s. in

teaching

Web)

dv = adt

permitted.

Dissemination

1 t + 5 ≤ dt 15

is

of

the

not

¢-

World

L30 s

L45 m>s

instructors

t

dv =

States

v

Wide

copyright

+ B A:

by

and

use

United

on

learning.

1 2 t + 5t - 75 ≤ m>s 30

for

work

student

the

v = ¢-

protected

the

(including

Thus, when v = 0,

this

assessing

is

work

solely

of

1 2 t¿ + 5t¿ - 75 30

and

work

0 = -

part

the

is

This

provided

integrity

of

Choosing the root t¿ 7 75 s,

any

courses

Ans. destroy

of

and

t¿ = 133.09 s = 133 s

sale

will

their

Also, the change in velocity is equal to the area under the a–t graph. Thus, ¢v = 0 =

L

adt

1 1 1 (3)(75) + B ¢ - t¿ + 5 ≤ (t¿ - 75) R 2 2 15

0 = -

1 2 t¿ + 5t¿ - 75 30

This equation is the same as the one obtained previously. The slope of the v–tgraph is zero when t = 75 s, which is the instant a = v t = 75 s = -






1 A 752 B + 5(75) - 75 = 112.5 m>s 30

dv = 0.Thus, dt

( 15 )t

5 t¿

SOLUTION

+ B A:

1

a

3

75

t(s)

12–51. continued The v–t graph is shown in Fig. a. s t Graph: Using the result of v, the equation of the s–t graph can be obtained by integrating the kinematic equation ds = vdt. For the time interval 0 … t 6 30 s, the initial condition s = 0 at t = 0 s will be used as the integration limit. Thus, + B A:

ds = vdt s

L0

t

ds =

s = a

L0

0.05t2 dt

1 3 t bm 60

When t = 30 s, 1 A 303 B = 450 m 60

s t = 30 s =

For the time interval 30 s 6 t … t¿ = 133.09 s, the initial condition is s = 450 m when t = 30 s. ds = vdt 1 2 t + 5t - 75 b dt 30

Web)

L30 s

a-

in

L450 m

or

t

ds =

laws

s

teaching

+ B A:

instructors

States

World

permitted.

Dissemination

Wide

copyright

5 1 s = a - t3 + t2 - 75t + 750b m 90 2

use

United

on

learning.

is

of

the

not

When t = 75 s and t¿ = 133.09 s,

work

student

the

by

and

1 5 A 753 B + A 752 B - 75(75) + 750 = 4500 m 90 2

the

(including

for

s t = 75 s = -

work

solely

of

protected

5 1 A 133.093 B + A 133.092 B - 75(133.09) + 750 = 8857 m 90 2

and

any

courses

part

the

is

This

The s–t graph is shown in Fig. b.

provided

integrity

of

and

work

this

assessing

is

s t = 133.09 s = -

sale

will

their

destroy

of

When t = 30 s, v = 45 m/s and s = 450 m. When t = 75 s, v = vmax = 112.5 m/s and s = 4500 m. When t = 133 s, v = 0 and s = 8857 m.






Ans.

*12–52. The snowmobile moves along a straight course according to the v –t graph. Construct the s–t and a–t graphs for the same 50-s time interval. When t = 0, s = 0.

v (m/s)

12

SOLUTION s t Graph: The position function in terms of time t can be obtained by applying 12 ds 2 . For time interval 0 s … t 6 30 s, v = v = t = a t b m>s. dt 30 5

t (s) 30

ds = vdt s

L0

t

ds =

2 tdt L0 5

1 s = a t2 b m 5 1 A 302 B = 180 m 5

or

s =

laws

At t = 30 s ,

in

teaching

Web)

For time interval 30 s
permitted.

Dissemination

Wide

copyright

ds = vdt instructors United

on

learning.

is

of

the

not

12dt L30 s

and

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L180 m

World

t

ds =

States

s

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by

s = (12t - 180) m

the

s = 12(50) - 180 = 420 m assessing

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a t Graph: The acceleration function in terms of time t can be obtained by applying dv 2 dv a = . For time interval 0 s ◊ t<30 s and 30 ss and a = = 0, respectively. dt






50

12–53. A two-stage missile is fired vertically from rest with the acceleration shown. In 15 s the first stage A burns out and the second stage B ignites. Plot the v -t and s-t graphs which describe the two-stage motion of the missile for 0 … t … 20 s.

a (m/s2)

SOLUTION

25

Since v = v–t graph.

L

B

18

a dt, the constant lines of the a–t graph become sloping lines for the

t (s) 15

The numerical values for each point are calculated from the total area under the a–t graph to the point. At t = 15 s,

v = (18)(15) = 270 m>s

At t = 20 s,

v = 270 + (25)(20 - 15) = 395 m>s

Since s =

v dt, the sloping lines of the v–t graph become parabolic curves for the

laws

or

s–t graph.

L

Wide

copyright

in

teaching

Web)

The numerical values for each point are calculated from the total area under the v–t graph to the point. 1 (15)(270) = 2025 m 2

s =

At t = 20 s,

s = 2025 + 270(20 - 15) +

is

of

the

not

instructors

States

World

permitted.

Dissemination

At t = 15 s,

for

work

student

the

by

and

use

United

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learning.

1 (395 - 270)(20 - 15) = 3687.5 m = 3.69 km 2

work

solely

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the

(including

Also:

and

work

this

assessing

is

0 … t … 15:

any

destroy

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and

v = v0 + ac t = 0 + 18t

courses

part

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provided

integrity

of

a = 18 m>s 2

will

their

1 a t2 = 0 + 0 + 9t2 2 c

sale

s = s0 + v0 t + When t = 15:

v = 18(15) = 270 m>s s = 9(15)2 = 2025 m = 2.025 km 15 … t … 20: a = 25 m>s 2 v = v0 + ac t = 270 + 25(t - 15) s = s0 + v0 t +

1 1 ac t2 = 2025 + 270(t - 15) + (25)(t - 15)2 2 2

When t = 20: v = 395 m>s s = 3687.5 m = 3.69 km © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





A

20

12–54. The dragster starts from rest and has an acceleration described by the graph. Determine the time t¿ for it to stop. Also, what is its maximum speed? Construct the v - t and s- t graphs for the time interval 0 … t … t¿ .

a (ft/s2)

80

SOLUTION

t¿

v– t Graph: For the time interval 0 … t 6 5 s, the initial condition is v = 0 when t = 0 s.

A

+ :

5 1 1

B

a⫽⫺t⫹5

dv = adt v

t

dv =

L0

L0

80dt

v = (80t) ft>s The maximum speed occurs at the instant when the acceleration changes sign when t = 5 s. Thus, vmax = v|t = 5 s = 80(5) = 400 ft>s

Web)

laws

or

Ans.

Wide World

permitted.

Dissemination

dv = adt

is

of

the

not

instructors

States

+ B A:

copyright

in

teaching

For the time interval 5 6 t … t¿ , the initial condition is v = 400 ft>s when t = 5 s.

on

learning.

t

and

( -t + 5)dt

for

work

student

L5 s

by

L400 ft > s

the

dv =

use

United

v

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solely

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protected

the

(including

t2 + 5t + 387.5b ft>s 2

assessing

is

v = a-

part

the

is and

any

courses

t¿ 2 + 5t¿ + 387.5 2

This

0 = -

provided

integrity

of

and

work

this

Thus when v = 0,

t¿ = 33.28 s = 33.3 s

sale

will

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Choosing the positive root,

Ans.

Also, the change in velocity is equal to the area under the a- t graph. Thus ¢v =

L

adt

1 0 = 80(5) + e [( - t¿ + 5)(t¿ - 5)] f 2 0 = -

t¿ 2 + 5t¿ + 387.5 2

This quadratic equation is the same as the one obtained previously. The v-t graph is shown in Fig. a.






t (s)

12–54. continued s–t Graph: For the time interval 0 … t 6 5 s, the initial condition is s = 0 when t = 0 s. + B A:

ds = vdt s

t

ds =

80dt L0 s = (40t2) ft

L0

When t = 5 s, s|t = 5 s = 40(52) = 1000 ft For the time interval 5 s 6 t … t¿ = 45 s, the initial condition is s = 1000 ft when t = 5 s. + B A:

ds = vdt t

ds =

L1000 ft

L5 s

t2 + 5t + 387.5b dt 2

t3 5 + t2 + 387.5t - 979.17bft 6 2 laws

s = a-

a-

or

s

in

teaching

Web)

When t = t¿ = 33.28 s, 33.283 5 + (33.282) + 387.5(33.28) - 979.17 = 8542 ft 6 2 of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

s|t = 33.28 s = -

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on

learning.

is

The s –t graph is shown in Fig. b.






12–55. A race car starting from rest travels along a straight road and for 10 s has the acceleration shown. Construct the v–t graph that describes the motion and find the distance traveled in 10 s.

a (m/s2) 6

6 a=

SOLUTION

6

dv = adt t

L0

v = a

or

1 A 63 B = 12.0 m>s, 18

teaching

Web)

v =

1 3 t b m>s 18

laws

At t = 6 s,

1 2 t dt L0 6

dv =

Wide

copyright

in

For time interval 6 s 6 t … 10 s,

instructors

States

World

permitted.

Dissemination

dv = adt the

is and

use

United

on

learning.

6dt by

L6s

the

L12.0m>s

not

t

dv =

of

v

the

(including

for

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v = (6t - 24) m>s

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solely

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protected

v = 6(10) - 24 = 36.0 m>s

assessing

is

At t = 10 s,

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

ds Position: The position in terms of time t can be obtained by applying v = . dt For time interval 0 s … t 6 6 s,

L0

t

ds = s = ¢

sale

will

their

destroy

of

ds = vdt s

1 3 t dt L0 18 1 4 t ≤m 72 1 A 64 B = 18.0 m. 72

When t = 6 s, v = 12.0 m>s and s = For time interval 6 s 6 t … 10 s,

ds = vdt s

t

dv =

L18.0 m

L6s

(6t - 24)dt

s = A 3t2 - 24t + 54 B m When t = 10 s, v = 36.0 m>s and s = 3 102 - 24(10) + 54 = 114 m






Ans.

t2 t (s)

v -t Graph: The velocity function in terms of time t can be obtained by applying dv formula a = . For time interval 0 s … t 6 6 s, dt

v

1 — 6

10

*12–56. The v–t graph for the motion of a car as it moves along a straight road is shown. Draw the a–t graph and determine the maximum acceleration during the 30-s time interval. The car starts from rest at s = 0.

v (ft/s)

60 v 40

SOLUTION

t

30

0.4t2

v

For t 6 10 s: t (s)

v = 0.4t2 a =

10

dv = 0.8t dt

At t = 10 s: a = 8 ft>s2 For 10 6 t … 30 s: laws in

teaching

Web)

dv = 1 dt

Dissemination

Wide

copyright

a =

or

v = t + 30

permitted.

Ans.

sale

will

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destroy

of

and

any

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is

This

provided

integrity

of

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is

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the

(including

for

work

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the

by

and

use

United

on

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is

of

the

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amax = 8 ft>s2






30

12–57. The v–t graph for the motion of a car as it moves along a straight road is shown. Draw the s–t graph and determine the average speed and the distance traveled for the 30-s time interval. The car starts from rest at s = 0.

v (ft/s)

60 v 40

SOLUTION

t

30

0.4t2

v

For t 6 10 s, t (s)

v = 0.4t2

10

ds = v dt s

L0

t

ds =

0.4t2 dt

L0

s = 0.1333t3 At t = 10 s, laws

or

s = 133.3 ft

in

teaching

Web)

For 10 6 t 6 30 s,

permitted.

Dissemination

Wide

copyright

v = t + 30

is

of

the

not

instructors

States

World

ds = v dt

on

learning.

t United

s

and

use

1t + 302 dt

for

work

student

L10

by

L133.3

the

ds =

is

work

solely

of

protected

the

(including

s = 0.5t2 + 30t - 216.7

part

the

is

This

s = 1133 ft

provided

integrity

of

and

work

this

assessing

At t = 30 s,

destroy

of

and

any

courses

1133 ¢s = = 37.8 ft>s ¢t 30

Ans.

sale

will

their

(vsp)Avg =

sT = 1133 ft = 1.13(103) ft When t = 0 s, s = 133 ft. When t = 30 s, s = sI = 1.33 (103) ft






Ans.

30

12–58. The jet-powered boat starts from rest at s = 0 and travels along a straight line with the speed described by the graph. Construct the s -t and a -t graph for the time interval 0 … t … 50 s.

v (m/s)

v ⫽ 4.8 (10 ⫺3)t 3 75 v ⫽ ⫺3t ⫹ 150

SOLUTION t (s)

s–t Graph: The initial condition is s = 0 when t = 0. + B A:

25

ds = vdt s

L0

t

ds =

L0

4.8(10-3)t3 dt

s = [1.2110-3)t4 ]m At t = 25 s, s|t = 25 s = 1.2(10-3)(254) = 468.75 m

ds = vdt

permitted.

Dissemination

Wide

copyright

+ B A:

in

teaching

Web)

laws

or

For the time interval 25 s 6 t … 50 s, the initial condition s = 468.75 m when t = 25 s will be used.

instructors

States

the

not

(- 3t + 150) dt

on

learning.

is

of

L25 s

by

and

use

L468.75 m

World

t

ds =

United

s

solely

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the

(including

for

work

student

the

3 s = a - t2 + 150t - 2343.75b m 2

this

assessing

is

work

When t = 50 s,

any sale

a–t Graph: For the time interval 0 … t 6 25 s,

will

their

destroy

of

and

The s -t graph is shown in Fig. a.

courses

part

the

is

This

provided

a =

dv d = [4.8(10-3)t3] = (0.0144t2) m > s2 dt dt

When t = 25 s, a|t = 25 s = 0.0144(252) m > s2 = 9 m > s2 For the time interval 25 s 6 t … 50 s, a =

dv d = (- 3t + 150) = - 3 m > s2 dt dt

The a -t graph is shown in Fig. b. When t = 25 s, a = amax = 9 m > s2 and s = 469 m. When t = 50 s, s = 1406 m. © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





integrity

of

and

work

3 s|t = 50 s = - (502) + 150(50) - 2343.75 = 1406.25 m 2

50

12–59. An airplane lands on the straight runway, originally traveling at 110 ft> s when s = 0. If it is subjected to the decelerations shown, determine the time t¿ needed to stop the plane and construct the s–t graph for the motion.

a (ft/s2) 5 –3

SOLUTION

–8

v0 = 110 ft>s ¢ v = 1 a dt 0 - 110 = - 3(15 - 5) - 8(20 - 15) - 3(t¿ - 20) t¿ = 33.3 s

Ans.

= 550 ft

5s

st=

15s

= 1500 ft

st=

20s

= 1800 ft

st=

33.3s

laws

or

st=

sale

will

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of

and

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is

This

provided

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of

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work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

= 2067 ft






15

20

t'

t (s)

*12–60. v (m/s)

A car travels along a straight road with the speed shown by the v -t graph. Plot the a -t graph.

6

1

v⫽— 5 t

1

v ⫽ ⫺— 3 (t ⫺ 48)

SOLUTION t (s)

a–t Graph: For 0 … t 6 30 s, v =

1 t 5

a =

dv 1 = = 0.2 m > s2 dt 5

30

For 30 s 6 t … 48 s

teaching

Web)

laws

or

1 v = - (t - 48) 3 1 dv = - (1) = - 0.333 m > s2 a = dt 3

sale

will

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destroy

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and

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is

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of

and

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this

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is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

Using these results, a-t graph shown in Fig. a can be plotted.






48

12–61. v (m/s)

A car travels along a straight road with the speed shown by the v–t graph. Determine the total distance the car travels until it stops when t = 48 s. Also plot the s–t and a–t graphs.

6 v

1 — 5 t

v

1 — 3 (t

48)

SOLUTION For 0 … t … 30 s, t (s) 30

1 v = t 5 a =

dv 1 = dt 5

ds = v dt t

L0

1 2 t 10 in

teaching

Web)

s =

1 t dt L0 5 or

ds =

laws

s

s = 90 m,

Dissemination

Wide

copyright

When t = 30 s,

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

1 v = - (t - 48) 3

work

student

the

by

and

dv 1 = dt 3

the

(including

for

a =

assessing

is

work

solely

of

protected

ds = v dt

this

integrity

of

and

work

1 - 1t - 482dt 3 L30

any

sale

will

their

destroy

of

and

1 s = - t2 + 16t - 240 6

courses

part

the

is

This

L90

t

ds =

provided

s

When t = 48 s, s = 144 m

Ans.

Also, from the v–t graph ¢s =






L

v dt

s-0 =

1 1621482 = 144 m 2

Ans.

48

12–62. A motorcyclist travels along a straight road with the velocity described by the graph. Construct the s - t and a -t graphs.

v (ft/s) 150 v 20 t 50

50 v 2 t2

SOLUTION

t (s) 5

s–t Graph: For the time interval 0 … t 6 5 s, the initial condition is s = 0 when t = 0. + B A:

ds = vdt s

t

ds =

L0

2

L0

2t dt

2 s = a t3 b ft 3

laws

or

When t = 5 s,

in

teaching

Web)

2 3 (5 ) = 83.33 ft = 83.3 ft and a = 20 ft>s2 3

Wide

copyright

s =

instructors

States United

on

learning.

is

of

the

not

ds = vdt

and

use

+ B A:

World

permitted.

Dissemination

For the time interval 5 s 6 t … 10 s, the initial condition is s = 83.33 ft when t = 5 s.

by

t

the

s

(20t - 50)dt

work (including

of

protected

the

L5 s

student

L83.33 ft

for

ds =

work

solely

t

this

= (10t2 - 50t) 2

assessing

s

is

s2

5s

integrity

of

and

work

83.33 ft

any their

destroy

of

and

When t = 10 s,

courses

part

the

is

This

provided

s = (10t2 - 50t + 83.33) ft

sale

will

s|t = 10 s = 10(102) - 50(10) + 83.33 = 583 ft The s -t graph is shown in Fig. a. a–t Graph: For the time interval 0 … t 6 5 s, + B A:

a =

dv d (2t2) = (4t) ft>s 2 = dt dt

When t = 5 s, a = 4(5) = 20 ft>s2 For the time interval 5 s 6 t … 10 s, + B A:

a =

d dv = (20t - 50) = 20 ft > s2 dt dt

The a -t graph is shown in Fig. b.






10

12–63. The speed of a train during the first minute has been recorded as follows: t (s) 0 20 40 60 0 16 21 24 v (m>s) Plot the v-t graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled.

SOLUTION The total distance traveled is equal to the area under the graph. 1 1 1 (20)(16) + (40 - 20)(16 + 21) + (60 - 40)(21 + 24) = 980 m 2 2 2

Ans.

sale

will

their

destroy

of

and

any

courses

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is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

sT =






*12–64. A man riding upward in a freight elevator accidentally drops a package off the elevator when it is 100 ft from the ground. If the elevator maintains a constant upward speed of 4 ft>s, determine how high the elevator is from the ground the instant the package hits the ground. Draw the v–t curve for the package during the time it is in motion. Assume that the package was released with the same upward speed as the elevator.

SOLUTION For package: (+ c )

v2 = v20 + 2ac(s2 - s0) v2 = (4)2 + 2( - 32.2)( 0 - 100) v = 80.35 ft>s T

(+ c ) v = v0 + act -80.35 = 4 + ( - 32.2)t

laws

or

t = 2.620 s

teaching

Web)

For elevator: in

s2 = s0 + vt

Wide

copyright

(+ c )

not

instructors

States

World

permitted.

Dissemination

s = 100 + 4(2.620)

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

s = 110 ft






12–65. Two cars start from rest side by side and travel along a straight road. Car A accelerates at 4 m>s2 for 10 s and then maintains a constant speed. Car B accelerates at 5 m>s2 until reaching a constant speed of 25 m/s and then maintains this speed. Construct the a–t, v–t, and s–t graphs for each car until t = 15 s. What is the distance between the two cars when t = 15 s?

SOLUTION Car A: v = v0 + ac t vA = 0 + 4t At t = 10 s,

vA = 40 m>s

s = s0 + v0t +

1 (4)t2 = 2t2 2

t 7 10 s,

ds = v dt

teaching

sA = 200 m

Dissemination

Wide

copyright

in

At t = 10 s,

Web)

laws

or

sA = 0 + 0 +

1 2 at 2 c

permitted. not

instructors

States

World

40 dt

the

is

L10

United

on

learning.

L200

t

ds =

of

sA

work

student

the

by

and

use

sA = 40t - 200

the

(including

for

sA = 400 m is

work

solely

of

protected

At t = 15 s,

work

this

assessing

Car B:

any

destroy

of

and

vB = 0 + 5t

courses

part

the

is

This

provided

integrity

of

and

v = v0 + a c t

their

25 = 5s 5

will

t =

sale

When vB = 25 m/s,

s = s0 + v0t + sB = 0 + 0 +

1 2 at 2 c

1 (5)t2 = 2.5t2 2

When t = 10 s, vA = (vA)max = 40 m/s and sA = 200 m. When t = 5 s, sB = 62.5 m. When t = 15 s, sA = 400 m and sB = 312.5 m.






12–65. continued At t = 5 s,

sB = 62.5 m

t 7 5 s,

ds = v dt sB

L62.5

t

ds =

L5

25 dt

sB - 62.5 = 25t - 125 sB = 25t - 62.5 When t = 15 s,

sB = 312.5

Distance between the cars is ¢s = sA - sB = 400 - 312.5 = 87.5 m

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Car A is ahead of car B.






12–66. A two-stage rocket is fired vertically from rest at s = 0 with an acceleration as shown. After 30 s the first stage A burns out and the second stage B ignites. Plot the v–t and s–t graphs which describe the motion of the second stage for 0 … t … 60 s.

a (m/s2)

B A

15

9

SOLUTION

a

0.01t2

For 0 … t … 30 s v

l

dv =

L0

0.01 t2 dt

L0

30

v = 0.00333t3 When t = 30 s, v = 90 m>s For 30 s … t … 60 s

L30

15dt laws

L90

l

dv =

or

v

in

teaching

Web)

v = 15t - 360 v = 540 m>s

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

When t = 60 s,






60

t (s)

12–67. A two-stage rocket is fired vertically from rest at s = 0 with an acceleration as shown. After 30 s the first stage A burns out and the second stage B ignites. Plot the s–t graph which describes the motion of the second stage for 0 … t … 60 s.

a (m/s2)

B A

15

9 a ⫽ 0.01t 2

SOLUTION v– t Graph: When t = 0, v = 0. For 0 … t … 30 s,

A+cB

30

dv = a dt v

L0 v2

t

dv =

v

0.01t2dt

L0

0.01 3 t t 2 3 0

= 0

laws

or

v = {0.003333t3} m>s

in

teaching

Web)

When t = 30 s, v = 0.003333(303) = 90 m>s

Wide Dissemination

World

permitted.

dv = a dt

is

of

the

not

instructors

States

A+cB

copyright

For 30 s 6 t … 60 s,

on

learning.

t use

United

v

and

15 dt

by

for

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L30 s

the

L90 m>s

dv =

the

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t

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= 15t 2

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v

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v2

30 s

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assessing

is

90 m>s

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When t = 60 s, v = 15(60) - 360 = 540 m>s

of

and

any

courses

v = {15t - 360} m>s

This

provided

integrity

of

and

work

v - 90 = 15t - 450

sale

s–t Graph: When t = 0, s = 0. For 0 … t … 30 s,

A+cB

ds = vdt s

t

ds =

L0

L0

0.003333t3dt

s

s 2 = 0.0008333t4 2 0

t 0

4

s = {0.0008333 t } m When t = 30 s, s = 0.0008333(304) = 675 m For 30 s 6 t … 60 s,

A+cB

ds = vdt s

L675 m © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





t

ds =

L30 s

(15t - 360) dt

60

t (s)

12–67. continued s2

s 675 m

= (7.5t2 - 360t) 2

t 30 s

s - 675 = (7.5t2 - 360t) - [7.5(302) - 360(30)] s = {7.5t2 - 360t + 4725} m When t = 60 s, s = 7.5(602) - 360(60) + 4725 = 10 125 m

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States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Using these results, the s–t graph shown in Fig. a can be plotted.






*12–68. a (m/s2)

The a–s graph for a jeep traveling along a straight road is given for the first 300 m of its motion. Construct the v –s graph. At s = 0, v = 0.

2

SOLUTION

200

a s Graph: The function of acceleration a in terms of s for the interval 0 m … s 6 200 m is a - 0 2 - 0 = s - 0 200 - 0

a = (0.01s) m>s2

For the interval 200 m 6 s … 300 m, a - 2 0 - 2 = s - 200 300 - 200

a = (-0.02s + 6) m>s2

laws

or

v s Graph: The function of velocity v in terms of s can be obtained by applying vdv = ads. For the interval 0 m ◊ s<200 m,

in

teaching

Web)

vdv = ads s

0.01sds

permitted.

Dissemination

L0

instructors

States

World

L0

vdv =

Wide

copyright

v

use

United

on

learning.

is

of

the

not

v = (0.1s) m>s At s = 200 m,

for

work

student

the

by

and

v = 0.100(200) = 20.0 m>s

is

work

solely

of

protected

the

(including

For the interval 200 m 6 s … 300 m,

work

this

assessing

vdv = ads

integrity

of provided

( - 0.02s + 6)ds part

the

is

and

any

L200 m

courses

L20.0 m>s

and

s

vdv =

This

v

destroy

of

A 2 -0.02s2 + 12s - 1200 B m>s sale

will

their

v = At s = 300 m,






v = 2 -0.02(3002) + 12(300) - 1200 = 24.5 m>s

300

s (m)

12–69. The v–s graph for the car is given for the first 500 ft of its motion. Construct the a–s graph for 0 … s … 500 ft. How long does it take to travel the 500-ft distance? The car starts at s = 0 when t = 0.

v (ft/s)

60 v = 0.1s + 10

SOLUTION a – s Graph: The acceleration a in terms of s can be obtained by applying vdv = ads. a = v

10

dv = (0.1s + 10)(0.1) = (0.01s + 1) ft>s2 ds

500

At s = 0 and s = 500 ft, a = 0.01(0) + 1 = 1.00 ft>s2 and a = 0.01(500) + 1 = 6.00 ft>s2, respectively. Position: The position s in terms of time t can be obtained by applying v =

ds . dt

ds v

dt =

laws

or

a s = 0 = 100 ft >s2

Wide

copyright

in

teaching

Web)

a s = 500ft = 6.00 ft >s2

World

permitted.

Dissemination

ds L0 0.1s + 10

is

of

the

not

instructors

L0

s

dt =

States

t

the

by

and

use

United

on

learning.

t = 10ln (0.01s + 1)

work

student

t = 10ln [0.01(500) + 1] = 17.9 s

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This

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is

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the

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When s = 500 ft,






Ans.

s (ft)

12–70. The boat travels along a straight line with the speed described by the graph. Construct the s–t and a -s graphs. Also, determine the time required for the boat to travel a distance s = 400 m if s = 0 when t = 0.

v (m/s)

80

SOLUTION s t Graph: For 0 … s 6 100 m, the initial condition is s = 0 when t = 0 s. + B A:

v

ds dt = v t

v2 20

s (m) 100

s = A t2 B m When s = 100 m, 100 = t2

t = 10 s

For 100 m 6 s … 400 m, the initial condition is s = 100 m when t = 10 s.

laws

or

ds v

Web)

ds

teaching

s

dt =

in

dt = t

L10 s

assessing

is

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the

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work

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use

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on

learning.

is

of

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not

instructors

States

World

permitted.

Dissemination

Wide

copyright

L100 m 0.2s s t - 10 = 5ln 100 s t - 2 = ln 5 100 s et>5 - 2 = 100 s et>5 = 100 e2 provided

integrity

of

and

work

this

s = A 13.53et>5 B m

and

any

courses

part

the

is

This

When s = 400 m,

sale

t = 16.93 s = 16.9 s

will

their

destroy

of

400 = 13.53et>5

The s–t graph is shown in Fig. a. a s Graph: For 0 m … s 6 100 m, a = v

dv = A 2s1>2 B A s - 1>2 B = 2 m>s2 ds

For 100 m 6 s … 400 m, a = v

dv = (0.2s)(0.2) = 0.04s ds

When s = 100 m and 400 m, a s = 100 m = 0.04(100) = 4 m>s2 a s = 400 m = 0.04(400) = 16 m>s2 The a–s graph is shown in Fig. b.






4s

s

ds dt = L0 2s1>2 L0 t = s1>2

+ B A:

0.2s

Ans.

400

12–71. The v -s graph of a cyclist traveling along a straight road is shown. Construct the a -s graph.

v (ft/s)

15 v ⫽ ⫺0.04 s ⫹ 19

v ⫽ 0.1s ⫹ 5 5

SOLUTION s (ft)

a–s Graph: For 0 … s 6 100 ft, + B A:

a = v

100

dv = A 0.1s + 5 B A 0.1 B = A 0.01s + 0.5 B ft>s2 ds

Thus at s = 0 and 100 ft a ƒ s = 0 = 0.01 A 0 B + 0.5 = 0.5 ft>s2

teaching

Web)

laws

or

a ƒ s = 100 ft = 0.01 A 100 B + 0.5 = 1.5 ft>s2

permitted.

Dissemination

World

A - 0.04s + 19 B A - 0.04 B = A 0.0016s - 0.76 B ft>s2

is

of

the

not

dv = ds

instructors

a = v

States

+ B A:

Wide

copyright

in

For 100 ft 6 s … 350 ft,

by

and

use

United

on

learning.

Thus at s = 100 ft and 350 ft

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

a ƒ s = 350 ft = 0.0016 A 350 B - 0.76 = - 0.2 ft>s2

work

solely

of

protected

the

(including

for

work

student

the

a ƒ s = 100 ft = 0.0016 A 100 B - 0.76 = - 0.6 ft>s2

sale

will

their

destroy

of

and

any

courses

The a -s graph is shown in Fig. a.

Thus at s = 0 and 100 ft a ƒ s = 0 = 0.01 A 0 B + 0.5 = 0.5 ft>s2 a ƒ s = 100 ft = 0.01 A 100 B + 0.5 = 1.5 ft>s2 At s = 100 ft, a changes from amax = 1.5 ft>s2 to a min = - 0.6 ft>s2.






350

*■ 12–72. a (ft/s2)

The a–s graph for a boat moving along a straight path is given. If the boat starts at s = 0 when v = 0, determine its speed when it is at s = 75 ft, and 125 ft, respectively. Use Simpson’s rule with n = 100 to evaluate v at s = 125 ft.

a

5

6( s

10)5/3

5

SOLUTION Velocity: The velocity v in terms of s can be obtained by applying vdv = ads. For the interval 0 ft … s 6 100 ft,

100

vdv = ads v

L0

s

vdv =

5ds

L0

v = 210s = ft>s At s = 75 ft, v = 210(75) = 27.4 ft>s

Ans.

At s = 100 ft,

Ans. or

v = 210(100) = 31.62 ft>s

teaching

Web)

laws

For the interval 100 ft 6 s … 125 ft,

Wide

copyright

in

vdv = ads v

States

World

permitted.

35 + 61 2s - 1025>34ds

the

not

instructors use

United

on

learning.

is

L100 ft

of

L31.62 ft>s

Dissemination

125 ft

vdv =

student

the

by

and

Evaluating the integral on the right using Simpson’s rule, we have

assessing

is

work

solely

of

protected

the

(including

for

work

v2 v ` = 201.032 2 31.62 ft/s

Ans.

provided

integrity

of

and

work

this

At s = 125 ft,

sale

will

their

destroy

of

and

any

courses

part

the

is

This

v = 37.4 ft>s






s (ft)

12–73. The position of a particle is defined by r = 55 cos 2t i + 4 sin 2t j6 m, where t is in seconds and the arguments for the sine and cosine are given in radians. Determine the magnitudes of the velocity and acceleration of the particle when t = 1 s. Also, prove that the path of the particle is elliptical.

SOLUTION Velocity: The velocity expressed in Cartesian vector form can be obtained by applying Eq. 12–7. v =

dr = {-10 sin 2ti + 8 cos 2tj} m>s dt

When t = 1 s, v = - 10 sin 2(1)i + 8 cos 2(1)j = { -9.093i - 3.329j} m>s. Thus, the magnitude of the velocity is v = 2v2x + v2y = 2(- 9.093)2 + ( -3.329)2 = 9.68 m>s

Ans.

laws

or

Acceleration: The acceleration expressed in Cartesian vector from can be obtained by applying Eq. 12–9.

in

teaching

Web)

dv = { -20 cos 2ti - 16 sin 2tj} m>s2 dt

Wide

copyright

a =

on

Ans.

the

by

and

use

United

a = 2a2x + a2y = 28.3232 + ( -14.549)2 = 16.8 m>s2

learning.

is

of

the

not

instructors

States

World

the

(including

for

work

student

Traveling Path: Here, x = 5 cos 2t and y = 4 sin 2t. Then,

(1)

(2)

destroy

of

and

any

courses

part

the

is

This

y2 = sin2 2t 16

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

x2 = cos2 2t 25

sale

will

their

Adding Eqs (1) and (2) yields

y2 x2 + = cos2 2t + sin2 2t 25 16 However, cos2 2t + sin2 2t = 1. Thus, y2 x2 + = 1 25 16






(Equation of an Ellipse) (Q.E.D.)

permitted.

Dissemination

When t = 1 s, a = -20 cos 2(1)i - 16 sin 2(1)j = {8.323i - 14.549j} m>s2. Thus, the magnitude of the acceleration is

12–74. The velocity of a particle is v = 53i + (6 - 2t)j6 m>s, where t is in seconds. If r = 0 when t = 0, determine the displacement of the particle during the time interval t = 1 s to t = 3 s.

SOLUTION Position: The position r of the particle can be determined by integrating the kinematic equation dr = vdt using the initial condition r = 0 at t = 0 as the integration limit. Thus, dr = vdt t

r

L0

dr =

L0

C 3i + (6 - 2t)j D dt

r = c 3ti + A 6t - t2 B j d m When t = 1 s and 3 s,

laws

or

r t = 1 s = 3(1)i + C 6(1) - 12 D j = [3i + 5j] m>s

in

teaching

Web)

r t = 3 s = 3(3)i + C 6(3) - 32 D j = [9i + 9j] m>s

Dissemination

Wide

copyright

Thus, the displacement of the particle is

is

of

the

not

instructors

States

World

permitted.

¢r = r t = 3 s - r t = 1 s

by

and

use

United

on

learning.

= (9i + 9j) - (3i + 5j)

sale

will

their

destroy

of

and

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courses

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the

is

This

provided

integrity

of

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work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

= {6i + 4j} m






Ans.

12–75. A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration of a = 56ti + 12t2k6 ft>s2. Determine the particle’s position (x, y, z) at t = 1 s.

SOLUTION Velocity: The velocity expressed in Cartesian vector form can be obtained by applying Eq. 12–9. dv = adt v

L0

t

dv =

16ti + 12t2k2 dt

L0

v = {3t2i + 4t3k} ft/s Position: The position expressed in Cartesian vector form can be obtained by applying Eq. 12–7. laws

or

dr = vdt Web)

t teaching

r

in

13t2i + 4t3k2 dt

Wide

L0

copyright

Lr1

dr =

not

instructors

States

World

permitted.

Dissemination

r - (3i + 2j + 5k) = t3i + t4k

and

use

United

on

learning.

is

of

the

r = {(t3 + 3) i + 2j + (t4 + 5)k} ft

(including

for

work

student

the

by

When t = 1 s, r = (13 + 3)i + 2j + (14 + 5)k = {4i + 2j + 6k} ft.

is

work

solely

of

protected

the

The coordinates of the particle are

Ans.

sale

will

their

destroy

of

and

any

courses

part

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is

This

provided

integrity

of

and

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this

assessing

(4 ft, 2 ft, 6 ft)






*12–76. The velocity of a particle is given by v = 516t2i + 4t3j + (5t + 2)k6 m>s, where t is in seconds. If the particle is at the origin when t = 0, determine the magnitude of the particle’s acceleration when t = 2 s. Also, what is the x, y, z coordinate position of the particle at this instant?

SOLUTION Acceleration: The acceleration expressed in Cartesian vector form can be obtained by applying Eq. 12–9. a =

dv = {32ti + 12t2j + 5k} m>s2 dt

When t = 2 s, a = 32(2)i + 12 A 22 B j + 5k = {64i + 48j + 5k} m>s2. The magnitude of the acceleration is a = 2a2x + a2y + a2z = 2642 + 482 + 52 = 80.2 m>s2

Ans.

or

Position: The position expressed in Cartesian vector form can be obtained by applying Eq. 12–7.

teaching

Web)

laws

dr = v dt in

2 3 A 16t i + 4t j + (5t + 2)k B dt

Wide

L0

permitted.

dr =

Dissemination

L0

copyright

t

r

the

not

instructors

States

World

5 16 3 t i + t4j + a t2 + 2tb k d m 3 2

and

use

United

on

learning.

is

of

r = c

(including

for

work

student

the

by

When t = 2 s,

the

16 3 5 A 2 B i + A 24 B j + c A 22 B + 2(2) d k = {42.7i + 16.0j + 14.0k} m. 3 2 work

this

assessing

is

work

solely

of

protected

r =

part

the

is

This

provided

integrity

of

and

Thus, the coordinate of the particle is

Ans.

sale

will

their

destroy

of

and

any

courses

(42.7, 16.0, 14.0) m






12–77. The car travels from A to B, and then from B to C, as shown in the figure. Determine the magnitude of the displacement of the car and the distance traveled.

2 km

A B

3 km

SOLUTION Displacement:

¢r = {2i - 3j} km ¢r = 222 + 32 = 3.61 km

Ans.

Distance Traveled:

C

d = 2 + 3 = 5 km

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






12–78. A car travels east 2 km for 5 minutes, then north 3 km for 8 minutes, and then west 4 km for 10 minutes. Determine the total distance traveled and the magnitude of displacement of the car. Also, what is the magnitude of the average velocity and the average speed?

SOLUTION Total Distance Traveled and Displacement: The total distance traveled is s = 2 + 3 + 4 = 9 km

Ans.

and the magnitude of the displacement is ¢r = 2(2 - 4)2 + 32 = 3.606 km = 3.61 km

Ans.

Average Velocity and Speed: The total time is ¢t = 5 + 8 + 10 = 23 min = 1380 s The magnitude of average velocity is Ans. laws

or

vavg

3.606 A 103 B ¢r = = 2.61 m>s = ¢t 1380

teaching in

9 A 103 B s = = 6.52 m>s ¢t 1380

Wide

copyright

permitted.

Dissemination

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

A vsp B avg =

Web)

and the average speed is






12–79. A car traveling along the straight portions of the road has the velocities indicated in the figure when it arrives at points A, B, and C. If it takes 3 s to go from A to B, and then 5 s to go from B to C, determine the average acceleration between points A and B and between points A and C.

y vC x

vB B

SOLUTION vA = 20 i

vA

vB = 21.21 i + 21.21j

20 m/s

A

vC = 40i aAB =

21.21 i + 21.21 j - 20 i ¢v = ¢t 3

aAB = { 0.404 i + 7.07 j } m>s2 40 i - 20 i ¢v = ¢t 8

or

aAC =

Ans.

laws

aAC = { 2.50 i } m>s2

sale

will

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destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

Ans.






45

30 m/s

C

40 m/s

*12–80. A particle travels along the curve from A to B in 2 s. It takes 4 s for it to go from B to C and then 3 s to go from C to D. Determine its average speed when it goes from A to D.

y D

B

5m

15 m C

SOLUTION

10 m

1 1 sT = (2p)(10)) + 15 + (2p(5)) = 38.56 4 4 sT 38.56 = 4.28 m>s = tt 2 + 4 + 3

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

vsP =

x

A






12–81. The position of a crate sliding down a ramp is given by x = (0.25t3) m, y = (1.5t2) m, z = (6 - 0.75t5>2) m, where t is in seconds. Determine the magnitude of the crate’s velocity and acceleration when t = 2 s.

SOLUTION Velocity: By taking the time derivative of x, y, and z, we obtain the x, y, and z components of the crate’s velocity. d # vx = x = A 0.25t3 B = A 0.75t2 B m>s dt d # vy = y = A 1.5t2 B = A 3t B m>s dt

A - 1.875t3>2 B m>s or

d # vz = z = A 6 - 0.75t5>2 B = dt

teaching

Web)

vz = - 1.875 A 2 B 3>2 = - 5.303 m/s in

vy = 3(2) = 6 m>s

not

instructors

States

World

permitted.

Dissemination

Thus, the magnitude of the crate’s velocity is

Wide

copyright

vx = 0.75 A 22 B = 3 m>s

laws

When t = 2 s,

Ans.

the

by

and

use

United

on

learning.

is

of

the

v = 2vx 2 + vy 2 + vz 2 = 232 + 62 + ( - 5.303)2 = 8.551 ft>s = 8.55 ft

of

protected

the

(including

for

work

student

Acceleration: The x, y, and z components of the crate’s acceleration can be obtained by taking the time derivative of the results of vx, vy, and vz, respectively.

part

the

is

any

courses and

A - 2.815t1>2 B m>s2 will

destroy

of

their

d # az = vz = A - 1.875t3>2 B = dt

sale

d # ay = vy = (3t) = 3 m>s2 dt

This

provided

integrity

of

and

work

this

assessing

is

work

solely

d # ax = vx = A 0.75t2 B = (1.5t) m>s2 dt

When t = 2 s, ax = 1.5(2) = 3 m>s2

ay = 3 m>s2

az = - 2.8125 A 21>2 B = - 3.977 m>s2

Thus, the magnitude of the crate’s acceleration is a = 2ax 2 + ay 2 + az 2 = 232 + 32 + ( - 3.977)2 = 5.815 m>s2 = 5.82 m>s






Ans.

12–82. A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by y2 = [120(103)x] m. If the 1 x component of acceleration is ax = a t2 b m>s2, where t is 4 in seconds, determine the magnitude of the rocket’s velocity and acceleration when t = 10 s.

SOLUTION Position: The parameter equation of x can be determined by integrating ax twice with respect to t. L

dvx =

L

vx

t

dvx =

L0

axdt

1 3 t b m>s 12 or

vx = a

1 2 t dt L0 4

L

vxdt

teaching

Web)

dx =

laws

L

in

1 3 t dt L0 12

Wide Dissemination

L0

t

dx =

copyright

x

the

not

instructors

States

World

permitted.

1 4 t bm 48 use

United

on

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y = A 50t2 B m

1 4 t b 48 work

y2 = 120 A 103 B a

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Substituting the result of x into the equation of the path,

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Velocity:

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courses

d # vy = y = A 50t2 B = A 100t B m>s dt When t = 10 s, vx =

1 A 103 B = 83.33 m>s 12

vy = 100(10) = 1000 m>s

Thus, the magnitude of the rocket’s velocity is v = 2vx 2 + vy2 = 283.332 + 10002 = 1003 m>s

Ans.

Acceleration: # d ay = vy = (100t) = 100 m>s2 dt When t = 10 s, ax =

1 A 102 B = 25 m>s2 4

Thus, the magnitude of the rocket’s acceleration is a = 2ax 2 + ay2 = 2252 + 1002 = 103 m>s2 © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

12–83. y

The particle travels along the path defined by the parabola y = 0.5x2. If the component of velocity along the x axis is vx = 15t2 ft>s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = 1 s. When t = 0, x = 0, y = 0.

y

0.5x2

SOLUTION dx . dt

Position: The x position of the particle can be obtained by applying the vx =

O

dx = vx dt x

L0

t

dx =

L0

5tdt

x = A 2.50t2 B ft Thus, y = 0.5 A 2.50t2 B 2 = A 3.125t4 B ft. At

t = 1 s, x = 2.5 A 12 B = 2.50 ft

and

y = 3.125 A 14 B = 3.125 ft. The particle’s distance from the origin at this moment is d = 2(2.50 - 0)2 + (3.125 - 0)2 = 4.00 ft

laws

or

Ans.

copyright

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# # Acceleration: Taking the first derivative of the path y = 0.5x2, we have y = xx. The second derivative of the path gives Wide

$ # $ y = x2 + xx States

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permitted.

Dissemination

(1)

use

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learning.

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# $ $ However, x = vx, x = ax and y = ay. Thus, Eq. (1) becomes

and

ay = v2x + xax

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dvx = 5 ft>s 2, and x = 2.50 ft . Then, from When t = 1 s, vx = 5(1) = 5 ft>s ax = dt Eq. (2)

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ay = 52 + 2.50(5) = 37.5 ft>s2

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x

*12–84. The motorcycle travels with constant speed v0 along the path that, for a short distance, takes the form of a sine curve. Determine the x and y components of its velocity at any instant on the curve.

y v0 y

π x) c sin ( –– L x

c L

SOLUTION y = c sin a

p xb L

p p # # ca cos xb x y = L L vy =

p p c vx a cos x b L L

v20 = v2y + v2x p 2 p cb cos2 a xb R L L

vx = v0 B 1 + a

-2 p 2 p cb cos2 a xb R L L

laws

or

v20 = v2x B 1 + a

1

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Ans. 1 Dissemination

-2 v0 pc p p 2 p vy = acos xb B 1 + a cb cos2 a x b R L L L L

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c L

12–85. A particle travels along the curve from A to B in 1 s. If it takes 3 s for it to go from A to C, determine its average velocity when it goes from B to C.

y

B

20 m

SOLUTION (rAC - rAB) 40i - (20i + 20j) ¢r = = = {10i - 10j} m>s ¢t ¢t 2

Ans.

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laws

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vang =

C

A

Time from B to C is 3 - 1 = 2 s






x

12–86. y

When a rocket reaches an altitude of 40 m it begins to travel along the parabolic path 1y - 4022 = 160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy = 180 m>s, determine the magnitudes of the rocket’s velocity and acceleration when it reaches an altitude of 80 m.

(y

40)2

160x

SOLUTION vy = 180 m>s 40 m

(y - 40)2 = 160 x 2(y - 40)vy = 160vx

x

(1)

2(80 - 40)(180) = 160vx vx = 90 m>s v = 2902 + 1802 = 201 m>s

dt

= 0 or

d vy

laws

ay =

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From Eq. 1,

Dissemination

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copyright

2 v2y + 2(y - 40)ay = 160 ax

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2(180)2 + 0 = 160 ax

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ax = 405 m>s2

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a = 405 m>s2






Ans.

12–87. Pegs A and B are restricted to move in the elliptical slots due to the motion of the slotted link. If the link moves with a constant speed of 10 m/s, determine the magnitude of the velocity and acceleration of peg A when x = 1 m.

y

A

v

Velocity: The x and y components of the peg’s velocity can be related by taking the first time derivative of the path’s equation.

B x2 4

x2 + y2 = 1 4 1 # # (2xx) + 2yy = 0 4 1 # # xx + 2yy = 0 2 or 1 xv + 2yvy = 0 2 x

(1)

laws

or

At x = 1 m, teaching

Web)

23 m 2 Dissemination

Wide

copyright

y =

in

(1)2 + y2 = 1 4

not

instructors

States

World

permitted.

Here, vx = 10 m>s and x = 1. Substituting these values into Eq. (1),

is

of

the

1 23 (1)(10) + 2 ¢ ≤ vy = 0 2 2

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vy = -2.887 m>s = 2.887 m>s T

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Thus, the magnitude of the peg’s velocity is

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v = 2vx 2 + vy 2 = 2102 + 2.8872 = 10.4 m>s

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part

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Acceleration: The x and y components of the peg’s acceleration can be related by taking the second time derivative of the path’s equation.

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of

1 # # ## ## # # (xx + xx) + 2(yy + yy) = 0 2 1 #2 ## ## # A x + xx B + 2 A y 2 + yy B = 0 2 or 1 A v 2 + xax B + 2 A vy 2 + yay B = 0 2 x Since vx is constant, ax = 0. When x = 1 m, y =

(2) 23 m, vx = 10 m>s, and 2

vy = -2.887 m>s. Substituting these values into Eq. (2), 23 1 a d = 0 A 102 + 0 B + 2 c ( - 2.887)2 + 2 2 y ay = - 38.49 m>s2 = 38.49 m>s2 T Thus, the magnitude of the peg’s acceleration is a = 2ax 2 + ay 2 = 202 + ( -38.49)2 = 38.5 m>s2 © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





D

C

SOLUTION

Ans.

y2

1

x 10 m/s

*12–88. The van travels over the hill described by y = (- 1.5(10–3) x2 + 15) ft. If it has a constant speed of 75 ft>s, determine the x and y components of the van’s velocity and acceleration when x = 50 ft.

y 15 ft

y

( 1.5 (10 3) x2 x

SOLUTION

100 ft

Velocity: The x and y components of the van’s velocity can be related by taking the first time derivative of the path’s equation using the chain rule. y = -1.5 A 10 - 3 B x2 + 15 # # y = -3 A 10 - 3 B xx or vy = -3 A 10 - 3 B xvx When x = 50 ft, vy = - 3 A 10 - 3 B (50)vx = - 0.15vx

(1)

The magnitude of the van’s velocity is v = 2vx 2 + vy 2

laws

or

(2)

in

teaching

Web)

Substituting v = 75 ft>s and Eq. (1) into Eq. (2),

permitted.

Dissemination

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copyright

75 = 2vx 2 + ( - 0.15vx)2

on

learning.

is

of

the

instructors

States

Ans.

not

World

vx = 74.2 ft>s ;

the

by

and

use

United

Substituting the result of nx into Eq. (1), we obtain

Ans.

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(including

for

work

student

vy = -0.15(- 74.17) = 11.12 ft>s = 11.1 ft>s c

work

this

assessing

is

work

solely

Acceleration: The x and y components of the van’s acceleration can be related by taking the second time derivative of the path’s equation using the chain rule.

and

any

courses

part

the

is

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provided

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and

$ # # ## y = - 3 A 10 - 3 B (xx + xx) destroy

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ay = - 3 A 10 - 3 B A vx 2 + xax B When x = 50 ft, vx = - 74.17 ft>s. Thus, ay = -3 A 10 - 3 B c ( - 74.17)2 + 50ax d ay = -(16.504 + 0.15ax)

(3)

Since the van travels with a constant speed along the path,its acceleration along the tangent of the path is equal to zero. Here, the angle that the tangent makes with the horizontal at dy x = 50 ft is u = tan - 1 ¢ ≤ 2 = tan - 1 c -3 A 10 - 3 B x d 2 = tan - 1(- 0.15) = -8.531°. dx x = 50 ft x = 50 ft Thus, from the diagram shown in Fig. a, ax cos 8.531° - ay sin 8.531° = 0

(4)

Solving Eqs. (3) and (4) yields






ax = -2.42 ft>s = 2.42 ft>s2 ;

Ans.

ay = -16.1 ft>s = 16.1 ft>s2 T

Ans.

15) ft

12–89. It is observed that the time for the ball to strike the ground at B is 2.5 s. Determine the speed vA and angle uA at which the ball was thrown.

vA uA

A

1.2 m 50 m

SOLUTION Coordinate System: The x–y coordinate system will be set so that its origin coincides with point A. x-Motion: Here, (vA)x = vA cos uA, xA = 0, xB = 50 m, and t = 2.5 s. Thus, + B A:

xB = xA + (vA)xt 50 = 0 + vA cos uA(2.5) vA cos uA = 20

(1)

y-Motion: Here, (vA)y = vA sin uA, = - 9.81 m>s2. Thus,

yB = -1.2 m,

ay = -g

and

or

yA = 0 ,

Web)

laws

1 a t2 2 y

teaching

yB = yA + (vA)y t +

in

A+cB

(2)

learning.

is

of

the

not

instructors

States

World

vA sin uA = 11.7825

by

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vA = 23.2 m>s

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the

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uA = 30.5°






Ans.

permitted.

Dissemination

Wide

copyright

1 ( -9.81) A 2.52 B 2

- 1.2 = 0 + vA sin uA (2.5) +

B

12–90. Determine the minimum initial velocity v0 and the corresponding angle u0 at which the ball must be kicked in order for it to just cross over the 3-m high fence.

v0

3m

u0

6m

SOLUTION Coordinate System: The x - y coordinate system will be set so that its origin coincides with the ball’s initial position. x-Motion: Here, (v0)x = v0 cos u, x0 = 0, and x = 6 m. Thus, + B A:

x = x0 + (v0)xt 6 = 0 + (v0 cos u)t t =

6 v0 cos u

(1)

Web)

laws

teaching

Wide

copyright

World

permitted.

Dissemination

1 (-9.81)t2 2

not of

the

3 = 0 + v0 (sin u) t +

in

1 a t2 2 y instructors

y = y0 + (v0)y t +

States

A+cB

or

y-Motion: Here, (v0)x = v0 sin u, ay = - g = - 9.81 m > s2, and y0 = 0. Thus,

is

3 = v0 (sin u) t - 4.905t2

the

by

and

use

United

on

learning.

(2)

the

(including

for

work

student

Substituting Eq. (1) into Eq. (2) yields

work

solely

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58.86 B sin 2u - cos2 u

(3)

integrity

of

and

work

this

assessing

is

v0 =

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provided

From Eq. (3), we notice that v0 is minimum when f(u) = sin 2u - cos2 u is df(u) = 0 maximum. This requires du df(u) = 2 cos 2u + sin 2u = 0 du tan 2u = - 2 2u = 116.57° u = 58.28° = 58.3°

Ans.

Substituting the result of u into Eq. (2), we have (v0)min =






58.86 = 9.76 m>s B sin 116.57° - cos2 58.28°

Ans.

12–91. During a race the dirt bike was observed to leap up off the small hill at A at an angle of 60° with the horizontal. If the point of landing is 20 ft away, determine the approximate speed at which the bike was traveling just before it left the ground. Neglect the size of the bike for the calculation. A

20 ft

SOLUTION + )s = s + v t (: 0 0 20 = 0 + vA cos 60° t ( + c ) s = s0 + v0 +

1 a t2 2 c

0 = 0 + vA sin 60° t +

1 ( - 32.2) t2 2

Solving

or

t = 1.4668 s laws

vA = 27.3 ft s

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

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this

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is

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solely

of

protected

the

(including

for

work

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the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

Ans.






60°

*12–92. The girl always throws the toys at an angle of 30° from point A as shown.Determine the time between throws so that both toys strike the edges of the pool B and C at the same instant.With what speed must she throw each toy?

A

30° 1m B

2.5 m

SOLUTION

4m

To strike B: + )s = s + v t (: 0 0 2.5 = 0 + vA cos 30° t (+ c ) s = s0 + v0 t +

1 2 a t 2 c

0.25 = 1 + vA siv 30° t -

1 (9.81)t2 2

laws

or

Solving

teaching

Web)

t = 0.6687 s in

(vA)B = 4.32 m>s

Dissemination

Wide

copyright

Ans.

not

instructors

States

World

permitted.

To strike C:

and

use

United

on

learning.

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of

the

+ )s = s + v t (: 0 0

for

work

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the

by

4 = 0 + vA cos 30° t

work

solely

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protected

the

(including

1 2 a t 2 c

integrity

of

and provided

part

the

is and

any

courses

1 (9.81)t2 2

This

0.25 = 1 + vA siv 30° t -

work

this

assessing

is

(+ c ) s = s0 + v0 t +

(vA)C = 5.85 m>s

sale

t = 0.790 s

will

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destroy

of

Solving

Ans.

Time between throws: ¢t = 0.790 s - 0.6687 s = 0.121 s






Ans.

C 0.25 m

12–93. The player kicks a football with an initial speed of v0 = 90 ft>s. Determine the time the ball is in the air and the angle u of the kick. v0 u A

126 ft

SOLUTION Coordinate System: The x – y coordinate system will be set with its origin coinciding with starting point of the football. x-motion: Here, x0 = 0, x = 126 ft, and (v0)x = 90 cos u + B A:

x = x0 + (v0)x t 126 = 0 + (90 cos u) t t =

126 90 cos u

(1)

laws

teaching

Web)

1 a t2 2 y

in

y = y0 + (v0)y t +

Wide

copyright

A+cB

or

y-motion: Here, y0 = y = 0, (v0)y = 90 sin u, and ay = - g = - 32.2 ft. Thus,

World

permitted.

Dissemination

1 ( -32.2)t2 2

not

instructors

States

O = 0 + (90 sin u)t +

of

the

O = (90 sin u)t - 16.1t2

and

use

United

on

learning.

is

(2)

for

work

student

the

by

Substitute Eq. (1) into (2) yields

work

solely

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protected

the

(including

2 126 126 ≤ - 16.1 ¢ ≤ 90 cos u 90 cos u

integrity

of

and provided

the part any

courses

(3) destroy

of

O = 126 sin u cos u - 31.556

is

This

126 sin u 31.556 cos u cos2 u

and

O =

work

this

assessing

is

O = 90 sin u ¢

sale

will

their

Using the trigonometry identity sin 2u = 2 sin u cos u, Eq. (3) becomes 63 sin 2u = 31.556 sin 2u = 0.5009 2u = 30.06 or 149.94 u = 15.03° = 15.0° or u = 74.97° = 75.0°

Ans.

If u = 15.03°, t =

126 = 1.45 s 90 cos 15.03°

Ans.

126 = 5.40 s 90 cos 74.97°

Ans.

If u = 74.97°, t = Thus,

u = 15.0°, t = 1.45 s u = 75.0°, t = 5.40 s






12–94. From a videotape, it was observed that a pro football player kicked a football 126 ft during a measured time of 3.6 seconds. Determine the initial speed of the ball and the angle u at which it was kicked. v0

θ A

126 ft

SOLUTION + B A:

s = s0 + v0 t 126 = 0 + (v0)x (3.6) (v0)x = 35 ft>s

A+cB

s = s0 + v0 t +

1 2 a t 2 c

O = 0 + (v0)y (3.6) +

1 (- 32.2)(3.6)2 2

or

(v0)y = 57.96 ft>s v0 = 2(35)2 + (57.96)2 = 67.7 ft>s

Ans.

Wide

= 58.9°

sale

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this

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is

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solely

of

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the

(including

for

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student

the

by

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United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

57.96 35

copyright

u = tan - 1

in

teaching

Web)

laws

Ans.






12–95. A projectile is given a velocity v0 at an angle f above the horizontal. Determine the distance d to where it strikes the sloped ground. The acceleration due to gravity is g.

y

v0 φ

SOLUTION + B A:

d cos u = 0 + v0 (cos f) t s = s0 + v0 t +

1 2 a t 2 c

d sin u = 0 + v0 (sin f)t +

1 ( - g) t2 2

Thus, or

d cos u 1 d cos u 2 b - ga b v0 cos f 2 v0 cos f

teaching

Web)

laws

d sin u = v0 sin fa

in

gd cos2 u 2v20 cos2 f

Dissemination

Wide

copyright

sin u = cos u tan f -

not

instructors

States

World

permitted.

2v20 cos2 f

the

g cos2 u use

United

on

learning.

is

of

d = (cos u tan f - sin u)

by

and

v 20 sin 2f - 2 tan u cos2 f g cos u

work

student

the

Ans.

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the

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for

d =






x d

s = s0 + v0 t

A+cB

θ

*12–96. A projectile is given a velocity v0 . Determine the angle f at which it should be launched so that d is a maximum. The acceleration due to gravity is g.

y

v0 φ

SOLUTION + B A:

d cos u = 0 + v0 (cos f) t sy = s0 + v0 t +

1 2 at 2 c

d sin u = 0 + v0 (sin f)t +

1 ( - g)t2 2

Thus, or

d cos u 1 d cos u 2 b - ga b v0 cos f 2 v0 cos f

teaching

Web)

laws

d sin u = v0 sin f a

in

gd cos2 u 2v 20 cos2 f

Dissemination

Wide

copyright

sin u = cos u tan f -

not

instructors

States

World

permitted.

2v 20 cos2 f

the

g cos2 u use

United

on

learning.

is

of

d = (cos u tan f - sin u)

of

protected

the

(including

for

work

student

the

by

and

v 20 A sin 2f - 2 tan u cos2 f B g cos u

d =

any sale

sin 2f tan u + 1 = 0 cos 2f

will

their

destroy

of

and

cos 2f + tan u sin 2f = 0

courses

part

the

is

This

provided

integrity

of

and

work

d(d) v 20 = C cos 2f(2) - 2 tan u(2 cos f)(- sin f) D = 0 df g cos u

this

assessing

is

work

solely

Require:

tan 2f = - ctn u 1 tan - 1 (- ctn u) 2

f =






x d

sx = s0 + v0 t

A+cB

θ

Ans.

12–97. Determine the maximum height on the wall to which the firefighter can project water from the hose, if the speed of the water at the nozzle is vC = 48 ft>s.

A

vC = 48 ft/s

SOLUTION

h

θ C 3 ft

(+ c ) v = v0 + ac t

B 30 ft

0 = 48 sin u - 32.2 t + )s = s + v t (: 0 0 30 = 0 + 48 (cos u)(t) 48 sin u = 32.2

30 48 cos u

sin u cos u = 0.41927

laws

or

sin 2u = 0.83854

in

teaching

Web)

u = 28.5°

Dissemination

Wide

copyright

t = 0.7111 s

the

not

instructors

States

World

permitted.

1 a t2 2 c use

United

on

learning.

is

of

(+ c ) s = s0 + v0 t +

work

student

the

by

and

1 ( - 32.2)(0.7111)2 2

the

(including

for

h - 3 = 0 + 48 sin 28.5° (0.7111) +

sale

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is

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provided

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assessing

is

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of

protected

h = 11.1 ft






Ans.

■

12–98.

Determine the smallest angle u, measured above the horizontal, that the hose should be directed so that the water stream strikes the bottom of the wall at B. The speed of the water at the nozzle is vC = 48 ft>s. A vC

48 ft/s

SOLUTION + ) (:

h

u C 3 ft

s = s0 + v0 t

B 30 ft

30 = 0 + 48 cos u t 30 48 cos u s = s0 + v0 t +

1 a t2 2 c

0 = 3 + 48 sin u t +

or

2 48 sin u (30) 30 - 16.1 a b 48 cos u 48 cos u

Web) in

0 = 3 +

1 ( -32.2)t2 2 laws

(+ c)

teaching

t =

Dissemination

Wide

copyright

0 = 3 cos2 u + 30 sin u cos u - 6.2891

is

of

the

not

instructors

States

World

permitted.

3 cos2 u + 15 sin 2u = 6.2891

by

and

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Solving

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and

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is

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protected

the

(including

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work

student

the

u = 6.41° or 77.9 °






Ans.

12–99. Measurements of a shot recorded on a videotape during a basketball game are shown. The ball passed through the hoop even though it barely cleared the hands of the player B who attempted to block it. Neglecting the size of the ball, determine the magnitude vA of its initial velocity and the height h of the ball when it passes over player B.

C 30

vA

B h

A 7 ft

25 ft

SOLUTION + ) (:

s = s0 + v0t 30 = 0 + vA cos 30° tAC

(+ c )

s = s0 + v0t +

1 2 at 2 c

10 = 7 + vA sin 30° tAC -

1 (32.2)(t2AC) 2

Solving vA = 36.73 = 36.7 ft>s

or

Ans.

teaching

Web)

laws

tAC = 0.943 s + ) (:

Wide

copyright

in

s = s0 + v0t

not

instructors

States

World

permitted.

Dissemination

25 = 0 + 36.73 cos 30° tAB

United

on

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is

of

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1 2 act 2

and

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s = s0 + v0t +

the

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the

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for

work

student

1 (32.2)(t2AB) 2

h = 7 + 36.73 sin 30° tAB -

courses

part

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This

any

Ans.

sale

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of

h = 11.5 ft

and

tAB = 0.786 s

provided

integrity

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this

Solving






5 ft

10 ft

*12–100. It is observed that the skier leaves the ramp A at an angle uA = 25° with the horizontal. If he strikes the ground at B, determine his initial speed vA and the time of flight tAB.

vA

uA A 4m

3

5 4

SOLUTION + B A:

100 m

s = v0 t 4 100 a b = vA cos 25°tAB 5

A+cB

s = s0 + v0 t +

B

1 ac t2 2

1 3 -4 - 100 a b = 0 + vA sin 25°tAB + ( - 9.81)t2AB 5 2

Ans.

tAB = 4.54 s

Ans.

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States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

vA = 19.4 m>s

or

Solving,






12–101. It is observed that the skier leaves the ramp A at an angle uA = 25° with the horizontal. If he strikes the ground at B, determine his initial speed vA and the speed at which he strikes the ground.

vA

uA A 4m

3

5 4

100 m

SOLUTION Coordinate System: x - y coordinate system will be set with its origin to coincide with point A as shown in Fig. a.

B

4 x-motion: Here, xA = 0, xB = 100 a b = 80 m and (vA)x = vA cos 25°. 5 + B A:

xB = xA + (vA)xt 80 = 0 + (vA cos 25°)t 80 t = vA cos 25°

or

(1)

the

not

instructors

States

1 a t2 2 y

on

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is

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yB = yA + (vA)y t +

use

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A+ c B

World

permitted.

Dissemination

Wide

copyright

in

teaching

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laws

3 y-motion: Here, yA = 0, yB = - [4 + 100 a b ] = - 64 m and (vA)y = vA sin 25° 5 and ay = - g = - 9.81 m>s2.

work

student

the

by

and

1 (- 9.81)t2 2

(including

for

- 64 = 0 + vA sin 25° t +

(2)

assessing

is

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solely

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protected

the

4.905t2 - vA sin 25° t = 64

provided

integrity

of

and

work

this

Substitute Eq. (1) into (2) yieldS

2 80 80 ≤ = vA sin 25° ¢ ≤ = 64 vA cos 25° yA cos 25°

2 80 ≤ = 20.65 vA cos 25°

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4.905 ¢

80 = 4.545 vA cos 25° vA = 19.42 m>s = 19.4 m>s Substitute this result into Eq. (1), t =






80 = 4.54465 19.42 cos 25°

Ans.

12–101. continued Using this result,

A+ c B

(vB)y = (vA)y + ay t = 19.42 sin 25° + ( -9.81)(4.5446) = - 36.37 m>s = 36.37 m>s T

And + B A:

(vB)x = (vA)x = vA cos 25° = 19.42 cos 25° = 17.60 m>s :

Thus, vB = 2(vB)2x + (vB)2y = 236.372 + 17.602 Ans.

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Dissemination

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laws

or

= 40.4 m>s






12–102. A golf ball is struck with a velocity of 80 ft>s as shown. Determine the distance d to where it will land. vA

80 ft/s B A

45

10 d

SOLUTION Horizontal Motion: The horizontal component of velocity is (v0)x = 80 cos 55° = 45.89 ft>s.The initial and final horizontal positions are (s0)x = 0 and sx = d cos 10°, respectively. + B A:

sx = (s0)x + (v0)x t d cos 10° = 0 + 45.89t

(1)

Vertical Motion: The vertical component of initial velocity is (v0)y = 80 sin 55° = 65.53 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = d sin 10°, respectively. 1 (a ) t2 2 cy 1 d sin 10° = 0 + 65.53t + ( - 32.2)t2 2

laws

or

sy = (s0)y + (v0)y t +

in

teaching

Web)

(2) copyright

(+ c )

permitted.

Dissemination

Wide

Solving Eqs. (1) and (2) yields Ans.

of

the

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instructors

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d = 166 ft

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the

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work

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by

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United

on

learning.

is

t = 3.568 s






12–103. The ball is thrown from the tower with a velocity of 20 ft/s as shown. Determine the x and y coordinates to where the ball strikes the slope. Also, determine the speed at which the ball hits the ground.

y 20 ft/s 5

4

3

SOLUTION

80 ft 1

Assume ball hits slope. + B A:

2

s = s0 + v0 t x = 0 +

3 (20)t = 12t 5

x 20 ft

A+cB

s = s0 + v0 t + y = 80 +

1 2 a t 2 c

4 1 (20)t + ( -32.2)t2 = 80 + 16t - 16.1t2 5 2

teaching

Web)

laws

or

Equation of slope: y - y1 = m(x - x1) in

1 (x - 20) 2

Dissemination

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copyright

y - 0 =

learning.

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States

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y = 0.5x - 10

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Thus,

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student

the

80 + 16t - 16.1t2 = 0.5(12t) - 10

assessing

is

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protected

the

16.1t2 - 10t - 90 = 0

provided

integrity

of

and

work

this

Choosing the positive root:

and

any

courses

part

the

is

This

t = 2.6952 s

Ans.

Since 32.3 ft 7 20 ft, assumption is valid.

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destroy

of

x = 12(2.6952) = 32.3 ft

y = 80 + 16(2.6952) - 16.1(2.6952)2 = 6.17 ft 3 (20) = 12 ft>s 5

+ B A:

vx = (v0)x =

A+cB

vy = (v0)y + act =

v =

4 (20) + (- 32.2)(2.6952) = -70.785 ft>s 5

(12)2 + (-70.785)2 = 71.8 ft s






Ans.

Ans.

*12–104. The projectile is launched with a velocity v0. Determine the range R, the maximum height h attained, and the time of flight. Express the results in terms of the angle u and v0. The acceleration due to gravity is g.

y

v0 u

h x R

SOLUTION + B A:

s = s0 + v0 t R = 0 + (v0 cos u)t s = s0 + v0 t +

1 a t2 2 c

0 = 0 + (v0 sin u) t +

1 R (g) ¢ ≤ 2 v0 cos u laws

0 = v0 sin u -

1 ( - g)t2 2

or

A+cB

teaching

Web)

v20 sin 2u g

in

Ans. Wide

copyright

R =

the

not

instructors

States

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Dissemination

v20 (2 sin u cos u) R = v0 cos u v0 g cos u United

on

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t =

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2v0 sin u g

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the

v2 = v20 + 2ac(s - s0) is

A+cB

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=

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is sale

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v20 sin2 u 2g

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h =

provided

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assessing

0 = (v0 sin u)2 + 2(- g)(h - 0)






Ans.

12–105. Determine the horizontal velocity vA of a tennis ball at A so that it just clears the net at B. Also, find the distance s where the ball strikes the ground.

vA B

s

21 ft

SOLUTION Vertical Motion: The vertical component of initial velocity is (v0)y = 0. For the ball to travel from A to B, the initial and final vertical positions are (s0)y = 7.5 ft and sy = 3 ft, respectively.

A+cB

sy = (s0)y + (v0)y t + 3 = 7.5 + 0 +

1 (a ) t2 2 cy

1 ( -32.2)t 21 2

t1 = 0.5287 s

laws

or

For the ball to travel from A to C, the initial and final vertical positions are (s0)y = 7.5 ft and sy = 0, respectively. Web) teaching

1 ( -32.2)t22 2

Wide

permitted.

Dissemination

0 = 7.5 + 0 +

1 (a ) t2 2 cy

in

sy = (s0)y + (v0)y t +

copyright

A+cB

on

learning.

is

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t2 = 0.6825 s

protected

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Horizontal Motion: The horizontal component of velocity is (v0)x = vA. For the ball to travel from A to B, the initial and final horizontal positions are (s0)x = 0 and sx = 21 ft, respectively. The time is t = t1 = 0.5287 s. work

solely

of

sx = (s0)x + (v0)x t

assessing

is

+ B A;

provided

integrity

of

and

work

this

21 = 0 + vA (0.5287)

Ans.

and

any

courses

part

the

is

This

vA = 39.72 ft>s = 39.7 ft>s

sale

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For the ball to travel from A to C, the initial and final horizontal positions are (s0)x = 0 and sx = (21 + s) ft, respectively. The time is t = t2 = 0.6825 s. + B A;

sx = (s0)x + (v0)x t 21 + s = 0 + 39.72(0.6825) s = 6.11 ft






7.5 ft 3 ft

C

Ans.

A

12–106. The ball at A is kicked with a speed vA = 8 ft > s and at an angle uA = 30°. Determine the point (x, -y) where it strikes the ground. Assume the ground has the shape of a parabola as shown.

y vA A

θA

x –y

SOLUTION

B y = –0.04x 2

(vA)x = 8 cos 30° = 6.928 ft>s

x

(vA)y = 8 sin 30° = 4 ft > s + )s = s + v t (: 0 0 x = 0 + 6.928 t ( + c ) s = s0 + v0 t +

1 2 a t 2 c

1 ( -32.2)t2 2

(2) or

y = 0 + 4t +

(1)

in

teaching

Web)

laws

y = - 0.04 x2

Wide

copyright

From Eqs. (1) and (2):

instructors

States

World

permitted.

Dissemination

y = 0.5774 x - 0.3354 x2

use

United

on

learning.

is

of

the

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- 0.04 x2 = 0.5774x - 0.3354 x2

for

work

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by

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0.2954 x2 = 0.5774x

Ans.

is

work

solely

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the

(including

x = 1.95 ft

and

work

this

assessing

Thus,

integrity

of

y = - 0.04(1.954)2 = - 0.153 ft

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will

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and

any

courses

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is

This

provided

Ans.






12–107. The ball at A is kicked such that uA = 30°. If it strikes the ground at B having coordinates x = 15 ft, y = - 9 ft, determine the speed at which it is kicked and the speed at which it strikes the ground.

y vA A

θA

x –y

SOLUTION

B y = –0.04x 2

+ )s = s + v t (: 0 0

x

15 = 0 + vA cos 30° t ( + c ) s = s0 + v0 t +

1 2 a t 2 c

-9 = 0 + vA sin 30° t +

1 ( - 32.2)t2 2

vA = 16.5 ft>s

Ans.

t = 1.047 s

laws

or

+ ) (v ) = 16.54 cos 30° = 14.32 ft>s (: B x

in

teaching

Web)

(+ c ) v = v0 + ac t

Dissemination

Wide

copyright

(vB)y = 16.54 sin 30° + ( - 32.2)(1.047)

of

the

not

instructors

States

World

permitted.

= - 25.45 ft>s

Ans.

United

on

learning.

is

(14.32)2 + (- 25.45)2 = 29.2 ft s

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vB =






*12–108. The man at A wishes to throw two darts at the target at B so that they arrive at the same time. If each dart is thrown with a speed of 10 m>s, determine the angles uC and uD at which they should be thrown and the time between each throw. Note that the first dart must be thrown at uC 1 7uD2, then the second dart is thrown at uD .

5m uC uD

A

s = s0 + v0t 5 = 0 + (10 cos u) t

(+ c )

(1)

v = v0 + act

-10 sin u = 10 sin u - 9.81 t t =

2(10 sin u) = 2.039 sin u 9.81

From Eq. (1), laws

or

5 = 20.39 sin u cos u in

teaching

Web)

sin 2u = 2 sin u cos u

Wide

copyright

Since

States

World

permitted.

Dissemination

sin 2u = 0.4905 instructors

The two roots are uD = 14.7° United

on

learning.

is

of

the

not

Ans. Ans.

work

student

the

by

and

use

uC = 75.3°

of

protected

the

(including

for

From Eq. (1): tD = 0.517 s

work

this

assessing

is

work

solely

tC = 1.97 s

sale

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of

and

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part

the

is

This

provided

integrity

of

and

So that ¢t = tC - tD = 1.45 s






D B

SOLUTION + ) (:

C

Ans.

12–109. A boy throws a ball at O in the air with a speed v0 at an angle u1. If he then throws another ball with the same speed v0 at an angle u2 6 u1, determine the time between the throws so that the balls collide in mid air at B.

B O

u1

u2

y

SOLUTION Vertical Motion: For the first ball, the vertical component of initial velocity is (v0)y = v0 sin u1 and the initial and final vertical positions are (s0)y = 0 and sy = y, respectively. (+ c)

sy = (s0)y + (v0)y t +

x

1 (a ) t2 2 cy

y = 0 + v0 sin u1t1 +

1 ( -g)t21 2

(1)

For the second ball, the vertical component of initial velocity is (v0)y = v0 sin u2 and the initial and final vertical positions are (s0)y = 0 and sy = y, respectively. 1 (a ) t2 2 cy

y = 0 + v0 sin u2t2 +

1 ( -g)t22 2

(2)

or

sy = (s0)y + (v0)y t +

laws

(+ c)

the

not

instructors

States

sx = (s0)x + (v0)x t

learning.

is

of

+ B A:

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

Horizontal Motion: For the first ball, the horizontal component of initial velocity is (v0)x = v0 cos u1 and the initial and final horizontal positions are (s0)x = 0 and sx = x, respectively.

(3)

student

the

by

and

use

United

on

x = 0 + v0 cos u1 t1

is

work

solely

of

protected

the

(including

for

work

For the second ball, the horizontal component of initial velocity is (v0)x = v0 cos u2 and the initial and final horizontal positions are (s0)x = 0 and sx = x, respectively. this

assessing

sx = (s0)x + (v0)x t

(4)

and

any

courses

part

the

is

This

x = 0 + v0 cos u2 t2

provided

integrity

of

and

work

+ B A:

cos u1 t cos u2 1

sale

t2 =

will

their

destroy

of

Equating Eqs. (3) and (4), we have

(5)

Equating Eqs. (1) and (2), we have v0 t1 sin u1 - v0 t2 sin u2 =

1 g A t21 - t22 B 2

(6)

Solving Eq. [5] into [6] yields t1 = t2 =

2v0 cos u2 sin(u1 - u2) g(cos2 u2 - cos2 u1) 2v0 cos u1 sin(u1 - u2) g(cos2 u2 - cos2u1)

Thus, the time between the throws is ¢t = t1 - t2 = =






2v0 sin(u1 - u2)(cos u2 - cos u1) g(cos2 u2 - cos2 u1) 2v0 sin (u1 - u2) g(cos u2 + cos u1)

Ans.

12–110. Small packages traveling on the conveyor belt fall off into a l-m-long loading car. If the conveyor is running at a constant speed of vC = 2 m>s, determine the smallest and largest distance R at which the end A of the car may be placed from the conveyor so that the packages enter the car.

vc

2 m/s

30 3m A

B

SOLUTION R

Vertical Motion: The vertical component of initial velocity is (v0)y = 2 sin 30° = 1.00 m>s. The initial and final vertical positions are (s0)y = 0 and sy = 3 m, respectively.

A+TB

sy = (s0)y + (v0)y t + 3 = 0 + 1.00(t) +

1 (a ) t2 2 cy

1 (9.81) A t2 B 2

Choose the positive root t = 0.6867 s

or

Horizontal Motion: The horizontal component of velocity is (v0)x = 2 cos 30° = 1.732 m>s and the initial horizontal position is (s0)x = 0. If sx = R, then + B A:

teaching

Web)

laws

sx = (s0)x + (v0)x t in

R = 0 + 1.732(0.6867) = 1.19 m

Dissemination

Wide

copyright

Ans.

United

on

learning.

is

of

the

sx = (s0)x + (v0)x t

and

use

+ B A:

not

instructors

States

World

permitted.

If sx = R + 1, then

for

work

student

the

by

R + 1 = 0 + 1.732(0.6867)

Ans.

assessing

is

work

solely

of

protected

the

(including

R = 0.189 m

Ans.

sale

will

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destroy

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and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

Thus, Rmin = 0.189 m, Rmax = 1.19 m






1m

12–111. The fireman wishes to direct the flow of water from his hose to the fire at B. Determine two possible angles u1 and u2 at which this can be done. Water flows from the hose at vA = 80 ft>s.

A u vA 20 ft B

SOLUTION + B A:

s = s0 + v0 t 35 ft

35 = 0 + (80)(cos u )t

A+cB

s = s0 + v0 t +

1 2 act 2

-20 = 0 - 80 (sin u)t +

1 (-32.2)t 2 2

Thus, or

0.1914 0.4375 t + 16.1 ¢ ≤ cos u cos2 u

laws

20 = 80 sin u

Wide

copyright

in

teaching

Web)

20 cos2 u = 17.5 sin 2u + 3.0816 Dissemination

Solving, (below the horizontal)

u2 = 85.2°

(above the horizontal)

permitted.

u1 = 24.9°

is

of

the

not

instructors

States

World

Ans.

sale

will

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destroy

of

and

any

courses

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is

This

provided

integrity

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and

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this

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is

work

solely

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protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

Ans.






*12–112. The baseball player A hits the baseball at vA = 40 ft>s and uA = 60° from the horizontal. When the ball is directly overhead of player B he begins to run under it. Determine the constant speed at which B must run and the distance d in order to make the catch at the same elevation at which the ball was hit.

vA = 40 ft/s

θA

A

15 ft

SOLUTION Vertical Motion: The vertical component of initial velocity for the football is (v0)y = 40 sin 60° = 34.64 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = 0, respectively. (+ c )

sy = (s0)y + (v0)y t + 0 = 0 + 34.64t +

1 (a ) t2 2 cy

1 ( -32.2) t2 2

t = 2.152 s

in

teaching

Web)

laws

or

Horizontal Motion: The horizontal component of velocity for the baseball is (v0)x = 40 cos 60° = 20.0 ft>s. The initial and final horizontal positions are (s0)x = 0 and sx = R, respectively. sx = (s0)x + (v0)x t

Dissemination

Wide

copyright

+ ) (:

is

of

the

not

instructors

States

World

permitted.

R = 0 + 20.0(2.152) = 43.03 ft

by

and

use

United

on

learning.

The distance for which player B must travel in order to catch the baseball is Ans.

the

(including

for

work

student

the

d = R - 15 = 43.03 - 15 = 28.0 ft

work

this

assessing

is

work

solely

of

protected

Player B is required to run at a same speed as the horizontal component of velocity of the baseball in order to catch it. integrity

of

and

vB = 40 cos 60° = 20.0 ft s

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

Ans.






B

C

vA

d

12–113. The man stands 60 ft from the wall and throws a ball at it with a speed v0 = 50 ft>s. Determine the angle u at which he should release the ball so that it strikes the wall at the highest point possible. What is this height? The room has a ceiling height of 20 ft.

v0

50 ft/s u

h

5 ft 60 ft

SOLUTION vx = 50 cos u s = s0 + v0t x = 0 + 50 cos u t (+ c)

(1)

v = v0 + act vy = 50 sin u - 32.2 t

(+ c)

s = s0 + v0t +

(2)

1 2 at 2 c

y = 0 + 50 sin u t - 16.1 t2

(3) laws

or

+ ) (:

teaching

Web)

v2 = n20 + 2ac(s - s0) in

(+ c)

(4) is

of

the

not

instructors

States

World

v2y = 2500 sin2 u - 64.4 s

by

and

use

United

on

learning.

Require vy = 0 at s = 20 - 5 = 15 ft

(including

for

work

student

the

0 = 2500 sin2 u - 64.4 (15)

Ans.

assessing

is

work

solely

of

protected

the

u = 38.433° = 38.4°

part

the

is

destroy

of

and

From Eq. (1)

sale

will

their

t = 0.9652 s

any

courses

0 = 50 sin 38.433° - 32.2 t

This

provided

integrity

of

and

work

this

From Eq. (2)

x = 50 cos 38.433°(0.9652) = 37.8 ft Time for ball to hit wall From Eq. (1), 60 = 50(cos 38.433°)t t = 1.53193 s From Eq. (3) y = 50 sin 38.433°(1.53193) - 16.1(1.53193)2 y = 9.830 ft h = 9.830 + 5 = 14.8 ft






Ans.

permitted.

Dissemination

Wide

copyright

v2y = (50 sin u)2 + 2(-32.2)(s - 0)

20

12–114. A car is traveling along a circular curve that has a radius of 50 m. If its speed is 16 m>s and is increasing uniformly at 8 m>s2, determine the magnitude of its acceleration at this instant.

SOLUTION v = 16 m>s at = 8 m>s2 r = 50 m an =

(16)2 v2 = = 5.12 m>s2 r 50

a = 2(8)2 + (5.12)2 = 9.50 m>s2

sale

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on

learning.

is

of

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not

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States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






12–115. Determine the maximum constant speed a race car can have if the acceleration of the car cannot exceed 7.5 m>s2 while rounding a track having a radius of curvature of 200 m.

SOLUTION Acceleration: Since the speed of the race car is constant, its tangential component of acceleration is zero, i.e., at = 0. Thus, a = an = 7.5 =

v2 r

v2 200

v = 38.7 m>s

sale

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and

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is

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provided

integrity

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of

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the

(including

for

work

student

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by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






*12–116. A car moves along a circular track of radius 250 ft such that its speed for a short period of time, 0 … t … 4 s, is v = 31t + t22 ft>s, where t is in seconds. Determine the magnitude of its acceleration when t = 3 s. How far has it traveled in t = 3 s?

SOLUTION v = 3(t + t 2) at =

dv = 3 + 6t dt

When t = 3 s, an =

at = 3 + 6(3) = 21 ft>s2

[3(3 + 32)]2 = 5.18 ft>s2 250

a = 2(21)2 + (5.18)2 = 21.6 ft>s2

Ans.

3 or

31t + t22dt

in

teaching

Web)

3 32 t + t3 ` 2 0

Wide

Ans.

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and

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is

This

provided

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of

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student

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on

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is

of

the

not

instructors

States

World

¢s = 40.5 ft






permitted.

Dissemination

¢s =

L0

laws

ds =

copyright

L

12–117. A car travels along a horizontal circular curved road that has a radius of 600 m. If the speed is uniformly increased at a rate of 2000 km>h2, determine the magnitude of the acceleration at the instant the speed of the car is 60 km>h.

SOLUTION at = ¢

2 1h 2000 km 1000 m ba b = 0.1543 m>s2 ba 2 1 km 3600 s h

v = a an =

1h 60 km 1000 m ba ba b = 16.67 m>s h 1 km 3600 s

v2 16.672 = 0.4630 m>s2 = r 600

a = 2a2t + a2n = 20.15432 + 0.46302 = 0.488 m>s2

sale

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destroy

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and

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student

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on

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of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






12–118. The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m>s. For a short distance from s = 0, its # speed is increased by v = 10.05s2 m>s2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.

. v v

50 m

SOLUTION v dv = at ds v

L4

10

v dv =

L0

0.5v 2 - 8 =

0.05s ds

0.05 (10)2 2

v = 4.583 = 4.58 m>s an =

Ans.

v 2 (4.583)2 = 0.420 m>s2 = r 50 laws

or

at = 0.05(10) = 0.5 m>s2

Web)

a = 2(0.420)2 + (0.5)2 = 0.653 m>s2

will

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the

(including

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student

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by

and

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on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Ans.






(0.05s) m/s2 4 m/s

12–119. The automobile is originally at rest at s = 0. If its speed is # increased by v = 10.05t22 ft>s2, where t is in seconds, determine the magnitudes of its velocity and acceleration when t = 18 s.

300 ft s

240 ft

SOLUTION at = 0.05t2 v

t

dv =

L0

0.05 t2 dt

L0

v = 0.0167 t3 s

L0

t

ds =

L0

0.0167 t3 dt

s = 4.167(10 - 3) t4 s = 437.4 ft laws

or

When t = 18 s,

in

teaching

Web)

Therefore the car is on a curved path.

Dissemination

Wide

Ans.

copyright

v = 0.0167(183) = 97.2 ft>s

the

not

instructors

States

World

permitted.

(97.2)2 = 39.37 ft>s2 240

on

learning.

is

of

an =

the

by

and

use

United

at = 0.05(182) = 16.2 ft/s2

the

(including

for

work

student

a = 2(39.37)2 + (16.2)2

sale

will

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destroy

of

and

any

courses

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the

is

This

provided

integrity

of

and

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assessing

is

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solely

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protected

a = 42.6 ft>s2






Ans.

*12–120. The automobile is originally at rest s = 0. If it then starts to # increase its speed at v = 10.05t22 ft>s2, where t is in seconds, determine the magnitudes of its velocity and acceleration at s = 550 ft.

300 ft s

240 ft

SOLUTION The car is on the curved path. at = 0.05 t2 v

t

dv =

L0

0.05 t2 dt

L0

v = 0.0167 t3 s

L0

t

ds =

L0

0.0167 t3 dt

or

s = 4.167(10 - 3) t4

teaching

Web)

laws

550 = 4.167(10-3) t4

Dissemination

Wide

copyright

in

t = 19.06 s

not

instructors

States

World

permitted.

So that

use

United

on

learning.

is

of

the

v = 0.0167(19.06)3 = 115.4

and

v = 115 ft>s

for

work

student

the

by

Ans.

work

solely

of

protected

the

(including

(115.4)2 = 55.51 ft>s2 240

assessing

is

an =

part

the

is sale

will

their

destroy

of

and

any

courses

a = 2(55.51)2 + (18.17)2 = 58.4 ft>s2

This

provided

integrity

of

and

work

this

at = 0.05(19.06)2 = 18.17 ft>s2






Ans.

12–121. When the roller coaster is at B, it has a speed of 25 m>s, which is increasing at at = 3 m>s2. Determine the magnitude of the acceleration of the roller coaster at this instant and the direction angle it makes with the x axis.

y

y

1 x2 100

A

s

SOLUTION

B

Radius of Curvature:

x

1 2 y = x 100 dy 1 = x dx 50 1 50

dx2

=

2 3>2 1 xb R 50

5

2 1 2 50

2

= 79.30 m x = 30 m in

2

d2y

B1 + a

or

r =

dy 2 3>2 b R dx

Web)

B1 + a

laws

=

dx2

teaching

d2y

30 m

Dissemination

Wide

copyright

Acceleration:

not

instructors

States

World

permitted.

# a t = v = 3 m>s2

by

and

use

United

on

learning.

is

of

the

vB 2 252 = 7.881 m>s2 = r 79.30

the

an =

the

(including

for

work

student

The magnitude of the roller coaster’s acceleration is

Ans.

assessing

is

work

solely

of

protected

a = 2at 2 + an 2 = 232 + 7.8812 = 8.43 m>s2

dy 1 2 ≤ = tan-1 c A 30 B d = 30.96°. dx x = 30 m 50

part

the

is

This

provided

integrity

of

and

work

this

The angle that the tangent at B makes with the x axis is f = tan-1 ¢

and

any

courses

As shown in Fig. a, an is always directed towards the center of curvature of the path. Here, destroy

of

an 7.881 b = 69.16°. Thus, the angle u that the roller coaster’s acceleration makes b = tan-1 a at 3 sale

will

their

a = tan-1 a

with the x axis is u = a - f = 38.2° b






Ans.

12–122. If the roller coaster starts from rest at A and its speed increases at at = (6 – 0.06s) m>s2, determine the magnitude of its acceleration when it reaches B where sB = 40 m.

y

y

1 x2 100

A

s

SOLUTION

B

Velocity: Using the initial condition v = 0 at s = 0,

x

v dv = at ds v

L0

s

vdv =

L0

A 6 - 0.06s B ds

30 m

v = a 212s - 0.06s2 b m>s

(1)

Thus, vB = 412 A 40 B - 0.06 A 40 B 2 = 19.60 m>s laws

or

Radius of Curvature: 1 2 x 100 dy 1 = x dx 50

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

y =

United

on

learning.

is

of

the

1 50

and

use

=

the

dx2

by

d2y

work

student

the

(including

for

of

= 79.30 m this

assessing

2

5

work

solely

protected

is

2 1 2 50

integrity

of

dx2

=

x = 30 m

and

any

courses

part

the

is

This

provided

2

d2y

2 3>2 1 xb R 50

B1 + a

and

r =

dy 2 3>2 b R dx

work

B1 + a

destroy

of

Acceleration: sale

will

their

# a t = v = 6 - 0.06(40) = 3.600 m>s2 an =

19.602 v2 = 4.842 m>s2 = r 79.30

The magnitude of the roller coaster’s acceleration at B is a = 2at 2 + an 2 = 23.6002 + 4.8422 = 6.03 m>s2






Ans.

12–123. The speedboat travels at a constant speed of 15 m> s while making a turn on a circular curve from A to B. If it takes 45 s to make the turn, determine the magnitude of the boat’s acceleration during the turn.

A r

SOLUTION Acceleration: During the turn, the boat travels s = vt = 15(45) = 675 m. Thus, the 675 s radius of the circular path is r = = m. Since the boat has a constant speed, p p at = 0. Thus, v2 = r

152 = 1.05 m>s2 675 a b p

Ans.

sale

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is

This

provided

integrity

of

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this

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is

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solely

of

protected

the

(including

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work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

a = an =






B

*12–124. The car travels along the circular path such that its speed is increased by a t = (0.5et) m>s2, where t is in seconds. Determine the magnitudes of its velocity and acceleration after the car has traveled s = 18 m starting from rest. Neglect the size of the car.

s

SOLUTION v

L0

t

dv =

L0

0.5e t dt ρ

v = 0.5(e t - 1) t

18

L0

ds = 0.5

L0

(e t - 1)dt

18 = 0.5(e t - t - 1) Solving, t = 3.7064 s v = 0.5(e 3.7064 - 1) = 19.85 m>s = 19.9 m>s # at = v = 0.5e t ƒ t = 3.7064 s = 20.35 m>s2

in

teaching

Web)

laws

or

Ans.

19.852 v2 = 13.14 m>s2 = r 30

World

permitted.

Dissemination

Wide

copyright

an =

States

a = 2a2t + a2n = 220.352 + 13.142 = 24.2 m>s2

sale

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destroy

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and

any

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is

This

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on

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is

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not

instructors

Ans.






30 m

18 m

12–125. The car passes point A with a speed of 25 m>s after which its speed is defined by v = (25 - 0.15s) m>s. Determine the magnitude of the car’s acceleration when it reaches point B, where s = 51.5 m.

y y

B 16 m

Velocity: The speed of the car at B is vB = C 25 - 0.15 A 51.5 B D = 17.28 m>s Radius of Curvature: 1 2 x 625

dy = -3.2 A 10-3 B x dx = - 3.2 A 10-3 B

dx2

or laws

4

2 - 3.2 A 10-3 B 2

Web)

= 324.58 m

teaching

=

2

Wide

d2y

2 3>2

c 1 + a -3.2 A 10-3 B x b d

x = 50 m instructors

States

World

permitted.

2

3>2

in

r =

dy 2 b dx

Dissemination

B1 + a

copyright

dx2

B

d2y

on

learning.

is

of

the

not

Acceleration:

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

vB 2 17.282 = 0.9194 m>s2 = r 324.58 dv = A 25 - 0.15s B A - 0.15 B = A 0.225s - 3.75 B m>s2 at = v ds an =

integrity

of

and

work

this

assessing

When the car is at B A s = 51.5 m B

and

any

courses

part

the

is

This

provided

a t = C 0.225 A 51.5 B - 3.75 D = - 2.591 m>s2

their

destroy

of

Thus, the magnitude of the car’s acceleration at B is sale

will

a = 2a2t + a2n = 2( - 2.591)2 + 0.91942 = 2.75 m>s2






Ans.

1 2 x 625 s A x

SOLUTION

y = 16 -

16

12–126. y

If the car passes point A with a speed of 20 m>s and begins to increase its speed at a constant rate of at = 0.5 m>s2, determine the magnitude of the car’s acceleration when s = 100 m.

y

B 16 m

Velocity: The speed of the car at C is vC 2 = vA 2 + 2a t (sC - sA) vC 2 = 202 + 2(0.5)(100 - 0) vC = 22.361 m>s Radius of Curvature: 1 2 x 625

laws

teaching

Web)

= -3.2 A 10-3 B

Wide

copyright

dx2

in

d2y

or

dy = -3.2 A 10-3 B x dx

permitted. United

on

learning.

is

of

the

not

instructors

= 312.5 m use

2

Dissemination

World States

4

- 3.2 A 10-3 B

x=0

and

dx2

=

for

work

student

2

d2y

2 3>2

c1 + a -3.2 A 10-3 B x b d

by

r =

dy 2 3>2 b R dx

the

B1 + a

of

protected

the

(including

Acceleration: work this

assessing

is

integrity

of

and

work

provided

vC 2 22.3612 = = 1.60 m>s2 r 312.5

their

destroy

The magnitude of the car’s acceleration at C is

of

and

any

courses

part

the

is

This

an =

solely

# a t = v = 0.5 m>s

sale

will

a = 2a2t + a2n = 20.52 + 1.602 = 1.68 m>s2






Ans.

1 2 x 625 s A x

SOLUTION

y = 16 -

16

12–127. A train is traveling with a constant speed of 14 m/s along the curved path. Determine the magnitude of the acceleration of the front of the train, B, at the instant it reaches point A 1y = 02.

y (m)

10 m

( —y )

x = 10 e 15

x (m)

SOLUTION

A vt = 14 m/s

y

x = 10e(15) y = 15 lna

x b 10

dy 10 1 15 = 15 a b a b = x x dx 10 d2y

= -

dx2

15 x2

At x = 10, laws

Web)

= 39.06 m

Wide

|- 0.15|

teaching

=

World

permitted.

dx2

2

in

d2y

3

Dissemination

2

C 1 + (1.5)2 D 2

copyright

r =

dy 2 2 b R dx

or

3

B1 + a

and

use

United

on

learning.

is

of

the

not

instructors

States

dv = 0 dt

at =

work

student

the

by

(14)2 v2 = = 5.02 m s2 r 39.06

sale

will

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destroy

of

and

any

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is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

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the

(including

for

an = a =






Ans.

B

*12–128. When a car starts to round a curved road with the radius of curvature of 600 ft, it is traveling at 75 ft>s . If the car’s # speed begins to decrease at a rate of v = ( - 0.06t2) ft>s2 , determine the magnitude of the acceleration of the car when it has traveled a distance of s = 700 ft. s

SOLUTION Velocity: Using the initial condition v = 75 ft>s when t = 0 s, L

dt =

L

atdt

v

t

dv =

Lv = 75 ft>s

L0

-0.06t2dt

v = 175 - 0.02t3) ft>s

or

Position: Using the initial condition s = 0 at t = 0 s, laws

ds = vdt L0

in

teaching

Web)

175 - 0.02t32dt

Wide

L0

t

ds =

copyright

s

of

the

not

instructors

States

World

permitted.

Dissemination

s = 375t - 0.005t44 ft

and

use

United

on

learning.

is

At s = 700 ft,

for

work

student

the

by

700 = 75t - 0.005t4

solely

of

protected

the

(including

Solving the above equation by trial and error,

and

destroy

of

and

sale

will

v2 552 = = 5.042 ft>s2 r 600

their

an =

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

t = 10 s and t = 20 s. Pick the first solution. # Acceleration: When t = 10 s, at = v = - 0.0611022 = - 6 ft>s2 v = 75 - 0.0211032 = 55 ft>s

Thus, the magnitude of the truck’s acceleration is a = 2at 2 + an2 = 21-622 + 5.0422 = 7.84 ft>s2






Ans.

r ⫽ 600 ft

12–129. When the motorcyclist is at A, he increases his speed along # the vertical circular path at the rate of v = 10.3t2 ft>s2, where t is in seconds. If he starts from rest at A, determine the magnitudes of his velocity and acceleration when he reaches B.

300 ft

A

60°

300 ft

SOLUTION v

L0

B

t

dv =

L0

0.3tdt

v = 0.15t2 s

L0

t

ds =

L0

0.15t2 dt

s = 0.05t3 When s = p3 (300) ft,

p 3 (300)

= 0.05t3

t = 18.453 s

v = 0.15(18.453)2 = 51.08 ft>s = 51.1 ft>s

or

Ans.

in

teaching

Web)

laws

# at = v = 0.3t|t = 18.453 s = 5.536 ft>s2 v2 51.082 = = 8.696 ft>s2 r 300 States

World

permitted.

Dissemination

Wide

copyright

an =

the

Ans. is

on

learning. United

and

use by

work

student

the

the

(including

for

work

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of

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is

integrity

of

and

work

provided

any

courses

part

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is

This

destroy

of

and

will

their






not

instructors

(5.536)2 + (8.696)2 = 10.3 ft s2

of

a2t + a2n =

a =

12–130. When the motorcyclist is at A, he increases his speed along # the vertical circular path at the rate of v = (0.04s) ft>s2 where s is in ft. If he starts at vA = 2 ft>s where s = 0 at A, determine the magnitude of his velocity when he reaches B. Also, what is his initial acceleration?

300 ft

A

SOLUTION

v dv =

L

at ds

v

s

v dv =

L2

L0

0.04s ds s

v

v2 2 = 0.0252 2 2 2 0

teaching

Web)

laws

or

v2 - 2 = 0.0252 2

Wide

copyright

in

v2 = 0.0452 + 4 = 0.041s2 + 1002

permitted.

Dissemination

v = 0.2 2s2 + 100

use

United

on

learning.

is

of

the

not

instructors

States

World

At B, s = ru = 300 A p3 B = 100p ft. Thus

and

v2

Ans. student

the

by

= 0.2 2(100p22 + 100 = 62.9 ft>s

(including

for

work

s = 100p ft

is

work

solely

of

protected

the

Acceleration: At t = 0, s = 0, and v = 2.

integrity

of

and

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# at = v = 0.04 s part

the

is

= 0

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provided

at 2

an 2

= s=0

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1222

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v2 r

courses

s=0

an =

= 0.01333 ft>s2

a = 21022 + 10.0133322 = 0.0133 ft>s2






300 ft

B

# Velocity: At s = 0, v = 2. Here, ac = v = 0.045. Then L

60⬚

12–131. At a given instant the train engine at E has a speed of 20 m>s and an acceleration of 14 m>s2 acting in the direction shown. Determine the rate of increase in the train’s speed and the radius of curvature r of the path.

v 75 a

r

SOLUTION Ans.

at = 14 cos 75° = 3.62 m>s2 an = 14 sin 75° an =

(20)2 r Ans.

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permitted.

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in

teaching

Web)

laws

or

r = 29.6 m






14 m/s2

E

20 m/s

*12–132. v

Car B turns such that its speed is increased by (at)B = (0.5et) m>s2, where t is in seconds. If the car starts from rest when u = 0°, determine the magnitudes of its velocity and acceleration when the arm AB rotates u = 30°. Neglect the size of the car.

B

SOLUTION Velocity: The speed v in terms of time t can be obtained by applying a =

5m

dv . dt

A

dv = adt v

L0

t

dv =

0.5et dt

L0

v = 0.5 A et - 1 B

(1)

30° pb = 2.618 m. 180° The time required for the car to travel this distance can be obtained by applying When u = 30°, the car has traveled a distance of s = ru = 5a

or

ds . dt

laws

v =

teaching in

Wide

copyright

0.5 A e - 1 B dt

permitted.

Dissemination

t

World

L0

instructors

L0

States

t

ds =

Web)

ds = vdt 2.618 m

and

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on

learning.

is

of

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not

2.618 = 0.5 A et - t - 1 B by

t = 2.1234 s

(including

for

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student

the

Solving by trial and error

is

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solely

of

protected

the

Substituting t = 2.1234 s into Eq. (1) yields

v = 0.5 A e2.1234 - 1 B = 3.680 m>s = 3.68 m>s provided

integrity

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this

assessing

Ans.

of

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any

courses

part

the

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This

Acceleration: The tangential acceleration for the car at t = 2.1234 s is at = 0.5e2.1234 = 4.180 m>s2. To determine the normal acceleration, apply Eq. 12–20. will

their

destroy

3.6802 v2 = = 2.708 m>s2 r 5 sale

an =

The magnitude of the acceleration is a = 2a2t + a2n = 24.1802 + 2.7082 = 4.98 m>s2






Ans.

u

12–133. v

Car B turns such that its speed is increased by (at)B = (0.5et) m>s2, where t is in seconds. If the car starts from rest when u = 0°, determine the magnitudes of its velocity and acceleration when t = 2 s. Neglect the size of the car.

B

SOLUTION

5m

dv Velocity: The speed v in terms of time t can be obtained by applying a = . dt

A

dv = adt v

L0

t

dv =

L0

t

0.5e dt

v = 0.5 A et - 1 B When t = 2 s,

v = 0.5 A e2 - 1 B = 3.195 m>s = 3.19 m>s

Ans.

laws

or

Acceleration: The tangential acceleration of the car at t = 2 s is at = 0.5e2 = 3.695 m>s2. To determine the normal acceleration, apply Eq. 12–20.

in

teaching

Web)

v2 3.1952 = 2.041 m>s2 = r 5 Dissemination

Wide

copyright

an =

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instructors

States

World

permitted.

The magnitude of the acceleration is Ans.

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the

a = 2a2t + a2n = 23.6952 + 2.0412 = 4.22 m>s2






u

12–134. A boat is traveling along a circular curve having a radius of 100 ft. If its speed at t = 0 is 15 ft/s and is increasing at # v = 10.8t2 ft>s2, determine the magnitude of its acceleration at the instant t = 5 s.

SOLUTION v

L15

5

dv =

L0

0.8tdt

v = 25 ft>s an =

v2 252 = = 6.25 ft>s2 r 100

At t = 5 s,

# at = v = 0.8(5) = 4 ft>s2 a2t + a2n =

42 + 6.252 = 7.42 ft s2

Ans.

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copyright

in

teaching

Web)

laws

or

a =






12–135. A boat is traveling along a circular path having a radius of 20 m. Determine the magnitude of the boat’s acceleration when the speed is v = 5 m>s and the rate of increase in the # speed is v = 2 m>s2.

SOLUTION at = 2 m>s2 an =

y2 52 = = 1.25 m>s2 r 20

a = 2a2t + a2n = 222 + 1.252 = 2.36 m>s2

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Dissemination

Wide

copyright

in

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laws

or

Ans.






*■12–136. Starting from rest, a bicyclist travels around a horizontal circular path, r = 10 m, at a speed of v = 10.09t2 + 0.1t2 m>s, where t is in seconds. Determine the magnitudes of his velocity and acceleration when he has traveled s = 3 m.

SOLUTION s

L0

t

ds =

L0

10.09t2 + 0.1t2dt

s = 0.03t3 + 0.05t2 When s = 3 m,

3 = 0.03t3 + 0.05t2

Solving, t = 4.147 s

or

ds = 0.09t2 + 0.1t dt laws

v =

v = 0.09(4.147)2 + 0.1(4.147) = 1.96 m>s

Wide

1.962 v2 = 0.3852 m>s2 = r 10

permitted.

an =

Dissemination

dv = 0.18t + 0.1 ` = 0.8465 m>s2 dt t = 4.147 s

the

by

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Ans.

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(including

for

work

student

a = 2a2t + a2n = 2(0.8465)2 + (0.3852)2 = 0.930 m>s2






Ans.

12–137. A particle travels around a circular path having a radius of 50 m. If it is initially traveling with a speed of 10 m> s and its # speed then increases at a rate of v = 10.05 v2 m>s2, determine the magnitude of the particle’s acceleraton four seconds later.

SOLUTION Velocity: Using the initial condition v = 10 m>s at t = 0 s, dt =

dv a

t

L0

v

dt =

t = 20 ln

L10 m>s

dv 0.05v

v 10

laws

or

v = (10et>20) m>s

in

teaching

Web)

When t = 4 s,

instructors

States

World

permitted.

Dissemination

Wide

copyright

v = 10e4>20 = 12.214 m>s

and

use

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on

learning.

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of

the

not

Acceleration: When v = 12.214 m>s (t = 4 s),

for

work

student

the

by

at = 0.05(12.214) = 0.6107 m>s2

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solely

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the

(including

(12.214)2 v2 = = 2.984 m>s2 r 50

this

assessing

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an =

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is

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work

Thus, the magnitude of the particle’s acceleration is

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a = 2at2 + an2 = 20.61072 + 2.9842 = 3.05 m>s2






Ans.

12–138. When the bicycle passes point A, it has a speed of 6 m> s, # which is increasing at the rate of v = 10.52 m>s2. Determine the magnitude of its acceleration when it is at point A.

y

y ⫽ 12 ln (

x ) 20

A x 50 m

SOLUTION Radius of Curvature: y = 12 ln a

x b 20

dy 1 1 12 = 12 a ba b = x dx x>20 20

` dx2

=

`-

12 ` x2

4

= 226.59 m

Web)

d2y

x = 50 m

Wide

`

12 2 3>2 ≤ R x

B1 + ¢

World

permitted.

Dissemination

r =

dy 2 3>2 b R dx

laws

B1 + a

or

12 x2

teaching

= -

in

dx2

copyright

d2y

learning.

is

of

the

not

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States

Acceleration:

the

by

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on

at = v = 0.5 m>s2

the

(including

for

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student

v2 62 = = 0.1589 m>s2 r 226.59 is

work

solely

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protected

an =

integrity

of

and

work

this

assessing

The magnitude of the bicycle’s acceleration at A is

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and

any

courses

part

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is

This

provided

a = 2at2 + an2 = 20.52 + 0.15892 = 0.525 m>s2






Ans.

12–139. The motorcycle is traveling at a constant speed of 60 km> h. Determine the magnitude of its acceleration when it is at point A.

y

y2 ⫽ 2x A

x 25 m

SOLUTION Radius of Curvature: y = 22x1>2 dy 1 = 22x - 1>2 dx 2 1 22x - 3>2 4

r =

d2y

` dx2

2

1 2

B 1 + ¢ 22x - 1>2 ≤ R =

3>2 or

4

1 ` - 22x - 3>2 ` 4

= 364.21 m x = 25 m Dissemination

Wide

copyright

`

dy 2 3>2 b R dx

Web)

B1 + a

laws

= -

teaching

dx2

in

d2y

is

of

the

not

instructors

States

World

permitted.

Acceleration: The speed of the motorcycle at a is

by

and

use

United

on

learning.

km 1000 m 1h ≤¢ ≤¢ ≤ = 16.67 m>s h 1 km 3600 s

for

work

student

the

v = ¢ 60

work

solely

of

protected

the

(including

v2 16.672 = = 0.7627 m>s2 r 364.21

this

assessing

is

an =

part

the

is

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provided

integrity

of

and

work

Since the motorcycle travels with a constant speed, at = 0. Thus, the magnitude of the motorcycle’s acceleration at A is

sale

will

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destroy

of

and

any

courses

a = 2at 2 + an2 = 202 + 0.76272 = 0.763 m>s2






Ans.

*12–140. The jet plane travels along the vertical parabolic path. When it is at point A it has a speed of 200 m>s, which is increasing at the rate of 0.8 m>s2. Determine the magnitude of acceleration of the plane when it is at point A.

y y ⫽ 0.4x2

A

SOLUTION y = 0.4x2

10 km

dy = 0.8x ` = 4 dx x = 5 km d2y dx2 r =

= 0.8

x 5 km

[1 + (4)2]3/2 = 87.62 km 0.8

at = 0.8 m/s2 or

(0.200)2 = 0.457(10 - 3) km>s2 87.62

teaching

Web)

laws

an =

Wide

copyright

in

an = 0.457 km/s2

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of

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Ans.






permitted.

Dissemination

a = 210.822 + 10.45722 = 0.921 m>s2

12–141. The ball is ejected horizontally from the tube with a speed of 8 m>s. Find the equation of the path, y = f1x2, and then find the ball’s velocity and the normal and tangential components of acceleration when t = 0.25 s.

y vA A

SOLUTION vx = 8 m>s + ) (:

s = v0t x = 8t

A+cB

s = s0 + v0 t + y = 0 + 0 +

1 2 a t 2 c

1 ( - 9.81)t2 2

y = -4.905t2

in

teaching

Web)

laws

or

x 2 y = -4.905 a b 8

Ans. Dissemination

Wide

(Parabola)

copyright

y = -0.0766x2

of

the

not

instructors

States

World

permitted.

v = v0 + act

the

by

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use

United

on

learning.

is

vy = 0 - 9.81t

the

(including

for

work

student

When t = 0.25 s,

assessing

is

work

solely

of

protected

vy = - 2.4525 m>s

Ans.

part

the

is

any

courses

will

ay = 9.81 m>s2

sale

ax = 0

their

destroy

of

2.4525 b = 17.04° 8

and

u = tan - 1 a

This

provided

integrity

of

and

work

this

v = 31822 + 12.452522 = 8.37 m>s

an = 9.81 cos 17.04° = 9.38 m>s2

Ans.

at = 9.81 sin 17.04° = 2.88 m>s2

Ans.






8 m/s x

12–142. A toboggan is traveling down along a curve which can be approximated by the parabola y = 0.01x2. Determine the magnitude of its acceleration when it reaches point A, where its speed is vA = 10 m>s, and it is increasing at the rate of # vA = 3 m>s2.

y

y = 0.01x2

A 36 m

SOLUTION

x 60 m

Acceleration: The radius of curvature of the path at point A must be determined d2y dy first. Here, = 0.02x and 2 = 0.02, then dx dx r =

[1 + (dy>dx)2]3>2 |d2y>dx2|

=

[1 + (0.02x)2]3>2 2 = 190.57 m |0.02| x = 60 m

To determine the normal acceleration, apply Eq. 12–20. an =

v2 102 = = 0.5247 m>s2 r 190.57

Web)

laws

or

# Here, at = vA = 3 m>s. Thus, the magnitude of acceleration is

teaching

32 + 0.52472 = 3.05 m s2

Ans. in

a2t + a2n =

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Dissemination

Wide

copyright

a =






12–143. A particle P moves along the curve y = 1x2 - 42 m with a constant speed of 5 m>s. Determine the point on the curve where the maximum magnitude of acceleration occurs and compute its value.

SOLUTION y = (x2 - 4) at =

dv = 0, dt

To obtain maximum a = an, r must be a minimum. This occurs at: x = 0,

y = -4 m

Ans.

laws

= 2 Web)

dx2

teaching

d2y

in

dy ` = 2x = 0; dx x = 0

or

Hence,

Wide

permitted.

Dissemination

World the

not

instructors

States

is on by

and

`

learning.

dx2

of

d2y

United

`

3

[1 + 0]2 1 = = 020 2 use

rmin =

dy 2 2 b d dx

copyright

3

c1 + a

the

(including

for

work

student

the

v2 52 = 1 = 50 m>s2 rmin 2

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(a)max = (an)max =






Ans.

*12–144. The Ferris wheel turns such that the speed of the passengers is # increased by v = 14t2 ft>s2, where t is in seconds. If the wheel starts from rest when u = 0°, determine the magnitudes of the velocity and acceleration of the passengers when the wheel turns u = 30°.

40 ft u

SOLUTION v

t

dv =

L0

L0

4tdt

v = 2t2 s

t

ds =

2t2 dt L0 2 s = 3 t3 p When s = (40) ft, 6 L0

p 2 (40) = t 3 6 3

t = 3.1554 s

v = 2(3.1554)2 = 19.91 ft>s = 19.9 ft>s

or

Ans.

in

teaching

Web)

laws

# at = v = 4t 0 t = 3.1554 s = 12.62 ft>s2

permitted.

Dissemination

Wide

copyright

19.912 v2 = = 9.91 ft>s2 r 40 instructors

States

World

an =

not

a = 2at 2 + an2 = 212.622 + 9.912 = 16.0 ft>s2

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Ans.






12–145. If the speed of the crate at A is 15 ft> s, which is increasing at # a rate v = 3 ft>s2, determine the magnitude of the acceleration of the crate at this instant.

y y ⫽ 1 x2 16

A

x

SOLUTION

10 ft

Radius of Curvature: y =

1 2 x 16

dy 1 = x dx 8 d2y 1 = 8 dx2 Thus, or Web)

laws

teaching

4

1 ` ` 8

in

= 32.82 ft

Wide

` dx2

=

x = 10 ft

not

instructors

States

World

permitted.

`

d2y

2 3>2

1 8

B1 + ¢ x≤ R

Dissemination

r =

dy 2 3>2 b R dx

copyright

B1 + a

and

use

United

on

learning.

is

of

the

Acceleration:

for

work

student

the

by

# at = v = 3ft>s2

work

solely

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the

(including

v2 152 = = 6.856 ft>s2 r 32.82

this

assessing

is

an =

the

is

This

provided

integrity

of

and

work

The magnitude of the crate’s acceleration at A is

sale

will

their

destroy

of

and

any

courses

part

a = 2at 2 + an2 = 232 + 6.8562 = 7.48 ft>s2






Ans.

12–146. The race car has an initial speed vA = 15 m>s at A. If it increases its speed along the circular track at the rate a t = 10.4s2 m>s2, where s is in meters, determine the time needed for the car to travel 20 m. Take r = 150 m. r

SOLUTION

s A

n dn at = 0.4s = ds a ds = n dn n

s

0.4s ds =

L0

L15

n dn

0.4s2 s n2 n ` = ` 2 0 2 15

laws

or

n2 225 0.4s2 = 2 2 2

in

teaching

Web)

n2 = 0.4s2 + 225 ds = 20.4s2 + 225 dt States

World

permitted.

Dissemination

Wide

copyright

n =

not the of

on

learning.

is

dt United

and

L0

work the

(including

for

of solely

= 0.632 456t

work

s

assessing

1n (s + 2s2 + 562.5) `

student

= 0.632 456t protected

ds

L0 2s2 + 562.5

is

s

the

by

L0 20.4s2 + 225

=

instructors

t

ds

use

s

work

this

0

any will sale

t = 1.21 s

their

destroy

of

and

At s = 20 m,

courses

part

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is

This

provided

integrity

of

and

1n (s + 2s2 + 562.5) - 3.166 196 = 0.632 456t






Ans.

12–147. A boy sits on a merry-go-round so that he is always located at r = 8 ft from the center of rotation. The merry-go-round is originally at rest, and then due to rotation the boy’s speed is increased at 2 ft>s2. Determine the time needed for his acceleration to become 4 ft>s2.

SOLUTION a = 2a2n + a2t at = 2 v = v0 + act v = 0 + 2t an =

(2t)2 v2 = r 8

4 =

(2)2 + a

or

A

(2t)2 2 b 8

teaching

Web)

laws

16 t4 64 in

16 = 4 +

Wide

copyright

t = 2.63 s

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the

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permitted.

Dissemination

Ans.






*12–148. A particle travels along the path y = a + bx + cx 2, where a, b, c are constants. If the speed of the particle is constant, v = v0, determine the x and y components of velocity and the normal component of acceleration when x = 0.

SOLUTION y = a + bx + cx2 # # # y = bx + 2 c x x $ $ # $ y = b x + 2 c (x)2 + 2 c xx # # y = bx

When x = 0, # # v20 + x 2 + b2 x 2

vy =

v0

Ans.

21 + b2 v0 b

Ans.

Dissemination

Wide

copyright

in

teaching

Web)

laws

21 + b2 v20 an = r

or

# vx = x =

dy 2 2 C 1 + A dx B D

World the

not

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States United

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2

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of

dx

use

d2y

the

`

by

r =

permitted.

3

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this

assessing

= 2c






integrity

of

and

part

the

is

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courses and

destroy

of

will

2 c v20

(1 + b2)3>2 2c

(1 + b2)3>2

sale

an =

r =

their

At x = 0,

This

provided

dx2

work

d2y

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solely

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the

(including

for

work

student

dy = b + 2c x dx

Ans.

12–149. The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively. Determine in t = 2 s, (a) the displacement along the path of each particle, (b) the position vector to each particle, and (c) the shortest distance between the particles.

y

SOLUTION (a)

sA = 0.7(2) = 1.40 m

Ans.

sB = 1.5(2) = 3 m

Ans.

5m

B

1.40 uA = = 0.280 rad. = 16.04° 5

(b)

uB =

A vB

1.5 m/s

vA

3 = 0.600 rad. = 34.38° 5

For A

laws

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x = 5 sin 16.04° = 1.382 = 1.38 m

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y = 5(1 - cos 16.04°) = 0.1947 = 0.195 m

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rA = {1.38i + 0.195j} m

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x = -5 sin 34.38° = -2.823 = -2.82 m

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y = 5(1 -cos 34.38°) = 0.8734 = 0.873 m

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rB = { - 2.82i + 0.873j} m

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¢r = rB - rA = {-4.20i + 0.678j} m

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¢r = 31 - 4.2022 + 10.67822 = 4.26 m






x

O 0.7 m/s

12–150. The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively. Determine the time when they collide and the magnitude of the acceleration of B just before this happens.

y

SOLUTION st = 2p(5) = 31.4159 m

5m

sA = 0.7 t B

sB = 1.5 t

A vB

Require

0.7 t + 1.5 t = 31.4159 t = 14.28 s = 14.3 s

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(1.5)2 v2B = 0.45 m>s2 = r 5

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x

O vA

sA + sB = 31.4159

aB =

1.5 m/s

0.7 m/s

12–151. The position of a particle traveling along a curved path is s = (3t3 - 4t2 + 4) m, where t is in seconds. When t = 2 s, the particle is at a position on the path where the radius of curvature is 25 m. Determine the magnitude of the particle’s acceleration at this instant.

SOLUTION Velocity: v =

d 3 A 3t - 4t2 + 4 B = A 9t2 - 8t B m>s dt

When t = 2 s, v ƒ t = 2 s = 9 A 22 B - 8122 = 20 m>s Acceleration: or

dv d 2 = A 9t - 8t B = A 18t - 8 B m>s2 ds dt laws

at =

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at ƒ t = 2 s = 18(2) - 8 = 28 m>s2 in

202 = 16 m>s2 25

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a = 2at 2 + an2 = 2282 + 162 = 32.2 m>s2






Ans.

*12–152. If the speed of the box at point A on the track is 30 ft>s # which is increasing at the rate of v = 5 ft>s2, determine the magnitude of the acceleration of the box at this instant.

y

y ⫽ 0.004x2 ⫹10

A

10 ft

x

SOLUTION

50 ft

Radius of Curvature: y = 0.004x2 + 10 dy = 0.008x dx d2y dx2

= 0.008

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Acceleration: for

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v2 302 = = 5.763 ft>s2 r 156.17 # at = v = 5 ft>s2

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a = 2at 2 + an2 = 252 + 5.7632 = 7.63 ft>s2






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■

12–153.

A go-cart moves along a circular track of radius 100 ft such that its speed for a short period of time, 0 … t … 4 s, is 2 v = 6011 - e -t 2 ft>s. Determine the magnitude of its acceleration when t = 2 s. How far has it traveled in t = 2 s? Use Simpson’s rule with n = 50 to evaluate the integral.

SOLUTION v = 60(1 - e - t ) 2

at =

dv 2 2 = 60( -e - t )(-2t) = 120 t e - t dt

at|t = 2 = 120(2)e - 4 = 4.3958 v|t = 2 = 60(1 - e - 4) = 58.9011 an =

(58.9011)2 = 34.693 100

a = 2(4.3958)2 + (34.693)2 = 35.0 m>s2 6011 - e - t 2 dt

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s = 67.1 ft

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12–154.

The ball is kicked with an initial speed vA = 8 m>s at an angle uA = 40° with the horizontal. Find the equation of the path, y = f(x), and then determine the normal and tangential components of its acceleration when t = 0.25 s.

y

vA = 8 m/s uA

40

A

x

x

SOLUTION Horizontal Motion: The horizontal component of velocity is (v0)x = 8 cos 40° = 6.128 m>s and the initial horizontal and final positions are (s0)x = 0 and sx = x, respectively. + B A:

sx = (s0)x + (y0)x t x = 0 + 6.128t

(1)

or

Vertical Motion: The vertical component of initial velocity is (v0)y = 8 sin 40° = 5.143 m>s. The initial and final vertical positions are (s0)y = 0 and sy = y, respectively.

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sy = (s0)y + (v0)y t +

1 ( -9.81) A t2 B 2

(2)

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y = 0 + 5.143t +

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Eliminate t from Eqs (1) and (2),we have

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y = {0.8391x - 0.1306x2} m = {0.839x - 0.131x2} m for

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Acceleration: When t = 0.25 s, from Eq. (1), x = 0 + 6.128(0.25) = 1.532 m. Here, dy dy = 0.8391 - 0.2612x. At x = 1.532 m, = 0.8391 - 0.2612(1.532) = 0.4389 dx dx and the tangent of the path makes an angle u = tan-1 0.4389 = 23.70° with the x axis. The magnitude of the acceleration is a = 9.81 m>s2 and is directed downward. From the figure, a = 23.70°. Therefore, Ans.

an = a cos a = 9.81 cos 23.70° = 8.98 m>s2

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at = - a sin a = - 9.81 sin 23.70° = - 3.94 m>s2






y

12–155. The race car travels around the circular track with a speed of 16 m>s. When it reaches point A it increases its speed at at = (43 v1>4) m>s2, where v is in m>s. Determine the magnitudes of the velocity and acceleration of the car when it reaches point B. Also, how much time is required for it to travel from A to B?

y A

200 m

SOLUTION

B

x

4 1 at = v4 3 dv = at dt dv =

4 1 v 4 dt 3

v

L16 3

t

dv

0.75

=

1

v4

L0

dt

v

v4 16 = t 3

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v4 - 8 = t 4

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ds = v dt

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7 3 (t + 8)3 2 7 0

s =

7 3 (t + 8)3 - 54.86 7

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s =

7 3 p (200) = 100p = (t + 8)3 - 54.86 2 7

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t = 10.108 s = 10.1 s 4

v = (10.108 + 8)3 = 47.551 = 47.6 m>s at = an =

1 4 (47.551)4 = 3.501 m>s2 3

(47.551)2 v2 = 11.305 m>s2 = r 200

a = 2(3.501)2 + (11.305)2 = 11.8 m>s2






Ans.

Ans.

*12–156. A particle P travels along an elliptical spiral path such that its position vector r is defined by r = 52 cos10.1t2i + 1.5 sin10.1t2j + 12t2k6 m, where t is in seconds and the arguments for the sine and cosine are given in radians. When t = 8 s, determine the coordinate direction angles a, b, and g, which the binormal axis to the osculating plane makes with the x, y, and z axes. Hint: Solve for the velocity vP and acceleration a P of the particle in terms of their i, j, k components. The binormal is parallel to vP * a P . Why?

z

P r

SOLUTION

y

rP = 2 cos (0.1t)i + 1.5 sin (0.1t)j + 2tk # vP = r = - 0.2 sin (0.1t)i + 0.15 cos (0.1t)j + 2k

x

$ aP = r = -0.02 cos 10.1t2i - 0.015 sin 10.1t2j When t = 8 s,

or

vP = -0.2 sin (0.8 rad)i + 0.15 cos (0.8 rad)j + 2k = -0.143 47i + 0.104 51j + 2k

in

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aP = -0.02 cos (0.8 rad)i - 0.015 sin (0.8 rad)j = -0.013 934i - 0.010 76 j

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copyright

Since the binormal vector is perpendicular to the plane containing the n–t axis, and ap and vp are in this plane, then by the definition of the cross product,

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j k 0.104 51 2 3 = 0.021 52i - 0.027 868j + 0.003k -0.010 76 0 (including

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b = 210.0215222 + 1- 0.02786822 + 10.00322 = 0.035 338

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i b = vP * aP = 3 -0.14 347 - 0.013 934

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ub = 0.608 99i - 0.788 62j + 0.085k

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a = cos - 1(0.608 99) = 52.5°

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b = cos - 1(-0.788 62) = 142°

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g = cos - 1(0.085) = 85.1°

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Ans.

Note: The direction of the binormal axis may also be specified by the unit vector ub¿ = - ub, which is obtained from b¿ = ap * vp. For this case, a = 128°, b = 37.9°, g = 94.9°






Ans.

12–157. The motion of a particle is defined by the equations x = (2t + t2) m and y = (t2) m, where t is in seconds. Determine the normal and tangential components of the particle’s velocity and acceleration when t = 2 s.

SOLUTION Velocity: Here, r =

E A 2 t + t 2 B i + t2 j F m.To determine the velocity v, apply Eq. 12–7. v =

dr = {(2 + 2t) i + 2tj } m>s dt

When t = 2 s, v = [2 + 2(2)]i + 2(2)j = {6i + 4j} m>s. Then v = 262 + 42 = 7.21 m>s. Since the velocity is always directed tangent to the path, vn = 0

and

vt = 7.21 m>s

The velocity v makes an angle u = tan-1

Ans.

4 = 33.69° with the x axis. 6 laws

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Acceleration: To determine the acceleration a, apply Eq. 12–9.

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dv = {2i + 2j} m>s2 dt

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a =

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a = 222 + 22 = 2.828 m>s2

2 = 45.0° with the x axis. From the 2 the

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The acceleration a makes an angle f = tan-1

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figure, a = 45° - 33.69 = 11.31°. Therefore,

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an = a sin a = 2.828 sin 11.31° = 0.555 m>s2

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at = a cos a = 2.828 cos 11.31° = 2.77 m>s2






12–158. The motorcycle travels along the elliptical track at a constant speed v. Determine the greatest magnitude of the acceleration if a 7 b.

y

b

x2 a2 x

a

SOLUTION b 2a2 - x2, we have a

Acceleration: Differentiating twice the expression y = dy bx = dx a2a2 - x2 d2y dx2

= -

ab (a2 - x2)3>2

The radius of curvature of the path is

dx2

=

2

2-

2 3>2

bx

≤ R

a2a2 - x2 ab

c1 + =

2

(a2 - x2)3>2

3>2 b2x2 d a2(a2 - x2) ab

(1)

(a2 - x2)3>2 in

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2

d2y

B1 + ¢ -

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r =

dy 2 3>2 b d dx

laws

c1 + a

permitted.

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copyright

To have the maximum normal acceleration, the radius of curvature of the path must be a minimum. By observation, this happens when y = 0 and x = a. When x : a, of

the

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3>2 3>2 b2x2 b2x2 b3x3 b2x2 . 7 7 1. Then, c1 + 2 2 d :c 2 2 d = 3 2 2 2 2 2 a (a - x ) a (a - x ) a (a - x ) a (a - x2)3>2

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Substituting this value into Eq. [1] yields r =

the

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b2 3 x . At x = a, a4

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b2 3 b2 a B = 4 A a a

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To determine the normal acceleration, apply Eq. 12–20.

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v2 a v2 = 2 = 2 v2 r b >a b destroy

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(an)max =

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Since the motorcycle is traveling at a constant speed, at = 0. Thus, amax = (an)max =






a 2 v b2

Ans.

y2 b2

1

12–159. A particle is moving along a circular path having a radius of 4 in. such that its position as a function of time is given by u = cos 2t, where u is in radians and t is in seconds. Determine the magnitude of the acceleration of the particle when u = 30°.

SOLUTION When u =

p 6

rad,

p 6

= cos 2t

t = 0.5099 s

# du = - 2 sin 2t 2 u = = - 1.7039 rad>s dt t = 0.5099 s $ d2u = - 2.0944 rad>s2 u = 2 = - 4 cos 2t 2 dt t = 0.5099 s r = 4

# r = 0

$ r = 0

or

# $ ar = r - ru2 = 0 - 4(-1.7039)2 = - 11.6135 in.>s2 $ ## au = ru + 2ru = 4( - 2.0944) + 0 = - 8.3776 in.>s2 ( -11.6135)2 + ( -8.3776)2 = 14.3 in. s2

Ans.

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a2r + a2u =

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*12–160. A particle travels around a limaçon, defined by the equation r = b - a cos u, where a and b are constants. Determine the particle’s radial and transverse components of velocity and acceleration as a function of u and its time derivatives.

SOLUTION r = b - a cos u # # r = a sin uu $ # # r = a cos uu2 + a sin uu # # vr = r = a sin uu

Ans.

# vu = r u = (b - a cos u)u # # $ # $ ar = r - r u2 = a cos uu2 + a sin uu - (b - a cos u)u2 $ # = (2a cos u - b) u2 + a sin u u $ $ # # # # au = ru + 2 r u = (b - a cos u)u + 2 a a sin uu bu

Ans.

laws

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$ # = (b - a cos u)u + 2au2 sin u

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12–161. If a particle’s position is described by the polar coordinates r = 411 + sin t2 m and u = 12e -t2 rad, where t is in seconds and the argument for the sine is in radians, determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s.

SOLUTION When t = 2 s, r = 4(1 +sin t) = 7.637 # r = 4 cos t = -1.66459 $ r = -4 sin t = -3.6372 u = 2 e-t # u = - 2 e - t = - 0.27067 $ u = 2 e-t = 0.270665 # vr = r = -1.66 m>s # vu = ru = 7.637(-0.27067) = - 2.07 m>s # $ ar = r - r1u22 = - 3.6372 - 7.6371- 0.2706722 = - 4.20 m>s2 $ ## au = ru + 2ru = 7.63710.2706652 + 21- 1.6645921- 0.270672 = 2.97 m>s2

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Dissemination

Ans.

12–162. An airplane is flying in a straight line with a velocity of 200 mi>h and an acceleration of 3 mi>h2. If the propeller has a diameter of 6 ft and is rotating at an angular rate of 120 rad>s, determine the magnitudes of velocity and acceleration of a particle located on the tip of the propeller.

SOLUTION vPl = ¢

200 mi 5280 ft 1h ≤¢ ≤¢ ≤ = 293.3 ft>s h 1 mi 3600 s

aPl = ¢

3 mi 5280 ft 1h 2 ≤ ¢ ≤ ¢ ≤ = 0.001 22 ft>s2 1 mi 3600 s h2

vPr = 120(3) = 360 ft>s v = 2v2Pl + v2Pr = 2(293.3)2 + (360)2 = 464 ft>s (360)2 v2Pr = = 43 200 ft>s2 r 3

a = 2a2Pl + a2Pr = 2(0.001 22)2 + (43 200)2 = 43.2(103) ft>s2

Ans.

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aPr =

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12–163. A car is traveling along the circular curve of radius r = 300 ft. At the instant shown, its angular rate of $ rotation is # u = 0.4 rad>s, which is increasing at the rate of u = 0.2 rad>s2. Determine the magnitudes of the car’s velocity and acceleration at this instant.

A

r = 300 ft

.

θ = 0.4 rad/s θ = 0.2 rad/s2 ..

SOLUTION

θ

Velocity: Applying Eq. 12–25, we have # # vr = r = 0 vu = ru = 300(0.4) = 120 ft>s Thus, the magnitude of the velocity of the car is v = 2v2r + v2u = 202 + 1202 = 120 ft>s

Ans.

laws

or

Acceleration: Applying Eq. 12–29, we have # $ ar = r - ru2 = 0 - 300 A 0.42 B = - 48.0 ft>s2 $ ## au = ru + 2ru = 300(0.2) + 2(0)(0.4) = 60.0 ft>s2

teaching

Web)

Thus, the magnitude of the acceleration of the car is in

(-48.0)2 + 60.02 = 76.8 ft s2

Ans.

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a2r + a2u =

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*12–164. A velocity of $ # radar gun at O rotates with the angular 2 u = 0.1 rad>s and angular acceleration of u = 0.025 rad>s , at the instant u = 45°, as it follows the motion of the car traveling along the circular road having a radius of r = 200 m. Determine the magnitudes of velocity and acceleration of the car at this instant. r ⫽ 200 m u O

SOLUTION Time Derivatives: Since r is constant, # $ r = r = 0 Velocity: # vr = r = 0 # vu = ru = 200(0.1) = 20 m>s Thus, the magnitude of the car’s velocity is v = 2vr2 + vu2 = 202 + 202 = 20 m>s

laws

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Acceleration:

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# # ar = r - ru2 = 0 - 200(0.12) = - 2 m>s2 $ # # au = r u + 2ru = 200(0.025) + 0 = 5 m>s2

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Thus, the magnitude of the car’s acceleration is

and

a = 2ar2 + au2 = 2(- 2)2 + 52 = 5.39 m>s2

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12–165. If a particle moves along a path such that r = 12 cos t2 ft and u = 1t>22 rad, where t is in seconds, plot the path r = f1u2 and determine the particle’s radial and transverse components of velocity and acceleration.

SOLUTION r = 2 cos t t 2

u =

# r = - 2 sin t # 1 u = 2

$ r = - 2 cos t

$ u = 0 Ans.

# 1 vu = ru = (2 cos t ) a b = cos t 2

Ans.

#2 1 2 5 $ ar = r - ru = -2 cos t - (2cos t) a b = - cos t 2 2

Ans.

$ 1 ## au = ru + 2ru = 2 cos t102 + 21- 2 sin t2 a b = - 2 sin t 2

Ans.

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laws

or

# vr = r = - 2 sin t






12–166. If a particle’s position is described by the polar coordinates r = 12 sin 2u2 m and u = 14t2 rad, where t is in seconds, determine the radial and transverse components of its velocity and acceleration when t = 1 s.

SOLUTION When t = 1 s, u = 4t = 4 # u = 4 $ u = 0 r = 2 sin 2u = 1.9787 # # r = 4 cos 2u u = -2.3280 $ # $ r = -8 sin 2u(u)2 + 8 cos 2u u = -126.638 # vr = r = -2.33 m>s # vu = ru = 1.9787(4) = 7.91 m>s # $ ar = r - r(u)2 = -126.638 - (1.9787)(4)2 = - 158 m>s2 $ ## au = ru + 2 ru = 1.9787(0) + 2( - 2.3280)(4) = - 18.6 m>s2

Web)

laws

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Ans.

States

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or

Ans.

12–167. The car travels along the circular curve having a radius r = # 400 ft. At the instant shown, its angular rate of rotation is $ u = 0.025 rad>s, which is decreasing at the rate u = - 0.008 rad>s2. Determine the radial and transverse components of the car’s velocity and acceleration at this instant and sketch these components on the curve.

r . u

SOLUTION r = 400 # u = 0.025

# r = 0

$ r = 0

u = -0.008

# vr = r = 0 # vu = ru = 400(0.025) = 10 ft>s # $ ar = r - r u2 = 0 - 400(0.025)2 = - 0.25 ft>s2 $ # # au = r u + 2 r u = 400(- 0.008) + 0 = -3.20 ft>s2

Ans. Ans. Ans.

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laws

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Ans.






400 ft

*12–168. The car travels along the circular curve of radius r = 400 ft with a constant speed of v = 30 ft>s. Determine the angular # rate of rotation u of the radial line r and the magnitude of the car’s acceleration. r . u

SOLUTION r = 400 ft # vr = r = 0

# r = 0

$ r = 0

# # vu = r u = 400 a u b

# 2 v = 6(0)2 + a 400 u b = 30 # u = 0.075 rad>s $ u = 0 # $ ar = r - r u2 = 0 - 400(0.075)2 = - 2.25 ft>s2 $ # # au = r u + 2 r u = 400(0) + 2(0)(0.075) = 0

laws

or

Ans.

a = 2( -2.25)2 + (0)2 = 2.25 ft>s2

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Ans.






400 ft

12–169. The time rate of change of acceleration is referred to as the jerk, which is often used as a means of measuring passenger # discomfort. Calculate this vector, a, in terms of its cylindrical components, using Eq. 12–32.

SOLUTION $ $ # ## $ a = a r -ru 2 bur + a ru + 2ru buu + z uz #$ # # $# $ ### ### $ ## $ #$ $# #$ ## # $# a = ar - ru2 - 2ruu b ur + a r - ru2 bu r + a r u + ru + 2ru + 2r u b uu + a ru + 2ru buu + z uz + zu z # # # But, ur = uuu uu = - uur

# uz = 0

Substituting and combining terms yields #$ # $# # ### #$ $# ### # a = a r - 3ru 2 - 3ruu b ur + a 3r u + ru + 3ru - ru3 b uu + az buz

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Ans.






12–170. A particle is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by u = sin 3t, where u is in radians, the argument for the sine are in radians , and t is in seconds. Determine the acceleration of the particle at u = 30°. The particle starts from rest at u = 0°.

SOLUTION r = 6 in.,

# r = 0,

$ r = 0

u = sin 3t # u = 3 cos 3t $ u = -9 sin 3t At u = 30°, 30° p = sin 3t 180°

laws

or

t = 10.525 s

(including

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Thus, # u = 2.5559 rad>s $ u = -4.7124 rad>s2 # $ ar = r - r u2 = 0 - 6(2.5559)2 = -39.196 $ ## au = r u + 2ru = 6( - 4.7124) + 0 = - 28.274

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a = 2( - 39.196)2 + ( - 28.274)2 = 48.3 in.>s2






Ans.

12–171. The slotted link is pinned # at O, and as a result of the constant angular velocity u = 3 rad>s it drives the peg P for a short distance along the spiral guide r = 10.4u2 m, where u is in radians. Determine the radial and transverse components of the velocity and acceleration of P at the instant u = p>3 rad.

0.5 m

P r

· u

r

3 rad/s

SOLUTION # u = 3 rad>s

At u =

p , 3

u

r = 0.4 u # # r = 0.4 u $ $ r = 0.4 u

O

r = 0.4189 # r = 0.4(3) = 1.20 $ r = 0.4(0) = 0

# v = r = 1.20 m>s # vu = r u = 0.4189(3) = 1.26 m>s # $ ar = r - ru2 = 0 - 0.4189(3)2 = - 3.77 m>s2 $ ## au = r u + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2

laws

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Ans.

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Dissemination

copyright

Ans.

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0.4 u

12–172. Solve Prob. # 12–171 if the slotted link has . an angular # # acceleration u = 8 rad>s 2 when u = 3 rad>s at u = p>3 rad.

SOLUTION # u = 3 rad/s

u =

0.5 m

r = 0.4 u # # r = 0.4 u $ $ r = 0.4 u

r · u

p 3

O

r = 0.4189 # r = 1.20 $ r = 0.4(8) = 3.20 laws

or

# vr = r = 1.20 m>s # vu = r u = 0.4189(3) = 1.26 m>s # $ ar = r - ru2 = 3.20 - 0.4189(3)2 = - 0.570 m>s2 $ ## au = r u + 2 ru = 0.4189(8) + 2(1.20)(3) = 10.6 m>s2

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permitted.

Ans. World

Dissemination

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sale Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

r

3 rad/s u

# u = 3 $ u = 8


P

0.4 u

12–173. The slotted link is pinned # at O, and as a result of the constant angular velocity u = 3 rad>s it drives the peg P for a short distance along the spiral guide r = 10.4u2 m, where u is in radians. Determine the velocity and acceleration of the particle at the instant it leaves the slot in the link, i.e., when r = 0.5 m.

0.5 m

P r

· u

SOLUTION

u

r = 0.4 u # # r = 0.4 u $ $ r = 0.4 u # u = 3 $ u = 0

O

At r = 0.5 m,

laws

or

0.5 = 1.25 rad 0.4

u =

in

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# r = 1.20 Dissemination

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$ r = 0

permitted.

# vr = r = 1.20 m>s # vu = r u = 0.5(3) = 1.50 m>s # $ ar = r - r(u)2 = 0 - 0.5(3)2 = - 4.50 m>s2 $ ## au = ru + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2

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Ans.

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Ans.

for protected is

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r

3 rad/s

Ans.

0.4u

12–174. A particle moves in the x–y plane such that its position is defined by r = 52ti + 4t2j6 ft, where t is in seconds. Determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s.

SOLUTION r = 2ti + 4 t2j|t = 2 = 4i + 16j v = 2i + 8 tj|t = 2 = 2i + 16j a = 8j u = tan - 1 a

16 b = 75.964° 4

v = 2(2)2 + (16)2 = 16.1245 ft>s 16 b = 82.875° 2 or

f = tan - 1 a

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a = 8 ft>s2

Wide

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in

f - u = 6.9112°

instructors

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Ans. not

vu = 16.1245 sin 6.9112° = 1.94 ft>s use

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d = 90° - u = 14.036°

Ans.

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ar = 8 cos 14.036° = 7.76 ft>s2

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au = 8 sin 14.036° = 1.94 ft>s2






Ans.

permitted.

Dissemination

vr = 16.1245 cos 6.9112° = 16.0 ft>s

12–175. A particle P moves along the spiral path r = 110>u2 ft, where u is in radians. If it maintains a constant speed of v = 20 ft>s, determine the magnitudes vr and vu as functions of u and evaluate each at u = 1 rad.

v P r

r = 10 –– θ

SOLUTION 10 u

r =

# 10 # r = - a 2 bu u

102 # 2 102 # b u + a 2 b u2 4 u u

(20)2 = a

# 102 b (1 + u2)u2 u4 laws

(20)2 = a

or

# 2 # Since v2 = r2 + a ru b

# Thus, u =

in

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2u2

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21 + u2

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# 2u2 10 20u vu = ru = a b £ ≥ = 2 u 21 + u 21 + u2

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When u = 1 rad

Ans.

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≤ = - 14.1 ft>s

destroy

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22

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20

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vr = ¢ -

vu =






20 2

= 14.1 ft s

Ans.

not

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States

World

10 20 2u2 # vr = r = - a 2 b £ ≥ = 2 u 21 + u 21 + u2

θ

*12–176. The driver of the car maintains a constant speed of 40 m>s. Determine the angular velocity of the camera tracking the car when u = 15°.

r

(100 cos 2u) m

SOLUTION

u

Time Derivatives: r = 100 cos 2u # # r = (-200 (sin 2 u)u ) m>s At u = 15°, r u = 15° = 100 cos 30° = 86.60 m # # # r u = 15° = -200 sin 30°u = - 100u m>s Velocity: Referring to Fig. a, vr = - 40 cos f and vu = 40 sin f. laws

or

# vr = r

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# -40 cos f = - 100u Dissemination

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(1)

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# vu = ru

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# 40 sin f = 86.60u

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f = 40.89° # u = 0.3024 rad>s = 0.302 rad>s

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Ans.






12–177. When u = 15°, the car has a speed of 50 m>s which is increasing at 6 m>s2. Determine the angular velocity of the camera tracking the car at this instant.

r

(100 cos 2u) m

SOLUTION

u

Time Derivatives: r = 100 cos 2u # # r = ( - 200 (sin 2u) u ) m>s $ # $ r = -200 C (sin 2u) u + 2 (cos 2u) u 2 D m>s2 At u = 15°, r u = 15° = 100 cos 30° = 86.60 m # # # r u = 15° = -200 sin 30°u = - 100u m>s $ # # r u = 15° = -200 C sin 30° u + 2 cos 30° u2 D =

$

#

laws

or

A -100u - 346.41u2 B m>s2

Dissemination

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Velocity: Referring to Fig. a, vr = - 50 cos f and vu = 50 sin f. Thus, # vr = r # -50 cos f = -100u

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# vu = ru

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# 50 sin f = 86.60u

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f = 40.89° # u = 0.378 rad>s

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12–178. The small washer slides down the cord OA. When it is at the midpoint, its speed is 200 mm>s and its acceleration is 10 mm>s2. Express the velocity and acceleration of the washer at this point in terms of its cylindrical components.

z A v, a

700 mm z O

SOLUTION

u 2

2

x

OB = 2(400) + (300) = 500 mm 2

vr = (200) a

2

500 b = 116 mm>s 860.23

vu = 0 vz = (200) a

700 b = 163 mm>s 860.23

Thus, v = { -116ur - 163uz} mm>s

Ans. or

500 b = 5.81 860.23

laws

ar = 10 a

in

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au = 0

permitted.

Dissemination

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700 b = 8.14 860.23 States

World

az = 10 a

instructors

Thus, a = { -5.81ur - 8.14uz} mm>s2 United

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r

y 300 mm

OA = 2(400) + (300) + (700) = 860.23 mm 2

400 mm

12–179. A block moves outward along the slot in the platform with # a speed of r = 14t2 m>s, where t is in seconds. The platform rotates at a constant rate of 6 rad/s. If the block starts from rest at the center, determine the magnitudes of its velocity and acceleration when t = 1 s.

·

θ = 6 rad/s r

θ

SOLUTION # r = 4t|t = 1 = 4 # $ u = 6 u = 0 1

L0

$ r = 4

1

dr =

L0

4t dt

r = 2t2 D 10 = 2 m

# # v = 3 A r B 2 + A ru B 2 = 2 (4)2 + [2(6)]2 = 12.6 m>s # $ $ ## a = r - ru2 2 + ru + 2 ru 2 = [4 - 2(6)2 ]2 + [0 + 2(4)(6)]2

Ans. Ans.

= 83.2 m s

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2






*12–180. Pin P is constrained to move along the curve defined by the lemniscate r = (4 sin 2u) ft. If the slotted arm OA rotates counterclockwise with a constant angular velocity of # u = 1.5 rad>s, determine the magnitudes of the velocity and acceleration of peg P when u = 60°.

A r P O

SOLUTION Time Derivatives: r = 4 sin 2u # u = 1.5 rad>s $ u = 0

# # r = (8(cos 2u)u ) ft>s # $ $ r = 8[(cos 2u)u - 2 sin 2u(u )2] ft>s2 When u = 60°, r|u = 60° = 4 sin 120° = 3.464 ft

laws

or

# r |u = 60° = 8 cos 120°(1.5) = - 6 ft>s

in

teaching

Web)

$ r|u = 60° = 8[0 - 2 sin 120°(1.52)] = - 31.18 ft>s2

Dissemination

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copyright

Velocity:

World

permitted.

# vu = ru = 3.464(1.5) = 5.196 ft>s of

the

not

instructors

States

# vr = r = - 6 ft>s

and

use

United

on

learning.

is

Thus, the magnitude of the peg’s velocity is by

v = 2vr2 + vu2 = 2( - 6)2 + 5.1962 = 7.94 ft>s for

work

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Acceleration:

part

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solely

# $ ar = r - ru2 = - 31.18 - 3.464(1.52) = - 38.97 ft>s2 $ ## au = ru + 2ru = 0 + 2( -6)(1.5) = - 18 ft>s2 and

any

courses

Thus, the magnitude of the peg’s acceleration is

sale

will

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destroy

of

a = 2ar2 + au2 = 2( -38.97)2 + ( -18)2 = 42.9 ft>s2






Ans.

u

r ⫽ (4 sin 2u) ft

12–181. Pin P is constrained to move along the curve defined by the lemniscate r = (4 sin 2u) ft. If the angular position of the slotted arm OA is defined by u = 13t 3>22 rad, determine the magnitudes of the velocity and acceleration of the pin P when u = 60°.

A r P O

SOLUTION Time Derivatives: r = 4 sin 2u # # r = (8(cos 2u)u ) ft>s # # $ r = 8[(cos 2u)u - 2 (sin 2u)u 2] ft>s2 When u = 60° =

p rad, 3 p = 3t 3>2 3

teaching

Web)

laws

or

t = 0.4958 s

States

World

permitted.

Dissemination

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in

Thus, the angular velocity and angular acceleration of arm OA when p u = rad (t = 0.4958 s) are 3

by

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# 9 u = t1>2 ` = 3.168 rad>s 2 t = 0.4958 s

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$ 9 u = t1>2 ` = 3.196 rad>s2 4 t = 0.4958 s

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the

is

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provided

integrity

of

and

work

r|u = 60° = 4 sin 120° = 3.464 ft and

any

courses

part

# r|u = 60° = 8 cos 120°(3.168) = - 12.67 ft>s

sale

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of

$ r|u = 60° = 8[cos 120°(3.196) - 2 sin 120°(3.1682)] = - 151.89 ft>s2 Velocity: # vr = r = - 12.67 ft>s

# vu = ru = 3.464(3.168) = 10.98 ft>s

Thus, the magnitude of the peg’s velocity is v = 2vr2 + vu2 = 2(- 12.67)2 + 10.982 = 16.8 ft>s

Ans.

Acceleration: # $ ar = r - ru2 = - 151.89 - 3.464(3.1682) = - 186.67 ft>s2 $ ## au = ru + 2ru = 3.464(3.196) + 2( - 12.67)(3.168) = - 69.24 ft>s2 Thus, the magnitude of the peg’s acceleration is a = 2ar2 + au2 = 2(- 186.67)2 + (- 69.24)2 = 199 ft>s2






Ans.

u

r ⫽ (4 sin 2u) ft

12–182. A cameraman standing at A is following the movement of a race car, B, which is traveling around a curved track at # a constant speed of 30 m>s. Determine the angular rate u at which the man must turn in order to keep the camera directed on the car at the instant u = 30°.

vB BB r r

2020 mm AA

r = 2(20) cos u = 40 cos u # # r = -(40 sin u ) u # # v = r u r + r u uu # # v = 2(r)2 + (r u)2 # # (30)2 = ( - 40 sin u)2 (u)2 + (40 cos u)2 (u)2 # (30)2 = (40)2 [sin2 u + cos2 u](u)2 # 30 u = = 0.75 rad>s 40

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Dissemination

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laws

or

Ans.





2020 mm

uu 2020 mm

SOLUTION


30 m/s

2020 mm

12–183. The slotted arm AB drives pin C through the spiral groove described by # the equation r = a u. If the angular velocity is constant at u, determine the radial and transverse components of velocity and acceleration of the pin.

B C

r

SOLUTION

# $ Time Derivatives: Since u is constant, then u = 0. # $ # $ r = au r = au r = au = 0

u

A

Velocity: Applying Eq. 12–25, we have # # vr = r = au # # vu = ru = auu

Ans. Ans.

Acceleration: Applying Eq. 12–29, we have # # # $ ar = r - ru2 = 0 - auu2 = - auu2 $ # # # # # au = ru + 2r u = 0 + 2(au)(u) = 2au2

or

Ans.

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permitted.

Dissemination

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copyright

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laws

Ans.






*12–184.

The slotted arm AB drives pin C through the spiral groove described by the equation r = (1.5 u) ft, where u is in radians. If the arm starts from # rest when u = 60° and is driven at an angular velocity of u = (4t) rad>s, where t is in seconds, determine the radial and transverse components of velocity and acceleration of the pin C when t = 1 s.

B C

r

SOLUTION

# $ Time Derivatives: Here, u = 4 t and u = 4 rad>s2. # $ $ # r = 1.5 u = 1.5 (4) = 6 ft>s2 r = 1.5 u r = 1.5 u = 1.5(4t) = 6t u

Velocity: Integrate the angular rate, 1 Then, r = b (6t2 + p) r ft. 2

At

t

du =

Lp3

r =

t = 1 s,

A

1 (6t2 + p) rad . 3

4tdt, we have u =

L0

u

1 C 6(12) + p D = 4.571 ft, r = 6(1) = 6.00 ft>s # 2

laws

or

# and u = 4(1) = 4 rad>s. Applying Eq. 12–25, we have # vr = r = 6.00 ft>s # vu = ru = 4.571 (4) = 18.3 ft>s

permitted.

Dissemination

copyright

Ans.

Wide

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Ans.

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# $ ar = r - ru2 = 6 - 4.571 A 42 B = - 67.1 ft>s2 $ # # au = r u + 2ru = 4.571(4) + 2(6) (4) = 66.3 ft>s2

is

of

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States

World

Acceleration: Applying Eq. 12–29, we have






Ans.

12–185. If the slotted arm AB rotates# counterclockwise with a constant angular velocity of u = 2 rad>s, determine the magnitudes of the velocity and acceleration of peg P at u = 30°. The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB.

D

P r ⫽ (4 sec u) ft

u

A

C

SOLUTION

4 ft

Time Derivatives: r = 4 sec u # u = 2 rad>s $ u = 0

# # r = (4 secu(tanu)u ) ft>s # # # $ $ r = 4 [secu(tanu)u + u (sec u (sec2u)u + tan u secu(tan u)u )] # # = 4[secu(tanu)u + u2(sec3u + tan2u secu)] ft>s2 When u = 30°,

laws

or

r|u = 30° = 4 sec 30° = 4.619 ft

in

teaching

Web)

# r |u = 30° = (4 sec30° tan30°)(2) = 5.333 ft>s

Dissemination

Wide

copyright

$ r |u = 30° = 4[0 + 2 2(sec3 30° + tan2 30° sec 30°)] = 30.79 ft>s2

not

instructors

States

World

permitted.

Velocity:

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# vu = ru = 4.619(2) = 9.238 ft>s

# vr = r = 5.333 ft>s

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Ans.

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protected

v = 2vr2 + vu2 = 25.3332 + 9.2382 = 10.7 ft>s

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Acceleration:

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# $ ar = r - ru2 = 30.79 - 4.619(2 2) = 12.32 ft>s2 $ ## au = ru + 2ru = 0 + 2(5.333)(2) = 21.23 ft>s2

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a = 2ar2 + au2 = 212.32 2 + 21.232 = 24.6 ft>s2






Ans.

12–186. The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB. When u = 30°, the angular velocity and angular acceleration of arm AB are $ # u = 2 rad>s and u = 3 rad>s2 , respectively. Determine the magnitudes of the velocity and acceleration of the peg P at this instant.

D

P r ⫽ (4 sec u) ft

u

A

C

SOLUTION

4 ft

Time Derivatives: r = 4 sec u # u = 2 rad>s $ u = 3 rad>s2

# # r = (4 secu(tanu)u ) ft>s # # # $ $ r = 4[secu(tanu)u + u (secusec2uu + tanu secu(tanu)u )] # $ = 4[secu(tanu)u + u 2(sec3u° + tan2u°secu°)] ft>s2 When u = 30°,

laws

or

r|u = 30° = 4 sec 30° = 4.619 ft

in

teaching

Web)

# r|u = 30° = (4 sec 30° tan 30°)(2) = 5.333 ft>s

Dissemination

Wide

copyright

$ r|u = 30° = 4[(sec 30° tan 30°)(3) + 22(sec3 30° + tan2 30° sec 30°)] = 38.79 ft>s2 not

instructors

States

World

permitted.

Velocity:

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# vu = ru = 4.619(2) = 9.238 ft>s

# vr = r = 5.333 ft>s

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v = 2vr2 + vu2 = 25.3332 + 9.2382 = 10.7 ft>s

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# $ ar = r - ru2 = 38.79 - 4.619(22) = 20.32 ft>s2 $ ## au = ru + 2ru = 4.619(3) + 2(5.333)(2) = 35.19 ft>s2

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a = 2ar2 + au2 = 220.322 + 35.192 = 40.6 ft>s2






Ans.

12–187. If the circular plate # rotates clockwise with a constant angular velocity of u = 1.5 rad>s, determine the magnitudes of the velocity and acceleration of the follower rod AB when u = 2>3p rad.

u

A

B

1/2

r ⫽ (10 ⫹ 50 u

SOLUTION Time Derivaties: r = A 10 + 50u1>2 B mm # # r = 25u-1>2u mm>s # # 1 $ r = 25 c u-1>2u - u-3>2u2 d mm>s2 2 2p rad, 3 or

2p 1>2 b d = 82.36 mm 3

Web)

3

= c 10 + 50a

laws

2p

in

r|u =

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1 2p -3>2 $ 2p r|u = = 25 c 0 - a b 11.522 d = - 9.279 mm>s2 3 2 3

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

2p -1>2 # 2p r|u = = 25 a b 11.52 = 25.91 mm>s 3 3

is

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Velocity: The radial component gives the rod’s velocity. assessing

# vr = r = 25.9 mm>s provided

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Acceleration: The radial component gives the rod’s acceleration. # $ ar = r - ru2 = - 9.279 - 82.36 A 1.52 B = - 195 mm>s2






Ans.

) mm

r

*12–188. When u = 2>3p rad, the angular velocity and angular # u = 1.5 rad>s and acceleration of the circular plate are $ u = 3 rad>s2, respectively. Determine the magnitudes of the velocity and acceleration of the rod AB at this instant.

u

A

B

1/2

r ⫽ (10 ⫹ 50 u

SOLUTION Time Derivatives: r = (10 + 50u1>2) mm # # r = 25u - 1>2u mm>s #2 $ 1 $ 2 r = 25 cu - 1>2 u - u - 3>2u d mm>s 2

or

2p rad, 3 laws

When u =

teaching

Web)

2p 1>2 b d = 82.36 mm 3

in

Wide

= c 10 + 50a

Dissemination

2p 3

copyright

r|u =

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2p - 1>2 1 2p - 3>2 $ 2p (3) - a b (1.52) d = 42.55 mm>s2 r|u = = 25 c a b 3 3 2 3

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2p - 1>2 # 2p r|u = = 25 a b (1.5) = 25.91 mm>s 3 3

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# v = r = 25.9 mm>s

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) mm

r

12–189. The box slides down the helical ramp with a constant speed of v = 2 m>s. Determine the magnitude of its acceleration. The ramp descends a vertical distance of 1 m for every full revolution. The mean radius of the ramp is r = 0.5 m.

0.5 m

SOLUTION Velocity: The inclination angle of the ramp is f = tan-1

L 1 = tan-1 B R = 17.66°. 2pr 2p(0.5)

Thus, from Fig. a, vu = 2 cos 17.66° = 1.906 m>s and vz = 2 sin 17.66° = 0.6066 m>s. Thus, # vu = ru # 1.906 = 0.5u # u = 3.812 rad>s

Wide

copyright

in

teaching

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laws

or

# $ # $ Acceleration: Since r = 0.5 m is constant, r = r = 0. Also, u is constant, then u = 0. Using the above results, # $ ar = r - r u2 = 0 - 0.5(3.812)2 = - 7.264 m>s2 $ ## au = r u + 2ru = 0.5(0) + 2(0)(3.812) = 0

States

World

permitted.

Dissemination

Since vz is constant az = 0. Thus, the magnitude of the box’s acceleration is instructors

( -7.264)2 + 02 + 02 = 7.26 m>s2

the

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ar 2 + au 2 + az 2 =

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a =






12–190. The box slides down the helical ramp which is defined by r = 0.5 m, u = (0.5t3) rad, and z = (2 – 0.2t2) m, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the box at the instant u = 2p rad.

0.5 m

SOLUTION Time Derivatives: r = 0.5 m # $ r = r = 0 # u = A 1.5t2 B rad>s

$ u = (3t) rad>s2

z = 2 - 0.2t2 $ z = - 0.4 m>s2

# z = ( -0.4t) m>s

or

When u = 2p rad, t = 2.325 s

teaching

Web)

laws

2p = 0.5t3

in

Thus,

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States

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Dissemination

Wide

copyright

# u t = 2.325 s = 1.5(2.325)2 = 8.108 rad>s $ u t = 2.325 s = 3(2.325) = 6.975 rad>s2 the

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# z t = 2.325 s = - 0.4(2.325) = - 0.92996 m>s

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$ z t = 2.325 s = - 0.4 m>s2

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# vz = z = - 0.92996 m>s

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# vr = r = 0 # vu = ru = 0.5(8.108) = 4.05385 m>s

this

assessing

is

Velocity:

Thus, the magnitude of the box’s velocity is v = 2vr 2 + vu 2 + vz 2 = 202 + 4.053852 + ( -0.92996)2 = 4.16 m>s

Ans.

Acceleration: # $ a r = r - r u2 = 0 - 0.5(8.108)2 = - 32.867 m>s2 $ # # a u = r u + 2r u = 0.5(6.975) + 2(0)(8.108)2 = 3.487 m>s2 $ az = z = - 0.4 m>s2 Thus, the magnitude of the box’s acceleration is a =






ar 2 + au 2 + az 2 =

( -32.867)2 + 3.4872 + ( -0.4)2 = 33.1 m>s2 Ans.

12–191. For a short distance the train travels along a track having the shape of a spiral, r = 11000>u2 m, where u is in radians. If it maintains a constant speed v = 20 m>s, determine the radial and transverse components of its velocity when u = 19p>42 rad.

r u

r

SOLUTION 1000 u

r =

1000 # # r = - 2 u u Since # # v2 = (r)2 + (r u)2

or

# (1 + u2)(u)2

Web)

u4

laws

(1000)2

in

(20)2 =

(1000)2 # 2 (1000)2 # 2 (u) + (u) 4 u u2

teaching

(20)2 =

Dissemination

Wide

copyright

Thus,

permitted.

# u =

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States

World

0.02u2

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21 + u2

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At u =

integrity

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Ans.






sale

# 1000 (0.140) = 19.8 m>s vu = r u = (9p/4)

will

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# vr = r = -2.80 m>s

and

-1000 # r = (0.140) = -2.80 (9p/4)2

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is

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the

# u = 0.140

Ans.

1000 u

12–192. For a short distance the train travels along a track having the shape of a spiral, r = 11000>u2 m, where u is in radians. # If the angular rate is constant, u = 0.2 rad>s, determine the radial and transverse components of its velocity and acceleration when u = 19p>42 rad.

r u

r

SOLUTION # u = 0.2 $ u = 0 r =

1000 u

# # r = -1000(u - 2) u # $ $ r = 2000(u - 3)(u)2 - 1000(u - 2) u 9p 4 or

When u =

teaching

Web)

laws

r = 141.477

Wide

copyright

in

# r = -4.002812 States

World

permitted.

Dissemination

$ r = 0.226513

not

instructors

# vr = r = -4.00 m>s # vu = r u = 141.477(0.2) = 28.3 m>s # $ ar = r - r(u)2 = 0.226513 - 141.477(0.2)2 = -5.43 m>s2 $ ## au = r u + 2ru = 0 + 2( -4.002812)(0.2) = - 1.60 m>s2

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Ans.






Ans.

1000 u

12–193. A particle moves along an Archimedean spiral r = (8u) ft, # where u is given in radians. If u = 4 rad>s (constant), determine the radial and transverse components of the particle’s velocity and acceleration at the instant u = p>2 rad. Sketch the curve and show the components on the curve.

y

r

(8 u) ft

r u

SOLUTION

# $ Time Derivatives: Since u is constant, u = 0. p r = 8u = 8a b = 4p ft 2

x

# # r = 8 u = 8(4) = 32.0 ft>s

$ $ r = 8u = 0

Velocity: Applying Eq. 12–25, we have # v r = r = 32.0 ft>s # vu = ru = 4p (4) = 50.3 ft>s

Ans. Ans.

or

Acceleration: Applying Eq. 12–29, we have # $ ar = r - ru2 = 0 - 4p A 42 B = - 201 ft>s2 $ # # au = ru + 2r u = 0 + 2(32.0)(4) = 256 ft>s2

teaching

Web)

laws

Ans.

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of

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the

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for

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on

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is

of

the

not

instructors

States

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permitted.

Dissemination

Wide

copyright

in

Ans.






12–194. Solve Prob. $12–193 if the particle has an angular # acceleration u = 5 rad>s2 when u = 4 rad>s at u = p 2 rad.

y

r

(8 u) ft

r u

SOLUTION

x

Time Derivatives: Here, p r = 8u = 8 a b = 4p ft 2 $ $ r = 8 u = 8(5) = 40 ft>s2

# # r = 8 u = 8(4) = 32.0 ft>s

Velocity: Applying Eq. 12–25, we have # vr = r = 32.0 ft>s # vu = r u = 4p(4) = 50.3 ft>s

Ans. Ans.

laws

or

Acceleration: Applying Eq. 12–29, we have # $ ar = r - r u2 = 40 - 4p A 42 B = - 161 ft>s2 $ # # au = r u + 2r u = 4p(5) + 2(32.0)(4) = 319 ft>s2

sale

will

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and

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is

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of

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is

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of

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the

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for

work

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of

the

not

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permitted.

Dissemination

copyright

Ans.

Wide

in

teaching

Web)

Ans.






12–195. The arm of the robot has a length of r = 3 ft grip A moves along the path z = 13 sin 4u2 ft, where u is in radians. If u = 10.5t2 rad, where t is in seconds, determine the magnitudes of the grip’s velocity and acceleration when t = 3 s.

A

z r u

SOLUTION u = 0.5 t # u = 0.5 $ u = 0

r = 3

z = 3 sin 2t

# r = 0

# z = 6 cos 2t

$ r = 0

$ z = - 12 sin 2t

At t = 3 s, z = -0.8382 # z = 5.761

or

$ z = 3.353

teaching

Web)

laws

vr = 0

Wide

copyright

in

vu = 3(0.5) = 1.5

Ans.

United

on

learning.

is

of

the

v = 2(0)2 + (1.5)2 + (5.761)2 = 5.95 ft>s

student

the

by

and

use

ar = 0 - 3(0.5)2 = -0.75

protected

the

(including

for

work

au = 0 + 0 = 0

this

assessing

is

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of

az = 3.353

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and

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a = 2(- 0.75)2 + (0)2 + (3.353)2 = 3.44 ft>s2






Ans.

not

instructors

States

World

permitted.

Dissemination

vz = 5.761

*12–196. For a short time the arm of the robot is extending at a # constant rate such that r = 1.5 ft>s when r = 3 ft, z = 14t22 ft, and u = 0.5t rad, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the grip A when t = 3 s.

A

z r u

SOLUTION u = 0.5 t rad # u = 0.5 rad>s $ u = 0

r = 3 ft

z = 4 t2 ft

# r = 1.5 ft>s

# z = 8 t ft>s

$ r = 0

$ z = 8 ft>s2

At t = 3 s, r = 3

z = 36

# r = 1.5

# z = 24

$ r = 0

$ z = 8 or

u = 1.5 # u = 0.5 $ u = 0

teaching

Web)

laws

vr = 1.5

Wide

copyright

in

vu = 3(0.5) = 1.5

instructors

States

World

permitted.

Dissemination

vz = 24

Ans.

United

on

learning.

is

of

the

not

v = 2(1.5)2 + (1.5)2 + (24)2 = 24.1 ft>s

student

the

by

and

use

ar = 0 - 3(0.5)2 = - 0.75

protected

the

(including

for

work

au = 0 + 2(1.5)(0.5) = 1.5

this

assessing

is

work

solely

of

az = 8

sale

will

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of

and

any

courses

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integrity

of

and

work

a = 2( -0.75)2 + (1.5)2 + (8)2 = 8.17 ft>s2






Ans.

12–197. The partial surface of the cam is that of a logarithmic spiral r = (40e0.05u) mm, where u is in #radians. If the cam is rotating at a constant angular rate of u = 4 rad>s, determine the magnitudes of the velocity and acceleration of the follower rod at the instant u = 30°. u

SOLUTION r = 40e0.05 u # # r = 2e 0.05u u

·

u

# 2 $ $ r = 0.1e 0.0 5 u a u b + 2e 0.05 u u p 6

u =

# u = -4 $ u = 0 r = 40e 0.05A 6 B = 41.0610 or

p

teaching

Web)

laws

p # r = 2e 0.05A 6 B (-4) = -8.2122

Wide

copyright

in

p $ r = 0.1e 0.05A 6 B (-4) 2 + 0 = 1.64244

of

the

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Ans.

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and

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is

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solely

of

protected

the

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work

student

the

by

and

use

United

on

learning.

is

Ans.






permitted.

Dissemination

# v = r = -8.2122 = 8.21 mm >s # $ a = r - r u2 = 1.642 44 - 41.0610(-4) 2 = - 665.33 = -665 mm>s 2

4 rad/s

r

40e0.05u

12–198. Solve if the cam has an angular# acceleration $ Prob. 12–197, 2 of u = 2 rad>s when its angular velocity is u = 4 rad>s at # u = 30°.

u

SOLUTION ·

r = 40e0.05u # # r = 2e0.05uu

u ⫽ 4 rad/s

# $ $ r = 0.1e0.05u A u B 2 + 2e0.05uu

u =

p p $ r = 0.1e0.05A 6 B( -4)2 + 2e0.05A 6 B(- 2) = - 2.4637

# v = r = 8.2122 = 8.21 mm>s

p 6

# u = -4 $ u = -2

# $ a = r - ru 2 = - 2.4637 - 41.0610(- 4)2 = - 659 mm>s2 or

r = 40e0.05A 6 B = 41.0610

teaching

Web)

laws

p

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p # r = 2e0.05AA 6 B ( -4) = - 8.2122






r ⫽ 40e0.05u

12–199. If the end of the cable at A is pulled down with a speed of 2 m>s, determine the speed at which block B rises. D C 2 m/s

A

SOLUTION Position-Coordinate Equation: Datum is established at fixed pulley D. The position of point A, block B and pulley C with respect to datum are sA, sB, and sC respectively. Since the system consists of two cords, two position-coordinate equations can be derived. (sA - sC) + (sB - sC) + sB = l1

(1)

sB + sC = l2

(2)

B

Eliminating sC from Eqs. (1) and (2) yields sA + 4sB = l1 = 2l2 laws

or

Time Derivative: Taking the time derivative of the above equation yields

Web)

(3) Wide

copyright

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teaching

vA + 4vB = 0

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permitted.

Dissemination

Since vA = 2 m>s, from Eq. (3) the

not

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2 + 4vB = 0

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vB = - 0.5 m>s = 0.5 m>s c






Ans.

*12–200. The motor at C pulls in the cable with an acceleration aC = (3t2) m>s2, where t is in seconds. The motor at D draws in its cable at aD = 5 m>s2. If both motors start at the same instant from rest when d = 3 m, determine (a) the time needed for d = 0, and (b) the relative velocity of block A with respect to block B when this occurs.

D B A d=3m

SOLUTION For A: sA + (sA - sC) = l 2vA = vC 2aA = aC = - 3t2 aA = - 1.5t2 = 1.5t2 vA = 0.5t3

:

: : laws

or

sA = 0.125 t4

teaching

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; instructors

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vB = 5t

;

copyright

aB = 5 m>s2

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0.125u2 + 2.5u = 3 part

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t = 1.0656 = 1.07 s

and

The positive root is u = 1.1355. Thus,

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work

Set u = t2

vA = 0.5(1.0656)3 = 0.6050 vB = 5(1.0656) = 5.3281 m>s vA = vB + vA>B 0.6050i = - 5.3281i + vA>B i vA B = 5.93 m s






:

Ans.

C

12–201. The crate is being lifted up the inclined plane using the motor M and the rope and pulley arrangement shown. Determine the speed at which the cable must be taken up by the motor in order to move the crate up the plane with a constant speed of 4 ft>s.

B

A C

SOLUTION Position-Coordinate Equation: Datum is established at fixed pulley B. The position of point P and crate A with respect to datum are sP and sA, respectively. 2sA + (sA - sP) = l 3sA - sP = 0 Time Derivative: Taking the time derivative of the above equation yields 3vA - vP = 0

(1)

Since vA = 4 ft>s, from Eq. [1] or

3(4) - vP = 0

laws

vP = 12 ft>s

Ans.

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(+ )






M

12–202. Determine the time needed for the load at B to attain a speed of 8 m>s, starting from rest, if the cable is drawn into the motor with an acceleration of 0.2 m>s2.

A vA

SOLUTION 4 sB + sA = l

B

4 nB = - vA 4 aB = - aA 4 aB = -0.2 aB = -0.05 m>s2 (+T)

vB = (vB)0 + aB t -8 = 0 - (0.05)(t) t = 160 s

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Ans.






vB

12–203. Determine the displacement of the log if the truck at C pulls the cable 4 ft to the right. B

SOLUTION 2sB + (sB - sC) = l 3sB - sC = l 3¢sB - ¢sC = 0 Since ¢sC = -4, then 3¢sB = -4 ¢sB = -1.33 ft = 1.33 ft :

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Ans.






C

*12–204. Determine the speed of cylinder A, if the rope is drawn towards the motor M at a constant rate of 10 m>s.

SOLUTION Position Coordinates: By referring to Fig. a, the length of the rope written in terms of the position coordinates sA and sM is A

3sA + sM = l

h

Time Derivative: Taking the time derivative of the above equation,

A+TB

3vA + vM = 0

Here, vM = 10 m>s. Thus, 3vA + 10 = 0 vA = -3.33 m>s = 3.33 m>s c

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Ans.






M

12–205. If the rope is drawn toward the motor M at a speed of vM = (5t3>2) m>s, where t is in seconds, determine the speed of cylinder A when t = 1 s.

SOLUTION Position Coordinates: By referring to Fig. a, the length of the rope written in terms of the position coordinates sA and sM is A

3sA + sM = l

M

h

Time Derivative: Taking the time derivative of the above equation,

A+TB

3vA + vM = 0

Here, vM = A 5t3>2 B m>s. Thus, 3vA + 5t3>2 = 0 or

5 5 vA = ¢ - t3>2 ≤ m>s = ¢ t3>2 ≤ m>s 2 = 1.67 m>s 3 3 t=1 s

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Ans.






12–206. If the hydraulic cylinder H draws in rod BC at 2 ft>s, determine the speed of slider A.

A B

C

H

SOLUTION 2sH + sA = l 2vH = -vA 2(2) = -vA vA = -4 ft>s = 4 ft>s ;

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12–207. If block A is moving downward with a speed of 4 ft>s while C is moving up at 2 ft>s, determine the speed of block B.

SOLUTION

B C

sA + 2sB + sC = l

A

vA + 2vB + vC = 0 4 + 2vB - 2 = 0 vB = -1 ft>s = 1 ft>s c

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*12–208. If block A is moving downward at 6 ft>s while block C is moving down at 18 ft>s, determine the speed of block B.

SOLUTION

B C

sA + 2sB + sC = l

A

vA + 2vB + vC = 0 6 + 2vB + 18 = 0 vB = -12 ft/s = 12 ft>s c

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Ans.






12–209. Determine the displacement of the block B if A is pulled down 4 ft.

C

SOLUTION

A

2 sA + 2 sC = l1 B

¢sA = - ¢sC sB - sC + sB = l2 2 ¢sB = ¢sC Thus, 2 ¢sB = - ¢sA

or

2 ¢sB = -4 ¢sB = - 2 ft = 2 ft c

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laws

Ans.






12–210. The pulley arrangement shown is designed for hoisting materials. If BC remains fixed while the plunger P is pushed downward with a speed of 4 ft>s, determine the speed of the load at A. B C P

SOLUTION 5 sB + (sB - sA) = l 6 sB - sA = l 6 vB - vA = 0 6(4) = vA vA = 24 ft>s

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4 ft/s A

12–211. Determine the speed of block A if the end of the rope is pulled down with a speed of 4 m>s.

4 m/s B

SOLUTION A

Position Coordinates: By referring to Fig. a, the length of the cord written in terms of the position coordinates sA and sB is sB + sA + 2(sA - a) = l sB + 3sA = l + 2a Time Derivative: Taking the time derivative of the above equation, vB + 3vA = 0 or

(+ T )

in

teaching

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laws

Here, vB = 4 m>s. Thus, vA = - 133 m>s = 1.33 m>s c

Ans.

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4 + 3vA = 0






*12–212. The cylinder C is being lifted using the cable and pulley system shown. If point A on the cable is being drawn toward the drum with a speed of 2 m>s, determine the speed of the cylinder.

A vA

SOLUTION l = sC + (sC - h) + (sC - h - sA)

C

l = 3sC - 2h - sA s

0 = 3vC - vA vA -2 = = -0.667 m>s = 0.667 m>s c 3 3

Ans.

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vC =






12–213. The man pulls the boy up to the tree limb C by walking backward at a constant speed of 1.5 m>s. Determine the speed at which the boy is being lifted at the instant xA = 4 m. Neglect the size of the limb. When xA = 0, yB = 8 m, so that A and B are coincident, i.e., the rope is 16 m long.

C

yB 8m B A

SOLUTION Position-Coordinate Equation: Using the Pythagorean theorem to determine lAC, we have lAC = 2x2A + 82. Thus, l = lAC + yB

xA

16 = 2x2A + 82 + yB yB = 16 - 2x2A + 64

(1)

or

dxA Time Derivative: Taking the time derivative of Eq. (1) and realizing that yA = dt dyB and yB = , we have dt laws

dyB xA dxA = 2 dt 2xA + 64 dt xA yB = yA 2 2xA + 64

(2)

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yB =

by

(1.5) = - 0.671 m>s = 0.671 m>s c

Ans. work

24 + 64

student

the

4

the

(including

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for

yB = -

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use

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is

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the

not

At the instant xA = 4 m, from Eq. [2]

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Note: The negative sign indicates that velocity yB is in the opposite direction to that of positive yB.






12–214. The man pulls the boy up to the tree limb C by walking backward. If he starts from rest when xA = 0 and moves backward with a constant acceleration a A = 0.2 m>s2, determine the speed of the boy at the instant yB = 4 m. Neglect the size of the limb.When xA = 0, yB = 8 m, so that A and B are coincident, i.e., the rope is 16 m long.

C

yB 8m B A

SOLUTION Position-Coordinate Equation: Using the Pythagorean theorem to determine lAC, we have lAC = 2x2A + 82. Thus, l = lAC + yB

xA

16 = 2x2A + 82 + yB yB = 16 - 2x2A + 64

(1)

or

dxA Time Derivative: Taking the time derivative of Eq. (1) Where vA = and dt dyB vB = , we have dt laws

dyB xA dxA = 2 dt 2xA + 64 dt xA vA vB = 2 2xA + 64

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vB =

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At the instant yB = 4 m, from Eq. (1), 4 = 16 - 2x2A + 64, xA = 8.944 m. The velocity of the man at that instant can be obtained.

the

(including

for

work

student

v2A = (v0)2A + 2(ac)A C sA - (s0)A D

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v2A = 0 + 2(0.2)(8.944 - 0)

the

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vA = 1.891 m>s

destroy

(1.891) = -1.41 m>s = 1.41 m>s c

28.9442 + 64

Ans.

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8.944

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vB = -

of

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courses

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Substitute the above results into Eq. (2) yields

Note: The negative sign indicates that velocity vB is in the opposite direction to that of positive yB.






12–215. The roller at A is moving upward with a velocity of vA = 3 ft>s and has an acceleration of aA = 4 ft>s2 when sA = 4 ft. Determine the velocity and acceleration of block B at this instant.

A

3 ft

SOLUTION

B

sB + 3 1sA22 + 32 = l - 12 # 1 # sB + C (sA)2 + 32 D (2sA) sA = 0 2 # - 12 # sB + C s2A + 9 D asA sA b = 0 3 1 1 $ # # $ sB - C (sA)2 + 9 D - 2 as2A s2A b + C s2A + 9 D - 2 a s2A b + C s2A + 9 D - 2 asAsA b = 0

# $ At sA = 4 ft, sA = 3 ft>s, sA = 4 ft>s2

or

1 # sB + a b (4)(3) = 0 5 laws

vB = -2.4 ft>s = 2.40 ft>s :

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1 3 1 1 ## s B - a b 14221322 + a b 1322 + a b 142142 = 0 5 5 5

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Ans.






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States

aB = -3.85 ft>s2 = 3.85 ft>s2 :

sA

4 ft

*12–216. The girl at C stands near the edge of the pier and pulls in the rope horizontally at a constant speed of 6 ft>s. Determine how fast the boat approaches the pier at the instant the rope length AB is 50 ft.

6 ft/s

xC

C

A

8 ft B xB

SOLUTION The length l of cord is 2(8)2 + x2B + xC = l Taking the time derivative: # # 1 [(8)2 + x2B] - 1/2 2 xBxB + xC = 0 2

(1)

# xC = 6 ft/s When AB = 50 ft,

teaching

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laws

or

xB = 2(50)2 - (8)2 = 49.356 ft

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From Eq. (1)

is

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Dissemination

# 1 [(8)2 + (49.356)2] - 1/2 2(49.356)(xB) + 6 = 0 2 # xB = - 6.0783 = 6.08 ft>s ;

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Ans.






12–217. The crate C is being lifted by moving the roller at A downward with a constant speed of vA = 2 m>s along the guide. Determine the velocity and acceleration of the crate at the instant s = 1 m. When the roller is at B, the crate rests on the ground. Neglect the size of the pulley in the calculation. Hint: Relate the coordinates xC and xA using the problem geometry, then take the first and second time derivatives.

4m B xA xC 4m

SOLUTION

s

xC + 2x2A + (4)2 = l # # 1 xC + (x2A + 16) - 1/2(2xA)(xA) = 0 2 1 # 2 $ # 2 $ xC - (x2A + 16) - 3/2 (2x2A)(xA ) + (x2A + 16) - 1/2 (xA) + (x2A + 16) - 1/2 (xA)(xA) = 0 2 l = 8 m , and when s = 1 m , xC = 3 m laws

or

xA = 3 m teaching

Web)

# vA = xA = 2 m>s

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$ aA = x A = 0

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vC + [(3)2 + 16] - 1/2 (3)(2) = 0

Ans.

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vC = - 1.2 m>s = 1.2 m>s c

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the

aC - [(3)2 + 16] - 3/2 (3)2(2)2 + [(3)2 + 16] - 1/2 (2)2 + 0 = 0

Ans.

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aC = - 0.512 m>s2 = 0.512 m>s2 c






A C

12–218. The man can row the boat in still water with a speed of 5 m>s. If the river is flowing at 2 m>s, determine the speed of the boat and the angle u he must direct the boat so that it travels from A to B.

B

vw

5 m/s u

SOLUTION

A

Solution I

25 m

Vector Analysis: Here, the velocity vb of the boat is directed from A to B. Thus, f = tan - 1 a

50 b = 63.43°. The magnitude of the boat’s velocity relative to the 25

flowing river is vb>w = 5 m>s. Expressing vb, vw, and vb/w in Cartesian vector form, we have vb = vb cos 63.43i + vb sin 63.43j = 0.4472vb i + 0.8944vb j, vw = [2i] m>s, and vb>w = 5 cos ui + 5 sin uj. Applying the relative velocity equation, we have vb = vw + vb>w

or

0.4472vb i + 0.8944vb j = 2i + 5 cos ui + 5 sin uj

teaching

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laws

0.4472vb i + 0.8944vb j = (2 + 5 cos u)i + 5 sin uj

(1)

instructors

States

World

Dissemination

0.4472vb = 2 + 5 cos u

(2)

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0.8944vb = 5 sin u

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u = 84.4°

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vb = 5.56 m>s

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Solution II

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52 = 22 + vb 2 - 2(2)(vb) cos 63.43°

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This

Scalar Analysis: Referring to the velocity diagram shown in Fig. a and applying the law of cosines,

vb 2 - 1.789vb - 21 = 0 vb =

-(-1.789) ; 2( -1.789)2 - 4(1)(- 21) 2(1)

Choosing the positive root, vb = 5.563 m>s = 5.56 m>s

Ans.

Using the result of vb and applying the law of sines, sin (180° - u) sin 63.43° = 5.563 5 u = 84.4°






Ans.

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Equating the i and j components, we have

50 m

2 m/s

12–219. Vertical motion of the load is produced by movement of the piston at A on the boom. Determine the distance the piston or pulley at C must move to the left in order to lift the load 2 ft. The cable is attached at B, passes over the pulley at C, then D, E, F, and again around E, and is attached at G.

A

6 ft/s

B

C

D G

E F

SOLUTION 2 sC + 2 s F = l 2 ¢sC = - 2 ¢sF ¢sC = - ¢sF ¢sC = -( - 2 ft) = 2 ft

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Dissemination

Wide

copyright

in

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laws

or

Ans.






12–220. If block B is moving down with a velocity vB and has an acceleration a B , determine the velocity and acceleration of block A in terms of the parameters shown.

sA A

h vB, aB B

SOLUTION l = sB + 3s2B + h2 1 # # 0 = sB + (s2A + h2) - 1/2 2sA sA 2 # - sB(s2A + h2)1/2 # vA = sA = sA vA = - vB (1 + a

h 2 1/2 b ) sA

Ans.

laws

or

h 2 h 2 1 # # # aA = vA = - vB(1 + a b )1/2 - vB a b (1 + a b ) - 1/2(h2)( - 2)(sA) - 3sA sA 2 sA h 2 1/2 vAvBh2 h 2 - 1/2 b ) + (1 + a b ) sB sA s3A

in

teaching

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learning.

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permitted.

Dissemination

Wide

copyright

aA = - aB(1 + a






12–221. Collars A and B are connected to the cord that passes over the small pulley at C. When A is located at D, B is 24 ft to the left of D. If A moves at a constant speed of 2 ft>s to the right, determine the speed of B when A is 4 ft to the right of D.

C

10 ft B

A D

SOLUTION l = 312422 + 11022 + 10 = 36 ft 311022 + s2B + 3 11022 + s2A = 36 1 1 1 1 # # (100 + s2B)- 2 (2s B sB) + (100 + s2A)- 2 (2sAsA) = 0 2 2 1 # sA sA 100 + s2B 2 # ba b sB = - a sB 100 + s2A

At sA = 4,

or

311022 + s2B + 3 11022 + 1422 = 36

in

teaching

Web)

laws

sB = 23.163 ft

Dissemination

Wide

copyright

Thus, 1

permitted.

4(2) 100 + (23.163)2 2 # b = - 0.809 ft>s = 0.809 ft>s : sB = - a ba 23.163 100 + 42

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Ans.

2 ft/s

12–222. Two planes, A and B, are flying at the same altitude. If their velocities are vA = 600 km>h and vB = 500 km>h such that the angle between their straight-line courses is u = 75°, determine the velocity of plane B with respect to plane A.

A vA B

u

SOLUTION

vB

vB = vA + vB/A 75°

[500 ; ] = [600 c u ] + vB/A + ) (;

500 = -600 cos 75° + (vB/A)x (vB/A)x = 655.29 ;

(+ c)

0 = -600 sin 75° + (vB/A)y (vB/A)y = 579.56 c 1vB>A2 = 31655.2922 +1579.5622 vB/A = 875 km/h

Web)

laws

or

Ans. teaching

579.56 b = 41.5° b 655.29

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Dissemination

Wide

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u = tan-1 a






12–223. At the instant shown, cars A and B are traveling at speeds of 55 mi/h and 40 mi/h, respectively. If B is increasing its speed by 1200 mi>h2, while A maintains a constant speed, determine the velocity and acceleration of B with respect to A. Car B moves along a curve having a radius of curvature of 0.5 mi.

vB = 40 mi/h B A vA = 55 mi/h

SOLUTION vB = -40 cos 30°i + 40 sin 30°j = {-34.64i + 20j} mi>h vA = { -55i } mi>h vB>A = nB - nA = ( - 34.64i + 20j) - (- 55i) = {20.36i + 20j} mi>h vB>A = 220.362 + 202 = 28.5 mi>h

Ans.

20 = 44.5° 20.36

Ans. (aB)t = 1200 mi>h2

or

v2A 402 = = 3200 mi>h2 r 0.5

teaching

Web)

(aB)n =

a

laws

u = tan - 1

Wide

copyright

in

aB = (3200 cos 60° - 1200 cos 30°)i + (3200 sin 60° + 1200 sin 30°)j

instructors

States

World

permitted.

Dissemination

= {560.77i + 3371.28j} mi>h2

United

on

learning.

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aA = 0

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= {560.77i + 3371.28j} - 0 = {560.77i + 3371.28j} mi>h2

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aB>A = aB - aA

Ans.

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3371.28 = 80.6° 560.77

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u = tan - 1

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aB>A = 2(560.77)2 + (3371.28)2 = 3418 mi>h2






30°

12–224. At the instant shown, car A travels along the straight portion of the road with a speed of 25 m>s. At this same instant car B travels along the circular portion of the road with a speed of 15 m>s. Determine the velocity of car B relative to car A.

15 r

A

SOLUTION

B

vA = [25 cos 30° i - 25 sin 30° j] m>s = [21.65i - 12.5j] m>s vB = [15 cos 15° i - 15 sin 15° j] m>s = [14.49i - 3.882j] m>s Applying the relative velocity equation, vB = vA + vB>A 14.49i - 3.882j = 21.65i - 12.5j + vB>A

laws

or

vB>A = [- 7.162i + 8.618j] m>s teaching

Web)

Thus, the magnitude of vB/A is given by in

vB>A = 2( -7.162)2 + 8.6182 = 11.2 m>s

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permitted.

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Wide

copyright

Ans.

The direction angle uv of vB/A measured down from the negative x axis, Fig. b is learning.

is

8.618 b = 50.3° d 7.162

on United

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sale or transmission in any form or by any means, electronic, mechanical,



Ans.

use by

the

uv = tan - 1 a


C 30

Velocity: Referring to Fig. a, the velocity of cars A and B expressed in Cartesian vector form are


200 m

15

12–225. An aircraft carrier is traveling forward with a velocity of 50 km>h. At the instant shown, the plane at A has just taken off and has attained a forward horizontal air speed of 200 km>h, measured from still water. If the plane at B is traveling along the runway of the carrier at 175 km>h in the direction shown, determine the velocity of A with respect to B.

B 15

SOLUTION

50 km/h

vB = vC + vB>C v B = 50i + 175 cos 15°i + 175 sin 15°j = 219.04i + 45.293j vA = vB + vA>B 200i = 219.04i + 45.293j + (vA>B)xi + (vA>B)y j 200 = 219.04 + (vA>B)x 0 = 45.293 + (vA>B)y (vA>B)x = - 19.04

Web)

laws

or

(vA>B)y = -45.293

teaching

vA>B = 2( -19.04)2 + ( -45.293)2 = 49.1 km>h

Wide

copyright

in

Ans. Dissemination

45.293 b = 67.2° d 19.04

World

permitted.

Ans.

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u = tan - 1 a






A

12–226. A car is traveling north along a straight road at 50 km>h. An instrument in the car indicates that the wind is directed toward the east. If the car’s speed is 80 km>h, the instrument indicates that the wind is directed toward the north-east. Determine the speed and direction of the wind.

SOLUTION Solution I Vector Analysis: For the first case, the velocity of the car and the velocity of the wind relative to the car expressed in Cartesian vector form are vc = [50j] km>h and vW>C = (vW>C)1 i. Applying the relative velocity equation, we have vw = vc + vw>c vw = 50j + (vw>c)1 i vw = (vw>c)1i + 50j

(1)

laws

or

For the second case, vC = [80j] km>h and vW>C = (vW>C)2 cos 45°i + (vW>C)2 sin 45° j. Applying the relative velocity equation, we have

Wide

copyright

in

teaching

Web)

vw = vc + vw>c

permitted.

Dissemination

vw = 80j + (vw>c)2 cos 45°i + (vw>c)2 sin 45° j States

World

vw = (vw>c)2 cos 45° i + C 80 + (vw>c)2 sin 45° D j United

on

learning.

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the

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(2)

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by

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use

Equating Eqs. (1) and (2) and then the i and j components,

(3)

the

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work

student

(vw>c)1 = (vw>c)2 cos 45°

(4)

this

assessing

is

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solely

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protected

50 = 80 + (vw>c)2 sin 45°

part

the

is

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(vw>c)1 = - 30 km>h destroy

of

and

any

courses

(vw>c)2 = -42.43 km>h

sale

will

their

Substituting the result of (vw>c)1 into Eq. (1), vw = [ -30i + 50j] km>h Thus, the magnitude of vW is vw = 2( - 30)2 + 502 = 58.3 km>h

Ans.

and the directional angle u that vW makes with the x axis is u = tan - 1 a






50 b = 59.0° b 30

Ans.

12–227. Two boats leave the shore at the same time and travel in the directions shown. If vA = 20 ft>s and vB = 15 ft>s, determine the velocity of boat A with respect to boat B. How long after leaving the shore will the boats be 800 ft apart?

vA A B 30 O

SOLUTION vA = vB + vA>B -20 sin 30°i + 20 cos 30°j = 15 cos 45°i + 15 sin 45°j + vA>B vA>B = {-20.61i + 6.714j} ft>s vA>B = 2( -20.61)2 + ( +6.714)2 = 21.7 ft>s u = tan - 1 (

Ans.

6.714 ) = 18.0° b 20.61

Ans.

(800)2 = (20 t)2 + (15 t)2 - 2(20 t)(15 t) cos 75° t = 36.9 s

or

Ans. laws

Also 800 800 = 36.9 s = vA>B 21.68

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Dissemination

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t =






45

vB

*12–228. At the instant shown, the bicyclist at A is traveling at 7 m/s around the curve on the race track while increasing his speed at 0.5 m>s2. The bicyclist at B is traveling at 8.5 m/s along the straight-a-way and increasing his speed at 0.7 m>s2. Determine the relative velocity and relative acceleration of A with respect to B at this instant.

vB = 8.5 m/s B 50 m

vA = vB + vA>B [7 |R ] = [8.5 : ] + [(vA>B)x : ] + [(vA>B)y T] 40° 7 sin 40° = 8.5 + (vA>B)x

(+ T )

7 cos 40° = (vA>B)y

Thus, (vA>B)x = 4.00 m>s ;

or

(vA>B)y = 5.36 m>s T

teaching

Web)

laws

(vA>B) = 2(4.00)2 + (5.36)2 vA>B = 6.69 m>s

Wide

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Ans.

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72 = 0.980 m>s2 50

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(aA)n =

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the

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aA = aB + aA>B work

= [0.7 : ] + [(aA>B)x : ] + [(aA>B)y T] assessing

this

40°

is

[0.980] ud 40° + [0.5] |R

provided

integrity

of

and

work

- 0.980 cos 40° + 0.5 sin 40° = 0.7 + (aA>B)x any of

and

(aA>B)x = 1.129 m>s2 ;

courses

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is

This

(+ : )

destroy

0.980 sin 40° + 0.5 cos 40° = (aA>B)y sale

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(+ T )

(aA>B)y = 1.013 m>s2 T (aA>B) = 2(1.129)2 + (1.013)2 aA>B = 1.52 m>s2 u = tan - 1






1.013 1.129

Ans. = 41.9° d

Ans.

permitted.

Dissemination

5.36 b = 53.3° d 4.00

States

u = tan - 1 a

50 m 40°

SOLUTION

+ ) (:

vA = 7 m/s

A

12–229. Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 90 ft>s and is increasing its speed at the rate of 15 ft>s2, whereas B has a speed of 105 ft>s and is decreasing its speed at 25 ft>s2. Determine the relative velocity and relative acceleration of car A with respect to car B at this instant.

vA A B vB rA

300 ft

60 rB

SOLUTION vA = vB + vA>B - 90i = -105 sin 30° i + 105 cos 30°j + vA>B vA>B = 5- 37.5i - 90.93j6 ft>s vA/B = 2( -37.5)2 + (- 90.93)2 = 98.4 ft>s

Ans.

90.93 b = 67.6° d 37.5

Ans.

u = tan - 1 a

aA = aB + aA>B j = 25 cos 60°i - 25 sin 60°j - 44.1 sin 60°i - 44.1 cos 60°j + aA>B teaching

Web)

300

or

19022

laws

-15i -

Wide

copyright

in

aA>B = {10.69i + 16.70j} ft>s2 aA>B = 2(10.69)2 + (16.70)2 = 19.8 ft>s2 instructors

States

World

permitted.

Dissemination

Ans. of

the

not

16.70 b = 57.4° a 10.69

United

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u = tan - 1 a






250 ft

12–230. The two cyclists A and B travel at the same constant speed v. Determine the speed of A with respect to B if A travels along the circular track, while B travels along the diameter of the circle.

v A r

f

B v

SOLUTION vA = v sin ui + v cos uj

vB = vi

vA>B = vA - vB = (v sin ui + v cos uj) - vi = (v sin u - v)i + v cos uj vA>B = 2(v sin u - v)2 + (v cos u)2

laws

or

= 22v2 - 2v2 sin u = v 22(1 - sin u)

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Ans.






u

12–231. At the instant shown, cars A and B travel at speeds of 70 mi>h and 50 mi>h, respectively. If B is increasing its speed by 1100 mi>h2, while A maintains a constant speed, determine the velocity and acceleration of B with respect to A.Car B moves along a curve having a radius of curvature of 0.7 mi.

A vA

Relative Velocity: vB = vA + vB>A 50 sin 30°i + 50 cos 30°j = 70j + vB>A vB>A = {25.0i - 26.70j} mi>h Thus, the magnitude of the relative velocity vB/A is yB>A = 225.02 + ( -26.70)2 = 36.6 mi>h

Ans.

laws

or

The direction of the relative velocity is the same as the direction of that for relative acceleration. Thus 26.70 = 46.9° c 25.0

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Ans. Wide

copyright

u = tan - 1

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of

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aB = aA + aB>A

the

(including

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work

student

(1100 sin 30° + 3571.43 cos 30°)i + (1100 cos 30° - 3571.43 sin 30°)j = 0 + aB>A

assessing

is

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solely

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protected

aB>A = {3642.95i - 833.09j} mi>h2

provided

integrity

of

and

work

this

Thus, the magnitude of the relative velocity aB/A is

Ans.

and

any

courses

part

the

is

This

aB>A = 23642.952 + (- 833.09)2 = 3737 mi>h2

their

destroy

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And its direction is




will

833.09 = 12.9° c 3642.95 sale

f = tan - 1

Ans.

permitted.

Dissemination

Relative Acceleration: Since car B is traveling along a curve, its normal y2B 502 acceleration is (aB)n = = 3571.43 mi>h2. Applying Eq. 12–35 gives = r 0.7


vB 30

SOLUTION


70 mi/h B

50 mi/h

12–232. At the instant shown, cars A and B travel at speeds of 70 mi>h and 50 mi>h, respectively. If B is decreasing its speed at 1400 mi>h2 while A is increasing its speed at 800 mi>h2, determine the acceleration of B with respect to A. Car B moves along a curve having a radius of curvature of 0.7 mi. A vA

Relative Acceleration: Since car B is traveling along a curve, its normal acceleration v2B 502 = is (aB)n = = 3571.43 mi>h2. Applying Eq. 12–35 gives r 0.7 aB = aA + aB>A (3571.43 cos 30° - 1400 sin 30°)i + ( -1400 cos 30° - 3571.43 sin 30°)j = 800j + aB>A aB>A = {2392.95i - 3798.15j} mi>h2 Thus, the magnitude of the relative acc. aB/A is aB>A = 22392.952 + (- 3798.15)2 = 4489 mi>h2

laws

or

Ans.

copyright

in

teaching

Web)


Ans. World

permitted.

Dissemination

Wide

3798.15 = 57.8° c 2392.95

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f = tan - 1





vB 30

SOLUTION


70 mi/h B

50 mi/h

12–233. A passenger in an automobile observes that raindrops make an angle of 30° with the horizontal as the auto travels forward with a speed of 60 km/h. Compute the terminal (constant)velocity vr of the rain if it is assumed to fall vertically.

vr va = 60 km/h

SOLUTION vr = va + vr>a -vr j = - 60i + vr>a cos 30°i - vr>a sin 30°j + ) (:

0 = - 60 + vr>a cos 30°

(+ c)

- vr = 0 - vr>a sin 30° vr>a = 69.3 km>h vr = 34.6 km h

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Dissemination

Wide

copyright

in

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laws

or

Ans.






12–234. A man can swim at 4 ft/s in still water. He wishes to cross the 40-ft-wide river to point B, 30 ft downstream. If the river flows with a velocity of 2 ft/s, determine the speed of the man and the time needed to make the crossing. Note: While in the water he must not direct himself toward point B to reach this point. Why?

30 ft B vr = 2 ft/s

A

SOLUTION Relative Velocity: vm = vr + vm>r 3 4 n i + vm j = 2i + 4 sin ui + 4 cos uj 5 m 5

(1)

4 v = 4 cos u 5 m

(2) teaching

Web)

laws

3 v = 2 + 4 sin u 5 m

or

Equating the i and j components, we have

Wide

copyright

in


States

World

permitted.

Dissemination

u = 13.29° instructors

vm = 4.866 ft>s = 4.87 ft>s

and

use

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Ans.

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solely

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protected

sAB 2402 + 302 = = 10.3 s vb 4.866

this

assessing

is

t =

for

work

student

the

by

Thus, the time t required by the boat to travel from points A to B is

sale

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is

This

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integrity

of

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work

In order for the man to reached point B, the man has to direct himself at an angle u = 13.3° with y axis.






40 ft

12–235. The ship travels at a constant speed of vs = 20 m>s and the wind is blowing at a speed of vw = 10 m>s, as shown. Determine the magnitude and direction of the horizontal component of velocity of the smoke coming from the smoke stack as it appears to a passenger on the ship.

vs

20 m/s

30 vw

10 m/s

45 y

x

SOLUTION Solution I Vector Analysis: The velocity of the smoke as observed from the ship is equal to the velocity of the wind relative to the ship. Here, the velocity of the ship and wind expressed in Cartesian vector form are vs = [20 cos 45° i + 20 sin 45° j] m>s = [14.14i + 14.14j] m>s and vw = [10 cos 30° i - 10 sin 30° j] = [8.660i - 5j] m>s. Applying the relative velocity equation, vw = vs + vw>s 8.660i - 5j = 14.14i + 14.14j + vw>s

or

vw>s = [- 5.482i - 19.14j] m>s

Web)

laws

Thus, the magnitude of vw/s is given by

teaching

vw = 2(-5.482)2 + ( -19.14)2 = 19.9 m>s

Wide

copyright

in

Ans.

instructors

States

World

permitted.

Dissemination

and the direction angle u that vw/s makes with the x axis is of

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19.14 b = 74.0° d 5.482

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Using the result of vw/s and applying the law of sines, sin f sin 75° = 10 19.91

f = 29.02°

Thus, u = 45° + f = 74.0° d






Ans.

*12–236. Car A travels along a straight road at a speed of 25 m>s while accelerating at 1.5 m>s2. At this same instant car C is traveling along the straight road with a speed of 30 m>s while decelerating at 3 m>s2. Determine the velocity and acceleration of car A relative to car C.

1.5 m/s r

100 m

15 m/s

Velocity: The velocity of cars A and C expressed in Cartesian vector form are vA = [ -25 cos 45°i - 25 sin 45°j] m>s = [- 17.68i - 17.68j] m>s vC = [-30j] m>s Applying the relative velocity equation, we have vA = vC + vA>C -17.68i - 17.68j = - 30j + vA>C

laws

or

vA>C = [ -17.68i + 12.32j] m>s

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Acceleration: The acceleration of cars A and C expressed in Cartesian vector form are

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aA = [ -1.5 cos 45°i - 1.5 sin 45°j] m>s2 = [ -1.061i - 1.061j] m>s2

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aC = [3j] m>s2

-1.061i - 1.061j = 3j + aA>C aA>C = [ -1.061i - 4.061j] m>s2 Thus, the magnitude of aA/C is given by aA>C = 2( -1.061)2 + (- 4.061)2 = 4.20 m>s2

Ans.

and the direction angle ua that aA/C makes with the x axis is ua = tan - 1 a






4.061 b = 75.4° d 1.061

3 m/s2

30

SOLUTION

12.32 b = 34.9° b 17.68

45

2

2 m/s2

uv = tan - 1 a

A

25 m/s

Ans.

B

C

30 m/s

12–237. Car B is traveling along the curved road with a speed of 15 m>s while decreasing its speed at 2 m>s2. At this same instant car C is traveling along the straight road with a speed of 30 m>s while decelerating at 3 m>s2. Determine the velocity and acceleration of car B relative to car C.

r

100 m

15 m/s

Velocity: The velocity of cars B and C expressed in Cartesian vector form are vB = [15 cos 60° i - 15 sin 60° j] m>s = [7.5i - 12.99j] m>s vC = [-30j] m>s Applying the relative velocity equation, vB = vC + vB>C 7.5i - 12.99j = - 30j + vB>C

or

vB>C = [7.5i + 17.01j] m>s

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laws

Thus, the magnitude of vB/C is given by vB>C = 27.52 + 17.012 = 18.6 m>s

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Acceleration: The normal component of car B’s acceleration is (aB)n =

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(aB)n = [2.25 cos 30° i + 2.25 sin 30° j] = [1.9486i + 1.125j] m>s2 aC = [3j] m>s2 Applying the relative acceleration equation, aB = aC + aB>C (- 1i + 1.732j) + (1.9486i + 1.125j) = 3j + aB>C aB>C = [0.9486i - 0.1429j] m>s2 Thus, the magnitude of aB/C is given by aB>C = 20.94862 + ( -0.1429)2 = 0.959 m>s2

Ans.

and the direction angle ua that aB/C makes with the x axis is ua = tan - 1 a






0.1429 b = 8.57° c 0.9486

3 m/s2

30

SOLUTION

=

45

1.5 m/s2

2 m/s2

uv = tan - 1 a

A

25 m/s

Ans.

B

C

30 m/s

12–238. At a given instant the football player at A throws a football C with a velocity of 20 m/s in the direction shown. Determine the constant speed at which the player at B must run so that he can catch the football at the same elevation at which it was thrown. Also calculate the relative velocity and relative acceleration of the football with respect to B at the instant the catch is made. Player B is 15 m away from A when A starts to throw the football.

A

15 m

Ball: + )s = s + v t (: 0 0 sC = 0 + 20 cos 60° t v = v0 + ac t

- 20 sin 60° = 20 sin 60° - 9.81 t t = 3.53 s

laws

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sC = 35.31 m teaching

Web)

Player B:

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vC = vB + vC>B

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(vC)y = 20 sin 60° = 17.32 m>s T

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(vC)x = 20 cos 60° = 10 m>s :

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- 17.32 = (vC>B)y (vC>B)x = 4.25 m>s : (vC>B)y = 17.32 m>s T vC>B = 2(4.25)2 + (17.32)2 = 17.8 m>s

Ans.

17.32 b = 76.2° 4.25

Ans.

u = tan - 1 a

c

aC = aB + aC>B - 9.81 j = 0 + aC>B aC B = 9.81 m s2 T






60° B

SOLUTION

(+ c )

C

20 m/s

Ans.

12–239. Both boats A and B leave the shore at O at the same time. If A travels at vA and B travels at vB, write a general expression to determine the velocity of A with respect to B.

A B

SOLUTION

O

Relative Velocity: vA = vB + vA>B vA j = vB sin ui + vB cos uj + vA>B vA>B = - vB sin ui + (vA - vB cos u)j Thus, the magnitude of the relative velocity vA>B is vA>B = 2( - vB sin u)2 + (vA - vB cos u)2 = 2vA2 + vB2 - 2vA vB cos u

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u = tan-1






13–1. z

The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t2i - 4tj - 1k6 lb, and F3 = 5- 2ti6 lb, where t is in seconds. Determine the distance the ball is from the origin 2 s after being released from rest.

F2

y F3

x

SOLUTION ©F = ma;

(2i + 6j - 2tk) + (t2i - 4tj - 1k) - 2ti - 6k = ¢

6 ≤ (axi + ay j + azk) 32.2

Equating components:

¢

6 ≤ a = t2 - 2t + 2 32.2 x

¢

6 ≤ a = - 4t + 6 32.2 y

6 ≤ a = -2t - 7 32.2 z

¢

Since dv = a dt, integrating from n = 0, t = 0, yields 6 t3 - t2 + 2t ≤ vx = 32.2 3

¢

6 ≤ v = - 2t2 + 6t 32.2 y

¢

6 ≤ v = - t2 - 7t 32.2 z laws

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¢

6 t3 7t2 ≤ sz = - 32.2 3 2

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6 2t3 + 3t2 ≤ sy = 32.2 3

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6 t4 t3 + t2 ≤ sx = 32.2 12 3

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Since ds = v dt, integrating from s = 0, t = 0 yields

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sz = -89.44 ft

the

sy = 35.78 ft

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sx = 14.31 ft,

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(14.31)2 + (35.78)2 + (- 89.44)2 = 97.4 ft

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s =






Ans.

F1

13–2. The 10-lb block has an initial velocity of 10 ft>s on the smooth plane. If a force F = 12.5t2 lb, where t is in seconds, acts on the block for 3 s, determine the final velocity of the block and the distance the block travels during this time.

v = 10 ft/s F = (2.5t) lb

SOLUTION + ©F = ma ; : x x

2.5t = ¢

10 ≤a 32.2

a = 8.05t dv = a dt n

L10

t

dv =

L0

8.05t dt

v = 4.025t2 + 10

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When t = 3 s, v = 46.2 ft>s

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ds = v dt

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s = 1.3417t3 + 10t

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s = 66.2 ft






Ans.

13–3. P

If the coefficient of kinetic friction between the 50-kg crate and the ground is mk = 0.3, determine the distance the crate travels and its velocity when t = 3 s. The crate starts from rest, and P = 200 N.

30⬚

SOLUTION Free-Body Diagram: The kinetic friction Ff = mkN is directed to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: Here, ay = 0. Thus, + c ©Fy = 0;

N - 50(9.81) + 200 sin 30° = 0 N = 390.5 N

+ ©Fx = max; :

200 cos 30° - 0.3(390.5) = 50a a = 1.121 m>s2

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v = v0 + act in

+ B A:

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Kinematics: Since the acceleration a of the crate is constant,

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1 (1.121) A 32 B = 5.04 m 2

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s = s0 + v0t +

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*13–4. P

If the 50-kg crate starts from rest and achieves a velocity of v = 4 m>s when it travels a distance of 5 m to the right, determine the magnitude of force P acting on the crate. The coefficient of kinetic friction between the crate and the ground is mk = 0.3.

30⬚

SOLUTION Kinematics: The acceleration a of the crate will be determined first since its motion is known. + ) (: v2 = v 2 + 2a (s - s ) c

0

0

42 = 02 + 2a(5 - 0) a = 1.60 m>s2 : Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.3N is required to be directed to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: or

N + P sin 30° - 50(9.81) = 50(0)

laws

+ c ©Fy = may;

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N = 490.5 - 0.5P

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Dissemination

Using the results of N and a, the

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P cos 30° - 0.3(490.5 - 0.5P) = 50(1.60)

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+ ©F = ma ; : x x

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P = 224 N






Ans.

13–5. The water-park ride consists of an 800-lb sled which slides from rest down the incline and then into the pool. If the frictional resistance on the incline is Fr = 30 lb, and in the pool for a short distance Fr = 80 lb, determine how fast the sled is traveling when s = 5 ft. 100 ft

SOLUTION + b a Fx = max;

800 a 32.2

800 sin 45° - 30 = a = 21.561 ft>s

s

2

v21 = v20 + 2ac(s - s0) v21 = 0 + 2(21.561)(10022 - 0)) v1 = = 78.093 ft>s - 80 =

800 a 32.2 or

+ ; a Fx = ma x;

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a = - 3.22 ft>s2

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v22 = v21 + 2ac(s2 - s1)

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Dissemination

v22 = (78.093)2 + 2(- 3.22)(5 - 0)

not

v2 = 77.9 ft>s

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100 ft

13–6. P

If P = 400 N and the coefficient of kinetic friction between the 50-kg crate and the inclined plane is mk = 0.25, determine the velocity of the crate after it travels 6 m up the plane. The crate starts from rest.

30°

SOLUTION

30°

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be directed down the plane to oppose the motion of the crate which is assumed to be directed up the plane. The acceleration a of the crate is also assumed to be directed up the plane, Fig. a. Equations of Motion: Here, ay ¿ = 0. Thus, ©Fy ¿ = may ¿;

N + 400 sin 30° - 50(9.81) cos 30° = 50(0) N = 224.79 N laws

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400 cos 30° - 50(9.81) sin 30° - 0.25(224.79) = 50a in

©Fx ¿ = may ¿;

or

Using the result of N,

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a = 0.8993 m>s2

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v2 = v0 2 + 2ac(s - s0)

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v2 = 0 + 2(0.8993)(6 - 0)

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v = 3.29 m>s






Ans.

13–7. P

If the 50-kg crate starts from rest and travels a distance of 6 m up the plane in 4 s, determine the magnitude of force P acting on the crate. The coefficient of kinetic friction between the crate and the ground is mk = 0.25.

30°

SOLUTION

30°

Kinematics: Here, the acceleration a of the crate will be determined first since its motion is known. s = s0 + v0t + 6 = 0 + 0 +

1 2 at 2 c

1 a(42) 2

a = 0.75 m>s2

in

teaching

Web)

laws

or

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be directed down the plane to oppose the motion of the crate which is directed up the plane, Fig. a.

Dissemination

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Equations of Motion: Here, ay ¿ = 0. Thus,

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permitted.

N + P sin 30° - 50(9.81) cos 30° = 50(0)

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©Fy ¿ = may ¿;

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N = 424.79 - 0.5P

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Using the results of N and a,

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P cos 30° - 0.25(424.79 - 0.5P) - 50(9.81) sin 30° = 50(0.75) assessing

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©Fx ¿ = max ¿;

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P = 392 N






*13–8. The speed of the 3500-lb sports car is plotted over the 30-s time period. Plot the variation of the traction force F needed to cause the motion.

v (ft/s) F

80 60

SOLUTION dv 60 Kinematics: For 0 … t 6 10 s. v = t = {6t} ft>s. Applying equation a = , 10 dt we have dv = 6 ft>s2 dt

a =

10

80 - 60 v - 60 = For 10 6 t … 30 s, , v = {t + 50} ft>s. Applying equation t - 10 30 - 10 dv a = , we have dt dv = 1 ft>s2 dt teaching

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a =

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Equation of Motion:

the

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3500 ≤ (1) = 109 lb 32.2

learning.

F = ¢

Ans.

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3500 ≤ (6) = 652 lb 32.2

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F = ¢

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Dissemination

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For 10 6 t … 30 s






Ans.

30

t (s)

13–9. p

The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. If the magnitude of P is increased until the crate begins to slide, determine the crate’s initial acceleration if the coefficient of static friction is ms = 0.5 and the coefficient of kinetic friction is mk = 0.3.

20

SOLUTION Equations of Equilibrium: If the crate is on the verge of slipping, Ff = ms N = 0.5N. From FBD(a), + c ©Fy = 0;

N + P sin 20° - 80(9.81) = 0

(1)

+ ©F = 0; : x

P cos 20° - 0.5N = 0

(2)

Solving Eqs.(1) and (2) yields P = 353.29 N

N = 663.97 N

N - 80(9.81) + 353.29 sin 20° = 80(0) teaching

Web)

+ c ©Fy = may ;

laws

or

Equations of Motion: The friction force developed between the crate and its contacting surface is Ff = mkN = 0.3N since the crate is moving. From FBD(b),

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permitted.

Dissemination

353.29 cos 20° - 0.3(663.97) = 80a

not

instructors

States

+ ©F = ma ; : x x

in

N = 663.97 N

Ans.

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a = 1.66 m>s2






13–10. p

The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. Determine the crate’s acceleration in t = 2 s if the coefficient of static friction is ms = 0.4, the coefficient of kinetic friction is mk = 0.3, and the towing force is P = (90t2) N, where t is in seconds.

20

SOLUTION Equations of Equilibrium: At t = 2 s, P = 90 A 22 B = 360 N. From FBD(a) + c ©Fy = 0;

N + 360 sin 20° - 80(9.81) = 0

+ ©F = 0; : x

360 cos 20° - Ff = 0

N = 661.67 N

Ff = 338.29 N

Since Ff 7 (Ff)max = ms N = 0.4(661.67) = 264.67 N, the crate accelerates. Equations of Motion: The friction force developed between the crate and its contacting surface is Ff = mkN = 0.3N since the crate is moving. From FBD(b), + c ©Fy = may ;

N - 80(9.81) + 360 sin 20° = 80(0) laws in

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360 cos 20° - 0.3(661.67) = 80a

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+ ©F = ma ; : x x

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N = 661.67 N

a = 1.75 m>s2

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Ans.






13–11. The safe S has a weight of 200 lb and is supported by the rope and pulley arrangement shown. If the end of the rope is given to a boy B of weight 90 lb, determine his acceleration if in the confusion he doesn’t let go of the rope. Neglect the mass of the pulleys and rope. S

SOLUTION B

Equation of Motion: The tension T developed in the cord is the same throughout the entire cord since the cord passes over the smooth pulleys. From FBD(a), + c ©Fy = may;

T- 90 = - a

90 ba 32.2 B

(1)

From FBD(b), 2T - 200 = - a

200 ba 32.2 S

(2) or

+ c ©Fy = may ;

teaching

Web)

laws

Kinematic: Establish the position-coordinate equation, we have

Dissemination

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copyright

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2sS +sB = l

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Taking time derivative twice yields the

2aS + aB = 0

(3)

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1+ T 2

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Solving Eqs.(1),(2), and (3) yields

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aB = -2.30 ft>s2 = 2.30 ft>s2 T = 96.43 lb

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aS = 1.15 ft>s2 T






Ans.

*13–12. The boy having a weight of 80 lb hangs uniformly from the bar. Determine the force in each of his arms in t = 2 s if the bar is moving upward with (a) a constant velocity of 3 ft>s, and (b) a speed of v = 14t22 ft>s, where t is in seconds.

SOLUTION (a) T = 40 lb

Ans.

(b) v = 4t 2 a = 8t + c a Fy = may ;

2T - 80 =

80 18t2 32.2

At t = 2 s. T = 59.9 lb

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Ans.






13–13. The bullet of mass m is given a velocity due to gas pressure caused by the burning of powder within the chamber of the gun. Assuming this pressure creates a force of F = F0 sin 1pt>t02 on the bullet, determine the velocity of the bullet at any instant it is in the barrel. What is the bullet’s maximum velocity? Also, determine the position of the bullet in the barrel as a function of time.

F F0

t0


F0 pt dv = a b sin a b m dt t0

v dv =

v = a

t

L0

a

F0 pt b sin a b dt m t0

v = -a

F0t0 pt t b cos a b d pm t0 0

F0t0 pt b a 1 - cos a b b pm t0 pt b = -1, or t = t0. t0

teaching

Web)

vmax occurs when cos a

Ans.

or

L0

pt b = ma t0

laws

a =

F0 sin a

Ans.

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2F0t0 pm

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vmax =

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the

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F0t0 pt b a 1 - cos a b b dt pm t0

is

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the

t0 F0t0 pt b at sin a b b p pm t0

student

s = a

for

t0 F0t0 pt t b ct sin a b d pm p t0 0 is

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s = a

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by

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L0

United

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t

13–14. The 2-Mg truck is traveling at 15 m> s when the brakes on all its wheels are applied, causing it to skid for a distance of 10 m before coming to rest. Determine the constant horizontal force developed in the coupling C, and the frictional force developed between the tires of the truck and the road during this time.The total mass of the boat and trailer is 1 Mg.

C

SOLUTION Kinematics: Since the motion of the truck and trailer is known, their common acceleration a will be determined first. + b a:

v2 = v0 2 + 2ac(s - s0) 0 = 152 + 2a(10 - 0) a = -11.25 m>s2 = 11.25 m>s2 ;

Free-Body Diagram:The free-body diagram of the truck and trailer are shown in Figs. (a) and (b), respectively. Here, F representes the frictional force developed when the truck skids, while the force developed in coupling C is represented by T. laws in

teaching

Web)

-T = 1000(-11.25)

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+ ©F = ma ; : x x

or

Equations of Motion:Using the result of a and referrning to Fig. (a),

T = 11 250 N = 11.25 kN instructors

States

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Ans.

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Using the results of a and T and referring to Fig. (b), by

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11 250 - F = 2000( -11.25) the

+ c ©Fx = max ;

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F = 33 750 N = 33.75 kN






Ans.

13–15. A freight elevator, including its load, has a mass of 500 kg. It is prevented from rotating by the track and wheels mounted along its sides.When t = 2 s, the motor M draws in the cable with a speed of 6 m> s, measured relative to the elevator. If it starts from rest, determine the constant acceleration of the elevator and the tension in the cable. Neglect the mass of the pulleys, motor, and cables.

M

SOLUTION 3sE + sP = l 3vE = - vP

A+TB

vP = vE + vP>E -3vE = vE + 6 vE = -

A+cB

6 = -1.5 m>s = 1.5 m>s c 4

v = v0 + ac t laws

or

1.5 = 0 + aE (2) aE = 0.75 m>s2 c

in

teaching

Web)

Ans.

4T - 500(9.81) = 500(0.75)

Dissemination

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permitted.

T = 1320 N = 1.32 kN

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Ans.






*13–16. The man pushes on the 60-lb crate with a force F. The force is always directed down at 30° from the horizontal as shown, and its magnitude is increased until the crate begins to slide. Determine the crate’s initial acceleration if the coefficient of static friction is ms = 0.6 and the coefficient of kinetic friction is mk = 0.3.

F 30⬚

SOLUTION Force to produce motion: + ©F = 0; : x

Fcos 30° - 0.6N = 0

+ c ©Fy = 0;

N - 60 - F sin 30° = 0 N = 91.80 lb

F = 63.60 lb

Since N = 91.80 lb, 63.60 cos 30° - 0.3(91.80) = a

60 ba 32.2

a = 14.8 ft>s2

Ans.

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+ ©F = ma ; : x x






13–17. The double inclined plane supports two blocks A and B, each having a weight of 10 lb. If the coefficient of kinetic friction between the blocks and the plane is mk = 0.1, determine the acceleration of each block.

B

A 60

SOLUTION Equation of Motion: Since blocks A and B are sliding along the plane, the friction forces developed between the blocks and the plane are (Ff)A = mk NA = 0.1 NA and (Ff)B = mk NB = 0.1NB. Here, aA = aB = a. Applying Eq. 13–7 to FBD(a), we have a + a Fy¿ = may¿; Q + a Fx¿ = max¿;

10 b (0) NA = 5.00 lb 32.2

NA - 10 cos 60° = a

10 ba 32.2

T + 0.1(5.00) - 10 sin 60° = - a

(1)

laws

or

From FBD(b),

teaching

T - 0.1(8.660) - 10 sin 30° = a

in

a + a Fx¿ = max¿;

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NB - 10 cos 30° = a

Web)

10 b (0) NB = 8.660 lb 32.2

Q + a Fy¿ = may¿;

(2) is

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10 ba 32.2

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Ans. (including

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a = 3.69 ft>s2

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the

T = 7.013 lb






30

13–18. A 40-lb suitcase slides from rest 20 ft down the smooth ramp. Determine the point where it strikes the ground at C. How long does it take to go from A to C?

A

20 ft

30 4 ft

SOLUTION + R ©Fx = m ax ;

40 sin 30° =

40 a 32.2

( +R)v2 = v20 + 2 ac(s - s0); v2B = 0 + 2(16.1)(20) vB = 25.38 ft>s ( + R) v = v0 + ac t ; laws

or

25.38 = 0 + 16.1 tAB in

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tAB = 1.576 s

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+ )s = (s ) + (v ) t (: x x 0 x 0

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R = 0 + 25.38 cos 30°(tBC)

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1 2 at 2 c

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4 = 0 + 25.38 sin 30° tBC +

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tBC = 0.2413 s provided

Ans. part

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This

R = 5.30 ft

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Total time = tAB + tBC = 1.82 s






C R

a = 16.1 ft>s2

( + T) sy = (sy)0 + (vy)0 t +

B

13–19. Solve Prob. 13–18 if the suitcase has an initial velocity down the ramp of vA = 10 ft>s and the coefficient of kinetic friction along AB is mk = 0.2.

A

20 ft

30⬚ 4 ft

SOLUTION +R©Fx = max;

40 40 sin 30° - 6.928 = a 32.2 a = 10.52 ft>s

v2B = (10)2 + 2(10.52)(20) vB = 22.82 ft>s (+R) v = v0 + ac t; 22.82 = 10 + 10.52 tAB laws

or

tAB = 1.219 s

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+ ) s = (s ) + (v ) t (: x x 0 x 0 Dissemination

R = 0 + 22.82 cos 30° (tBC)

permitted.

1 a t2 2 c 1 + (32.2)(tBC)2 2

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(+ T ) sy = (sy)0 + (vy)0 t +

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tBC = 0.2572 s

Ans.

this

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R = 5.08 ft

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Total time = tAB + tBC = 1.48 s






C R

2

( +R) v2 = v20 + 2 ac (s - s0);

4 = 0 + 22.82 sin 30° tBC

B

Ans.

*13–20. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = 13200t22 N, where t is in seconds. If the car has an initial velocity v1 = 2 m>s when t = 0, determine its velocity when t = 2 s.

M v1

SOLUTION

17

8

15

Q+ ©Fx¿ = max¿ ;

3200t2 - 40019.812a

8 b = 400a 17

a = 8t2 - 4.616

dv = adt v

2

dv =

L2

L0

18t 2 -4.6162 dt

v = 14.1 m>s

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Ans.






2 m/s

13–21. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = 13200t22 N, where t is in seconds. If the car has an initial velocity v1 = 2 m>s at s = 0 and t = 0, determine the distance it moves up the plane when t = 2 s.

M v1

SOLUTION

17

8

15

Q+ ©Fx¿ = max¿;

3200t2 - 40019.812 a

8 b = 400a 17

a = 8t2 - 4.616

dv = adt v

v =

L0

ds = 2.667t3 - 4.616t + 2 dt

s

2

ds =

L0

(2.667 t3 - 4.616t + 2) dt laws

L0

18t2 - 4.6162 dt

or

L2

t

dv =

s = 5.43 m

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Ans.






2 m/s

13–22. Determine the required mass of block A so that when it is released from rest it moves the 5-kg block B a distance of 0.75 m up along the smooth inclined plane in t = 2 s. Neglect the mass of the pulleys and cords.

E

C

D

SOLUTION

B

1 a t2, we have 2 c

Kinematic: Applying equation s = s0 + v0 t +

A 60

(a+ )

1 0.75 = 0 + 0 + aB A 22 B 2

aB = 0.375 m>s2

Establishing the position - coordinate equation, we have 2sA + (sA - sB) = l

3sA - sB = l

Taking time derivative twice yields 3aA - aB = 0

laws

or

(1)

in

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From Eq.(1), aA = 0.125 m>s2

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3aA - 0.375 = 0

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T - 5(9.81) sin 60° = 5(0.375) for

work

student

the

a + ©Fy¿ = may¿ ;

is

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solely

of

protected

the

(including

T = 44.35 N

provided

integrity

of

3(44.35) - 9.81mA = mA ( -0.125) part

the

is

This

+ c ©Fy = may ;

and

work

this

assessing

From FBD(a),

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Ans.

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mA = 13.7 kg






not

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States

World

Equation of Motion: The tension T developed in the cord is the same throughout the entire cord since the cord passes over the smooth pulleys. From FBD(b),

13–23.

The winding drum D is drawing in the cable at an accelerated rate of 5 m>s2. Determine the cable tension if the suspended crate has a mass of 800 kg.

D

SOLUTION sA + 2 sB = l aA = - 2 aB 5 = - 2 aB aB = - 2.5 m>s2 = 2.5 m>s2 c + c ©Fy = may ;

2T - 800(9.81) = 800(2.5) T = 4924 N = 4.92 kN

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*13–24. If the motor draws in the cable at a rate of v = (0.05s3>2) m>s, where s is in meters, determine the tension developed in the cable when s = 10 m. The crate has a mass of 20 kg, and the coefficient of kinetic friction between the crate and the ground is mk = 0.2.

s v M

SOLUTION Kinematics: Since the motion of the create is known, its acceleration a will be determined first. a = v

dv 3 = A 0.05s3>2 B c (0.05) a bs1>2 d = 0.00375s2 m>s2 ds 2

When s = 10 m, a = 0.00375(102) = 0.375 m>s2 : Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to oppose the motion of the crate which is to the right, Fig. a. laws in

teaching

Web)

N - 20(9.81) = 20(0)

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+ c ©Fy = may;

or

Equations of Motion: Here, ay = 0. Thus,

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permitted.

Dissemination

N = 196.2 N

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is

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Using the results of N and a, by

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T - 0.2(196.2) = 20(0.375)

student

the

+ ©F = ma ; : x x

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T = 46.7 N






Ans.

13–25. If the motor draws in the cable at a rate of v = (0.05t2) m>s, where t is in seconds, determine the tension developed in the cable when t = 5 s. The crate has a mass of 20 kg and the coefficient of kinetic friction between the crate and the ground is mk = 0.2.

s v M

SOLUTION Kinematics: Since the motion of the crate is known, its acceleration a will be determined first. a =

dv = 0.05(2t) = (0.1t) m>s2 dt

When t = 5 s, a = 0.1(5) = 0.5 m>s2 : Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to oppose the motion of the crate which is to the right, Fig. a. laws in

teaching

Web)

N - 20(9.81) = 0

Wide

copyright

+ c ©Fy = may;

or

Equations of Motion: Here, ay = 0. Thus,

instructors

States

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permitted.

Dissemination

N = 196.2 N

United

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Using the results of N and a, by

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T - 0.2(196.2) = 20(0.5)

student

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+ ©F = ma ; : x x

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T = 49.2 N






Ans.

13–26. The 2-kg shaft CA passes through a smooth journal bearing at B. Initially, the springs, which are coiled loosely around the shaft, are unstretched when no force is applied to the shaft. In this position s = s¿ = 250 mm and the shaft is at rest. If a horizontal force of F = 5 kN is applied, determine the speed of the shaft at the instant s = 50 mm, s¿ = 450 mm. The ends of the springs are attached to the bearing at B and the caps at C and A.

s¿

s

3 kN/m

kCB

SOLUTION FCB = kCBx = 3000x + ©F = ma ; ; x x

FAB = kABx = 2000x

5000 - 3000x - 2000x = 2a 2500 - 2500x = a

a dx - v dy v

0.2

(2500 - 2500x) dx =

v dv

or

2500(0.2)2 v2 ≤ = 2 2

teaching

Web)

2500(0.2) - ¢

L0

laws

L0

v = 30 m>s

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A

B

C

kAB

2 kN/m

F

5 kN

13–27. The 30-lb crate is being hoisted upward with a constant acceleration of 6 ft>s2. If the uniform beam AB has a weight of 200 lb, determine the components of reaction at the fixed support A. Neglect the size and mass of the pulley at B. Hint: First find the tension in the cable, then analyze the forces in the beam using statics.

y

5 ft A

B

x

SOLUTION 6 ft/s2

Crate: + c ©Fy = may ;

T - 30 = a

30 b(6) 32.2

T = 35.59 lb

Beam: + ©F = 0; : x a + c ©Fy = 0;

Ax = 35.6 lb

Ay - 200 - 35.59 = 0

Ans.

Ay = 236 lb

Ans.

MA = 678 lb # ft

MA - 200(2.5) - (35.59)(5) = 0

Ans.

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laws

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+ ©MA = 0;

- Ax + 35.59 = 0






*13–28. The driver attempts to tow the crate using a rope that has a tensile strength of 200 lb. If the crate is originally at rest and has a weight of 500 lb, determine the greatest acceleration it can have if the coefficient of static friction between the crate and the road is ms = 0.4, and the coefficient of kinetic friction is mk = 0.3.

30

SOLUTION Equilibrium: In order to slide the crate, the towing force must overcome static friction. + ©F = 0; : x

- T cos 30° +0.4N = 0

(1)

+ ©F = 0; :

N + T sin 30° -500 = 0

(2)

Solving Eqs.(1) and (2) yields: T = 187.6 lb

N = 406.2 lb

laws

or

Since T 6 200 lb, the cord will not break at the moment the crate slides.

teaching

Web)

After the crate begins to slide, the kinetic friction is used for the calculation. N = 400 lb

in

N + 200 sin 30° - 500 = 0

Wide

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Dissemination

500 a 32.2

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200 cos 30° - 0.3(400) =

States

+ ©F = ma ; : x x

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a = 3.43 ft>s2






Ans.

13–29. F (lb)

The force exerted by the motor on the cable is shown in the graph. Determine the velocity of the 200-lb crate when t = 2.5 s.

250 lb M t (s) 2.5

SOLUTION Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a. Equilibrium: For the crate to move, force F must overcome the weight of the crate. Thus, the time required to move the crate is given by A

+ c ©Fy = 0;

t = 2s

100t - 200 = 0

250 t = (100t) lb. By referring to Fig. a, 2.5

Equation of Motion: For 2 s 6 t 6 2.5 s, F =

+ c ©Fy = may;

100t - 200 =

200 a 32.2

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Kinematics: The velocity of the crate can be obtained by integrating the kinematic equation, dv = adt. For 2 s … t 6 2.5 s, v = 0 at t = 2 s will be used as the lower integration limit.Thus,

Web)

laws

or

a = (16.1t - 32.2) ft>s2

World

adt

not the on

learning.

is

L

States

dv =

instructors

L

of

(+ c )

United

t

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v

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(16.1t - 32.2)dt

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the

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for

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dv =

the

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2s

work

solely

= A 8.05t2 - 32.2t + 32.2 B ft>s part

the

is

When t = 2.5 s,

This

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v = A 8.05t2 - 32.2t B 2

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v = 8.05(2.52) - 32.2(2.5) + 32.2 = 2.01 ft>s






Ans.

13–30. F (N)

The force of the motor M on the cable is shown in the graph. Determine the velocity of the 400-kg crate A when t = 2 s.

2500 F ⫽ 625 t 2 M

2

SOLUTION Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a. Equilibrium: For the crate to move, force 2F must overcome its weight. Thus, the time required to move the crate is given by + c ©Fy = 0;

2(625t2) - 400(9.81) = 0

A

t = 1.772 s Equations of Motion: F = A 625t2 B N. By referring to Fig. a, + c ©Fy = may;

2 A 625t2 B - 400(9.81) = 400a

laws

or

a = (3.125t2 - 9.81) m>s2

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Kinematics: The velocity of the crate can be obtained by integrating the kinematic equation, dv = adt. For 1.772 s … t 6 2 s, v = 0 at t = 1.772 s will be used as the lower integration limit. Thus, adt

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dv =

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A 3.125t2 - 9.81 B dt

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L1.772 s

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L0

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dv =

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v = A 1.0417t3 - 9.81t B 2

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When t = 2 s,

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= A 1.0417t - 9.81t + 11.587 B m>s

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1.772 s

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v = 1.0417(23) - 9.81(2) + 11.587 = 0.301 m>s






Ans.

t (s)

13–31. The tractor is used to lift the 150-kg load B with the 24-mlong rope, boom, and pulley system. If the tractor travels to the right at a constant speed of 4 m> s, determine the tension in the rope when sA = 5 m. When sA = 0, sB = 0.

12 m

sB

SOLUTION

B A

12 - sB + 2s2A + (12)2 = 24 1 # -sB + A s2A + 144 B - 2 asAsA b = 0

sA

3 1 1 # 2 # $ $ -sB - A s2A + 144 B - 2 a sAsA b + A s2A + 144 B - 2 a s2A b + A s2A + 144 B - 2 asA sA b = 0

3 2

((5)2 + 144)

-

A s2A + 144 B 2 1

S

or

(4)2 + 0 1

((5)2 + 144)2

S = 1.0487 m>s2

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(5)2(4)2

$ # s2A + sAsA

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A s2A + 144 B

-

3 2

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aB = - C

# s2As2A

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$ sB = - C

T - 150(9.81) = 150(1.0487) States

World

permitted.

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+ c ©Fy = may ;

Ans.

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T = 1.63 kN






*13–32. The tractor is used to lift the 150-kg load B with the 24-mlong rope, boom, and pulley system. If the tractor travels to the right with an acceleration of 3 m>s2 and has a velocity of 4 m> s at the instant sA = 5 m, determine the tension in the rope at this instant. When sA = 0, sB = 0.

12 m

sB

SOLUTION

B A

12 = sB + 2s2A + (12)2 = 24 sA

3 1 # # -sB + A s2A + 144 B - 2 a 2sA sA b = 0 2 3 1 1 # 2 # $ $ -sB - A s2A + 144 B - 2 a sAsA b + A s2A + 144 B - 2 a s2A b + A s2A + 144 B - 2 asAsA b = 0

(5)2(4)2 3 2

((5)2 + 144)

-

A s2A + 144 B 2 1

S

(4)2 + (5)(3) 1

((5)2 + 144)2

or

3

S = 2.2025 m>s2

Web)

A s2A + 144 B 2

$ # s2A + sA sA

laws

-

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copyright

in

aB = - C

# s2A s2A

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$ sB = - C

T - 150(9.81) = 150(2.2025) States

World

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+ c ©Fy = may ;

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T = 1.80 kN

13–33. Each of the three plates has a mass of 10 kg. If the coefficients of static and kinetic friction at each surface of contact are ms = 0.3 and mk = 0.2, respectively, determine the acceleration of each plate when the three horizontal forces are applied.

18 N

D C

15 N

B A

SOLUTION Plates B, C and D + ©F = 0; : x

100 - 15 - 18 - F = 0 F = 67 N Fmax = 0.3(294.3) = 88.3 N 7 67 N

Plate B will not slip. aB = 0

Ans.

100 - 18 - F = 0

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Web)

laws

+ ©F = 0; : x

or

Plates D and C

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F = 82 N

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Fmax = 0.3(196.2) = 58.86 N 6 82N

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Slipping between B and C.

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Assume no slipping between D and C,

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100 - 39.24 - 18 = 20 ax is

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+ ©F = ma ; : x x

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Check slipping between D and C.

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ax = 2.138 m>s2 :

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courses

F - 18 = 10(2.138)

will sale

F = 39.38 N

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and

+ ©F = m a ; : x x

Fmax = 0.3(98.1) = 29.43 N 6 39.38 N Slipping between D and C. Plate C: + ©F = m a ; : x x

100 - 39.24 - 19.62 = 10 ac ac = 4.11 m>s2 :

Ans.

Plate D: + ©F = m a ; : x x

19.62 - 18 = 10 a D a D = 0.162m>s2 :






Ans.

100 N

13–34. Each of the two blocks has a mass m. The coefficient of kinetic friction at all surfaces of contact is m. If a horizontal force P moves the bottom block, determine the acceleration of the bottom block in each case.

B P

A (a) B

SOLUTION

P

A

Block A: + ©F = ma ; (a) ; x x

(b)

P - 3mmg = m aA aA =

P -3mg m

Ans.

(b) sB + sA = l aA = - aB

(1)

Block A: P - T - 3mmg = m aA

or

(2) laws

+ ©F = m a ; ; x x

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Block B: mmg - T = maB

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+ ©F = ma ; ; x x

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Subtract Eq.(3) from Eq.(2):

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P - 4mmg = m (aA - aB)

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aA =

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Use Eq.(1);






Ans.

13–35. The conveyor belt is moving at 4 m>s. If the coefficient of static friction between the conveyor and the 10-kg package B is ms = 0.2, determine the shortest time the belt can stop so that the package does not slide on the belt.

B

SOLUTION + ©F = m a ; : x x

0.2198.12 = 10 a a = 1.962 m>s2

+ 2v = v + a t 1: 0 c 4 = 0 + 1.962 t t = 2.04 s

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Ans.






*13–36. The 2-lb collar C fits loosely on the smooth shaft. If the spring is unstretched when s = 0 and the collar is given a velocity of 15 ft> s, determine the velocity of the collar when s = 1 ft.

15 ft/s s C

1 ft k

SOLUTION Fs = kx;

Fs = 4 A 21 + s2 - 1 B

+ ©F = ma ; : x x 1

-

L0

¢ 4s ds -

s

-4 A 21 + s2 - 1 B ¢ 4s ds

21 + s 1

- C 2s2 - 4 31 + s2 D 0 =

2

≤ =

v

L15

a

21 + s

2

2 dv b av b 32.2 ds

≤ = a

2 b v dv 32.2

1 A v2 - 152 B 32.2

v = 14.6 ft>s

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4 lb/ft

13–37. Cylinder B has a mass m and is hoisted using the cord and pulley system shown. Determine the magnitude of force F as a function of the cylinder’s vertical position y so that when F is applied the cylinder rises with a constant acceleration aB. Neglect the mass of the cord, pulleys, hook and chain.

d/2

d/2

F y

SOLUTION + c ©Fy = may ;

2F cos u - mg = maB

2F £

2y +

A d2 B 2

aB

y

A B

d 2 2

2y + 2

≥ - mg = maB

m(aB + g)24y2 + d2 4y

Ans.

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laws

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F =

y 2

where cos u =






B

13–38. vA

The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s speed is vA = 2.5 m>s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is mk = 0.3, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs. Take u = 30°.

2.5 m/s A 3m

u

SOLUTION Q +©Fy = may;

NC - 1219.812 cos 30° = 0 NC = 101.95 N

+ R©Fx = max ;

1219.812 sin 30° - 0.31101.952 = 12 aC aC = 2.356 m>s2

(+ R)

2 vB2 = vA + 2ac1sB - sA2

v2B = 12.522 + 212.356213 - 02 vB = 4.5152 = 4.52 m>s

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B

13–39. y

An electron of mass m is discharged with an initial horizontal velocity of v0. If it is subjected to two fields of force for which Fx = F0 and Fy = 0.3F0, where F0 is constant, determine the equation of the path, and the speed of the electron at any time t.

++++++++++++++

v0

+ ©F = ma ; : x x

F0 = max

+ c ©Fy = may;

0.3 F0 = may

Thus, vx

t

dvx =

Lv0

F0 dt L0 m

F0 t + v0 m vy t 0.3F0 dvy = dt L0 L0 m

vx =

0.3F0 t m

2 F0 0.3F0 2 t + v0 b + a tb m C m

Web)

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Dissemination

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F0 t + v0 b dt m

the

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a

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L0

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t

dx =

Ans.

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x

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1 21.09F20 t2 + 2F0tmv0 + m2v20 m

=

L0

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v =

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F0 t2 + v0 t 2m

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x =

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x =

F0 2m 1 2m a by + v0 a by2 2m 0.3F0 B 0.3F0

x =

y 1 2m + v0 a by 2 0.3 B 0.3F0






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t = a

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0.3F0 t dt L0 m

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dy =

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y

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vy =

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++++++++++++++

SOLUTION

x

*13–40. v

The engine of the van produces a constant driving traction force F at the wheels as it ascends the slope at a constant velocity v. Determine the acceleration of the van when it passes point A and begins to travel on a level road, provided that it maintains the same traction force.

A F u

SOLUTION Free-Body Diagram: The free-body diagrams of the van up the slope and on the level road are shown in Figs. a and b, respectively. Equations of Motion: Since the van is travelling up the slope with a constant velocity, its acceleration is a = 0. By referring to Fig. a, ©Fx ¿ = max ¿;

F - mg sin u = m(0) F = mg sin u

Since the van maintains the same tractive force F when it is on level road, from Fig. b, mg sin u = ma or

+ ©F = ma ; : x x

a = g sin u

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Ans.






13–41. The 2-kg collar C is free to slide along the smooth shaft AB. Determine the acceleration of collar C if (a) the shaft is fixed from moving, (b) collar A, which is fixed to shaft AB, moves downward at constant velocity along the vertical rod, and (c) collar A is subjected to a downward acceleration of 2 m>s2. In all cases, the collar moves in the plane.

B

45

A

SOLUTION (a) + b©Fx¿ = max¿ ;

219.812 sin 45° = 2aC

aC = 6.94 m>s2

(b) From part (a) a C>A = 6.94 m>s2 a C = a A + a C>A

Where a A = 0

= 6.94 m>s2 (c) aC = aA + aC>A = 2 + aC>A b √T laws

or

(1) aC>A = 5.5225 m>s2 b

Web)

219.812 sin 45° = 212 cos 45°+ aC>A2

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+ b©Fx¿ = max¿ ;

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From Eq.(1)

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aC = 2 + 5.52 25 = 3.905 + 5.905 ; b T T

Ans.

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aC = 23.9052 + 5.9052 = 7.08 m>s2

the

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5.905 = 56.5° ud 3.905

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u = tan-1






Ans.

C

13–42. The 2-kg collar C is free to slide along the smooth shaft AB. Determine the acceleration of collar C if collar A is subjected to an upward acceleration of 4 m>s2.

B

45⬚

A

SOLUTION + ©F = ma ; ; x x

N sin 45° = 2aC>AB sin 45° N = 2 aC>AB

+ c ©Fy = may ;

N cos 45° - 19.62 = 2142 - 2aC>AB cos 45° aC>AB = 9.76514 aC = aAB + aC>AB 1aC)x = 0 + 9.76514 sin 45° = 6.905 ; (aC)y = 4 - 9.76514 cos 45° = 2.905T or

aC = 216.90522 + 12.90522 = 7.49 m>s2

teaching

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Ans.

2.905 b = 22.8° ud 6.905

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Ans.

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u = tan-1 a






C

13–43. The coefficient of static friction between the 200-kg crate and the flat bed of the truck is ms = 0.3. Determine the shortest time for the truck to reach a speed of 60 km> h, starting from rest with constant acceleration, so that the crate does not slip.

SOLUTION Free-Body Diagram: When the crate accelerates with the truck, the frictional force Ff develops. Since the crate is required to be on the verge of slipping, Ff = msN = 0.3N. Equations of Motion: Here, ay = 0. By referring to Fig. a, + c ©Fy = may;

N - 200(9.81) = 200(0) N = 1962 N

+ ©F = ma ; : x x

-0.3(1962) = 200(- a)

or

a = 2.943 m>s2 ; km 1000 m 1h ba ba b = h 1 km 3600 s 16.67 m>s. Since the acceleration of the truck is constant, Dissemination

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Kinematics: The final velocity of the truck is v = a 60

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v = v0 + ac t

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16.67 = 0 + 2.943t

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t = 5.66 s






*13–44. When the blocks are released, determine their acceleration and the tension of the cable. Neglect the mass of the pulley.

SOLUTION Free-Body Diagram: The free-body diagram of blocks A and B are shown in Figs. b and c, respectively. Here, aA and aB are assumed to be directed downwards so that they are consistent with the positive sense of position coordinates sA and sB of blocks A and B, Fig. a. Since the cable passes over the smooth pulleys, the tension in the cable remains constant throughout.

A 10 kg B 30 kg

Equations of Motion: By referring to Figs. b and c, + c ©Fy = may;

2T - 10(9.81) = - 10aA

(1)

T - 30(9.81) = - 30aB

(2)

and

laws

or

+ c ©Fy = may;

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copyright

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Kinematics: We can express the length of the cable in terms of sA and sB by referring to Fig. a.

instructors

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2sA + sB = l

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The second derivative of the above equation gives

(3)

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2aA + aB = 0

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Solving Eqs. (1), (2), and (3) yields work

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aB = 7.546 m>s2 = 7.55 m>s2 T

Ans.

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aA = - 3.773 m>s2 = 3.77 m>s2 c

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T = 67.92 N = 67.9 N






Ans.

13–45. If the force exerted on cable AB by the motor is F = (100t3>2) N, where t is in seconds, determine the 50-kg crate’s velocity when t = 5 s. The coefficients of static and kinetic friction between the crate and the ground are ms = 0.4 and mk = 0.3, respectively. Initially the crate is at rest.

A

SOLUTION Free-Body Diagram: The frictional force Ff is required to act to the left to oppose the motion of the crate which is to the right. Equations of Motion: Here, ay = 0. Thus, N - 50(9.81) = 50(0)

+ c ©Fy = may;

N = 490.5 N Realizing that Ff = mkN = 0.3(490.5) = 147.15 N, 100t3>2 - 147.15 = 50a

laws

or

a = A 2t3>2 - 2.943 B m>s

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Equilibrium: For the crate to move, force F must overcome the static friction of Ff = msN = 0.4(490.5) = 196.2 N. Thus, the time required to cause the crate to be on the verge of moving can be obtained from.

Web)

+ c ©Fx = max;

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100t3>2 - 196.2 = 0

+ ©F = 0; : x

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Kinematics: Using the result of a and integrating the kinematic equation dv = a dt with the initial condition v = 0 at t = 1.567 as the lower integration limit, this work

provided

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A 2t3>2 - 2.943 B dt

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v = A 0.8t5>2 - 2.943t B 2

v = A 0.8t5>2 - 2.943t + 2.152 B m>s sale

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When t = 5 s, v = 0.8(5)5>2 - 2.943(5) + 2.152 = 32.16 ft>s = 32.2 ft>s






Ans.

B

13–46. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.

A

P

u B C

SOLUTION Require aA = aB = a Block A: + c ©Fy = 0; + ©F = ma ; ; x x

N cos u - mg = 0 N sin u = ma a = g tan u

P - N sin u = ma

teaching

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+ ©F = ma ; ; x x

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Block B:

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P - mg tan u = mg tan u P = 2mg tan u

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13–47. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not slip on B. The coefficient of static friction between A and B is ms. Neglect any friction between B and C.

A

C

SOLUTION Require aA = aB = a Block A:

N sin u + ms N cos u = ma N =

mg cos u - ms sin u sin u + ms cos u b cos u - ms sin u in

teaching

Web)

a = ga

or

+ ©F = ma ; ; x x

N cos u - ms N sin u - mg = 0

laws

+ c ©Fy = 0;

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copyright

Block B:

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permitted.

P - ms N cos u - N sin u = ma

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sin u + ms cos u b cos u - ms sin u

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sin u + ms cos u sin u + ms cos u b = mga b cos u - ms sin u cos u - ms sin u

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States

+ ©F = ma ; ; x x






P

u B

Ans.

*13–48. A parachutist having a mass m opens his parachute from an at-rest position at a very high altitude. If the atmospheric drag resistance is FD = kv2, where k is a constant, determine his velocity when he has fallen for a time t. What is his velocity when he lands on the ground? This velocity is referred to as the terminal velocity, which is found by letting the time of fall t : q .

FD

v

SOLUTION + T ©Fz = m az ;

mg - kv2 = m

v

m

dv dt

t

m dv dt = 2 L0 1mg - kv 2 L0 v

m dv = t mg k L0 - v2 k v mg + v m A k 1 T = t ln D mg k ¢ mg 2 - v A k A k 0 in

teaching

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mg + v mg A k k t ¢2 = ln mg m A k - v A k

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mg + v A k = mg - v A k

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mg

mg e 2t2 k - v e2t2mgk mg = + v A k A k

When t : q






vt =

mg A k

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2t2mg k

- 1 mg e v = C S mg A k e 2t2 k + 1

Ans.

Ans.

13–49. The smooth block B of negligible size has a mass m and rests on the horizontal plane. If the board AC pushes on the block at an angle u with a constant acceleration a0, determine the velocity of the block along the board and the distance s the block moves along the board as a function of time t. The block starts from rest when s = 0, t = 0.

a0 θ

A

B s

SOLUTION Q+ ©Fx = m ax;

0 = m aB sin f aB = aAC + aB>AC aB = a0 + aB>AC

Q+

aB sin f = - a0 sin u + aB>AC

Thus, 0 = m( -a0 sin u + aB>AC) laws

Web) teaching in

a0 sin u dt

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s =






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C

13–50. A projectile of mass m is fired into a liquid at an angle u0 with an initial velocity v0 as shown. If the liquid develops a frictional or drag resistance on the projectile which is proportional to its velocity, i.e., F = kv, where k is a constant, determine the x and y components of its position at any instant. Also, what is the maximum distance xmax that it travels?

y

v0

θ0 O


- kv cos u = m ax

+ c ©Fy = m ay ;

-m g - k v sin u = m ay

or -k

d2x dx = m 2 dt dt

-mg - k

dy d2y = m 2 dt dt

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-m v0 cos u0 e-(k>m)t + C3 k sale

x = y = -

mg mg m -(k>m)t t - (v0 sin u0 + )( )e k k k

When t = 0, x = y = 0, thus x = y = -

m v0 cos u0(1 - e-(k>m)t) k

Ans.

mgt mg m + (v0 sin u0 + )(1 - e-(k>m)t) k k k

Ans.

m v0 cos u0 k

Ans.

As t : q xmax =






x

13–51. The block A has a mass mA and rests on the pan B, which has a mass mB . Both are supported by a spring having a stiffness k that is attached to the bottom of the pan and to the ground. Determine the distance d the pan should be pushed down from the equilibrium position and then released from rest so that separation of the block will take place from the surface of the pan at the instant the spring becomes unstretched.

A B y d

SOLUTION For Equilibrium + c ©Fy = may;

Fs = (mA + mB)g yeq =

(mA + mB)g Fs = k k

Block: + c ©Fy = may ;

-mA g + N = mA a

Block and pan or

- (mA + mB)g + k(yeq + y) = (mA + mB)a

teaching

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laws

+ c ©Fy = may;

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-mAg + N mA + mB bg + y d = (mA + mB) a b mA k

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kd = - (mA + mB)g

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Ans.

k

*13–52. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5 m from the platform’s center. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is m = 0.2.

z

5m

SOLUTION Equation of Motion: Since the girl is on the verge of slipping, Ff = msN = 0.2N. Applying Eq. 13–8, we have ©Fb = 0 ; ©Fn = man ;

N - 15(9.81) = 0

N = 147.15 N

0.2(147.15) = 15 a

v2 b 5

v = 3.13 m>s

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Ans.






13–53. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block is given a speed of v = 10 m>s, determine the radius r of the circular path along which it travels.

r

B v

SOLUTION A

Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The tension in the cord is equal to the weight of cylinder A, i.e., T = 15(9.81) N = 147.15 N. Here, an must be directed towards the center of the circular path (positive n axis). v2 102 and referring to Fig. (a), = r r

Equations of Motion: Realizing that an = ©Fn = man;

147.15 = 2 a

102 b r

r = 1.36 m

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13–54. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block travels along a circular path of radius r = 1.5 m, determine the speed of the block.

r

B v

SOLUTION A

Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The tension in the cord is equal to the weight of cylinder A, i.e., T = 15(9.81) N = 147.15 N. Here, an must be directed towards the center of the circular path (positive n axis). v2 v2 and referring to Fig. (a), = r 1.5

Equations of Motion: Realizing that an = ©Fn = man;

147.15 = 2 a

2

v b 1.5

v = 10.5 m>s

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13–55. The 5-kg collar A is sliding around a smooth vertical guide rod. At the instant shown, the speed of the collar is v = 4 m>s, which is increasing at 3 m>s 2. Determine the normal reaction of the guide rod on the collar, and force P at this instant.

v ⫽ 4 m/s

30⬚ A 0.5 m

SOLUTION + ©F = ma ; : t t

P cos 30° = 5(3) P = 17.32 N = 17.3 N

+ T ©Fn = man;

Ans.

N + 5 A 9.81 B - 17.32 sin 30° = 5a

42 b 0.5

N = 119.61 N = 120 N T

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P

*13–56. Cartons having a mass of 5 kg are required to move along the assembly line at a constant speed of 8 m/s. Determine the smallest radius of curvature, r, for the conveyor so the cartons do not slip. The coefficients of static and kinetic friction between a carton and the conveyor are ms = 0.7 and mk = 0.5, respectively.

8 m/s

ρ

SOLUTION + c ©Fb = m ab ;

N - W = 0 N = W Fx = 0.7W

+ ©F = m a ; ; n n

0.7W =

W 82 ( ) 9.81 r

r = 9.32 m

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Ans.






13–57. The block B, having a mass of 0.2 kg, is attached to the vertex A of the right circular cone using a light cord. If the block has a speed of 0.5 m>s around the cone, determine the tension in the cord and the reaction which the cone exerts on the block and the effect of friction.

z

200 mm

B

400 mm

SOLUTION r 300 = ; 200 500 + Q©Fy = may;

300 mm

r = 120 mm = 0.120 m (0.5)2 4 3 T - 0.2(9.81)a b = B 0.2 ¢ ≤Ra b 5 0.120 5 T = 1.82 N

+ a©Fx = max;

Ans.

(0.5)2 3 4 NB - 0.2(9.81)a b = - B 0.2 ¢ ≤Ra b 5 0.120 5 NB = 0.844 N

or

Ans.

teaching

Web)

laws

Also, in

(0.5)2 3 4 b Ta b - NB a b = 0.2 a 5 5 0.120

permitted.

Dissemination

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copyright

+ ©F = ma ; : n n

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States

World

4 3 Ta b + NB a b - 0.2(9.81) = 0 5 5 use

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+ c ©Fb = 0;

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T = 1.82 N

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NB = 0.844 N






Ans.

A

13–58. The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is ms = 0.2. If the spool is located 0.25 m from A, determine the minimum constant speed the spool can have so that it does not slip down the rod.

z 5 4

S 0.25 m

A

SOLUTION 4 r = 0.25a b = 0.2 m 5 + ©F = m a ; ; n n

v2 3 4 b Ns a b - 0.2Ns a b = 2 a 5 5 0.2

+ c ©Fb = m ab;

4 3 Ns a b + 0.2Ns a b - 2(9.81) = 0 5 5 Ns = 21.3 N v = 0.969 m>s

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Ans.






3

13–59. z

The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is ms = 0.2. If the spool is located 0.25 m from A, determine the maximum constant speed the spool can have so that it does not slip up the rod.

5

S 0.25 m

A

SOLUTION 4 r = 0.25( ) = 0.2 m 5 + ©F = m a ; ; n n + c ©Fb = m ab ;

3 4 v2 Ns( ) + 0.2Ns( ) = 2( ) 5 5 0.2 4 3 Ns( ) - 0.2Ns( ) - 2(9.81) = 0 5 5 Ns = 28.85 N v = 1.48 m s

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3 4

*13–60. At the instant u = 60°, the boy’s center of mass G has a downward speed vG = 15 ft>s. Determine the rate of increase in his speed and the tension in each of the two supporting cords of the swing at this instant. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords.

u

10 ft

SOLUTION + R©Ft = mat ;

60 cos 60° =

60 a 32.2 t

2T - 60 sin 60° =

at = 16.1 ft>s2 60 152 a b 32.2 10

Ans.

T = 46.9 lb

Ans.

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laws

or

Q+ ©Fn = man ;

G






13–61. At the instant u = 60°, the boy’s center of mass G is momentarily at rest. Determine his speed and the tension in each of the two supporting cords of the swing when u = 90°. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords.

u

10 ft

SOLUTION

G

60 60 cos u = a 32.2 t

+R© t = mat;

Q+ ©Fn = man;

2T - 60 sin u =

v dn = a ds v

L0

at = 32.2 cos u

60 v2 a b 32.2 10

(1)

however ds = 10du

90°

v dn =

L60°

322 cos u du

v = 9.289 ft>s

or

Ans.

Ans.

Wide

T = 38.0 lb

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60 9.2892 a b 32.2 10

copyright

2T - 60 sin 90° =

in

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laws

From Eq. (1)






13–62. The 10-lb suitcase slides down the curved ramp for which the coefficient of kinetic friction is mk = 0.2. If at the instant it reaches point A it has a speed of 5 ft>s, determine the normal force on the suitcase and the rate of increase of its speed.

y

1 x2 y = –– 8 6 ft

A

SOLUTION

x

1 2 x 8

n =

dy 1 = tan u = x 2 = - 1.5 dx 4 x= -6 d2y

=

2

dx

u = - 56.31°

1 4 3

or Web)

(5)2 10 b¢ ≤ 32.2 23.436

N - 10 cos 56.31° = a

Dissemination

+Q©Fn = man ;

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2

= 23.436 ft laws

dx2

212 4

teaching

2

d2y

C 1 + (- 1.5)2 D 2 3

=

in

r =

dy 2 2 b R dx

copyright

B1 + a

N = 5.8783 = 5.88 lb

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10 a 32.2 t

- 0.2(5.8783) + 10 sin 56.31° =

the

by

+R©Ft = mat;

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Ans.

13–63. z

The 150-lb man lies against the cushion for which the coefficient of static friction is ms = 0.5. Determine the resultant normal and frictional forces the cushion exerts on him if, due to rotation about the z axis, he has a constant speed v = 20 ft>s. Neglect the size of the man.Take u = 60°.

8 ft G

SOLUTION + a a Fy = m1an2y ;

u

150 202 a b sin 60° N - 150 cos 60° = 32.2 8 N = 277 lb

+ b a Fx = m1an2x ;

Ans. 150 202 a b cos 60° 32.2 8

- F + 150 sin 60° = F = 13.4 lb

Ans.

Note: No slipping occurs

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Since ms N = 138.4 lb 7 13.4 lb






*13–64. z

The 150-lb man lies against the cushion for which the coefficient of static friction is ms = 0.5. If he rotates about the z axis with a constant speed v = 30 ft>s, determine the smallest angle u of the cushion at which he will begin to slip off.

8 ft G

SOLUTION + ©F = ma ; ; n n

150 1302 a b 0.5N cos u + N sin u = 32.2 8

+ c ©Fb = 0;

- 150 + N cos u - 0.5 N sin u = 0

u

2

N =

150 cos u - 0.5 sin u

10.5 cos u + sin u2150 1cos u - 0.5 sin u2

150 1302 a b 32.2 8 2

=

0.5 cos u + sin u = 3.49378 cos u - 1.74689 sin u u = 47.5°

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13–65. Determine the constant speed of the passengers on the amusement-park ride if it is observed that the supporting cables are directed at u = 30° from the vertical. Each chair including its passenger has a mass of 80 kg. Also, what are the components of force in the n, t, and b directions which the chair exerts on a 50-kg passenger during the motion?

4m

b

6m

u t

SOLUTION

n

+ ©F = m a ; ; n n

v2 ) T sin 30° = 80( 4 + 6 sin 30°

+ c ©Fb = 0;

T cos 30° - 8019.812 = 0 T = 906.2 N v = 6.30 m>s 16.3022 ) = 283 N Fn = 50( 7

Ans.

©Ft = m at;

Ft = 0

Ans.

©Fb = m ab ;

Fb - 490.5 = 0

or

Ans.

laws

©Fn = m an ;

Fb = 490 N

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13–66. The man has a mass of 80 kg and sits 3 m from the center of the rotating platform. Due to the rotation his speed is # increased from rest by v = 0.4 m>s2. If the coefficient of static friction between his clothes and the platform is ms = 0.3, determine the time required to cause him to slip.

3m 10 m

SOLUTION ©Ft = m at ;

Ft = 80(0.4) Ft = 32 N

©Fn = m an ;

Fn = (80)

v2 3

F = ms Nm = 2(Ft)2 + (Fn)2 0.3(80)(9.81) =

2

A

(32)2 + ((80) v )2 3

laws

or

v4 55 432 = 1024 + (6400)( ) 9

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v = 2.9575 m>s

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dv = 0.4 dt

at = v

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work

v = 0.4 t

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2.9575 = 0.4 t

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t = 7.39 s






Ans.

13–67. The vehicle is designed to combine the feel of a motorcycle with the comfort and safety of an automobile. If the vehicle is traveling at a constant speed of 80 km> h along a circular curved road of radius 100 m, determine the tilt angle u of the vehicle so that only a normal force from the seat acts on the driver. Neglect the size of the driver.

u

SOLUTION Free-Body Diagram: The free-body diagram of the passenger is shown in Fig. (a). Here, an must be directed towards the center of the circular path (positive n axis). 1h km 1000 m ba ba b h 1 km 3600 s = 22.22 m>s. Thus, the normal component of the passenger’s acceleration is given by 22.222 v2 = 4.938 m>s2. By referring to Fig. (a), = an = r 100 in

teaching

Web)

laws

or

Equations of Motion: The speed of the passenger is v = a80

9.81m cos u

N =

Wide

instructors

States

World

permitted.

Dissemination

N cos u - m(9.81) = 0

copyright

+ c ©Fb = 0;

United

on

learning.

is

of

the

not

9.81m sin u = m(4.938) cos u by

and

use

+ ©F = ma ; ; n n

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the

u = 26.7°






Ans.

*13–68. The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m> s, determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at the instant it reaches point A. Neglect the size of the car.

y

y

20 (1

x2 ) 6400

A

SOLUTION

80 m

dy d2y = - 0.00625x and 2 = - 0.00625. The slope angle u at point Geometry: Here, dx dx A is given by tan u =

dy 2 = - 0.00625(80) dx x = 80 m

u = -26.57°

and the radius of curvature at point A is r =

[1 + (dy>dx)2]3>2 2

2

|d y>dx |

=

[1 + (- 0.00625x)2]3>2 2 = 223.61 m | -0.00625| x = 80 m

in

teaching

Web)

laws

or

Equations of Motion: Here, at = 0. Applying Eq. 13–8 with u = 26.57° and r = 223.61 m, we have 800(9.81) sin 26.57° - Ff = 800(0)

Dissemination

Wide

copyright

©Ft = mat;

permitted.

Ans. of

the

not

instructors

States

World

Ff = 3509.73 N = 3.51 kN

student

the

and

United

on

learning.

is

92 b 223.61 use

800(9.81) cos 26.57° - N = 800 a by

©Fn = man;

work

N = 6729.67 N = 6.73 kN

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Ans.






x

13–69. The 0.8-Mg car travels over the hill having the shape of a parabola. When the car is at point A, it is traveling at 9 m> s and increasing its speed at 3 m>s2. Determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at this instant. Neglect the size of the car.

y

y

20 (1

x2 ) 6400

A

SOLUTION

80 m

dy d2y Geometry: Here, = - 0.00625x and 2 = - 0.00625. The slope angle u at point dx dx A is given by tan u =

dy 2 = - 0.00625(80) dx x = 80 m

u = -26.57°

and the radius of curvature at point A is r =

C 1 + (dy>dx)2 D 3>2 2

2

|d y>dx |

=

C 1 + ( -0.00625x)2 D 3>2

2

|- 0.00625|

= 223.61 m

x = 80 m

laws

or

Equation of Motion: Applying Eq. 13–8 with u = 26.57° and r = 223.61 m, we have in

teaching

Web)

800(9.81) sin 26.57° - Ff = 800(3)

Wide

copyright

©Ft = mat;

Ans.

States

World

permitted.

Dissemination

Ff = 1109.73 N = 1.11 kN the

not

instructors

92 ≤ 223.61

by

and

use

on

learning.

is

of

800(9.81) cos 26.57° - N = 800 a United

©Fn = man;

Ans.

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student

the

N = 6729.67 N = 6.73 kN






x

13–70. The package has a weight of 5 lb and slides down the chute. When it reaches the curved portion AB, it is traveling at 8 ft>s 1u = 0°2. If the chute is smooth, determine the speed of the package when it reaches the intermediate point C 1u = 30°2 and when it reaches the horizontal plane 1u = 45°2. Also, find the normal force on the package at C.

45° 8 ft/s

θ = 30° 20 ft A

B

SOLUTION + b ©Ft = mat ;

5 a 32.2 t

5 cos f =

at = 32.2 cos f + a©Fn = man ;

N - 5 sin f =

5 v2 ( ) 32.2 20

v dv = at ds v

Lg

f

v dv =

L45°

32.2 cos f (20 df)

in

teaching

Web)

laws

or

1 2 1 v - (8)2 = 644 (sin f - sin 45°) 2 2

Dissemination

Wide

copyright

At f = 45° + 30° = 75°,

permitted.

vC = 19.933 ft>s = 19.9 ft>s

the

not

instructors

States

World

Ans. is

of

Ans.

the

by

and

use

United

on

learning.

NC = 7.91 lb

the

(including

for

work

student

At f = 45° + 45° = 90°

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vB = 21.0 ft s






45°

Ans.

C

13–71. If the ball has a mass of 30 kg and a speed v = 4 m>s at the instant it is at its lowest point, u = 0°, determine the tension in the cord at this instant. Also, determine the angle u to which the ball swings and momentarily stops. Neglect the size of the ball.

u 4m

SOLUTION + c ©Fn = man;

T - 30(9.81) = 30 a

(4)2 b 4

T = 414 N +Q©Ft = mat;

Ans.

- 30(9.81) sin u = 30at at = - 9.81 sin u

at ds = v dv Since ds = 4 du, then u

L4

v dv or

L0

0

sin u(4 du) =

laws

-9.81

1 2

Wide

copyright

in

teaching

Web)

C 9.81(4)cos u D u0 = - (4)2

States

World

permitted.

Dissemination

39.24(cos u - 1) = - 8 instructors

u = 37.2°

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Ans.






*13–72. The ball has a mass of 30 kg and a speed v = 4 m>s at the instant it is at its lowest point, u = 0°. Determine the tension in the cord and the rate at which the ball’s speed is decreasing at the instant u = 20°. Neglect the size of the ball.

u 4m

SOLUTION +a©Fn = man;

T - 30(9.81) cos u = 30 a

+Q©Ft = mat;

- 30(9.81) sin u = 30at

v2 b 4

at = - 9.81 sin u at ds = v dv Since ds = 4 du, then u

u

9.81(4) cos u 2 =

v dv

1 1 (v)2 - (4)2 2 2

teaching

Web)

0

L4

or

L0

v

sin u (4 du) =

laws

-9.81

in

1 2 v 2

permitted.

Dissemination

Wide

copyright

39.24(cos u - 1) + 8 =

learning.

is

of

the

not

instructors

States

World

At u = 20°

the

by

and

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United

on

v = 3.357 m>s

Ans.

the

(including

for

work

student

at = -3.36 m>s2 = 3.36 m>s2 b

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T = 361 N






Ans.

13–73. Determine the maximum speed at which the car with mass m can pass over the top point A of the vertical curved road and still maintain contact with the road. If the car maintains this speed, what is the normal reaction the road exerts on the car when it passes the lowest point B on the road?

r

A

r

B r

SOLUTION Free-Body Diagram: The free-body diagram of the car at the top and bottom of the vertical curved road are shown in Figs. (a) and (b), respectively. Here, an must be directed towards the center of curvature of the vertical curved road (positive n axis). Equations of Motion: When the car is on top of the vertical curved road, it is required that its tires are about to lose contact with the road surface. Thus, N = 0. v2 v2 = Realizing that an = and referring to Fig. (a), r r + T ©Fn = man;

mg = m ¢

v2 ≤ r

v = 2gr

Ans.

Wide

copyright

in

teaching

Web)

laws

or

Using the result of v, the normal component of car acceleration is gr v2 an = = = g when it is at the lowest point on the road. By referring to Fig. (b), r r permitted.

Dissemination

N - mg = mg States

World

+ c ©Fn = man;

instructors

N = 2mg

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Ans.






r

13–74. If the crest of the hill has a radius of curvature r = 200 ft, determine the maximum constant speed at which the car can travel over it without leaving the surface of the road. Neglect the size of the car in the calculation. The car has a weight of 3500 lb.

v

r

SOLUTION T ©Fn = man;

3500 =

3500 v2 a b 32.2 200

v = 80.2 ft>s

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Dissemination

Wide

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laws

or

Ans.






200 ft

13–75. Bobs A and B of mass mA and mB (mA 7 mB) are connected to an inextensible light string of length l that passes through the smooth ring at C. If bob B moves as a conical pendulum such that A is suspended a distance of h from C, determine the angle u and the speed of bob B. Neglect the size of both bobs.

C

h

SOLUTION

A B

Free-Body Diagram: The free-body diagram of bob B is shown in Fig. a. The tension developed in the string is equal to the weight of bob A, i.e., T = mAg. Here, an must be directed towards the center of the horizontal circular path (positive n axis). Equations of Motion: The radius of the horizontal circular path is r = (l - h) sin u. vB2 v2 Thus, an = . By referring to Fig. a, = r (l - h) sin u + c ©Fb = 0; mAg cos u - mBg = 0 mB b mA vB 2 d (l - h) sin u

or

mAg sin u = mB c

teaching

Web)

+ ©F = ma ; ; n n

Ans.

laws

u = cos - 1 a

(1) Dissemination

Wide

in

mAg(l - h) sin u mB B

copyright

vB =

the

not

instructors

States

World

permitted.

2mA2 - mB2 . Substituting this value into Eq. (1), mA

and

use

United

on

learning.

is

of

From Fig. b, sin u =

work

student

the

by

mAg(l - h) 3mA2 - mB2 £ ≥ mB mA B

this

integrity

of

and

work

provided

g(l - h)(mA2 - mB2) mAmB B

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vB =






u

Ans.

*13–76. Prove that if the block is released from rest at point B of a smooth path of arbitrary shape, the speed it attains when it reaches point A is equal to the speed it attains when it falls freely through a distance h; i.e., v = 22gh.

B

h A

SOLUTION + R©Ft = mat;

mg sin u = mat

v dv = at ds = g sin u ds v

L0

at = g sin u However dy = ds sin u

h

v dv =

L0

g dy

v2 = gh 2 v = 22gh

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not

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permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Q.E.D.






13–77. The skier starts from rest at A(10 m, 0) and descends the smooth slope, which may be approximated by a parabola. If she has a mass of 52 kg, determine the normal force the ground exerts on the skier at the instant she arrives at point B. Neglect the size of the skier. Hint: Use the result of Prob. 13–76.

y 10 m

y

5m B

SOLUTION Geometry: Here,

dy d2y 1 1 = x and 2 = . The slope angle u at point B is given by dx 10 10 dx tan u =

dy 2 = 0 dx x = 0 m

u = 0°

and the radius of curvature at point B is

C 1 + (dy>dx) D

=

|d2y>dx2|

= 10.0 m or

x=0 m

teaching

Web)

Equations of Motion:

laws

r =

2 3>2 1 xb d 10 4 |1>10|

c1 + a

2 3>2

in

Wide

copyright

World

(1) not

on

learning.

is

of United is

work

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the

(including

for

work

student

the

by

and

use

Kinematics: The speed of the skier can be determined using v dv = at ds. Here, at must be in the direction of positive ds. Also, ds = 21 + (dy>dx)2dx 1 2 = 21 + 100 x dx

this

integrity

of

and

work

provided

the

is

This

part

x 1 2 100 x

≤ A 21 + any

courses

1 2 100 x dx

B

destroy

¢

L10 m 10 21 + sale

v2 = 9.81 m2>s2

will

their

L0

0

v dv = - 9.81

.

1 2 100 x

of

v

(+ )

assessing

1 x x. Then, sin u = 10 10 21 + and

Here, tan u =

Substituting v2 = 98.1 m2>s2, u = 0°, and r = 10.0 m into Eq.(1) yields N - 52(9.81) cos 0° = 52 a

98.1 b 10.0

N = 1020.24 N = 1.02 kN






Ans.

permitted.

Dissemination

y2 b r

N - 52(9.81) cos u = ma

the

+a©Fn = man;

at = -9.81 sin u

instructors

52(9.81) sin u = - 52at

States

+b©Ft = mat;

1 2 –– x 20

5 A

x

13–78. A spring, having an unstretched length of 2 ft, has one end attached to the 10-lb ball. Determine the angle u of the spring if the ball has a speed of 6 ft> s tangent to the horizontal circular path.

6 in.

A

u

SOLUTION Free-Body Diagram: The free-body diagram of the bob is shown in Fig. (a). If we denote the stretched length of the spring as l, then using the springforce formula, Fsp = ks = 20(l - 2) lb. Here, an must be directed towards the center of the horizontal circular path (positive n axis). Equations of Motion: The radius of the horizontal circular path is r = 0.5 + l sin u. 62 v2 Since an = = , by referring to Fig. (a), r 0.5 + l sin u

62 10 a b 32.2 0.5 + l sin u

(2) or

20(l - 2) sin u =

(1)

teaching

Web)

+ ©F = ma ; ; n n

20(l - 2) cos u - 10 = 0

laws

+ c ©Fb = 0;

Wide

copyright

in

Solving Eqs. (1) and (2) yields u = 31.26° = 31.3°

Ans.

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is

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the

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the

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United

on

learning.

is

of

the

instructors

l = 2.585 ft






not

States

World

permitted.

Dissemination

Ans.

k

20 lb>ft

13–79. u

The airplane, traveling at a constant speed of 50 m>s, is executing a horizontal turn. If the plane is banked at u = 15°, when the pilot experiences only a normal force on the seat of the plane, determine the radius of curvature r of the turn. Also, what is the normal force of the seat on the pilot if he has a mass of 70 kg.

r

SOLUTION + c a Fb = mab;

NP sin 15° - 7019.812 = 0 NP = 2.65 kN

+ ; a Fn = man;

NP cos 15° = 70 a

Ans. 502 b r

r = 68.3 m

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the

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Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






*13–80. A 5-Mg airplane is flying at a constant speed of 350 km> h along a horizontal circular path of radius r = 3000 m. Determine the uplift force L acting on the airplane and the banking angle u. Neglect the size of the airplane.

L u

r

SOLUTION Free-Body Diagram: The free-body diagram of the airplane is shown in Fig. (a). Here, an must be directed towards the center of curvature (positive n axis). Equations of Motion: The speed of the airplane is v = ¢ 350 = 97.22 m>s. Realizing that an =

1h km 1000 m ≤¢ ≤¢ ≤ h 1 km 3600 s

97.222 v2 = 3.151 m>s2 and referring to Fig. (a), = r 3000

+ c ©Fb = 0;

T cos u - 5000(9.81) = 0

(1)

+ ©F = ma ; ; n n

T sin u = 5000(3.151)

(2)

laws

or

Solving Eqs. (1) and (2) yields T = 51517.75 = 51.5 kN

teaching

Web)

Ans.

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on

learning.

is

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permitted.

Dissemination

Wide

copyright

in

u = 17.8°






13–81. A 5-Mg airplane is flying at a constant speed of 350 km> h along a horizontal circular path. If the banking angle u = 15°, determine the uplift force L acting on the airplane and the radius r of the circular path. Neglect the size of the airplane.

L u

r

SOLUTION Free-Body Diagram: The free-body diagram of the airplane is shown in Fig. (a). Here, an must be directed towards the center of curvature (positive n axis). 1h km 1000 m ba ba b h 1 km 3600 s

Equations of Motion: The speed of the airplane is v = a350 = 97.22 m>s. Realizing that an =

97.222 v2 and referring to Fig. (a), = r r

+ c ©Fb = 0;

L cos 15° - 5000(9.81) = 0 L = 50780.30 N = 50.8 kN

or

97.222 ≤ r

50780.30 sin 15° = 5000 ¢

laws

+ ©F = ma ; ; n n

Ans.

teaching

Web)

r = 3595.92 m = 3.60 km

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the

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Dissemination

Wide

copyright

in

Ans.






13–82. The 800-kg motorbike travels with a constant speed of 80 km > h up the hill. Determine the normal force the surface exerts on its wheels when it reaches point A. Neglect its size.

y A

y2

2x x

100 m

SOLUTION d2y dy 22 22 = and = - 3>2 . The angle that 2 1>2 dx dx 2x 4x the hill slope at A makes with the horizontal is Geometry: Here, y = 22x1>2. Thus,

u = tan - 1 a

dy 22 b2 = tan - 1 ¢ 1>2 ≤ 2 = 4.045° dx x = 100 m 2x x = 100 m

The radius of curvature of the hill at A is given by

dx2

2

= 2849.67 m

22

2-

2

4(1003>2)

x = 100 m

3>2

or

2(100 )

=

b R

Web)

dy

5

2

1>2

in

2

2

22

B1 + a

laws

rA =

dy 2 3>2 b R dx

teaching

B1 + a

States

World

permitted.

Dissemination

Wide

copyright

Free-Body Diagram: The free-body diagram of the motorcycle is shown in Fig. (a). Here, an must be directed towards the center of curvature (positive n axis).

work the

solely

of

protected

Thus, an =

(including

for

v2 22.222 = 0.1733 m>s2. By referring to Fig. (a), = rA 2849.67

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

Equations of Motion: The speed of the motorcycle is 1h km 1000 m ba ba b = 22.22 m>s v = a 80 h 1 km 3600 s

work

800(9.81)cos 4.045° - N = 800(0.1733) this

assessing

is

R+ ©Fn = man;

sale

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courses

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provided

integrity

of

and

work

N = 7689.82 N = 7.69 kN






Ans.

13–83. The ball has a mass m and is attached to the cord of length l. The cord is tied at the top to a swivel and the ball is given a velocity v0. Show that the angle u which the cord makes with the vertical as the ball travels around the circular path must satisfy the equation tan u sin u = v20>gl. Neglect air resistance and the size of the ball.

O

u l

SOLUTION

+ c ©Fb = 0; Since r = l sin u

T sin u = m a

v20 b r

T cos u - mg = 0 T = a

v0

mv20 l sin2 u

mv20 cos u b a 2 b = mg l sin u v20 gl

Q.E.D.

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the

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is

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World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

tan u sin u =

or

+ ©F = ma ; : n n






*13–84. The 5-lb collar slides on the smooth rod, so that when it is at A it has a speed of 10 ft> s. If the spring to which it is attached has an unstretched length of 3 ft and a stiffness of k = 10 lb>ft, determine the normal force on the collar and the acceleration of the collar at this instant.

y 10 ft/s A

1 x2 y 8 –– 2

SOLUTION y = 8 -

1 2 x 2 O

dy = tan u = x 2 = 2 dx x=2

x

u = 63.435° 2 ft

d2y

= -1

dx2

3

2

dx2

2

(1 + (- 2)2)2 = 11.18 ft | -1|

in

teaching

Web)

1 (2)2 = 6 2

Dissemination

Wide

copyright

y = 8 -

d2y

3

=

or

r =

dy 2 2 b R dx

laws

B1 + a

not

instructors

States

World

permitted.

OA = 2(2)2 + (6)2 = 6.3246

and

use

United

on

learning.

is

of

the

Fs = kx = 10(6.3246 - 3) = 33.246 lb

work

student

the

by

6 ; f = 71.565° 2

(10)2 5 ba b 32.2 11.18

solely

of

protected

the

(including

for

tan f =

work

5 cos 63.435° - N + 33.246 cos 45.0° = a

Ans.

and

any

courses

part

the

is

This

N = 24.4 lb

provided

integrity

of

and

work

this

assessing

is

+b©Fn = man;

destroy

of

5 sin 63.435° + 33.246 sin 45.0° = a

5 b at 32.2

sale

will

their

+ R©Ft = mat;

at = 180.2 ft>s2 an =

(10)2 v2 = 8.9443 ft>s2 = r 11.18

a = 2(180.2)2 + (8.9443)2 a = 180 ft>s2






Ans.

13–85. The spring-held follower AB has a weight of 0.75 lb and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.2 ft and z = 10.1 sin u2 ft. If the cam is rotating at a constant rate of 6 rad>s, determine the force at the end A of the follower when u = 90°. In this position the spring is compressed 0.4 ft. Neglect friction at the bearing C.

SOLUTION z = 0.1 sin 2u # # z = 0.2 cos 2uu # $ $ z = -0.4 sin 2uu2 + 0.2 cos 2uu

#

u = 6 rad>s

##

u = 0 $ z = -14.4 sin 2u

or

$ FA - 12(z + 0.3) = mz laws

a Fz = maz;

in

teaching

Web)

0.75 ( -14.4 sin 2u) 32.2

Dissemination

Wide

copyright

FA - 12(0.1 sin 2u + 0.3) =

is

of

the

not

instructors

States

World

permitted.

For u = 45°,

by

and

use

United

on

learning.

0.75 ( -14.4) 32.2

work

student

the

FA - 12(0.4) =

for

FA = 4.46 lb

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

Ans.






13–86. Determine the magnitude of the resultant force acting on a 5-kg particle at the instant t = 2 s, if the particle is moving along a horizontal path defined by the equations r = (2t + 10) m and u = (1.5t2 - 6t) rad, where t is in seconds.

SOLUTION r = 2t + 10|t = 2 s = 14 # r = 2 $ r = 0

teaching

Web)

laws

or

u = 1.5t2 - 6t # u = 3t - 6冷t = 2 s = 0 $ u = 3 # $ ar = r - ru2 = 0 - 0 = 0 $ ## au = ru + 2ru = 14(3) + 0 = 42

Fr = 5(0) = 0

©Fu = mau;

Fu = 5(42) = 210 N United

on

learning.

is

of

the

not

instructors

States

World

permitted.

©Fr = mar;

Dissemination

Wide

copyright

in

Hence,

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

F = 2(Fr)2 + (Fu)2 = 210 N






Ans.

13–87. The path of motion of a 5-lb particle in the horizontal plane is described in terms of polar coordinates as r = (2t + 1) ft and u = (0.5t2 - t) rad, where t is in seconds. Determine the magnitude of the resultant force acting on the particle when t = 2 s.

SOLUTION # $ r = 2 ft>s r = 0 # = 0 rad u = t - 1|t = 2 s = 1 rad>s

r = 2t + 1|t = 2 s = 5 ft

$ u = 1 rad>s2

u = 0.5t2 - t|t = 2 s # $ ar = r - ru2 = 0 - 5(1)2 = - 5 ft>s2 $ ## au = ru + 2ru = 5(1) + 2(2)(1) = 9 ft>s2 Fr =

5 ( - 5) = - 0.7764 lb 32.2

©Fu = mau;

Fu =

5 (9) = 1.398 lb 32.2 or

©Fr = mar;

F = 2F2r + F2u = 2(-0.7764)2 + (1.398)2 = 1.60 lb

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

Ans.






*13–88. A particle, having a mass of 1.5 kg, moves along a path defined by the equations r = 14 + 3t2 m, u = 1t2 + 22 rad, and z = 16 - t32 m, where t is in seconds. Determine the r, u, and z components of force which the path exerts on the particle when t = 2 s.

SOLUTION u = t2 + 2

# r = 3 m>s # u = 2t| t = 2 s = 4 rad>s

z = 6 - t3

z = - 3t2

r = 4 + 3t| t = 2 s = 10 m

##

$ r = 0 $ u = 2 rad>s2

#

##

z = - 6t| t = 2 s = - 12 m>s2

#

ar = r - r u 2 = 0 -10(4)2 = - 160 m>s2 $ ## au = ru + 2ru = 10(2) + 2(3)(4) = 44 m>s2

##

az = z = - 12 m>s2 Fr = 1.5(- 160) = - 240 N

Ans.

©Fu = mau;

Fu = 1.5(44) = 66 N

Ans.

©Fz = maz;

Fz - 1.5(9.81) = 1.5(- 12)

laws

or

©Fr = mar;

teaching

Web)

Fz = - 3.28 N

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

Ans.






13–89. Rod OA rotates # counterclockwise with a constant angular velocity of u = 5 rad>s. The double collar B is pinconnected together such that one collar slides over the rotating rod and the other slides over the horizontal curved rod, of which the shape is described by the equation r = 1.512 - cos u2 ft. If both collars weigh 0.75 lb, determine the normal force which the curved rod exerts on one collar at the instant u = 120°. Neglect friction.

A B r

SOLUTION ·

# $ Kinematic: Here, u = 5 rad>s and u = 0. Taking the required time derivatives at u = 120°, we have r = 1.5 (2 – cos ) ft.

r = 1.5(2 - cos u)|u = 120° = 3.75 ft # # r = 1.5 sin uu|u = 120° = 6.495 ft>s $ # $ r = 1.5(sin uu + cos uu 2)|u = 120° = - 18.75 ft>s2

teaching

Web)

laws

or

Applying Eqs. 12–29, we have # $ ar = r - ru2 = - 18.75 - 3.75(52) = - 112.5 ft>s2 $ # # au = r u + 2r u = 3.75(0) + 2(6.495)(5) = 64.952 ft>s2

Wide

copyright

in

Equation of Motion: The angle c must be obtained first. Dissemination

1.5(2 - cos u) r 2 = = 2.8867 dr>du 1.5 sin u u = 120°

World

permitted.

c = 70.89°

on

learning.

is

of

the

not

instructors

States

tan c =

the

by

and

use

United

Applying Eq. 13–9, we have

work

student

0.75 (-112.5) 32.2

the

(including

for

- N cos 19.11° =

is

work

solely

of

protected

a Fr = mar ;

Ans. integrity

of

and

work

this

assessing

N = 2.773 lb = 2.77 lb provided

0.75 (64.952) 32.2

FOA + 2.773 sin 19.11° =

destroy sale

will

their

FOA = 0.605 lb

of

and

any

courses

part

the

is

This

a Fu = mau ;






O = 5 rad/s

13–90. z

The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r = 1.5 m, u = 10.7t2 rad, and z = 1 -0.5t2 m, where t is in seconds. Determine the components of force Fr , Fu , and Fz which the slide exerts on him at the instant t = 2 s. Neglect the size of the boy.

u

SOLUTION r = 1.5 # $ r = r = 0 $ u = 0

u = 0.7t # u = 0.7

z

z = - 0.5t # z = - 0.5

$ z = 0

# $ ar = r - r(u)2 = 0 - 1.5(0.7)2 = - 0.735 $ ## au = ru + 2ru = 0 $ az = z = 0 Fr = 40( -0.735) = - 29.4 N

Ans.

©Fu = mau;

Fu = 0

Ans.

©Fz = maz;

Fz - 40(9.81) = 0

Wide

copyright

in

teaching

Web)

laws

or

©Fr = mar;

Fz = 392 N

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Ans.






r 1.5 m

13–91. The 0.5-lb particle is guided along the circular path using the slotted arm guide. If the arm has an$ angular velocity # u = 4 rad>s and an angular acceleration u = 8 rad>s2 at the instant u = 30°, determine the force of the guide on the particle. Motion occurs in the horizontal plane.

r

u 0.5 ft

SOLUTION

0.5 ft

r = 2(0.5 cos u) = 1 cos u # # r = - sin uu

# $ r = - cos uu2 - sin uu

# $ At u = 30°, u = 4 rad>s and u = 8 rad>s2 r = 1 cos 30° = 0.8660 ft

#

r = - sin 30° (4) = - 2 ft>s .. r = - cos 30° (4)2 - sin 30° (8) = - 17.856 ft>s2

Wide

copyright

in

teaching

Web)

laws

or

#2 $ ar = r - ru = - 17.856 - 0.8660(4)2 = - 31.713 ft>s2 $ ## au = ru + 2ru = 0.8660(8) + 2( -2)(4) = - 9.072 ft>s2 Dissemination

0.5 (- 31.713) 32.2

World

permitted.

N = 0.5686 lb learning.

is

of

the

not

instructors

- N cos 30° =

States

Q+ ©Fr = mar;

by

and

use

United

on

0.5 ( - 9.072) 32.2

for

work

student

F - 0.5686 sin 30° =

the

+a©Fu = mau;

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

F = 0.143 lb






Ans.

*13–92. Using a forked rod, a smooth cylinder C having a mass of 0.5 kg is forced to move along the vertical slotted path r = 10.5u2 m, where u is in radians. If the angular position of the arm is u = 10.5t22 rad, where t is in seconds, determine the force of the rod on the cylinder and the normal force of the slot on the cylinder at the instant t = 2 s. The cylinder is in contact with only one edge of the rod and slot at any instant.

C

r

SOLUTION r = 0.5u u = 0.5t2

# r = 0.5u # u = t

$ $ r = 0.5u $ u = 1

At t = 2 s, # u = 2 rad>2

u = 2 rad = 114.59° r = 1m

# r = 1 m>s

$ u = 1 rad>s2

$ r = 0.5 m>s2

0.5(2) r = c = 63.43° dr>du 0.5 # # ar = r - ru2 = 0.5 - 1(2)2 = - 3.5 $ ## au = ru + 2ru = 1(1) + 2(1)(2) = 5

Wide

copyright

in

teaching

Web)

laws

or

tan c =

NC cos 26.57° - 4.905 cos 24.59° = 0.5( - 3.5) States

World

permitted.

Dissemination

+ a©Fr = mar;

on

learning.

is

of

the

Ans.

by

and

use

United

F - 3.030 sin 26.57° + 4.905 sin 24.59° = 0.5(5) the

+ b© Fu = mau;

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

F = 1.81 N






Ans.

not

instructors

NC = 3.030 = 3.03 N

0.5 u

u

13–93. If arm OA .rotates with a constant clockwise angular velocity of u = 1.5 rad>s. determine the force arm OA exerts on the smooth 4-lb cylinder B when u = 45°. A

B r

SOLUTION

u u

Kinematics: Since the motion of cylinder B is known, ar and au will be determined 4 first. Here, = cos u or r = 4 sec u ft. The value of r and its time derivatives at the r instant u = 45° are

O

4 ft

r = 4 sec u |u = 45° = 4 sec 45° = 5.657 ft # # r = 4 sec u(tan u)u|u = 45° = 4 sec 45° tan 45°(1.5) = 8.485 ft>s $ # # # $ r = 4 C sec u(tan u)u + u A sec u sec2 uu + tan u sec u tan uu B D # $ #2 = 4 C sec u(tan u)u + sec3 uu2 + sec u tan2 uu D 2

u = 45°

laws

or

= 4 C sec 45° tan 45°(0) + sec 45°(1.5) + sec 45° tan2 45°(1.5)2 D

Web)

2

teaching

3

Dissemination

Wide

copyright

in

= 38.18 ft>s2

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Using the above time derivatives, # $ ar = r - ru 2 = 38.18 - 5.657 A 1.52 B = 25.46 ft>s2 $ ## au = ru - 2ru = 5.657(0) + 2(8.485)(1.5) = 25.46 ft>s2

4 (25.46) 32.2 the

is

This

integrity

of

and

work

N cos 45° - 4 cos 45° = provided

©Fr = mar;

this

assessing

is

work

solely

of

protected

the

(including

Equations of Motion: By referring to the free-body diagram of the cylinder shown in Fig. a,

and

any

courses

part

N = 8.472 lb destroy

of

FOA - 8.472 sin 45° - 4 sin 45° =

4 (25.46) 32.2

sale

will

their

©Fu = mau;

FOA = 12.0 lb






Ans.

13–94. The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangular spiral r = 1eu2 m, where u is in radians. Determine the tangential force F and the normal force N acting on the collar when u = 90°, if the force F maintains a constant angular motion # u = 2 rad>s.

F

SOLUTION

r

r = eu

u

# # r = euu # $ $ r = eu(u)2 + euu At u = 90° # u = 2 rad>s $ u = 0 r = 4.8105 laws

or

# r = 9.6210 in

teaching

Web)

$ r = 19.242

and by

work

student

(including

A B

= eu>eu = 1

the

r dr du

for

tan c =

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

# $ ar = r - r(u)2 = 19.242 - 4.8105(2)2 = 0 $ ## au = ru + 2ru = 0 + 2(9.6210)(2) = 38.4838 m>s2

is

work

solely

of

protected

the

c = 45°

-NC cos 45° + F cos 45° = 2(0)

+ ; a Fu = mau;

F sin 45° + NC sin 45° = 2(38.4838) and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

+ c a Fr = mar;

Ans.

F = 54.4 N






sale

will

their

destroy

of

NC = 54.4 N

Ans.

r eu

13–95. The ball has a mass of 2 kg and a negligible size. It is originally traveling around the horizontal circular path of radius r# 0 = 0.5 m such that the angular rate of rotation is u0 = 1 rad>s. If the attached cord ABC is drawn down through the hole at a constant speed of 0.2 m>s, determine the tension the cord exerts on the ball at the instant r = 0.25 m. Also, compute the angular velocity of the ball at this instant. Neglect the effects of friction between the ball and horizontal plane. Hint: First show that$ the equation of motion # in the u ## direction yields au = ru + 2ru = 11>r21d1r2u2>dt2 = 0. # When integrated, r2u = c, where the constant c is determined from the problem data.

A r

F

$ 1 d 2# ## 0 = m[ru + 2ru] = mc (r u) d = 0 r dt

Thus,

or

# d(r2u) = 0 # r2u = C

Dissemination

copyright

Ans.

Wide

in

teaching

Web)

laws

# (0.5)2(1) = C = (0.25)2u # u = 4.00 rad>s

World

permitted.

$ r = 0

not

instructors

States

# Since r = - 0.2 m>s,

the

by

and

use

United

on

learning.

is

of

the

$ # ar = r - r(u)2 = 0 - 0.25(4.00)2 = - 4 m>s2

work

student

-T = 2( - 4)

(including

for

a Fr = mar;

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

T = 8N






r0

0.2 m/s C

SOLUTION a Fu = mau;

u

B

Ans.

· u0

*13–96. A

The particle has a mass of 0.5 kg and is confined to move along the smooth horizontal slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when u = 30°. The rod is rotating with a constant angular velocity # u = 2 rad>s. Assume the particle contacts only one side of the slot at any instant.

· u

2 rad/s

0.5 m u

SOLUTION 0.5 = 0.5 sec u cos u # # r = 0.5 sec u tan uu # # # $ $ r = 0.5 E C (sec u tan uu) tan u + sec u(sec2 uu D u + sec u tan uu F # # $ = 0.5 C sec u tan2 uu + sec3 uu2 + sec u tan uu D

O

r =

# $ When u = 30°, u = 2 rad>s and u = 0 r = 0.5 sec 30° = 0.5774 m laws

or

# r = 0.5 sec 30° tan 30°(2) = 0.6667 m>s in

teaching

Web)

$ r = 0.5 C sec 30° tan2 30°(2)2 + sec3 30°(2)2 + sec 30° tan 30°(0) D Dissemination

Wide

copyright

= 3.849 m>s2

work the

(including

for

N cos 30° - 0.5(9.81) cos 30° = 0.5(1.540)

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

# $ ar = r - ru2 = 3.849 - 0.5774(2)2 = 1.540 m>s2 $ ## au = ru + 2ru = 0.5774(0) + 2(0.6667)(2) = 2.667 m>s2 Q+ ©Fr = mar;

Ans.

assessing

is

work

solely

of

protected

N = 5.79 N

this

F + 0.5(9.81) sin 30° - 5.79 sin 30° = 0.5(2.667) provided

integrity

of

and

work

+ R©Fu = mau;

sale

will

their

destroy

of

and

any

courses

part

the

is

This

F = 1.78 N






r

Ans.

13–97. Solve $ Problem 13–96# if the arm has an angular acceleration of u = 3 rad/s2 and u = 2 rad/s at this instant. Assume the particle contacts only one side of the slot at any instant.

A

· u

2 rad/s

r 0.5 m

u

SOLUTION 0.5 = 0.5 sec u cos u # # r = 0.5 sec u tan uu r =

O

# # # $ $ r = 0.5 E C (sec u tan uu) tan u + sec u(sec2 uu) D u + sec u tan uu F # # $ = 0.5 C sec u tan2 uu2 + sec3 uu2 + sec u tan uu D # $ When u = 30°, u = 2 rad>s and u = 3 rad>s2 r = 0.5 sec 30° = 0.5774 m laws

or

# r = 0.5 sec 30° tan 30°(2) = 0.6667 m>s

teaching in

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

= 4.849 m>s2 # $ ar = r - ru2 = 4.849 - 0.5774(2)2 = 2.5396 m>s2 $ ## au = ru + 2ru = 0.5774(3) + 2(0.6667)(2) = 4.3987 m>s2

Web)

$ r = 0.5 C sec 30° tan2 30°(2)2 + sec3 30°(2)2 + sec 30° tan 30°(3) D

work

student

N cos 30° - 0.5(9.81) cos 30° = 0.5(2.5396) the

(including

for

Q+ ©Fr = mar;

Ans.

assessing

is

work

solely

of

protected

N = 6.3712 = 6.37 N

this

F + 0.5(9.81) sin 30° - 6.3712 sin 30° = 0.5(4.3987) provided

integrity

of

and

work

+R©Fu = mau;

sale

will

their

destroy

of

and

any

courses

part

the

is

This

F = 2.93 N






Ans.

13–98. The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangular spiral r = 1eu2 m, where u is in radians. Determine the tangential force F and the normal force N acting on the collar when u =# 45°, if the force F maintains a constant angular motion u = 2 rad>s.

F

r

SOLUTION r = eu

θ

# # r = eu u # # $ r = eu(u)2 + eu u At u = 45° # u = 2 rad>s $ u = 0 r = 2.1933 laws

or

# r = 4.38656 in

teaching

Web)

$ r = 8.7731

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

# $ ar = r - r(u)2 = 8.7731 - 2.1933(2)2 = 0 $ # # au = r u + 2 r u = 0 + 2(4.38656)(2) = 17.5462 m>s2

the

(including

for

work

student

the

by

and

r = e u>e u = 1 dr a #b du solely

of

protected

tan c =

integrity

of

and

work

this

assessing

is

work

c = u = 45° provided

- NC cos 45° + F cos 45° = 2(0) destroy

of

and

any

courses

part

the

is

This

Q+ a Fr = mar ;

will

their

F sin 45° + NC sin 45° = 2(17.5462) sale

+ a a Fu = mau ;






N = 24.8 N

Ans.

F = 24.8 N

Ans.

r = eθ

13–99. For a short time, the 250-kg roller coaster car is traveling along the spiral track such that its position measured from the top of the track has components r = 8 m, u = 10.1t + 0.52 rad, and z = 1-0.2t2 m, where t is in seconds. Determine the magnitudes of the components of force which the track exerts on the car in the r, u, and z directions at the instant t = 2 s. Neglect the size of the car.

r=8m

SOLUTION # $ Kinematic: Here, r = 8 m, r = r = 0. Taking the required time derivatives at t = 2 s, we have # $ u = 0.100 rad>s u = 0 u = 0.1t + 0.5|t = 2s = 0.700 rad # z = - 0.200 m>s

z = - 0.2t|t = 2 s = - 0.400 m

$ z = 0

Applying Eqs. 12–29, we have $ $ ar = r - ru2 = 0 - 8(0.1002) = - 0.0800 m>s2 $ ## au = r u + 2 ru = 8(0) + 2(0)(0.200) = 0

teaching

Web)

laws

or

$ az = z = 0

Fr = 250( -0.0800) = - 20.0 N

©Fu = mau ;

Fu = 250(0) = 0

©Fz = maz ;

Fz - 250(9.81) = 250(0)

Ans.

States

World

permitted.

©Fr = mar ;

Dissemination

Wide

copyright

in

Equation of Motion:

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Fz = 2452.5 N = 2.45 kN

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Ans.






Ans.

*13–100. The 0.5-lb ball is guided along the vertical circular path r = 2rc cos# u using the arm OA. If the arm has an angular u = 0.4 rad>s and an angular acceleration velocity $ u = 0.8 rad>s2 at the instant u = 30°, determine the force of the arm on the ball. Neglect friction and the size of the ball. Set rc = 0.4 ft.

P

r

u O

SOLUTION r = 2(0.4) cos u = 0.8 cos u # # r = -0.8 sin uu # $ $ r = -0.8 cos uu2 - 0.8 sin uu # $ At u = 30°, u = 0.4 rad>s, and u = 0.8 rad>s2 r = 0.8 cos 30° = 0.6928 ft # r = - 0.8 sin 30°(0.4) = - 0.16 ft>s

Wide

copyright

in

teaching

Web)

laws

or

$ r = - 0.8 cos 30°(0.4)2 - 0.8 sin 30°(0.8) = - 0.4309 ft>s2 # $ ar = r - ru2 = -0.4309 - 0.6928(0.4)2 = - 0.5417 ft>s2 $ ## au = ru + 2ru = 0.6928(0.8) + 2( -0.16)(0.4) = 0.4263 ft>s2

World

N = 0.2790 lb

permitted.

Dissemination

0.5 (-0.5417) 32.2

learning.

is

of

the

not

instructors

N cos 30° - 0.5 sin 30° =

States

+Q©Fr = mar;

United

on

0.5 (0.4263) 32.2

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and

use

FOA + 0.2790 sin 30° - 0.5 cos 30° =

for

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the

a+ ©Fu = mau;

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FOA = 0.300 lb






Ans.

rc

A

13–101. The ball of mass m is guided along the vertical circular path r = 2rc cos u using # the arm OA. If the arm has a constant angular velocity u0, determine the angle u … 45° at which the ball starts to leave the surface of the semicylinder. Neglect friction and the size of the ball.

P

r

u O

SOLUTION r = 2rc cos u # # r = -2rc sin uu # $ $ r = - 2rc cos uu2 - 2rc sin uu # $ Since u is constant, u = 0. # # # # $ ar = r - ru2 = -2rc cos uu20 - 2rc cos uu20 = -4rc cos uu20 # - mg sin u = m(- 4rc cos uu20) # # 4rc u20 4rc u20 -1 tan u = u = tan ¢ ≤ g g

or

Ans.

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permitted.

Dissemination

Wide

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teaching

Web)

laws

+ Q©Fr = mar;






rc

A

13–102. Using a forked rod, a smooth cylinder P, having a mass of 0.4 kg, is forced to move along the vertical slotted path r = 10.6u2 m, where u is in radians. If the cylinder has a constant speed of vC = 2 m>s, determine the force of the rod and the normal force of the slot on the cylinder at the instant u = p rad. Assume the cylinder is in contact with only one edge of the rod and slot at any instant. Hint: To obtain the time derivatives necessary to compute the cylinder’s acceleration components ar and au, take the first and second time derivatives of r = 0.6u. Then, for further # information, use Eq. 12–26 to determine u. Also, take the # time derivative of Eq. 12–26, noting that vC = 0, to $ determine u.

P

r

SOLUTION $ $ # r = 0.6 u r = 0.6 u # # # # vr = r = 0.6u vu = ru = 0.6uu r = 0.6 u

2 # v2 = r 2 + a rub

2 laws

# uu2

Web)

$ u = -

teaching

1 + u2

World

= 1.011 rad>s

the

not

instructors

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learning.

is

2 0.621 + p2

States

# u =

of

At u = p rad,

permitted.

Dissemination

Wide

#$ # #$ 0 = 0.72u u + 0.36 a 2uu3 + 2u2u u b

or

0.6 21 + u2

in

# u =

copyright

# 2 # 2 22 = a 0.6u b + a 0.6uu b

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# r = 0.6(1.011) = 0.6066 m>s

United

$ (p)(1.011)2 = - 0.2954 rad>s2 u = 1 + p2 r = 0.6(p) = 0.6 p m

c = 72.34° sale

0.6u r = = u = p dr>du 0.6

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$ r = 0.6( -0.2954) = - 0.1772 m>s2 # $ ar = r - ru 2 = - 0.1772 - 0.6 p(1.011)2 = - 2.104 m>s2 $ ## au = ru + 2ru = 0.6p( - 0.2954) + 2(0.6066)(1.011) = 0.6698 m>s2 tan c =

+ ©F = ma ; ; r r

-N cos 17.66° = 0.4( - 2.104)

+ T ©Fu = mau ;

- F + 0.4(9.81) + 0.883 sin 17.66° = 0.4(0.6698) F = 3.92 N






u r

N = 0.883 N

Ans.

Ans.

0.6u

p

13–103. A ride in an amusement park consists of a cart which is supported by small wheels. Initially the cart is traveling in a circular path# of radius r0 = 16 ft such that the angular rate of rotation is u0 = 0.2 rad>s. If the attached cable OC is drawn # inward at a constant speed of r = - 0.5 ft>s, determine the tension it exerts on the cart at the instant r = 4 ft. The cart and its passengers have a total weight of 400 lb. Neglect the effects of friction. Hint: First show that the equation of .. .. motion in# the u direction yields # au = ru + 2ru = 11>r2 d1r2u2>dt = 0. When integrated, r2u = c, where the constant c is determined from the problem data.

r

u O

SOLUTION +Q©Fr = mar ;

-T = ¢

+a©Fu = mau ;

0 = ¢

# 400 $ ≤ ¢ r - ru 2 ≤ 32.2

(1)

$ 400 ## ≤ ¢ ru + 2ru ≤ 32.2

# 1 d From Eq. (2), ¢ ≤ ¢ r2u ≤ = 0 r dt

(2)

# r 2u = c

teaching

Web)

laws

or

# Since u0 = 0.2 rad>s when r0 = 16 ft, c = 51.2.

Wide

copyright

in

Hence, when r = 4 ft,

is

of

the

not

instructors

States

World

permitted.

Dissemination

# 51.2 u = ¢ 2 ≤ = 3.2 rad>s (4)

student

the

by

and

use

United

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learning.

$ Since r = - 0.5 ft>s, r = 0, Eq. (1) becomes

the

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400 a0 - (4)(3.2)2 b 32.2 is

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-T =

assessing

T = 509 lb

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C

· u0

*13–104. # The arm is rotating at a rate of u = 5 rad>s when $ u = 2 rad>s2 and u = 90°. Determine the normal force it must exert on the 0.5-kg particle if the particle is confined to move along the slotted path defined by the horizontal hyperbolic spiral ru = 0.2 m.

0.2 — u

r r u·

·· 5 rad/s, u u

SOLUTION p = 90° 2

u =

# u = 5 rad>s $ u = 2 rad>s2

Wide

copyright

in

teaching

Web)

laws

or

r = 0.2>u = 0.12732 m # # r = -0.2 u - 2 u = -0.40528 m>s # $ $ r = -0.2[- 2u - 3 (u)2 + u - 2 u] = 2.41801 # $ a r = r - r(u)2 = 2.41801 - 0.12732(5)2 = - 0.7651 m>s2 $ ## au = ru + 2 ru = 0.12732(2) + 2( -0.40528)(5) = - 3.7982 m>s2 Dissemination

0.2>u

World

permitted.

-0.2u - 2

not is

of

the

=

instructors

r dr (du )

States

tan c =

student

the

by

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United

on

learning.

p c = tan - 1(- ) = -57.5184° 2 Np cos 32.4816° = 0.5( - 0.7651)

+ ©F = ma ; ; u u

F + Np sin 32.4816° = 0.5( -3.7982) this

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for

work

+ c ©Fr = m ar ;

part

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of

and

work

NP = - 0.453 N F = - 1.66 N

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Ans.






90

2 rad/s2

13–105. The forked rod is used to move the smooth 2-lb particle around the horizontal path in the # shape of a limaçon, r = (2 + cos u) ft. If at all times u = 0.5 rad>s, determine the force which the rod exerts on the particle at the instant u = 90°. The fork and path contact the particle on only one side.

2 ft

u · u

SOLUTION r = 2 + cos u # # r = -sin uu $ # $ r = -cos uu2 - sin uu # $ At u = 90°, u = 0.5 rad>s, and u = 0 r = 2 + cos 90° = 2 ft # r = - sin 90°(0.5) = -0.5 ft>s

Wide

copyright

in

teaching

Web)

laws

or

$ r = - cos 90°(0.5)2 - sin 90°(0) = 0 # $ ar = r - ru2 = 0 - 2(0.5)2 = - 0.5 ft>s2 $ ## au = ru + 2ru = 2(0) + 2( -0.5)(0.5) = - 0.5 ft>s2 Dissemination

r 2 + cos u 2 = = -2 dr>du - sin u u = 90°

World

permitted.

c = - 63.43°

use

United

on

learning.

is

of

the

not

instructors

States

tan c =

and

2 (- 0.5) 32.2

by

N = 0.03472 lb work

student

the

- N cos 26.57° =

the

(including

for

+ c ©Fr = mar;

2 (-0.5) 32.2

work

solely

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protected

F - 0.03472 sin 26.57° =

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F = - 0.0155 lb

and

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+ ©F = ma ; ; u u






r

Ans.

3 ft

13–106. The forked rod is used to move the smooth 2-lb particle around the horizontal path in the # shape of a limaçon, r = (2 + cos u) ft. If at all times u = 0.5 rad>s, determine the force which the rod exerts on the particle at the instant u = 60°. The fork and path contact the particle on only one side.

2 ft

u · u

SOLUTION r = 2 + cos u # # r = -sin uu # $ $ r = -cos uu2 - sin uu # $ At u = 60°, u = 0.5 rad>s, and u = 0 r = 2 + cos 60° = 2.5 ft # r = - sin 60°(0.5) = -0.4330 ft>s

Wide

copyright

in

teaching

Web)

laws

or

$ r = - cos 60°(0.5)2 - sin 60°(0) = - 0.125 ft>s2 # $ ar = r - ru2 = - 0.125 - 2.5(0.5)2 = - 0.75 ft>s2 $ ## au = ru + 2ru = 2.5(0) + 2( - 0.4330)(0.5) = - 0.4330 ft>s2 Dissemination

r 2 + cos u 2 = = - 2.887 dr>du - sin u u = 60°

World

permitted.

c = -70.89°

use

United

on

learning.

is

of

the

not

instructors

States

tan c =

by

work

student

F - 0.04930 sin 19.11° =

the

+a©Fu

N = 0.04930 lb the

(including

for

- N cos 19.11° =

and

2 ( -0.75) 32.2

+ Q©Fr = mar;

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solely

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2 ( -0.4330) 32.2 work

this

assessing

is

= mau;

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F = - 0.0108 lb






r

Ans.

3 ft

13–107. The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a limaçon, r = (2 + cos u) ft. If u = (0.5t2) rad, where t is in seconds, determine the force which the rod exerts on the particle at the instant t = 1 s. The fork and path contact the particle on only one side.

2 ft

u · u

SOLUTION r = 2 + cos u # r = - sin uu # $ $ r = - cos uu2 - sin uu

u = 0.5t2 # u = t $ u = 1 rad>s2

$ At t = 1 s, u = 0.5 rad, u = 1 rad>s, and u = 1 rad>s2 r = 2 + cos 0.5 = 2.8776 ft # r = - sin 0.5(1) = - 0.4974 ft>s2

Wide

copyright

in

teaching

Web)

laws

or

$ r = - cos 0.5(1)2 - sin 0.5(1) = - 1.357 ft>s2 # $ ar = r - ru2 = - 1.375 - 2.8776(1)2 = - 4.2346 ft>s2 $ ## au = ru + 2ru = 2.8776(1) + 2( - 0.4794)(1) = 1.9187 ft>s2 Dissemination

r 2 + cos u 2 = = - 6.002 dr>du - sin u u = 0.5 rad

World

permitted.

c = - 80.54°

use

United

on

learning.

is

of

the

not

instructors

States

tan c =

by

work

student

F - 0.2666 sin 9.46° =

the

+a©Fu = mau;

N = 0.2666 lb the

(including

for

-N cos 9.46° =

and

2 ( -4.2346) 32.2

+ Q©Fr = mar;

work

this

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2 (1.9187) 32.2

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F = 0.163 lb






r

Ans.

3 ft

*13–108. The collar, which has a weight of 3 lb, slides along the smooth rod lying in the horizontal plane and having the shape of a parabola r = 4>11 - cos u2, where u is in radians and r is #in feet. If the collar’s angular rate is constant and equals u = 4 rad>s, determine the tangential retarding force P needed to cause the motion and the normal force that the collar exerts on the rod at the instant u = 90°.

P

r u

SOLUTION 4 1 - cos u

r =

# -4 sin u u

# r =

(1 - cos u)2 $ - 4 sin u u

$ r =

+

(1 - cos u)2

# - 4 cos u (u)2 (1 - cos u)2

# u = 4,

At u = 90°,

+

# 8 sin2 u u2 (1 - cos u)3

$ u = 0 r = 4 laws

or

# r = - 16 in

teaching

Web)

$ r = 128

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

$ ar = r - r(u)2 = 128 - 4(4)2 = 64 $ ## au = ru + 2 ru = 0 + 2(- 16)(4) = -128

(including

for

work

student

the

by

and

use

4 1 - cos u

r =

integrity

of

and

4 = -1 -4

the

=

part

2

- 4 sin u ° (1 - cos u)2 u = 90

provided

4 1 - cos u)

and

any

courses

dr (du )

=

is

r

This

tan c =

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the

-4 sin u dr = du (1 - cos u)2

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c = - 45° = 135°

3 (64) 32.2

+ c ©Fr = m ar ;

P sin 45° - N cos 45° =

+ © F = ma ; ; u u

- P cos 45° - N sin 45° =

3 ( -128) 32.2

Solving,






P = 12.6 lb

Ans.

N = 4.22 lb

Ans.

13–109. The smooth particle has a mass of 80 g. It is attached to an elastic cord extending from O to P and due to the slotted arm guide moves along the horizontal circular path r = 10.8 sin u2 m. If the cord has a stiffness k = 30 N>m and an unstretched length of 0.25 m, determine the force of the guide on the particle when u = 60°. The guide has a constant # angular velocity u = 5 rad>s.

P r

· u

0.4 m

SOLUTION

u

r = 0.8 sin u # # r = 0.8 cos u u

O

# $ $ r = -0.8 sin u (u)2 + 0.8 cos uu # $ u = 5, u = 0 r = 0.6928

At u = 60°,

# r = 2

laws

or

$ r = - 17.321

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

# $ ar = r - r(u)2 = -17.321 - 0.6928(5)2 = - 34.641 $ ## au = ru + 2 ru = 0 + 2(2)(5) = 20

World

Fs = 30(0.6928 - 0.25) = 13.284 N

a+ ©Fu = mau;

F - NP sin 30° = 0.08(20)

and

-13.284 + NP cos 30° = 0.08(-34.641)

the

(including

for

work

student

the

by

Q+ ©Fr = m ar;

use

United

on

learning.

is

of

the

not

instructors

States

Fs = ks;

this

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F = 7.67 N

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NP = 12.1 N






Ans.

5 rad/s

13–110. The smooth particle has a mass of 80 g. It is attached to an elastic cord extending from O to P and due to the slotted arm guide moves along the horizontal circular path r = (0.8 sin u) m. If the cord has a stiffness k = 30 N>m and an unstretched length of 0.25$ m, determine # the force of the guide on the particle when u = 2 rad>s2, u = 5 rad>s, and u = 60°.

P r

· u

0.4 m

SOLUTION r = 0.8 sin u

u

# # r = 0.8 cos u u

O

$ # $ r = -0.8 sin u (u)2 + 0.8 cos uu $ # u = 2 u = 5, At u = 60°,

r = 0.6928 # r = 2

or

$ r = -16.521

Dissemination

Wide

copyright

in

teaching

Web)

laws

# $ ar = r - r(u)2 = -16.521 - 0.6928(5)2 = - 33.841 $ ## au = r u + 2 ru = 0.6925(2) + 2(2)(5) = 21.386

World

permitted.

Fs = 30(0.6928 - 0.25) = 13.284 N

is

of

the

not

instructors

States

Fs = ks;

and

use

United

on

learning.

- 13.284 + NP cos 30° = 0.08( -33.841) by

Q+ ©Fr = m ar;

work

student

the

F - NP sin 30° = 0.08(21.386)

(including

for

+ a©Fu = mau;

assessing

is

work

solely

of

protected

the

F = 7.82 N

sale

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integrity

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this

NP = 12.2 N






Ans.

5 rad/s

13–111. A 0.2-kg spool slides down along a smooth rod. If # the rod has a constant angular rate of rotation u = 2 rad>s in the vertical plane, show that the equations of $ motion for the spool are r - 4r - 9.81 sin u = 0 and # 0.8r + Ns - 1.962 cos u = 0, where Ns is the magnitude of the normal force of the rod on the spool. Using the methods of differential equations, it can be shown that the solution of the first of these equations is # r = C1e -2t + C2e2t - 19.81>82 sin 2t. If r, r, and u are zero when t = 0, evaluate the constants C1 and C2 to determine r at the instant u = p>4 rad.

u

u

2 rad/s

SOLUTION

# # Kinematic: Here, u. = 2 rad>s and u = 0. Applying Eqs. 12–29, we have # $ $ $ ar = r - ru2 = r - r(22) = r - 4r $ ## # # au = ru + 2ru = r(0) + 2r(2) = 4r Equation of Motion: Applying Eq. 13–9, we have $ 1.962 sin u = 0.2(r - 4r) or

$ r - 4r - 9.81 sin u = 0

(Q.E.D.)

# 0.8r + Ns - 1.962 cos u = 0

(2)

instructors

States

World

Dissemination

(Q.E.D.)

not the of United

on

learning.

is

2dt, u = 2t. The solution of the differential and

use

L0

the

(including

for

work

student

the

L0 equation (Eq.(1)) is given by

1

by

# u =

u

Since u. = 2 rad>s, then

9.81 sin 2t 8

(3)

and

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is

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protected

r = C1 e - 2 t + C2 e2t -

part

the

is

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provided

integrity

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Thus,

(4)

will

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destroy

of

and

any

courses

9.81 # cos 2t r = - 2 C1 e - 2t + 2C2 e2t 4

(5)

9.81 # At t = 0, r = 0 . From Eq.(4) 0 = - 2 C1 (1) + 2C2 (1) 4

(6)

sale

At t = 0, r = 0 . From Eq.(3) 0 = C1 (1) + C2 (1) - 0

Solving Eqs. (5) and (6) yields C1 = -

9.81 16

C2 =

9.81 16

Thus, r = -

9.81 - 2t 9.81 2t 9.81 e e + sin 2t 16 16 8

9.81 -e - 2t + e2t a - sin 2tb 8 2 9.81 (sin h 2t - sin 2t) = 8 =

At u = 2t =






p , 4

r =

p p 9.81 asin h - sin 3 b = 0.198 m 8 4 4

Ans.

permitted.

Wide

copyright

in

teaching

Web)

# 1.962 cos u - Ns = 0.2(4r)

©Fu = mau;

(1)

laws

©Fr = mar ;

r

*13–112. The pilot of an airplane executes a vertical loop which in part follows the path of a cardioid, r = 600(1 + cos u) ft. If his speed at A ( u = 0°) is a constant vP = 80 ft>s , determine the vertical force the seat belt must exert on him to hold him to his seat when the plane is upside down at A. He weighs 150 lb.

A

r u

SOLUTION r = 600(1 + cos u)|u = 0° = 1200 ft # # r = -600 sin uu u = 0° = 0 $ # # $ r = -600 sin uu - 600 cos uu2 u = 0° = -600u2 # 2 # v2p = r2 + a ru b # 2 (80)2 = 0 + a1200u b

# u = 0.06667

# $ $ # 2vpvp = 2rr + 2 a ru b a ru + ru b

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

$ $ 0 = 0 + 0 + 2r2 uu u = 0 # $ ar = r - ru2 = -600(0.06667)2 - 1200(0.06667)2 = -8 ft>s2 $ ## au = ru + 2ru = 0 + 0 = 0 of

the

150 b ( -8) 32.2

on

learning. United

and work

student

the

the

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protected

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Ans. is

N = 113 lb

use

N - 150 = a

by

+ c ©Fr = mar;

600 (1 + cos u ) ft

13–113. The earth has an orbit with eccentricity e = 0.0821 around the d sun. Knowing that the earth’s minimum distance from the sun is 151.3(106) km, find the speed at which the earth travels when it is at this distance. Determine the equation in polar coordinates which describes the earth’s orbit about the sun.

SOLUTION e =

Ch2 GMS

e =

GMS 1 ¢1 ≤ (r0v0)2 GMS r0 r0 v20

y0 = =

where C =

GMS 1 ¢1 ≤ and h = r0 v0 r0 r0 v20 e = ¢

r0 v20 - 1≤ GMS

r0v20 = e + 1 GMS

GMS (e + 1) r0 B 66.73(10 - 12)(1.99)(1030)(0.0821 + 1) = 30818 m>s = 30.8 km>s B 151.3(109)

Ans.

teaching

Web)

laws

or

GMS GMS 1 1 = a1 bcos u + 2 2 2 r r0 r0 v0 r0v0

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

66.73(10 - 12)(1.99)(1030) 66.73(10 - 12)(1.99)(1030) 1 1 = b cos u + a1 r 151.3(109) 151.3(109)(30818)2 C 151.3(109) D 2 (30818)2 of

the

not

1 = 0.502(10 - 12) cos u + 6.11(10 - 12) r

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Ans.






13–114. A communications satellite is in a circular orbit above the earth such that it always remains directly over a point on the earth’s surface. As a result, the period of the satellite must equal the rotation of the earth, which is approximately 24 hours. Determine the satellite’s altitude h above the earth’s surface and its orbital speed.

SOLUTION The period of the satellite around r0 = h + re = C h + 6.378(106) D m is given by T =

circular

orbit

of

radius

2pr0 vs

24(3600) = vs =

the

2p C h + 6.378(106) D vs

2p C h + 6.378(106)

(1)

86.4(103)

teaching

Web)

laws

or

The velocity of the satellite orbiting around the circular orbit of radius r0 = h + re = C h + 6.378(106) D m is given by in

GMe C r0

Dissemination

Wide

copyright

yS =

World instructors

(2)

the

is on United

and

use by

for

work

student

the

Solving Eqs.(1) and (2), h = 35.87(106) m = 35.9 Mm

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(including

yS = 3072.32 m>s = 3.07 km>s Ans.






not

States

h + 6.378(106)

learning.

C

permitted.

66.73(10-12)(5.976)(1024)

of

yS =

13–115. The speed of a satellite launched into a circular orbit about the earth is given by Eq. 13–25. Determine the speed of a satellite launched parallel to the surface of the earth so that it travels in a circular orbit 800 km from the earth’s surface.

SOLUTION For a 800-km orbit v0 =

66.73(10 - 12)(5.976)(1024) B (800 + 6378)(103)

= 7453.6 m>s = 7.45 km>s

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in

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Web)

laws

or

Ans.






*13–116. A rocket is in circular orbit about the earth at an altitude of h = 4 Mm. Determine the minimum increment in speed it must have in order to escape the earth’s gravitational field.

h = 4 Mm

SOLUTION Circular Orbit: vC =

66.73(10 - 12)5.976(1024) GMe = = 6198.8 m>s A r0 B 4000(103) + 6378(103)

Parabolic Orbit: ve =

2(66.73)(10 - 12)5.976(1024) 2GMe = = 8766.4 m>s A r0 B 4000(103) + 6378(103)

¢v = ve - vC = 8766.4 - 6198.8 = 2567.6 m>s ¢v = 2.57 km>s

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(including

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by

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of

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permitted.

Dissemination

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copyright

in

teaching

Web)

laws

or

Ans.






13–117. Prove Kepler’s third law of motion. Hint: Use Eqs. 13–19, 13–28, 13–29, and 13–31.

SOLUTION From Eq. 13–19, GMs 1 = C cos u + r h2 For u = 0° and u = 180°, GMs 1 = C + rp h2

or

GMs 1 = -C + ra h2

teaching

Web)

laws

Eliminating C, from Eqs. 13–28 and 13–29,

World

permitted.

Dissemination

Wide

copyright

in

2GMs 2a = 2 b h2

on United

and

use

T2h2 4p2a2

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for

work

student

b2 =

by

p (2a)(b) h

the

T =

learning.

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of

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not

instructors

States

From Eq. 13–31,

any

courses

destroy

of

and

T2 = a






sale

4p2 b a3 GMs

will

their

GMs 4p2a3 = 2 2 Th h2

part

the

is

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provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

Thus,

Q.E.D.

13–118. The satellite is moving in an elliptical orbit with an eccentricity e = 0.25. Determine its speed when it is at its maximum distance A and minimum distance B from the earth. 2 Mm

A

SOLUTION Ch2 GMe

e =

GMe 1 ¢1 ≤ and h = r0 v0. r0 r0v20 GMe 1 ¢1 ≤ (r0 v0)2 GMe r0 r0 v20 e = ¢

v0 =

B

GMe (e + 1) r0

teaching

Web)

r0 v20 = e + 1 GMe

r0 v20 - 1≤ GMe or

e =

laws

where C =

Wide

copyright

permitted.

= 7713 m>s = 7.71 km>s

World

Ans.

States

C

Dissemination

66.73(10 - 12)(5.976)(1024)(0.25 + 1)

the

not

instructors

8.378(106)

by

work

solely is

(7713) = 4628 m>s = 4.63 km>s

13.96(106)

this

assessing

8.378(106)

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ra

vB =

and

rp

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vA =

of

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and

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8.378(106) r0 = = 13.96 A 106 B m 2GMe 2(66.73)(10 - 12)(5.976)(1024) - 1 - 1 r0 v0 8.378(106)(7713)2

on

learning.

is

of

vB = v0 =

in

where r0 = rp = 2 A 106 B + 6378 A 103 B = 8.378 A 106 B m.






Ans.

B

13–119. The elliptical orbit of a satellite orbiting the earth has an eccentricity of e = 0.45. If the satellite has an altitude of 6 Mm at perigee, determine the velocity of the satellite at apogee and the period.

P

A

SOLUTION 6 Mm

Here, rO = rP = 6(106) + 6378(103) = 12.378(106) m. h = rPvP h = 12.378(106)vP

(1)

and GMe 1 a1 b rP rPvP 2 12.378(106)

c1 -

12.378(106)vP 2

d

2.6027 vP 2

(2)

in

teaching

Web)

C = 80.788(10 - 9) -

66.73(10 - 12)(5.976)(1024)

or

1

C =

laws

C =

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permitted.

Dissemination

Ch2 GMe by

and

use

United

on

learning.

is

of

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States

e =

copyright

Using Eqs. (1) and (2),

66.73(10 - 12)(5.976)(1024)

the

(including

for

work

student

the

2 2.6027 d c 12.378(106)vP d 2 vP

assessing

is

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protected

0.45 =

c 80.788(10 - 9) -

the

is

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provided

integrity

of

and

work

this

vP = 6834.78 m>s

rP vP 2

any

courses

destroy

of

and

- 1

will

their

rP 2GMe

sale

ra =

part

Using the result of vP,

12.378(106) =

2(66.73)(10 - 12)(5.976)(1024) 12.378(106)(6834.782)

- 1

= 32.633(106) m Since h = rPvP = 12.378(106)(6834.78) = 84.601(109) m2>s is constant, rava = h 32.633(106)va = 84.601(109) va = 2592.50 m>s = 2.59 km>s

Ans.

Using the result of h, T =

= © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

p (r + ra) 2rPra h P p C 12.378(106) + 32.633(106) D 312.378(106)(32.633)(106) 84.601(109)





= 33 592.76 s = 9.33 hr

Ans.

*13–120. Determine the constant speed of satellite S so that it circles the earth with an orbit of radius r = 15 Mm. Hint: Use Eq. 13–1.

r ⫽ 15 Mm

SOLUTION F = G ms a y =

ms me r2

Also

S

y2s F = ms a b r

Hence

y20 ms me b = G r r2 me 5.976(1024) b = 5156 m>s = 5.16 km>s = 66.73(10 - 12) a r B 15(106)

Ans.

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the

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and

use

United

on

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is

of

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permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

A

G






13–121. The rocket is in free flight along an elliptical trajectory A¿A. The planet has no atmosphere, and its mass is 0.70 times that of the earth. If the rocket has an apoapsis and periapsis as shown in the figure, determine the speed of the rocket when it is at point A.

r B

A

6 Mm

SOLUTION Central-Force Motion: Use ra = M = 0.70Me, we have

r0 (2 GM>r0 y20 B -1

, with r0 = rp = 6 A 106 B m and

6(10)6

¢

2(66.73) (10-12) (0.7) [5.976(1024)] 6(106)y2P

≤ - 1 or

9 A 106 B =

laws

yA = 7471.89 m>s = 7.47 km>s

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solely

of

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the

(including

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work

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the

by

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on

learning.

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instructors

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permitted.

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copyright

in

teaching

Web)

Ans.






3 Mm

O

A¿

9 Mm

13–122. A satellite S travels in a circular orbit around the earth. A rocket is located at the apogee of its elliptical orbit for which e = 0.58. Determine the sudden change in speed that must occur at A so that the rocket can enter the satellite’s orbit while in free flight along the blue elliptical trajectory. When it arrives at B, determine the sudden adjustment in speed that must be given to the rocket in order to maintain the circular orbit.

120 Mm

B

S

10 Mm

SOLUTION GMe 1 11 2 [Eq. 13–21] and h = r0 v0 r0 r0 v20 [Eq. 13–20]. Substitute these values into Eq. 13–17 gives Central-Force Motion: Here, C =

1 GM A 1 - r2v2e B A r20v20 B 0 0 r0v02 r0 ch 2 e = = = -1 GMe GMe GMe

(1)

Rearrange Eq. (1) gives GMe 1 = 1 + e r0 v02

(2)

laws

or

Rearrange Eq. (2), we have teaching in

D

Web)

11 + e2GMe

Wide World

, we have

the

not

instructors

on

learning.

is

A 2 GMe >r0 v20 B - 1

permitted.

Dissemination

r0

States

Substitute Eq. (2) into Eq. 13–27, ra =

(3)

copyright

r0

of

v0 =

United

and

use

by

(4) work the

(including

B -1

1 - e bra 1 + e

student

2A

r0 = a

or

the

r0 1 1 + e

for

ra =

assessing

is

work

solely

of

protected

or the first elliptical orbit e = 0.58, from Eq. (4)

integrity

of

and

work

this

1 - 0.58 b C 12011062 D = 31.89911062 m 1 + 0.58 part

the

is

This

provided

1rp21 = r0 = a

their

destroy

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and

any

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Substitute r0 = 1rp21 = 31.89911062 m into Eq. (3) yields will

11 + 0.582166.732110 - 12215.9762110242 sale

1vp21 =

D

31.89911062

= 4444.34 m>s

Applying Eq. 13–20, we have 1va21 = a

rp ra

b 1vp21 = c

31.89911062 12011062

d14444.342 = 1181.41 m>s

When the rocket travels along the second elliptical orbit, from Eq. (4), we have 1011062 = a

1 - e b C 12011062 D 1 + e

e = 0.8462

Substitute r0 = 1rp22 = 1011062 m into Eq. (3) yields 1vp2 =






11 + 0.84622166.732110 - 12215.9672110242 D

101106)

= 8580.25 m>s

A

13–122. continued And in Eq. 13–20, we have (va22 = c

1rp22 1ra22

d 1vp22 = c

1011062 12011062

d 18580.252 = 715.02 m>s

For the rocket to enter into orbit two from orbit one at A, its speed must be decreased by ¢v = 1va21 - 1va22 = 1184.41 - 715.02 = 466 m>s

Ans.

If the rocket travels in a circular free-flight trajectory, its speed is given by Eq. 13–25.

vc =

66.73110 - 12215.9762110242 GMe = = 6314.89 m>s D r0 D 1011062

The speed for which the rocket must be decreased in order to have a circular orbit is ¢v = 1vp22 - ve = 8580.25 - 6314.89 = 2265.36 m>s = 2.27 km>s

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or

Ans.






13–123. An asteroid is in an elliptical orbit about the sun such that its periheliondistance is 9.3011092 km. If the eccentricity of the orbit is e = 0.073, determine the aphelion distance of the orbit.

SOLUTION rp = r0 = 9.3011092 km e =

GMs r0v20 1 ch2 = a1 b a b r0 GMs GMs r0v20

e = a

r0 v20 - 1b GMs

r0 v20 = e + 1 GMs

or laws

Web) teaching in

B -1

(2) Wide World

permitted.

Dissemination

9.301109211.0732 0.927

learning.

is

11 - e2

=

A

not

r01e + 12

r0 2 e + 1

the

- 1

=

copyright

r0 2GMs r0 v20

ra =

1 b e + 1

instructors

ra =

= a

States

r0v20

of

GMs

(1)

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the

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by

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on

ra = 10.811092 km






Ans.

*13–124. An elliptical path of a satellite has an eccentricity e = 0.130. If it has a speed of 15 Mm>h when it is at perigee, P, determine its speed when it arrives at apogee, A. Also, how far is it from the earth’s surface when it is at A? A

P

SOLUTION e = 0.130 vp = v0 = 15 Mm>h = 4.167 km>s e =

GMe r20 v20 Ch2 1 = ¢1 b ≤a 2 r0 GMe GMe r0v0

e = ¢

r0 v20 -1≤ GMe

r0 v20 = e +1 GMe or laws

teaching

Web)

v20 in

1.130166.732110 - 12215.9762110242

Wide

permitted.

Dissemination

C 4.16711032 D 2 States

World

=

(e + 1)GMe

copyright

r0 =

United

on

learning.

is

of

the

not

instructors

= 25.96 Mm

and

use

1 e + 1

by

(including

for

work

student

=

r0 v20

the

GMe

solely

of

protected

the

r0

work

B -1

this

A

2 e + 1

assessing

- 1

=

is

r0 2GMe r0v02

work

rA =

provided

integrity

of

and

r01e + 12 is

This

part

the

1 - e and

any

courses

rA =

0.870

will

their

destroy

of

25.961106211.1302 sale

=

= 33.7111062 m = 33.7 Mm vA = =

v0r0 rA 15125.96211062 33.7111062

= 11.5 Mm>h

Ans.

d = 33.7111062 - 6.37811062 = 27.3 Mm






Ans.

13–125. A satellite is launched with an initial velocity v0 = 2500 mi>h parallel to the surface of the earth. Determine the required altitude (or range of altitudes) above the earth’s surface for launching if the free-flight trajectory is to be (a) circular, (b) parabolic, (c) elliptical, and (d) hyperbolic. Take G = 34.4110-921lb # ft22>slug2, Me = 409110212 slug, the earth’s radius re = 3960 mi, and 1 mi = 5280 ft.

SOLUTION v0 = 2500 mi>h = 3.67(103) ft>s e =

(a)

1 =

C2h = 0 GMe

or C = 0

GMe r0 v20

GMe = 34.4(10 - 9)(409)(1021) = 14.07(1015) 14.07(1015) [3.67(1013)]2

= 1.046(109) ft

Wide

C2h = 1 GMe

on

learning.

is

of United

and

use by

student

the

(b)

the

not

instructors

States

World

permitted.

e =

Ans. Dissemination

1.047(109) - 3960 = 194(103) mi 5280

copyright

r =

in

teaching

Web)

v20

=

or

GMe

laws

r0 =

= 2.09(109) ft = 396(103) mi this

2(14.07)(1015)

work

integrity

[3.67(103)]2

of

=

and

v20

provided

2GMe

part

the

is

This

r0 =

assessing

is

work

solely

of

protected

the

(including

for

work

GMe 1 1 (r20 v20)a b ¢ 1 ≤ = 1 r0 GMe r0 v20

Ans.

sale

will

their

destroy

of

and

any

courses

r = 396(103) - 3960 = 392(103) mi

e 6 1

(c)

194(103) mi 6 r 6 392(103) mi

Ans.

e 7 1

(d)

r 7 392(103) mi






Ans.

13–126. A probe has a circular orbit around a planet of radius R and mass M. If the radius of the orbit is nR and the explorer is traveling with a constant speed v0, determine the angle u at which it lands on the surface of the planet B when its speed is reduced to kv0, where k 6 1 at point A.

B u A

SOLUTION

nR

When the probe is orbiting the planet in a circular orbit of radius rO = nR, its speed is given by GM GM = B rO B nR

vO =

The probe will enter the elliptical trajectory with its apoapsis at point A if its speed is decreased to va = kvO = k

GM at this point. When it lands on the surface of the B nR

planet, r = rB = R. 1 1 GM GM = a1 b cos u + 2 2 r rP rPvP2 rP vP 1 1 GM GM = a - 2 2 b cos u + 2 2 rP R rP vP rP vP laws

or

(1)

GM b = k 2nGMR is constant, A nR

Wide

copyright

in

teaching

Web)

Since h = rava = nR ak

States

World

permitted.

Dissemination

rPvP = h

(2)

by

and

use

United

on

learning.

is

of

the

not

instructors

k 2nGMR rP

vP =

rP 2GM - 1 rPvP2 rP nR = 2GM - 1 rPvP2 2nGMR vP 2 = rp(rp + nR)

(including

for

work

student

the

Also,

(3)

sale

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and

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of

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is

work

solely

of

protected

the

ra =

Solving Eqs.(2) and (3), rp =

k2n R 2 - k2

vp =

2 - k2 GM k B nR

Substituting the result of rp and vp into Eq. (1), 1 2 - k2 1 1 - 2 b cos u + 2 = a 2 R k nR k nR k nR u = cos - 1 a

k2n - 1 b 1 - k2

Here u was measured from periapsis.When measured from apoapsis, as in the figure then u = p - cos - 1 a






R

k2n - 1 b 1 - k2

Ans.

13–127. Upon completion of the moon exploration mission, the command module, which was originally in a circular orbit as shown, is given a boost so that it escapes from the moon’s gravitational field. Determine the necessary increase in velocity so that the command module follows a parabolic trajectory. The mass of the moon is 0.01230 Me.

A

SOLUTION

3 Mm

When the command module is moving around the circular orbit of radius r0 = 3(106) m, its velocity is vc =

66.73(10 - 12)(0.0123)(5.976)(1024) GMm = B r0 B 3(106)

= 1278.67 m>s The escape velocity of the command module entering into the parabolic trajectory is ve =

2(66.73)(10 - 12)(0.0123)(5.976)(1024) 2GMm = B r0 B 3(106) or

= 1808.31 m>s

teaching

Web)

laws

Thus, the required increase in the command module is ¢v = ve - vc = 1808.31 - 1278.67 = 529.64 m>s = 530 m>s

sale

will

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Ans.






*13–128. The rocket is traveling in a free-flight elliptical orbit about the earth such that e = 0.76 and its perigee is 9 Mm as shown. Determine its speed when it is at point B. Also determine the sudden decrease in speed the rocket must experience at A in order to travel in a circular orbit about the earth.

A

B

SOLUTION

9 Mm

GMe

1 a1 b [Eq. 13–21] and h = r0 v 0 r0 r0 v20 [Eq. 13–20] Substitute these values into Eq. 13–17 gives Central-Force Motion: Here C =

1 e A 1 - GM B 1 r20 v022 r20v20 r0v02 r0 ch2 e = = = - 1 GMe GMe GMe

(1)

Rearrange Eq.(1) gives GMe 1 = 1 + e r0 v02

laws

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(2)

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Rearrange Eq.(2), we have

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r0

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r0

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, we have by

12GMe >r0 v022 - 1

the

Substitute Eq.(2) into Eq. 13–27, ra =

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r0

the

B -1

(4)

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1 1 + e

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ra =

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this

Rearrange Eq.(4), we have

1 + e 1 + 0.76 br = a b C 911062 D = 66.011062 m 1 - e 0 1 - 0.76 destroy

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ra = a

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Substitute r0 = rp = 9 A 106 B m into Eq.(3) yields

vp =

(1 + 0.76)(66.73)(10 - 12)(5.976) (1024) D

9(106)

= 8830.82 m>s

Applying Eq. 13–20, we have va = a

rp ra

b np = B

911062 66.011062

R 18830.822 = 1204.2 m>s = 1.20 km>s

Ans.

If the rocket travels in a circular free-flight trajectory, its speed is given by Eq. 13–25. ve =

66.73110 - 12215.9762110242 GMe = = 6656.48 m>s D r0 D 911062

The speed for which the rocket must be decreased in order to have a circular orbit is ¢v = v p - v c = 8830.82 - 6656.48 = 2174.34 m>s = 2.17 km>s






Ans.

permitted.

B

copyright

11 + e2 GMe States

v0 =

13–129. A rocket is in circular orbit about the earth at an altitude above the earth’s surface of h = 4 Mm. Determine the minimum increment in speed it must have in order to escape the earth’s gravitational field.

h

SOLUTION Circular orbit: vC =

66.73(10 - 12)5.976(1024) GMe = = 6198.8 m>s B r0 B 4000(103) + 6378(103)

Parabolic orbit: ve =

2(66.73)(10 - 12)5.976(1024) 2GMe = = 8766.4 m>s B r0 B 4000(103) + 6378(103)

¢v = ve - vC = 8766.4 - 6198.8 = 2567.6 m>s ¢v = 2.57 km>s

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4 Mm

13–130. The satellite is in an elliptical orbit having an eccentricity of e = 0.15. If its velocity at perigee is vP = 15 Mm>h, determine its velocity at apogee A and the period of the satellite.

15 Mm/h P

SOLUTION Here, vP = c 15(106)

m 1h da b = 4166.67 m>s. h 3600 s

h = rPvP h = rP (4166.67) = 4166.67rp

(1)

and GMe 1 ¢1 ≤ rP rP vP 2

C =

66.73(10-12)(5.976)(1024) 1 B1 R rp rp(4166.672)

C =

22.97(106) 1 B1 R rP rP

e =

Ch2 GMe

or

C =

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66.73(10-12)(5.976)(1024)

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0.15 =

22.97(106) 1 B1 R (4166.67 rP)2 rP rP

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rP = 26.415(106) m

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- 1

2(66.73)(10-12)(5.976)(1024) 26.415(106)(4166.672)

will

26.415(106)

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rP vP 2

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rP 2GMe

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Using the result of rp

- 1

= 35.738(106) m Since h = rP vP = 26.415(106)(4166.672) = 110.06(109) m2>s is constant, rA vA = h 35.738(106)vA = 110.06(109) vA = 3079.71 m>s = 3.08 km>s

Ans.

Using the results of h, rA, and rP, T = =






p A r + rA B 2rP rA 6 P p C 26.415(106) + 35.738(106) D 226.415(106)(35.738)(106) 110.06(109)

= 54 508.43 s = 15.1 hr

Ans.

A

13–131. A rocket is in a free-flight elliptical orbit about the earth such that the eccentricity of its orbit is e and its perigee is r0. Determine the minimum increment of speed it should have in order to escape the earth’s gravitational field when it is at this point along its orbit.

SOLUTION To escape the earth’s gravitational field, the rocket has to make a parabolic trajectory. ParabolicTrajectory: ye =

A

2GMe r0

Elliptical Orbit: e =

Ch2 GMe

e =

GMe 1 ¢1 ≤ (r0 y0)2 GMe r0 r0 y20

GMe 1 ¢1 ≤ and h = r0 y0 r0 r0y20

in

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where C =

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Dissemination

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copyright

r0 y20 - 1b GMe

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e = a

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of

the

GMe (e + 1) r0

and

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B

by

y0 =

the

r0 y20 = e + 1 GMe

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GMe (e + 1) 2GMe GMe = a 22 - 21 + eb r0 r0 A A r0 solely

A

protected

¢y =

Ans.

*13–132. The rocket shown is originally in a circular orbit 6 Mm above the surface of the earth.It is required that it travel in another circular orbit having an altitude of 14 Mm.To do this,the rocket is given a short pulse of power at A so that it travels in free flight along the gray elliptical path from the first orbit to the second orbit.Determine the necessary speed it must have at A just after the power pulse, and at the time required to get to the outer orbit along the path AA¿. What adjustment in speed must be made at A¿ to maintain the second circular orbit?

A

O

6 Mm 14 Mm

SOLUTION r0

Central-Force Motion: Substitute Eq. 13–27, ra =

(2GM>r0v20) - 1

, with

ra = (14 + 6.378)(106) = 20.378(106) m and r0 = rp = (6 + 6.378)(106) = 12.378(106) m, we have 12.378(106) or

20.378(106) =

laws

2(66.73)(10 - 12)[5.976(1024)]

teaching

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≤ - 1

12.378(106)v2p

in

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permitted.

Dissemination

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copyright

vp = 6331.27 m>s

United

R (6331.27) = 3845.74 m>s

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20.378(106)

use

12.378(106)

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ra

≤ vp = B

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the

va = ¢

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Applying Eq. 13–20. we have

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p (r + ra)2rp ra h p

T =

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the

Eq. 13–20 gives h = rp vp = 12.378(106)(6331.27) = 78.368(109) m2>s. Thus, applying Eq. 13–31, we have

p [(12.378 + 20.378)(106)]212.378(20.378)(106) 78.368(109) sale

will

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destroy

of

=

= 20854.54 s The time required for the rocket to go from A to A¿ (half the orbit) is given by t =

T = 10427.38 s = 2.90 hr 2

Ans.

In order for the satellite to stay in the second circular orbit, it must achieve a speed of (Eq. 13–25) vc =

66.73(10 - 12)(5.976)(1024) GMe = = 4423.69 m>s = 4.42 km>s A r0 A 20.378(106)

Ans.

The speed for which the rocket must be increased in order to enter the second circular orbit at A¿ is ¢v = vc - va = 4423.69 - 3845.74 = 578 m s






Ans.

A'

14–1. The 20-kg crate is subjected to a force having a constant direction and a magnitude F = 100 N. When s = 15 m, the crate is moving to the right with a speed of 8 m/s. Determine its speed when s = 25 m. The coefficient of kinetic friction between the crate and the ground is mk = 0.25.

F 30°

SOLUTION Equation of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = mkN = 0.25N. Applying Eq. 13–7, we have + c a Fy = may ;

N + 100 sin 30° - 20(9.81) = 20(0) N = 146.2 N

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or

Principle of Work and Energy: The horizontal component of force F which acts in the direction of displacement does positive work, whereas the friction force Ff = 0.25(146.2) = 36.55 N does negative work since it acts in the opposite direction to that of displacement. The normal reaction N, the vertical component of force F and the weight of the crate do not displace hence do no work. Applying Eq.14–7, we have

Dissemination

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copyright

T1 + a U1 - 2 = T2 instructors

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25 m

on

learning.

is

of

the

not

100 cos 30° ds use

United

1 (20)(8 2) + 2 L15 m

by

1 (20)v2 2

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the

(including

the

36.55 ds =

for

L15 m

and

25 m

-

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v = 10.7 m s






Ans.

14–2. F (lb)

For protection, the barrel barrier is placed in front of the bridge pier. If the relation between the force and deflection of the barrier is F = (90(103)x1>2) lb, where x is in ft, determine the car’s maximum penetration in the barrier. The car has a weight of 4000 lb and it is traveling with a speed of 75 ft>s just before it hits the barrier.

F ⫽ 90(10)3 x1/2

x (ft)

SOLUTION Principle of Work and Energy: The speed of the car just before it crashes into the barrier is v1 = 75 ft>s. The maximum penetration occurs when the car is brought to a stop, i.e., v2 = 0. Referring to the free-body diagram of the car, Fig. a, W and N do no work; however, Fb does negative work. T1 + ©U1 - 2 = T2 1 4000 a b (752) + c 2 32.2 L0

xmax

90(103)x1>2dx d = 0

xmax = 3.24 ft

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Ans.






14–3. The crate, which has a mass of 100 kg, is subjected to the action of the two forces. If it is originally at rest, determine the distance it slides in order to attain a speed of 6 m>s. The coefficient of kinetic friction between the crate and the surface is mk = 0.2.

5

30

SOLUTION Equations of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = mk N = 0.2N. Applying Eq. 13–7, we have + c ©Fy = may;

3 N + 1000 a b - 800 sin 30° - 100(9.81) = 100(0) 5 N = 781 N

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Principle of Work and Energy: The horizontal components of force 800 N and 1000 N which act in the direction of displacement do positive work, whereas the friction force Ff = 0.2(781) = 156.2 N does negative work since it acts in the opposite direction to that of displacement. The normal reaction N, the vertical component of 800 N and 1000 N force and the weight of the crate do not displace, hence they do no work. Since the crate is originally at rest, T1 = 0. Applying Eq. 14–7, we have

of

the

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Dissemination

T1 + a U1-2 = T2

the

by

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learning.

is

4 1 0 + 800 cos 30°(s) + 1000 a bs - 156.2s = (100) A 62 B 5 2

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s = 1.35m






1000 N

800 N

Ans.

3 4

*14–4. The 2-kg block is subjected to a force having a constant direction and a magnitude F = 1300>11 + s22 N, where s is in meters. When s = 4 m, the block is moving to the left with a speed of 8 m>s. Determine its speed when s = 12 m. The coefficient of kinetic friction between the block and the ground is mk = 0.25.

s

SOLUTION + c ©Fy = 0;

NB = 2(9.81) +

150 1 + s

T1 + ©U1 - 2 = T2 12

12

300 1 1 150 (2)(8)2 - 0.25[2(9.81)(12 - 4)] - 0.25 ds + b ds cos 30° = (2)(v22 ) a 2 1 + s 1 + s 2 L4 L4 v22 = 24.76 - 37.5 lna

1 + 12 1 + 12 b + 259.81 lna b 1 + 4 1 + 4

v2 = 15.4 m>s

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Ans.






F 30

v

14–5. When a 7-kg projectile is fired from a cannon barrel that has a length of 2 m, the explosive force exerted on the projectile, while it is in the barrel, varies in the manner shown. Determine the approximate muzzle velocity of the projectile at the instant it leaves the barrel. Neglect the effects of friction inside the barrel and assume the barrel is horizontal.

F (MN) 15

10

5

SOLUTION 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

The work done is measured as the area under the force–displacement curve. This area is approximately 31.5 squares. Since each square has an area of 2.5 A 106 B (0.2), T1 + ©U1-2 = T2 0 + C (31.5)(2.5) A 106 B (0.2) D =

Ans.

(approx.)

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v2 = 2121 m>s = 2.12 km>s

1 (7)(v2)2 2






s (m)

14–6. The spring in the toy gun has an unstretched length of 100 mm. It is compressed and locked in the position shown. When the trigger is pulled, the spring unstretches 12.5 mm, and the 20-g ball moves along the barrel. Determine the speed of the ball when it leaves the gun. Neglect friction.

50 mm k

Principle of Work and Energy: Referring to the free-body diagram of the ball bearing shown in Fig. a, notice that Fsp does positive work. The spring has an initial and final compression of s1 = 0.1 - 0.05 = 0.05 m and s2 = 0.1 - (0.05 + 0.0125) = 0.0375 m. T1 + ©U1-2 = T2 1 1 1 0 + B ks1 2 - ks2 2 R = mvA 2 2 2 2

or

1 1 1 0 + B (2000)(0.05)2 - (2000)(0.03752) R = (0.02)vA 2 2 2 2 vA = 10.5 m>s

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Ans.





150 mm D

A B

SOLUTION


2 kN/m

14–7. As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which rests on the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x, determine the final speed of the block if it has an initial speed of 5 m>s and travels 10 m, both directed to the right and measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x¿ axis and moving at a constant velocity of 2 m>s relative to A. Hint: The distance the block travels will first have to be computed for observer B before applying the principle of work and energy.

A B

5 m/s 6N

10 m

Observer A: T1 + ©U1 - 2 = T2 1 1 (10)(5)2 + 6(10) = (10)v22 2 2 v2 = 6.08 m>s

or

Ans.

teaching

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laws

Observer B:

Wide

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F = ma

instructors

States

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permitted.

Dissemination

a = 0.6 m>s2

on

learning.

is

of

the

not

1 2 at 2 c

United

the

by

and

s = s0 + v0t +

use

+ B A:

the

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student

1 (0.6)t2 2 solely

of

protected

10 = 0 + 5t +

work

this

assessing

is

work

t2 + 16.67t - 33.33 = 0

any sale

Block moves 10 - 3.609 = 6.391 m

will

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destroy

of

and

At v = 2 m>s, s¿ = 2(1.805) = 3.609 m

courses

part

the

is

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provided

integrity

of

and

t = 1.805 s

Thus T1 + ©U1 - 2 = T2 1 1 (10)(3)2 + 6(6.391) = (10)v22 2 2 v2 = 4.08 m>s Note that this result is 2 m>s less than that observed by A.






x¿

2 m/s

SOLUTION

6 = 10a

x

Ans.

*14–8. If the 50-kg crate is subjected to a force of P = 200 N, determine its speed when it has traveled 15 m starting from rest. The coefficient of kinetic friction between the crate and the ground is mk = 0.3.

P

SOLUTION Free-Body Diagram: Referring to the free-body diagram of the crate, Fig. a, + c Fy = may;

N - 50(9.81) = 50(0)

N = 490.5 N

Thus, the frictional force acting on the crate is Ff = mkN = 0.3(490.5) = 147.15 N. Principle of Work and Energy: Referring to Fig. a, only P and Ff do work.The work of P will be positive, whereas Ff does negative work. T1 + g U1 - 2 = T2 0 + 200(15) - 147.15(15) =

1 (50)v2 2

v = 5.63 m>s

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Ans.






14–9. If the 50-kg crate starts from rest and attains a speed of 6 m>s when it has traveled a distance of 15 m, determine the force P acting on the crate. The coefficient of kinetic friction between the crate and the ground is mk = 0.3.

P

SOLUTION Free-Body Diagram: Referring to the free-body diagram of the crate, Fig. a, + c Fy = may;

N - 50(9.81) = 50(0)

N = 490.5 N

Thus, the frictional force acting on the crate is Ff = mkN = 0.3(490.5) = 147.15 N. Principle of Work and Energy: Referring to Fig. a, only P and Ff do work.The work of P will be positive, whereas Ff does negative work. T1 + gU1 - 2 = T2 0 + P(15) - 147.15(15) =

1 (50)(62) 2

P = 207 N

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Ans.






14–10. The 2-Mg car has a velocity of v1 = 100 km>h when the driver sees an obstacle in front of the car. If it takes 0.75 s for him to react and lock the brakes, causing the car to skid, determine the distance the car travels before it stops. The coefficient of kinetic friction between the tires and the road is mk = 0.25.

v1

SOLUTION Free-Body Diagram: The normal reaction N on the car can be determined by writing the equation of motion along the y axis. By referring to the free-body diagram of the car, Fig. a, + c ©Fy = may;

N - 2000(9.81) = 2000(0)

N = 19 620 N

Since the car skids, the frictional force acting on the car is Ff = mkN = 0.25(19620) = 4905N.

teaching

Web)

laws

or

Principle of Work and Energy: By referring to Fig. a, notice that only Ff does work, m 1h which is negative. The initial speed of the car is v1 = c 100(103) d a b = h 3600 s 27.78 m>s. Here, the skidding distance of the car is denoted as s¿ . Wide

copyright

in

T1 + ©U1-2 = T2

learning.

is

of

the

not

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permitted.

Dissemination

1 (2000)(27.782) + (-4905s¿) = 0 2

student

the

by

and

use

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on

s¿ = 157.31 m

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assessing

is

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solely

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the

(including

for

work

The distance traveled by the car during the reaction time is s– = v1t = 27.78(0.75) = 20.83 m. Thus, the total distance traveled by the car before it stops is

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and

work

s = s¿ + s– = 157.31 + 20.83 = 178.14 m = 178 m






Ans.

100 km/h

14–11. The 2-Mg car has a velocity of v1 = 100 km>h when the driver sees an obstacle in front of the car. It takes 0.75 s for him to react and lock the brakes, causing the car to skid. If the car stops when it has traveled a distance of 175 m, determine the coefficient of kinetic friction between the tires and the road.

v1

100 km/h

SOLUTION Free-Body Diagram: The normal reaction N on the car can be determined by writing the equation of motion along the y axis and referring to the free-body diagram of the car, Fig. a, + c ©Fy = may;

N - 2000(9.81) = 2000(0)

N = 19 620 N

Since the car skids, the frictional force acting on the car can be computed from Ff = mkN = mk(19 620).

in

teaching

Web)

laws

or

Principle of Work and Energy: By referring to Fig. a, notice that only Ff does work, 1h m which is negative. The initial speed of the car is v1 = c 100(103) d a b = h 3600 s 27.78 m>s. Here, the skidding distance of the car is s¿ .

permitted.

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T1 + ©U1-2 = T2

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1 (2000)(27.782) + C - mk(19 620)s¿ D = 0 2

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The distance traveled by the car during the reaction time is s– = v1t = 27.78(0.75) = 20.83 m. Thus, the total distance traveled by the car before it stops is

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s = s¿ + s–

mk = 0.255






Ans.

*14–12. Design considerations for the bumper B on the 5-Mg train car require use of a nonlinear spring having the loaddeflection characteristics shown in the graph. Select the proper value of k so that the maximum deflection of the spring is limited to 0.2 m when the car, traveling at 4 m>s, strikes the rigid stop. Neglect the mass of the car wheels.

F (N)

F

ks2

s (m)

SOLUTION B

0.2

1 ks2 ds = 0 (5000)(4)2— 2 L0 40 000 - k

(0.2)3 = 0 3

k = 15.0 MN>m2

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Ans.






14–13. The 2-lb brick slides down a smooth roof, such that when it is at A it has a velocity of 5 ft>s. Determine the speed of the brick just before it leaves the surface at B, the distance d from the wall to where it strikes the ground, and the speed at which it hits the ground.

y A

5 ft/s

15 ft

5

3 4

B

SOLUTION TA + ©UA-B = TB 30 ft

2 2 1 1 a b(5)2 + 2(15) = a b v2B 2 32.2 2 32.2 vB = 31.48 ft>s = 31.5 ft>s + b a:

Ans.

x d

s = s0 + v0t 4 d = 0 + 31.48 a bt 5 or

1 ac t2 2 in

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Web)

s = s0 + v0t -

laws

A+TB

States

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copyright

1 3 30 = 0 + 31.48 a bt + (32.2)t2 5 2

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16.1t2 + 18.888t - 30 = 0

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t = 0.89916 s is

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2 2 1 1 a b(5)2 + 2(45) = a b v2C 2 32.2 2 32.2

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TA + ©UA-C = TC

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Ans.

vC = 54.1 ft>s






Ans.

14–14. If the cord is subjected to a constant force of F = 300 N and the 15-kg smooth collar starts from rest at A, determine the velocity of the collar when it reaches point B. Neglect the size of the pulley.

200 mm B

C 30⬚ 200 mm F ⫽ 300 N

200 mm

SOLUTION

300 mm

Free-Body Diagram: The free-body diagram of the collar and cord system at an arbitrary position is shown in Fig. a.

A

Principle of Work and Energy: Referring to Fig. a, only N does no work since it always acts perpendicular to the motion. When the collar moves from position A to position B, W displaces vertically upward a distance h = (0.3 + 0.2) m = 0.5 m, while

force

F

displaces

a

distance

s = AC - BC = 20.72 + 0.42 -

of

20.2 + 0.2 = 0.5234 m . Here, the work of F is positive, whereas W does negative work. 2

2

TA + g UA - B = TB 1 (15)vB2 2

0 + 300(0.5234) + [- 15(9.81)(0.5)] = vB = 3.335 m>s = 3.34 m>s

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Ans.






14–15.

36

Barrier stopping force (kip)

The crash cushion for a highway barrier consists of a nest of barrels filled with an impact-absorbing material. The barrier stopping force is measured versus the vehicle penetration into the barrier. Determine the distance a car having a weight of 4000 lb will penetrate the barrier if it is originally traveling at 55 ft>s when it strikes the first barrel.

SOLUTION T1 + ©U1 - 2 = T2 1 4000 a b (55)2 - Area = 0 2 32.2

27 18 9 0

Area = 187.89 kip # ft 2(9) + (5 - 2)(18) + x(27) = 187.89 x = 4.29 ft 6 (15 - 5) ft

(O.K!)

Thus s = 5 ft + 4.29 ft = 9.29 ft

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2

5

10 15 20 25 Vehicle penetration (ft)

*14–16. Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is mk = 0.10.

SOLUTION Block A: + a©Fy = may;

A

NA - 60 cos 60° = 0

B

NA = 30 lb

60

FA = 0.1(30) = 3 lb Block B: +Q©Fy = may;

NB - 40 cos 30° = 0 or

NB = 34.64 lb

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laws

FB = 0.1(34.64) = 3.464 lb

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copyright

Use the system of both blocks. NA, NB, T, and R do no work.

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1 60 1 40 a b v2A + a b v2 2 32.2 2 32.2 B

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(0 + 0) + 60 sin 60°|¢sA| - 40 sin 30°|¢sB| - 3|¢sA| -3.464|¢sB| =

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2¢sA = - ¢sB

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2vA = -vB

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When |¢sB| = 2 ft, |¢sA| = 1 ft

Substituting and solving, vA = 0.771 ft>s vB = -1.54 ft>s






Ans.

30

14–17. If the cord is subjected to a constant force of F = 30 lb and the smooth 10-lb collar starts from rest at A, determine its speed when it passes point B. Neglect the size of pulley C.

y ⫽ 1 x2 2

y

C

B

4.5 ft F ⫽ 30 lb

A

SOLUTION

x 1 ft

Free-Body Diagram: The free-body diagram of the collar and cord system at an arbitrary position is shown in Fig. a. Principle of Work and Energy: By referring to Fig. a, only N does no work since it always acts perpendicular to the motion. When the collar moves from position A to position B, W displaces upward through a distance h = 4.5 ft, while force F displaces a distance of s = AC - BC = 262 + 4.52 - 2 = 5.5 ft. The work of F is positive, whereas W does negative work.

laws

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TA + gUA - B = TB

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1 10 a bv 2 2 32.2 B

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0 + 30(5.5) + [-10(4.5)] = vB = 27.8 ft>s

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3 ft

2 ft

14–18. The two blocks A and B have weights WA = 60 lb and WB = 10 lb. If the kinetic coefficient of friction between the incline and block A is mk = 0.2, determine the speed of A after it moves 3 ft down the plane starting from rest. Neglect the mass of the cord and pulleys. A

SOLUTION

5

Kinematics: The speed of the block A and B can be related by using position coordinate equation. sA + (sA - sB) = l 2¢sA - ¢sB = 0

4

2sA - sB = l

¢sB = 2¢sA = 2(3) = 6 ft 2vA - vB = 0

(1)

Equation of Motion: Applying Eq. 13–7, we have 60 4 (0) N - 60 a b = 5 32.2

N = 48.0 lb or

+ ©Fy¿ = may¿ ;

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3 1 1 0 + WA ¢ ¢sA ≤ - Ff ¢sA - WB ¢sB = mAv2A + mB v2B 5 2 2

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T1 + a U1 - 2 = T2

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3 1 60 1 10 60 B (3) R - 9.60(3) - 10(6) = ¢ ≤ v2A + ¢ ≤ v2B 5 2 32.2 2 32.2 1236.48 = 60v2A + 10v2B will sale

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vA = 3.52 ft>s vB = 7.033 ft s






Ans.

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Dissemination

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Principle of Work and Energy: By considering the whole system, WA which acts in the direction of the displacement does positive work. WB and the friction force Ff = mkN = 0.2(48.0) = 9.60 lb does negative work since they act in the opposite direction to that of displacement Here, WA is being displaced vertically (downward) 3 ¢s and WB is being displaced vertically (upward) ¢sB. Since blocks A and B are 5 A at rest initially, T1 = 0. Applying Eq. 14–7, we have

3

B

14–19. y

If the 10-lb block passes point A on the smooth track with a speed of vA = 5 ft>s, determine the normal reaction on the block when it reaches point B.

y ⫽ 1 x2 32

vA ⫽ 5 ft/s A 8 ft

SOLUTION Free-Body Diagram: The free-body diagram of the block at an arbitrary position is shown in Fig. a.

x B

Principle of Work and Energy: Referring to Fig. a, N does no work since it always acts perpendicular to the motion. When the block slides down the track from position A to position B, W displaces vertically downward h = 8 ft and does positive work. TA + g UA - B = TB 10 1 10 1 a b A 52 B + 10(8) = a bv 2 2 32.2 2 32.2 B laws

or

vB = 23.24 ft>s

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v2 . By referring to Fig. a, r

Dissemination

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copyright

Equation of Motion: Here, an =

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10 v2 ba b r 32.2

learning.

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N - 10 cos u = a

g Fn = man;

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10 v2 a b + 10 cos u 32.2 r

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(1)

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the

N =

solely

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protected

dy d2y 1 1 . The slope that the track at position B = x and 2 = dx 16 16 dx dx makes with the horizontal is uB = tan - 1 a b 2 = tan - 1(0) = 0°. The radius of dy x = 0 curvature of the track at position B

rB =

`

dy 2 3>2 b R dx

d2y dx2

`

B1 + ¢ =

`

2 3>2 1 x≤ R 16 sale

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Geometry: Here,

1 ` 16

4

= 16 ft x=0

Substituting u = uB = 0°, v = vB = 23.24 ft>s, and r = rB = 16 ft into Eq. (1),

NB =






10 23.242 c d + 10 cos 0° = 20.5 lb 32.2 16

Ans.

*14–20. The steel ingot has a mass of 1800 kg. It travels along the conveyor at a speed v = 0.5 m>s when it collides with the “nested” spring assembly. Determine the maximum deflection in each spring needed to stop the motion of the ingot. Take kA = 5 kN>m, kB = 3 kN>m.

0.5 m 0.45 m kB

A

SOLUTION

B

Assume both springs compress T1 + ©U1 - 2 = T2 1 1 1 (1800)(0.5)2 - (5000)s2 - (3000)(s - 0.05)2 = 0 2 2 2 225 - 2500 s2 - 1500(s2 - 0.1 s + 0.0025) = 0 s2 - 0.0375 s - 0.05531 = 0 s = 0.2547 m 7 0.05 m

sB = 0.205 m

Ans.

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laws

Ans.

or

(O.K!)

sA = 0.255 m






kA C

14–21. The steel ingot has a mass of 1800 kg. It travels along the conveyor at a speed v = 0.5 m>s when it collides with the “nested” spring assembly. If the stiffness of the outer spring is kA = 5 kN>m, determine the required stiffness kB of the inner spring so that the motion of the ingot is stopped at the moment the front, C, of the ingot is 0.3 m from the wall.

0.5 m 0.45 m kA

kB

A

SOLUTION

B

T1 + ©U1 - 2 = T2 1 1 1 (1800)(0.5)2 - (5000)(0.5 - 0.3)2 - (kB)(0.45 - 0.3)2 = 0 2 2 2 kB = 11.1 kN m

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C

14–22. The 25-lb block has an initial speed of v0 = 10 ft>s when it is midway between springs A and B. After striking spring B, it rebounds and slides across the horizontal plane toward spring A, etc. If the coefficient of kinetic friction between the plane and the block is mk = 0.4, determine the total distance traveled by the block before it comes to rest.

2 ft kA = 10 lb/in.

1 ft v0 = 10 ft/s

A

B

SOLUTION Principle of Work and Energy: Here, the friction force Ff = mk N = 0.4(25) = 10.0 lb. Since the friction force is always opposite the motion, it does negative work. When the block strikes spring B and stops momentarily, the spring force does negative work since it acts in the opposite direction to that of displacement. Applying Eq. 14–7, we have Tl + a U1 - 2 = T2 1 1 25 a b (10)2 - 10(1 + s1) - (60)s21 = 0 2 32.2 2 s1 = 0.8275 ft

Wide

copyright

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laws

or

Assume the block bounces back and stops without striking spring A. The spring force does positive work since it acts in the direction of displacement. Applying Eq. 14–7, we have

instructors

States

World

permitted.

Dissemination

T2 + a U2 - 3 = T3

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on

learning.

is

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not

1 (60)(0.82752) - 10(0.8275 + s2) = 0 2 by

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for

work

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the

s2 = 1.227 ft

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the

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Since s2 = 1.227 ft 6 2 ft, the block stops before it strikes spring A. Therefore, the above assumption was correct. Thus, the total distance traveled by the block before it stops is

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the

is

This

provided

sTot = 2s1 + s2 + 1 = 2(0.8275) + 1.227 + 1 = 3.88 ft

Ans.

kB = 60 lb/in.

14–23. The train car has a mass of 10 Mg and is traveling at 5 m> s when it reaches A. If the rolling resistance is 1> 100 of the weight of the car, determine the compression of each spring when the car is momentarily brought to rest.

300 kN/m 500 kN/m

A

SOLUTION Free-Body Diagram: The free-body diagram of the train in contact with the spring 1 [10 000(9.81)] = 981 N. is shown in Fig. a. Here, the rolling resistance is Fr = 100 The compression of springs 1 and 2 at the instant the train is momentarily at rest will be denoted as s1 and s2. Thus, the force developed in springs 1 and 2 are

A Fsp B 1 = k1s1 = 300(103)s1 and A Fsp B 2 = 500(103)s2. Since action is equal to

reaction,

A Fsp B 1 = A Fsp B 2

300(103)s1 = 500(103)s2 laws

or

s1 = 1.6667s2

Wide

copyright

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teaching

Web)

Principle of Work and Energy: Referring to Fig. a, W and N do no work, and Fsp and Fr do negative work.

instructors

States

World

permitted.

Dissemination

T1 + g U1 - 2 = T2

the

(including

for

work

student

the

by

1 1 C 300(103) D s12 f + e - C 500(103) D s22 f = 0 2 2

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1 (10 000)(52) + [- 981(30 + s1 + s2)] + 2 e-

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150(103)s12 + 250(103)s22 + 981(s1 + s2) - 95570 = 0

provided

integrity

of

and

work

this

Substituting Eq. (1) into Eq. (2),

and

any

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part

the

is

This

666.67(103)s22 + 2616s2 - 95570 = 0

sale

s2 = 0.3767 m = 0.377 m

will

their

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of

Solving for the positive root of the above equation,

Substituting the result of s2 into Eq. (1), s1 = 0.6278 m = 0.628 m






vA ⫽ 5 m/s

30 m

Ans.

*14–24. The 0.5-kg ball is fired up the smooth vertical circular track using the spring plunger. The plunger keeps the spring compressed 0.08 m when s = 0. Determine how far s it must be pulled back and released so that the ball will begin to leave the track when u = 135°.

B

135

SOLUTION

u s

Equations of Motion: A

©Fn = man;

y2B 0.5(9.81) cos 45° = 0.5 a b 1.5

y2B = 10.41 m2>s2

k

Principle of Work and Energy: Here, the weight of the ball is being displaced vertically by s = 1.5 + 1.5 sin 45° = 2.561 m and so it does negative work. The spring force, given by Fsp = 500(s + 0.08), does positive work. Since the ball is at rest initially, T1 = 0. Applying Eq. 14–7, we have

or

TA + a UA-B = TB

in

teaching

Web)

1 (0.5)(10.41) 2

500(s + 0.08) ds - 0.5(9.81)(2.561) =

copyright

L0

laws

s

0 +

Ans.

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Dissemination

Wide

s = 0.1789 m = 179 mm






500 N/m

1.5 m

14–25. The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed vB when he reaches B. Also, find the distance s to where he strikes the ground at C, if he makes the jump traveling horizontally at B. Neglect the skier’s size. He has a mass of 70 kg.

A 50 m B

4m

s

C

SOLUTION 30

TA + © UA - B = TB 0 + 70(9.81)(46) =

1 (70)(vB)2 2

vB = 30.04 m>s = 30.0 m>s + B A:

Ans.

s = s0 + v0 t s cos 30° = 0 + 30.04t s = s0 + v0 t +

1 2 a t 2 c laws

or

A+TB

in

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1 (9.81)t2 2

Wide

copyright

s sin 30° + 4 = 0 + 0 +

States

World

permitted.

Dissemination

Eliminating t,

use

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learning.

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s2 - 122.67s - 981.33 = 0

work

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Solving for the positive root

for

s = 130 m

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Ans.






14–26. The catapulting mechanism is used to propel the 10-kg slider A to the right along the smooth track. The propelling action is obtained by drawing the pulley attached to rod BC rapidly to the left by means of a piston P. If the piston applies a constant force F = 20 kN to rod BC such that it moves it 0.2 m, determine the speed attained by the slider if it was originally at rest. Neglect the mass of the pulleys, cable, piston, and rod BC.

A P B

F

SOLUTION 2 sC + sA = l 2 ¢ sC + ¢ sA = 0 2(0.2) = - ¢ sA -0.4 = ¢ sA T1 + ©U1 - 2 = T2 1 (10)(vA)2 2

vA = 28.3 m>s

Ans.

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laws

or

0 + (10 000)(0.4) =






C

14–27. Block A has a weight of 60 lb and B has a weight of 10 lb. Determine the distance A must descend from rest before it obtains a speed of 8 ft>s. Also, what is the tension in the cord supporting A? Neglect the mass of the cord and pulleys.

SOLUTION B

2 sA + sB = l 2 ¢ sA = - ¢ s B

A

2 vA = -vB For vA = 8 ft/s,

vB = - 16 ft/s

For the system: T1 + ©U1 - 2 = T2

or

1 60 1 10 a b(8)2 + a b ( -16)2 2 32.2 2 32.2

laws

[0 + 0] + [60(sA) - 10(2sA)] = sA = 2.484 = 2.48 ft

Wide

copyright

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Web)

Ans.

States

World

permitted.

Dissemination

For block A:

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T1 + ©U1 - 2 = T2 by

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1 60 a b (8)2 2 32.2

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0 + 60(2.484) - TA(2.484) =

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the

TA = 36.0 lb






Ans.

*14–28. The cyclist travels to point A, pedaling until he reaches a speed vA = 4 m>s. He then coasts freely up the curved surface. Determine how high he reaches up the surface before he comes to a stop. Also, what are the resultant normal force on the surface at this point and his acceleration? The total mass of the bike and man is 75 kg. Neglect friction, the mass of the wheels, and the size of the bicycle.

y

C

SOLUTION 1 2

x1/2

y1/2

B y

x

45

A

1 2

x + y = 2

4m

1 -1 1 1 dy x 2 + y-2 = 0 2 2 dx 1

dy -x - 2 = 1 dx y-2 T1 + ©U1 - 2 = T2 1 (75)(4)2 - 75(9.81)(y) = 0 2 y = 0.81549 m = 0.815 m

laws

or

Ans.

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x1>2 + (0.81549)1>2 = 2

permitted.

Dissemination

x = 1.2033 m

United

on

learning.

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not

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States

World

-(1.2033) - 1>2 dy = = - 0.82323 dx (0.81549) - 1>2

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use

tan u =

for

work

student

the

by

u = -39.46°

the

(including

Nb - 9.81(75) cos 39.46° = 0 is

work

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protected

Q+ ©Fn = m an ;

Ans. integrity

of

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work

this

assessing

Nb = 568 N provided

75(9.81) sin 39.46° = 75 at part

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any

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Ans.

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and

a = at = 6.23 m>s2






2

4m

x

14–29. The collar has a mass of 20 kg and slides along the smooth rod. Two springs are attached to it and the ends of the rod as shown. If each spring has an uncompressed length of 1 m and the collar has a speed of 2 m>s when s = 0, determine the maximum compression of each spring due to the backand-forth (oscillating) motion of the collar.

s

kA

50 N/m

kB

1m

1m 0.25 m

SOLUTION T1 + ©U1 - 2 = T2 1 1 1 (20)(2)2 - (50)(s)2 - (100)(s)2 = 0 2 2 2 s = 0.730 m

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100 N/m

14–30. A

The 30-lb box A is released from rest and slides down along the smooth ramp and onto the surface of a cart. If the cart is prevented from moving, determine the distance s from the end of the cart to where the box stops. The coefficient of kinetic friction between the cart and the box is mk = 0.6.

10 ft 4 ft

s C

B

SOLUTION Principle of Work and Energy: WA which acts in the direction of the vertical displacement does positive work when the block displaces 4 ft vertically. The friction force Ff = mk N = 0.6(30) = 18.0 lb does negative work since it acts in the opposite direction to that of displacement Since the block is at rest initially and is required to stop, TA = TC = 0. Applying Eq. 14–7, we have TA + a UA - C = TC 0 + 30(4) - 18.0s¿ = 0

s = 10 - s¿ = 3.33 ft

Ans.

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in

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laws

or

Thus,

s¿ = 6.667 ft






14–31. Marbles having a mass of 5 g are dropped from rest at A through the smooth glass tube and accumulate in the can at C. Determine the placement R of the can from the end of the tube and the speed at which the marbles fall into the can. Neglect the size of the can.

A

B 3m 2m

SOLUTION

C

TA + © UA-B = TB R

1 0 + [0.005(9.81)(3 - 2)] = (0.005)v2B 2 vB = 4.429 m>s

A+TB

s = s0 + v0t + 2 = 0 + 0 =

1 a t2 2 c

1 (9.81)t2 2 laws in

teaching

Web)

s = s0 + v0 t

Wide

copyright

+ b a:

or

t = 0.6386 s

R = 0 + 4.429(0.6386) = 2.83 m

permitted.

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on

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is

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TA + © UA-C = T1 by

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1 (0.005)v2C 2

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0 + [0.005(9.81)(3) =

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vC = 7.67 m>s






Ans.

*14–32. The cyclist travels to point A, pedaling until he reaches a speed vA = 8 m>s. He then coasts freely up the curved surface. Determine the normal force he exerts on the surface when he reaches point B. The total mass of the bike and man is 75 kg. Neglect friction, the mass of the wheels, and the size of the bicycle.

y

C x1/2

y1/2

B y

x

2

4m

SOLUTION 1 2

45

A

1 2

x + y = 2

4m

1 1 dy 1 -1 x 2 + y- 2 = 0 2 2 dx 1

dy - x- 2 = 1 dx y- 2 For y = x, 1

2 x2 = 2

laws

or

x = 1, y = 1 (Point B)

in

teaching

Web)

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permitted.

Dissemination

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copyright

dy = -1 dx instructors

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tan u =

use

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dy 1 1 1 1 3 1 = y2 a x - 2 b - x - 2 a b ay - 2 b a b 2 2 dx dx

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1 1 -3 1 1 y2 x 2 + a b 2 2 x

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For x = y = 1 dy = -1, dx r =

d2y dx2

= 1

[1 + ( -1)2]3>2 = 2.828 m 1

T1 + ©U1 - 2 = T2 1 1 (75)(8)2 - 75(9.81)(1) = (75)(v2B) 2 2 v2B = 44.38 Q+ ©Fn = man ;

NB - 9.81(75) cos 45° = 75 a NB = 1.70 kN






44.38 b 2.828 Ans.

x

14–33. The man at the window A wishes to throw the 30-kg sack on the ground. To do this he allows it to swing from rest at B to point C, when he releases the cord at u = 30°. Determine the speed at which it strikes the ground and the distance R.

A B

8m

8m u 16 m

C

SOLUTION TB + ©UB - C = TC

D

0 + 30(9.81)8 cos 30° =

R

1 (30)v2C 2

yC = 11.659 m>s TB + ©UB - D = TD 0 + 30(9.81)(16) =

1 (30) v2D 2

vD = 17.7 m>s

or

Ans.

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1 2 act 2

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( + T) s = s0 + v0t +

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16 = 8 cos 30° - 11.659 sin 30°t +

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t = 2.0784 s

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s = 24.985 m

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s = 8 sin 30° + 11.659 cos 30°(2.0784)

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work

+ )s = s + v t (: 0 0

Thus, R = 8 + 24.985 = 33.0 m

Ans.

Also, (vD)x = 11.659 cos 30° = 10.097 m>s (+ T) (vD)x = -11.659 sin 30° + 9.81(2.0784) = 14.559 m>s nD = 2(10.097)2 + (14.559) 2 = 7.7 m>s






Ans.

14–34. The spring bumper is used to arrest the motion of the 4-lb block, which is sliding toward it at v = 9 ft>s. As shown, the spring is confined by the plate P and wall using cables so that its length is 1.5 ft. If the stiffness of the spring is k = 50 lb>ft, determine the required unstretched length of the spring so that the plate is not displaced more than 0.2 ft after the block collides into it. Neglect friction, the mass of the plate and spring, and the energy loss between the plate and block during the collision.

1.5 ft.

5 ft.

SOLUTION T1 + ©U1 - 2 = T2 4 1 1 1 a b (9)2 - c (50)(s - 1.3)2 - (50)(s - 1.5)2 d = 0 2 32.2 2 2 0.20124 = s2 - 2.60 s + 1.69 - (s2 - 3.0 s + 2.25) 0.20124 = 0.4 s - 0.560 s = 1.90 ft

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laws

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Ans.






vA

P

k

A

14–35. The collar has a mass of 20 kg and is supported on the smooth rod. The attached springs are undeformed when d = 0.5 m. Determine the speed of the collar after the applied force F = 100 N causes it to be displaced so that d = 0.3 m. When d = 0.5 m the collar is at rest.

k' = 15 N/m F = 100 N 60°

SOLUTION

d

T1 + a U1 - 2 = T2

k = 25 N/m

1 (15)(0.5 - 0.3)2 2

0 + 100 sin 60°(0.5 - 0.3) + 196.2(0.5 - 0.3) 1 1 (25)(0.5 - 0.3)2 = (20)v2C 2 2

-

vC = 2.36 m>s

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*14–36. If the force exerted by the motor M on the cable is 250 N, determine the speed of the 100-kg crate when it is hoisted to s = 3 m. The crate is at rest when s = 0.

M

SOLUTION Kinematics: Expressing the length of the cable in terms of position coordinates sC and sP referring to Fig. a, s

3sC + (sC - sP) = l 4sC - sP = l

(1)

Using Eq. (1), the change in position of the crate and point P on the cable can be written as (+ T)

4¢sC - ¢sP = 0

Here, ¢sC = - 3 m. Thus, ¢sp = - 12 m = 12 m c

or

4(- 3) - ¢sP = 0

not

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Dissemination

T1 + g U1 - 2 = T2

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1 m v2 2 C

the

by

0 + T¢sP + [ - WC ¢sC] =

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1 (100)v2 2

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0 + 250(12) + [- 100(9.81)(3)] =

Ans.

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v = 1.07 m>s






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Principle of Work and Energy: Referring to the free-body diagram of the pulley system, Fig. b, F1 and F2 do no work since it acts at the support; however, T does positive work and WC does negative work.

Web)

laws

(+ T )

14–37. If the track is to be designed so that the passengers of the roller coaster do not experience a normal force equal to zero or more than 4 times their weight, determine the limiting heights hA and hC so that this does not occur. The roller coaster starts from rest at position A. Neglect friction.

A C

rC

20 m rB

hC B

SOLUTION Free-Body Diagram:The free-body diagram of the passenger at positions B and C are shown in Figs. a and b, respectively. Equations of Motion: Here, an =

v2 . The requirement at position B is that r

NB = 4mg. By referring to Fig. a, + c ©Fn = man;

4mg - mg = m ¢

vB 2 ≤ 15

or

vB 2 = 45g

teaching

Web)

laws

At position C, NC is required to be zero. By referring to Fig. b, in

vC 2 ≤ 20

Wide

permitted.

Dissemination

mg - 0 = m ¢

copyright

+ T ©Fn = man;

on

learning.

is

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States

World

vC 2 = 20g

of

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the

(including

for

work

student

the

by

and

use

United

Principle of Work and Energy: The normal reaction N does no work since it always acts perpendicular to the motion. When the rollercoaster moves from position A to B, W displaces vertically downward h = hA and does positive work.

this

assessing

is

work

solely

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part

the

is

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provided

integrity

of

and

work

TA + ©UA-B = TB

any

courses

1 m(45g) 2 sale

hA = 22.5 m

will

their

destroy

of

and

0 + mghA =

Ans.

When the rollercoaster moves from position A to C, W displaces vertically downward h = hA - hC = (22.5 - hC) m. TA + ©UA-B = TB 0 + mg(22.5 - hC) = hC = 12.5 m






1 m(20g) 2 Ans.

15 m

hA

14–38. The 150-lb skater passes point A with a speed of 6 ft>s. Determine his speed when he reaches point B and the normal force exerted on him by the track at this point. Neglect friction.

y

y2 B

25 ft

Free-Body Diagram:The free-body diagram of the skater at an arbitrary position is shown in Fig. a. Principle of Work and Energy: By referring to Fig. a, notice that N does no work since it always acts perpendicular to the motion. When the skier slides down the track from A to B, W displaces vertically downward h = yA - yB = 20 - C 2(25)1>2 D = 10 ft and does positive work. TA + ©UA-B = TB

or

1 150 1 150 ¢ ≤ A 62 B + C 150(10) D = ¢ ≤v 2 2 32.2 2 32.2 B vB = 26.08 ft>s = 26.1 ft>s

teaching

Web)

laws

Ans. in

v2 . By referring to Fig. a, r

not the

is on

150 v2 a b 32.2 r

learning.

N = 150 cos u -

of

150 v2 a b 32.2 r

United

150 cos u - N =

(1)

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the

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by

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use

R+ ©Fn = man ;

instructors

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Dissemination

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copyright

Equations of Motion:Here, an =

is

work

dy d2y 1 1 = 1>2 , and 2 = . The slope that the dx dx x 2x 3>2 dx track at position B makes with the horizontal is uB = tan-1 a b 2 dy x = 25 ft 1 = 11.31°. The radius of curvature of the track at position B is = tan a 1>2 b 2 x x = 25 ft given by sale

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Geometry: Here, y = 2x 1>2,

3>2

dy 2 3>2 B1 + a b R dx

2

d2y dx2

C1 + ¢

1 x 1>2

=

2

2-

2

≤ S

1 2x 3>2

6

= 265.15 ft

2 x = 25 ft

Substituting u = uB = 11.31°, v = vB = 26.08 ft>s, and r = rB = 265.15 ft into Eq. (1), NB = 150 cos 11.31° = 135 lb






20 ft x

SOLUTION

rB =

4x A

150 26.082 ¢ ≤ 32.2 265.15 Ans.

14–39. The 8-kg cylinder A and 3-kg cylinder B are released from rest. Determine the speed of A after it has moved 2 m starting from rest. Neglect the mass of the cord and pulleys.

SOLUTION Kinematics: Express the length of cord in terms of position coordinates sA and sB by referring to Fig. a 2sA + sB = l

A

(1) B

Thus (2)

2¢sA + ¢sB = 0

If we assume that cylinder A is moving downward through a distance of ¢sA = 2 m, Eq. (2) gives (+ T )

2(2) + ¢sB = 0

¢sB = - 4 m = 4 m c

laws

or

Taking the time derivative of Eq. (1), 2vA + vB = 0

teaching

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(3) in

(+ T )

Dissemination

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copyright

g T1 + gU1 - 2 = g T2

the

not

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World

permitted.

1 1 (8)vA2 + (3)vB2 2 2

learning.

is

of

0 + 8(2)9.81 - 3(4)9.81 =

for

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Positive net work on left means assumption of A moving down is correct. Since vB = - 2vA,

Ans. of

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vA = 1.98 m>s T

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vB = - 3.96 m>s = 3.96 m>s c






*14–40. Cylinder A has a mass of 3 kg and cylinder B has a mass of 8 kg. Determine the speed of A after it moves upwards 2 m starting from rest. Neglect the mass of the cord and pulleys.

SOLUTION a T1 + a U1 - 2 = a T2 A

1 1 (3)v2A + (8)v2B 2 2

0 + 2[F1 - 3(9.81)] + 4[8(9.81) - F2] =

B

Also, vB = 2vA, and because the pulleys are massless, F1 = 2F2. The F1 and F2 terms drop out and the work-energy equation reduces to 255.06 = 17.5v2A vA = 3.82 m>s

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Ans.






14–41. A 2-lb block rests on the smooth semicylindrical surface. An elastic cord having a stiffness k = 2 lb>ft is attached to the block at B and to the base of the semicylinder at point C. If the block is released from rest at A (u = 0°), determine the unstretched length of the cord so that the block begins to leave the semicylinder at the instant u = 45°. Neglect the size of the block.

k

2 lb/ft B

1.5 ft u

C

SOLUTION +b©Fn = man;

2 sin 45° =

v2 2 a b 32.2 1.5

v = 5.844 ft>s T1 + © U1-2 = T2 2 2 3p 2 1 1 1 (2) C p(1.5) - l0 D - (2) c (1.5) - l0 d - 2(1.5 sin 45°) = a b (5.844)2 2 2 4 2 32.2

0 +

l0 = 2.77 ft

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or

Ans.






A

14–42. The jeep has a weight of 2500 lb and an engine which transmits a power of 100 hp to all the wheels. Assuming the wheels do not slip on the ground, determine the angle u of the largest incline the jeep can climb at a constant speed v = 30 ft>s.

θ

SOLUTION P = FJv 100(550) = 2500 sin u(30) u = 47.2°

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Ans.






14–43. Determine the power input for a motor necessary to lift 300 lb at a constant rate of 5 ft> s. The efficiency of the motor is P = 0.65.

SOLUTION Power: The power output can be obtained using Eq. 14–10. P = F # v = 300(5) = 1500 ft # lb>s Using Eq. 14–11, the required power input for the motor to provide the above power output is power input =

1500 = 2307.7 ft # lb>s = 4.20 hp 0.65

Ans.

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=

power output P






*14–44. An automobile having a mass of 2 Mg travels up a 7° slope at a constant speed of v = 100 km>h. If mechanical friction and wind resistance are neglected, determine the power developed by the engine if the automobile has an efficiency P = 0.65.

7⬚

SOLUTION Equation of Motion: The force F which is required to maintain the car’s constant speed up the slope must be determined first. + ©Fx¿ = max¿;

F - 2(103)(9.81) sin 7° = 2(103)(0) F = 2391.08 N

100(103) m 1h b = 27.78 m>s. R * a h 3600 s The power output can be obtained using Eq. 14–10. Power: Here, the speed of the car is y = B

laws

or

P = F # v = 2391.08(27.78) = 66.418(103) W = 66.418 kW

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copyright

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Using Eq. 14–11, the required power input from the engine to provide the above power output is

World

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Dissemination

power output e

of

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power input =

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66.418 = 102 kW 0.65

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14–45. The Milkin Aircraft Co. manufactures a turbojet engine that is placed in a plane having a weight of 13000 lb. If the engine develops a constant thrust of 5200 lb, determine the power output of the plane when it is just ready to take off with a speed of 600 mi>h.

SOLUTION At 600 ms>h. 88 ft>s 1 b = 8.32 (103) hp 60 m>h 550

Ans.

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laws

or

P = 5200(600)a






14–46. To dramatize the loss of energy in an automobile, consider a car having a weight of 5000 lb that is traveling at 35 mi>h. If the car is brought to a stop, determine how long a 100-W light bulb must burn to expend the same amount of energy. 11 mi = 5280 ft.2

SOLUTION 5280 ft 1h 35 mi b * a b * a b = h 1 mi 3600 s

Energy: Here, the speed of the car is y = a 51.33 ft>s. Thus, the kinetic energy of the car is

1 2 1 5000 my = a b A 51.332 B = 204.59 A 103 B ft # lb 2 2 32.2

The power of the bulb is 73.73 ft # lb>s. Thus,

550 ft # lb>s 1 hp b * a b = 746 W 1 hp

or

204.59(103) U = 2774.98 s = 46.2 min = Pbulb 73.73

Ans.

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t =

Pbulb = 100 W * a

laws

U =






14–47. The escalator steps move with a constant speed of 0.6 m>s. If the steps are 125 mm high and 250 mm in length, determine the power of a motor needed to lift an average mass of 150 kg per step. There are 32 steps.

250 mm 125 mm v

0.6 m/s

SOLUTION Step height: 0.125 m The number of steps:

4 = 32 0.125

Total load: 32(150)(9.81) = 47 088 N 4 = 2 m, then 2

If load is placed at the center height, h =

laws

or

4 U = 47 088 a b = 94.18 kJ 2

2(32(0.25))2 + 42

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≤ = 0.2683 m>s in

4

permitted. World

94.18 U = = 12.6 kW t 7.454

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P = F # v = 47 088(0.2683) = 12.6 kW

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4m

*14–48. If the escalator in Prob. 14–47 is not moving, determine the constant speed at which a man having a mass of 80 kg must walk up the steps to generate 100 W of power—the same amount that is needed to power a standard light bulb.

250 mm 125 mm v

0.6 m/s

SOLUTION P =

(80)(9.81)(4) U1 - 2 = = 100 t t

n =

2(32(0.25))2 + 42 s = = 0.285 m>s t 31.4

t = 31.4 s

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4m

14–49. The 2-Mg car increases its speed uniformly from rest to 25 m>s in 30 s up the inclined road. Determine the maximum power that must be supplied by the engine, which operates with an efficiency of P = 0.8. Also, find the average power supplied by the engine.

1 10

SOLUTION Kinematics:The constant acceleration of the car can be determined from + B A:

v = v0 + ac t 25 = 0 + ac (30) ac = 0.8333 m>s2

Equations of Motion: By referring to the free-body diagram of the car shown in Fig. a, ©Fx¿ = max¿ ;

F - 2000(9.81) sin 5.711° = 2000(0.8333)

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F = 3618.93N

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permitted.

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(Pout)max = F # vmax = 3618.93(25) = 90 473.24 W

in

Power: The maximum power output of the motor can be determined from

by

Pout 90473.24 = 113 091.55 W = 113 kW = e 0.8

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14–50. The crate has a mass of 150 kg and rests on a surface for which the coefficients of static and kinetic friction are ms = 0.3 and mk = 0.2, respectively. If the motor M supplies a cable force of F = (8t2 + 20) N, where t is in seconds, determine the power output developed by the motor when t = 5 s.

M

SOLUTION Equations of Equilibrium: If the crate is on the verge of slipping, Ff = ms N = 0.3N. From FBD(a), + c ©Fy = 0;

N - 150(9.81) = 0

+ ©F = 0; : x

0.3(1471.5) - 3 A 8 t2 + 20 B = 0

N = 1471.5 N t = 3.9867 s

Equations of Motion: Since the crate moves 3.9867 s later, Ff = mk N = 0.2N. From FBD(b), + c ©Fy = may ; + ©F = ma ; : x x

N - 150(9.81) = 150 (0)

N = 1471.5 N or

0.2 (1471.5) - 3 A 8 t2 + 20 B = 150 ( -a)

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a = A 0.160 t2 - 1.562 B m>s2

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Kinematics:Applying dy = adt, we have

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Power: At t = 5 s, F = 8 A 52 B + 20 = 220 N. The power can be obtained using Eq. 14–10.

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P = F # v = 3 (220) (1.7045) = 1124.97 W = 1.12 kW






Ans.

14–51. The 50-kg crate is hoisted up the 30° incline by the pulley system and motor M. If the crate starts from rest and, by constant acceleration, attains a speed of 4 m>s after traveling 8 m along the plane, determine the power that must be supplied to the motor at the instant the crate has moved 8 m. Neglect friction along the plane. The motor has an efficiency of P = 0.74.

M

SOLUTION

30

Kinematics:Applying equation y2 = y20 + 2ac (s - s0), we have 42 = 02 + 2a(8 - 0)

a = 1.00 m>s2

Equations of Motion: + ©Fx¿ = max¿ ;

F - 50(9.81) sin 30° = 50(1.00)

F = 295.25 N

Power: The power output at the instant when y = 4 m>s can be obtained using Eq. 14–10.

or

P = F # v = 295.25 (4) = 1181 W = 1.181 kW

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power output e

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*14–52. The 50-lb load is hoisted by the pulley system and motor M. If the motor exerts a constant force of 30 lb on the cable, determine the power that must be supplied to the motor if the load has been hoisted s = 10 ft starting from rest. The motor has an efficiency of P = 0.76.

M

SOLUTION + c ©Fy = m ay ;

B

2(30) - 50 =

50 a 32.2 B

s

aB = 6.44 m>s2 ( + c) v2 = v20 + 2ac (s - s0) v2B = 0 + 2(6.44)(10 - 0) vB = 11.349 ft>s 2sB + sM = l

laws

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vM = -2( 11.349) = 22.698 ft>s

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in

Po = F # v = 30(22.698) = 680.94 ft # lb>s

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680.94 = 895.97 ft # lb>s 0.76 United

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Pi = 1.63 hp






Ans.

14–53. The 10-lb collar starts from rest at A and is lifted by applying a constant vertical force of F = 25 lb to the cord. If the rod is smooth, determine the power developed by the force at the instant u = 60°.

3 ft

4 ft

u F

SOLUTION

A

Work of F U1 - 2 = 25(5 - 3.464) = 38.40 lb # ft T1 + ©U1 - 2 = T2s 0 + 38.40 - 10(4 - 1.732) =

1 10 2 ( )v 2 32.2

v = 10.06 ft>s P = F # v = 25 cos 60°(10.06) = 125.76 ft # lb>s P = 0.229 hp

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14–54. The 10-lb collar starts from rest at A and is lifted with a constant speed of 2 ft>s along the smooth rod. Determine the power developed by the force F at the instant shown.

3 ft

4 ft

u F


A

4 F a b - 10 = 0 5 F = 12.5 lb

4 P = F # v = 12.5 a b (2) = 20 lb # ft>s 5 = 0.0364 hp

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14–55. The elevator E and its freight have a total mass of 400 kg. Hoisting is provided by the motor M and the 60-kg block C. If the motor has an efficiency of P = 0.6, determine the power that must be supplied to the motor when the elevator is hoisted upward at a constant speed of vE = 4 m>s.

M

C

SOLUTION vE

Elevator: E

Since a = 0, + c ©Fy = 0;

60(9.81) + 3T - 400(9.81) = 0 T = 1111.8 N

2sE + (sE - sP) = l 3vE = vP Since vE = -4 m>s,

or

(1111.8)(12) F # vP = = 22.2 kW e 0.6

Ans.

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laws

Pi =

vP = - 12 m>s






*14–56. The sports car has a mass of 2.3 Mg, and while it is traveling at 28 m/s the driver causes it to accelerate at 5 m>s2. If the drag resistance on the car due to the wind is FD = 10.3v22 N, where v is the velocity in m/s, determine the power supplied to the engine at this instant. The engine has a running efficiency of P = 0.68.

FD


F - 0.3v2 = 2.3(103)(5) F = 0.3v2 + 11.5(103)

At v = 28 m>s F = 11 735.2 N PO = (11 735.2)(28) = 328.59 kW PO 328.59 = 438 kW = e 0.68

Ans.

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Pi =






14–57. The sports car has a mass of 2.3 Mg and accelerates at 6 m>s2, starting from rest. If the drag resistance on the car due to the wind is FD = 110v2 N, where v is the velocity in m/s, determine the power supplied to the engine when t = 5 s. The engine has a running efficiency of P = 0.68.

FD


F - 10v = 2.3(103)(6) F = 13.8(103) + 10 v

+ )v = v + a t (: 0 c v = 0 + 6(5) = 30 m>s PO = F # v = [13.8(103) + 10(30)](30) = 423.0 kW PO 423.0 = 622 kW = e 0.68

Ans.

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Pi =






14–58. The block has a mass of 150 kg and rests on a surface for which the coefficients of static and kinetic friction are ms = 0.5 and mk = 0.4, respectively. If a force F = 160t22 N, where t is in seconds, is applied to the cable, determine the power developed by the force when t = 5 s. Hint: First determine the time needed for the force to cause motion.

F

SOLUTION + ©F = 0; : x

2F - 0.5(150)(9.81) = 0 F = 367.875 = 60t2 t = 2.476 s

+ ©F = m a ; : x x

2(60t2)- 0.4(150)(9.81) = 150ap ap = 0.8t2 - 3.924

dv = a dt

L2.476

A 0.8t2 - 3.924 B dt laws

L0

5

dv =

or

v

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5 0.8 3 b t - 3.924t ` = 19.38 m>s 3 2.476

Wide

copyright

v = a

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Dissemination

sP + (sP - sF) = l

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2vP = vF

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vF = 2(19.38) = 38.76 m>s

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F = 60(5)2 = 1500 N

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P = F # v = 1500(38.76) = 58.1 kW






Ans.

14–59. v

The rocket sled has a mass of 4 Mg and travels from rest along the horizontal track for which the coefficient of kinetic friction is mk = 0.20. If the engine provides a constant thrust T = 150 kN, determine the power output of the engine as a function of time. Neglect the loss of fuel mass and air resistance.

T


150(10)3 - 0.2(4)(10)3(9.81) = 4(10)3 a a = 35.54 m>s2

+ Bv = v + a t A: 0 c = 0 + 35.54t = 35.54t P = T # v = 150(10)3 (35.54t) = 5.33t MW

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*14–60. A loaded truck weighs 16(103) lb and accelerates uniformly on a level road from 15 ft>s to 30 ft>s during 4 s. If the frictional resistance to motion is 325 lb, determine the maximum power that must be delivered to the wheels.

SOLUTION a =

¢v 30 - 15 = = 3.75 ft>s2 ¢t 4

+ ©F = ma ; ; x x

F - 325 = ¢

16(103) ≤ (3.75) 32.2

F = 2188.35 lb 2188.35(30) = 119 hp 550

Ans.

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Pmax = F # vmax =






14–61. If the jet on the dragster supplies a constant thrust of T = 20 kN, determine the power generated by the jet as a function of time. Neglect drag and rolling resistance, and the loss of fuel. The dragster has a mass of 1 Mg and starts from rest.

T

SOLUTION Equations of Motion: By referring to the free-body diagram of the dragster shown in Fig. a, + ©F = ma ; : x x

20(103) = 1000(a)

a = 20 m>s2

Kinematics: The velocity of the dragster can be determined from + b a:

v = v0 + ac t v = 0 + 20 t = (20 t) m>s

Power: laws

or

P = F # v = 20(103)(20 t)

teaching

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= C 400(103)t D W

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14–62. An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete’s arm acting in the same direction as the force varies with time as shown in the second graph. Determine the power applied as a function of time and the work done in t = 0.3 s.

F (N)

800

t (s) 0.2

0.3

SOLUTION v (m/s)

For 0 … t … 0.2 F = 800 N v =

20 t = 66.67t 0.3

P =

F#v

20

t (s)

= 53.3 t kW

0.3

Ans.

For 0.2 … t … 0.3 F = 2400 - 8000 t laws

or

v = 66.67t P = F # v = 1160 t - 533t22 kW

in

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P dt instructors

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1160t - 533t 2 dt

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the

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53.3t dt +

by

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u =

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student

53.3 160 533 (0.2)2 + [(0.3)2 - (0.2)2] [(0.3)3 - (0.2)3] 2 2 3 is

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= 1.69 kJ






14–63. An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete’s arm acting in the same direction as the force varies with time as shown in the second graph. Determine the maximum power developed during the 0.3-second time period.

F (N)

800

t (s) 0.2

0.3

SOLUTION v (m/s)

See solution to Prob. 14–62. P = 160 t - 533 t2

20

dP = 160 - 1066.6 t = 0 dt

t (s) 0.3

t = 0.15 s 6 0.2 s Thus maximum occurs at t = 0.2 s Pmax = 53.3(0.2) = 10.7 kW

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*14–64. The 500-kg elevator starts from rest and travels upward with a constant acceleration a c = 2 m>s2. Determine the power output of the motor M when t = 3 s. Neglect the mass of the pulleys and cable.

M


3T - 500(9.81) = 500(2) E

T = 1968.33 N 3sE - sP = l 3 vE = vP When t = 3 s, (+ c ) v0 + ac t laws

or

vE = 0 + 2(3) = 6 m>s in

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vP = 3(6) = 18 m>s

permitted.

Dissemination

Wide

copyright

PO = 1968.33(18)

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PO = 35.4 kW






14–65. The 50-lb block rests on the rough surface for which the coefficient of kinetic friction is m k = 0.2. A force F = (40 + s 2) lb, where s is in ft, acts on the block in the direction shown. If the spring is originally unstretched (s = 0) and the block is at rest, determine the power developed by the force the instant the block has moved s = 1.5 ft.

F k

30


NB - A 40 + s2 B sin 30° - 50 = 0 NB = 70 + 0.5s2

T1 + ©U1 - 2 + T2 1.5

0 +

L0

A 40 + s2 B cos 30° ds -

1 2 (20)(1.5) - 0.2 2 L0

1.5

1 50 2 a bv 2 32.2 2

A 70 + 0.5s2 B ds =

0 + 52.936 - 22.5 - 21.1125 = 0.7764 v22 v2 = 3.465 ft>s laws

or

When s = 1.5 ft, in

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F = 40 + (1.5)2 = 42.25 lb Dissemination

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copyright

P = F # v = (42.25 cos 30°)(3.465)

permitted.

P = 126.79 ft # lb>s = 0.231 hp

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the © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





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Ans.

20 lb/ft

14–66. The girl has a mass of 40 kg and center of mass at G. If she is swinging to a maximum height defined by u = 60°, determine the force developed along each of the four supporting posts such as AB at the instant u = 0°. The swing is centrally located between the posts.

A 2m

␪ G

SOLUTION The maximum tension in the cable occurs when u = 0°. B

T1 + V1 = T2 + V2 0 + 40(9.81)(- 2 cos 60°) =

1 (40)v2 + 40(9.81)(- 2) 2

v = 4.429 m>s 4.4292 b 2

+ c ©Fy = 0;

2(2F) cos 30° - 784.8 = 0

T = 784.8 N F = 227 N

Ans.

or

T - 40(9.81) = (40)a

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+ c ©Fn = ma n;






30⬚

30⬚

14–67. Two equal-length springs are “nested” together in order to form a shock absorber. If it is designed to arrest the motion of a 2-kg mass that is dropped s = 0.5 m above the top of the springs from an at-rest position, and the maximum compression of the springs is to be 0.2 m, determine the required stiffness of the inner spring, kB, if the outer spring has a stiffness kA = 400 N>m.

s

A

SOLUTION

B

T1 + V1 = T2 + V2 1 1 (400)(0.2)2 + (kB)(0.2)2 2 2

0 + 0 = 0 - 2(9.81)(0.5 + 0.2) + kB = 287 N>m

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*14–68. The collar has a weight of 8 lb. If it is pushed down so as to compress the spring 2 ft and then released from rest 1h = 02, determine its speed when it is displaced h = 4.5 ft. The spring is not attached to the collar. Neglect friction. h

SOLUTION T1 + V1 = T2 + V2 0 +

k = 30 lb/ft

8 1 1 (30)(2)2 = a b v22 + 8(4.5) 2 2 32.2

v2 = 13.9 ft s

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14–69. The collar has a weight of 8 lb. If it is released from rest at a height of h = 2 ft from the top of the uncompressed spring, determine the speed of the collar after it falls and compresses the spring 0.3 ft. h

SOLUTION T1 + V1 = T2 + V2 0 + 0 =

k = 30 lb/ft

8 1 1 a b v22 - 8(2.3) + (30)(0.3)2 2 32.2 2

v2 = 11.7 ft s

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14–70. The 2-kg ball of negligible size is fired from point A with an initial velocity of 10 m/s up the smooth inclined plane. Determine the distance from point C to where it hits the horizontal surface at D. Also, what is its velocity when it strikes the surface?

B 10 m/s

1.5 m

A

C

SOLUTION

2m

Datum at A: TA + VA = TB + VB 1 1 (2)(10)2 + 0 = (2)(nB)2 + 2(9.81)(1.5) 2 2 vB = 8.401 m>s + B s = s + v t A: 0 0

or

4 d = 0 + 8.401 a b t 5

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1 2 at 2 c

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s = s0 + v0 t +

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1 3 -1.5 = 0 + 8.401 a b t + ( - 9.81)t2 5 2

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-4.905t2 + 5.040t + 1.5 = 0

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t = 1.269 s

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4 d = 8.401a b (1.269) = 8.53 m 5

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TA + VA = TD + VD

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Ans.

1 1 (2)(10)2 + 0 = (2)(nD)2 + 0 2 2 vD = 10 m s






Ans.

D d

14–71. y

The ride at an amusement park consists of a gondola which is lifted to a height of 120 ft at A. If it is released from rest and falls along the parabolic track, determine the speed at the instant y = 20 ft. Also determine the normal reaction of the tracks on the gondola at this instant. The gondola and passenger have a total weight of 500 lb. Neglect the effects of friction and the mass of the wheels.

A

1 x2 y ⫽ —– 260

SOLUTION y =

y ⫽ 20 ft

1 2 x 260

dy 1 = x dx 130 d2y 1 = dx2 130 At y = 120 - 100 = 20 ft x = 72.11 ft or

u = 29.02° teaching

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dy = 0.555 , dx

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tan u =

in

[1 + (0.555)2]3>2 = 194.40 ft 1 130 States

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500 v2 a b 32.2 194.40

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NG - 500 cos 29.02° = by

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1 500 2 a b v - 500(100) 2 32.2

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0 + 0 =

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v2 = 6440

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v = 80.2 ft>s

NG = 952 lb






Ans.

120 ft

x

*14–72. The 2-kg collar is attached to a spring that has an unstretched length of 3 m. If the collar is drawn to point B and released from rest, determine its speed when it arrives at point A.

k = 3 N/m 3m B

A

SOLUTION 4m

Potential Energy: The initial and final elastic potential energy are 1 1 (3) A 232 + 42 - 3 B 2 = 6.00 J and (3)(3 - 3)2 = 0, respectively.The gravitational 2 2 potential energy remains the same since the elevation of collar does not change when it moves from B to A. Conservation of Energy: TB + VB = TA + VA 1 (2) v2A + 0 2

vA = 2.45 m s

Ans.

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0 + 6.00 =






14–73. The 2-kg collar is attached to a spring that has an unstretched length of 2 m. If the collar is drawn to point B and released from rest, determine its speed when it arrives at point A. k ⫽ 3 N/m 3m B

A

SOLUTION

4m

Potential Energy: The stretches of the spring when the collar is at B and A are sB = 232 + 4 2 - 2 = 3 m and sA = 3 - 2 = 1 m, respectively. Thus, the elastic potential energy of the system at B and A are

(Ve)B =

(Ve)A =

1 1 ks 2 = (3)(32) = 13.5 J 2 B 2 1 1 ks 2 = (3)(12) = 1.5 J 2 A 2

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or

There is no change in gravitational potential energy since the elevation of the collar does not change during the motion.

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Conservation of Energy:

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TB + VB = TA + VA

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1 1 mvB2 + (Ve)B = mvA2 + (Ve)A 2 2

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1 (2)vA2 + 1.5 2

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0 + 13.5 =

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vA = 3.46 m>s






Ans.

14–74. The 0.5-lb ball is shot from the spring device shown. The spring has a stiffness k = 10 lb>in. and the four cords C and plate P keep the spring compressed 2 in. when no load is on the plate. The plate is pushed back 3 in. from its initial position. If it is then released from rest, determine the speed of the ball when it travels 30 in. up the smooth plane.

s

k C

P 30

SOLUTION Potential Energy: The datum is set at the lowest point (compressed position). 30 Finally, the ball is sin 30° = 1.25 ft above the datum and its gravitational 12 potential energy is 0.5(1.25) = 0.625 ft # lb. The initial and final elastic potential 2 + 3 2 2 2 1 1 (120)a b = 10.42 ft # lb and (120)a b = 1.667 ft # lb, energy are 2 12 2 12 respectively. Conservation of Energy:

laws

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©T1 + ©V1 = ©T2 + ©V2

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1 0.5 a b y2 + 0.625 + 1.667 2 32.2

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0 + 10.42 =

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y = 32.3 ft>s






14–75. The 0.5-lb ball is shot from the spring device shown. Determine the smallest stiffness k which is required to shoot the ball a maximum distance of 30 in. up the smooth plane after the spring is pushed back 3 in. and the ball is released from rest. The four cords C and plate P keep the spring compressed 2 in. when no load is on the plate.

s

k C

P 30

SOLUTION Potential Energy: The datum is set at the lowest point (compressed position). 30 sin 30° = 1.25 ft above the datum and its gravitational Finally, the ball is 12 potential energy is 0.5(1.25) = 0.625 ft # lb. The initial and final elastic potential 2 + 3 2 2 2 1 1 b = 0.08681k and (k) a b = 0.01389k, respectively. energy are (k) a 2 12 2 12 Conservation of Energy: ©T1 + ©V1 = ©T2 + ©V2

laws

or

0 + 0.08681k = 0 + 0.625 + 0.01389k k = 8.57 lb>ft

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*14–76. The roller coaster car having a mass m is released from rest at point A. If the track is to be designed so that the car does not leave it at B, determine the required height h. Also, find the speed of the car when it reaches point C. Neglect friction.

A B 7.5 m 20 m C

SOLUTION Equation of Motion: Since it is required that the roller coaster car is about to leave vB 2 vB 2 = the track at B, NB = 0. Here, an = . By referring to the free-body rB 7.5 diagram of the roller coaster car shown in Fig. a, ©Fn = ma n;

m(9.81) = m ¢

vB 2 ≤ 7.5

vB 2 = 73.575 m2>s2

or

Potential Energy: With reference to the datum set in Fig. b, the gravitational potential energy of the rollercoaster car at positions A, B, and C are A Vg B A = mghA = m(9.81)h = 9.81mh, A Vg B B = mghB = m(9.81)(20) = 196.2 m, teaching

Web)

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and A Vg B C = mghC = m(9.81)(0) = 0.

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Conservation of Energy: Using the result of vB 2 and considering the motion of the car from position A to B,

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TA + VA = TB + VB

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1 1 mvA 2 + A Vg B A = mvB 2 + A Vg B B 2 2

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1 m(73.575) + 196.2m 2

this

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0 + 9.81mh =

Ans.

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h = 23.75 m

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TB + VB = TC + VC

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Also, considering the motion of the car from position B to C,

1 1 mvB 2 + A Vg B B = mvC 2 + A Vg B C 2 2 1 1 m(73.575) + 196.2m = mvC 2 + 0 2 2 vC = 21.6 m>s






Ans.

h

14–77. A 750-mm-long spring is compressed and confined by the plate P, which can slide freely along the vertical 600-mm-long rods. The 40-kg block is given a speed of v = 5 m>s when it is h = 2 m above the plate. Determine how far the plate moves downwards when the block momentarily stops after striking it. Neglect the mass of the plate.

v ⫽ 5 m/s

A

h⫽2m

SOLUTION Potential Energy: With reference to the datum set in Fig. a, the gravitational potential energy of the block at positions (1) and (2) are A Vg B 1 = mgh1 = 40(9.81)(0) = 0

P

and A Vg B 2 = mgh2 = 40(9.81) C -(2 + y) D = C -392.4(2 + y) D , respectively. The compression of the spring when the block is at positions (1) and (2) are s1 = (0.75 - 0.6) = 0.15 m and s2 = s1 + y = (0.15 + y) m. Thus, the initial and final elastic potential energy of the spring are

k ⫽ 25 kN/m

A Ve B 1 = ks21 = (25)(103)(0.152) = 281.25 J 1 2

1 2

A Ve B 2 = ks22 = (25)(103)(0.15 + y)2 1 2

laws

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1 2

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T1 + V1 = T2 + V2

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by

1 (40)(52) + (0 + 281.25) = 0 + [ - 392.4(2 + y)] + 2 1 (25)(103)(0.15 + y)2 2 12500y2 + 3357.6y - 1284.8 = 0

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1 1 mv21 + c A Vg B 1 + A Ve B 1 d = mv22 + c A Vg B 2 + A Ve B 2 d 2 2

part

the

is

This

Solving for the positive root of the above equation, any

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y = 0.2133 m = 213 mm






600 mm

14–78. The 2-lb block is given an initial velocity of 20 ft/s when it is at A. If the spring has an unstretched length of 2 ft and a stiffness of k = 100 lb>ft, determine the velocity of the block when s = 1 ft.

2 ft B

vA = 20 ft/s A s

SOLUTION

k = 100 lb/ft

Potential Energy: Datum is set along AB. The collar is 1 ft below the datum when it is at C. Thus, its gravitational potential energy at this point is -2(1) = -2.00 ft # lb. 1 (100)(2 - 2)2 = 0 and The initial and final elastic potential energy are 2 1 (100) A 222 + 12 - 2 B 2 = 2.786 ft # lb, respectively. 2

C

Conservation of Energy: TA + VA = TC + VC 2 2 1 1 a b A 202 B + 0 = a b v2C + 2.786 + ( -2.00) 2 32.2 2 32.2 vC = 19.4 ft s

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14–79. The block has a weight of 1.5 lb and slides along the smooth chute AB. It is released from rest at A, which has coordinates of A(5 ft, 0, 10 ft). Determine the speed at which it slides off at B, which has coordinates of B(0, 8 ft, 0).

z

5 ft A

SOLUTION

10 ft

Datum at B:

B 8 ft

TA + VA = TB + VB 0 + 1.5(10) =

x

1 1.5 a b (vB)2 + 0 2 32.2

vB = 25.4 ft s

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y

*14–80. Each of the two elastic rubber bands of the slingshot has an unstretched length of 200 mm. If they are pulled back to the position shown and released from rest, determine the speed of the 25-g pellet just after the rubber bands become unstretched. Neglect the mass of the rubber bands. Each rubber band has a stiffness of k = 50 N>m.

50 mm 50 mm 240 mm

SOLUTION T1 + V1 = T2 + V2 1 1 0 + (2) a b(50)[2(0.05)2 + (0.240)2 - 0.2]2 = (0.025)v2 2 2 v = 2.86 m>s

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14–81. Each of the two elastic rubber bands of the slingshot has an unstretched length of 200 mm. If they are pulled back to the position shown and released from rest, determine the maximum height the 25-g pellet will reach if it is fired vertically upward. Neglect the mass of the rubber bands and the change in elevation of the pellet while it is constrained by the rubber bands. Each rubber band has a stiffness k = 50 N>m.

50 mm 50 mm 240 mm

SOLUTION T1 + V1 = T2 + V2 1 0 + 2a b (50)[2(0.05)2 + (0.240)2 - 0.2]2 = 0 + 0.025(9.81)h 2 h = 0.416 m = 416 mm

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14–82. If the mass of the earth is Me, show that the gravitational potential energy of a body of mass m located a distance r from the center of the earth is Vg = −GMem>r. Recall that the gravitational force acting between the earth and the body is F = G(Mem>r 2), Eq. 13–1. For the calculation, locate the datum at r : q. Also, prove that F is a conservative force.

r2

r

SOLUTION r1

The work is computed by moving F from position r1 to a farther position r2. Vg = - U = -

L

F dr r2

= -G Me m

Lr1

= - G Me ma

dr r2

1 1 - b r2 r1

or

As r1 : q , let r2 = r1, F2 = F1, then

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-G Me m r

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Vg :

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To be conservative, require

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G Me m 0 b ar 0r

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F = - § Vg = -

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-G Me m

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r2

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=






Q.E.D.

14–83. A rocket of mass m is fired vertically from the surface of the earth, i.e., at r = r1 . Assuming no mass is lost as it travels upward, determine the work it must do against gravity to reach a distance r2 . The force of gravity is F = GMem>r2 (Eq. 13–1), where Me is the mass of the earth and r the distance between the rocket and the center of the earth.

r2

r

SOLUTION F = G

r1

Mem r2 r2

F1-2 =

L

F dr = GMem

2

Lr1 r

1 1 - b r1 r2

Ans.

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= GMema

dr






*14–84. The firing mechanism of a pinball machine consists of a plunger P having a mass of 0.25 kg and a spring of stiffness k = 300 N>m. When s = 0, the spring is compressed 50 mm. If the arm is pulled back such that s = 100 mm and released, determine the speed of the 0.3-kg pinball B just before the plunger strikes the stop, i.e., s = 0. Assume all surfaces of contact to be smooth. The ball moves in the horizontal plane. Neglect friction, the mass of the spring, and the rolling motion of the ball.

s

B P k


1 1 1 1 (300)(0.1 + 0.05)2 = (0.25)(n2)2 + (0.3)(n2)2 + (300)(0.05)2 2 2 2 2

n2 = 3.30 m s

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300 N/m

14–85. A 60-kg satellite travels in free flight along an elliptical orbit such that at A, where rA = 20 Mm, it has a speed vA = 40 Mm>h. What is the speed of the satellite when it reaches point B, where rB = 80 Mm? Hint: See Prob. 14–82, where Me = 5.976(1024) kg and G = 66.73(10-12) m3>(kg # s2).

B vB

rB

80 Mm

rA

SOLUTION yA = 40 Mm>h = 11 111.1 m>s Since V = -

GMe m r

T1 + V1 = T2 + V2 66.73(10) - 12(5.976)(10)23(60) 66.73(10) - 12(5.976)(10)24(60) 1 1 2 (60)(11 111.1)2 (60)v = B 2 2 20(10)6 80(10)6 vB = 9672 m>s = 34.8 Mm>h

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Ans.






20 Mm

vA A

14–86. Just for fun, two 150-lb engineering students A and B intend to jump off the bridge from rest using an elastic cord (bungee cord) having a stiffness k = 80 lb>ft. They wish to just reach the surface of the river, when A, attached to the cord, lets go of B at the instant they touch the water. Determine the proper unstretched length of the cord to do the stunt, and calculate the maximum acceleration of student A and the maximum height he reaches above the water after the rebound. From your results, comment on the feasibility of doing this stunt.

120 ft

SOLUTION T1 + V1 = T2 + V2 0 + 2(150)(120) = 0 +

1 (80)(x)2 2

x = 30 ft Unstretched length of cord.

or

120 = l + 30 l = 90 ft

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Ans.

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When A lets go of B.

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T2 + V2 = T3 + V3

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1 (80)(30)2 = 0 + (150) h 2

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T2 + V2 = T3 + V3

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Thus, h 7 120 + 90 = 210 ft

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This is not possible since the 90 ft cord would have to stretch again, i.e., hmax = 120 + 90 = 210 ft.

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1 1 (80)(30)2 = 0 + 150 h + (80)[(h - 120) - 90]2 2 2 sale

0 +

36 000 = 150 h + 40(h2 - 420 h + 44 100) h2 - 416.25 h + 43 200 = 0 Choosing the root 7 210 ft h = 219 ft + c ©Fy = may;

Ans. 800(30) - 150 = a = 483 ft>s2

150 a 32.2 Ans.

It would not be a good idea to perform the stunt since a = 15 g which is excessive and A rises 219¿ - 120¿ = 99 ft above the bridge!






A

B

14–87. z

The 20-lb collar slides along the smooth rod. If the collar is released from rest at A, determine its speed when it passes point B. The spring has an unstretched length of 3 ft.

A 6 ft k ⫽ 20 lb/ft

SOLUTION

3 ft

rOA = {-2i + 3j + 6k} ft,

rOA = 7 ft

rOB = {4i} ft,

rOB = 4 ft

O 2 ft B

Put datum at x–y plane x

TA + VA = TB + VB 0 + (20 lb)(6 ft) +

1 1 20 (20 lb>ft)(7 ft - 3 ft)2 = a b v2B + 0 + 2 2 32.2

laws

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1 (20 lb>ft)(4 ft - 3 ft)2 2 vB = 29.5 ft>s

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Ans.






4 ft y

*14–88. Two equal-length springs having a stiffness kA = 300 N>m and kB = 200 N>m are “nested” together in order to form a shock absorber. If a 2-kg block is dropped from an at-rest position 0.6 m above the top of the springs, determine their deformation when the block momentarily stops.

0.6 m

B

SOLUTION Datum at initial position: T1 + V1 = T2 + V2 0 + 0 = 0 - 2(9.81)(0.6 + x) +

1 (300 + 200)(x)2 2

250x2 - 19.62x - 11.772 = 0 Solving for the positive root, x = 0.260 m

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A

14–89. When the 6-kg box reaches point A it has a speed of vA = 2 m>s. Determine the angle u at which it leaves the smooth circular ramp and the distance s to where it falls into the cart. Neglect friction.

vA = 2 m/s

20° A

θ

B

1.2 m

SOLUTION s

At point B: +b©Fn = man;

6(9.81) cos f = 6 a

n2B b 1.2

(1)

Datum at bottom of curve: TA + VA = TB + VB 1 1 (6)(2)2 + 6(9.81)(1.2 cos 20°) = (6)(vB)2 + 6(9.81)(1.2 cos f) 2 2 13.062 = 0.5v2B + 11.772 cos f

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(2)

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vB = 2.951 m>s

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(2.951)2 b = 42.29° 1.2(9.81)

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A+cB

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u = f - 20° = 22.3°

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-1.2 cos 42.29° = 0 - 2.951(sin 42.29°)t +

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4.905t2 + 1.9857t - 0.8877 = 0

s = 0 + (2.951 cos 42.29°)(0.2687) s = 0.587 m






Ans.

14–90. The Raptor is an outside loop roller coaster in which riders are belted into seats resembling ski-lift chairs. Determine the minimum speed v0 at which the cars should coast down from the top of the hill, so that passengers can just make the loop without leaving contact with their seats. Neglect friction, the size of the car and passenger, and assume each passenger and car has a mass m.

v0 r

h

SOLUTION Datum at ground: T1 + V1 = T2 + V2 1 2 1 mv + mgh = mv21 + mg2r 2 0 2 v1 = 2v20 + 2g(h-2r) + T ©Fn = man;

mg = ma

v21 b r

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v1 = 2gr

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g(5r - 2h)

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14–91. The Raptor is an outside loop roller coaster in which riders are belted into seats resembling ski-lift chairs. If the cars travel at v0 = 4 m>s when they are at the top of the hill, determine their speed when they are at the top of the loop and the reaction of the 70-kg passenger on his seat at this instant. The car has a mass of 50 kg. Take h = 12 m, r = 5 m. Neglect friction and the size of the car and passenger.

v0 r

h

SOLUTION Datum at ground: T1 + V1 = T2 + V2 1 1 (120)(4)2 + 120(9.81)(12) = (120)(v1)2 + 120(9.81)(10) 2 2 v1 = 7.432 m>s + T ©Fn = man;

Ans. (7.432)2 b 5

70(9.81) + N = 70 a N = 86.7 N

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*14–92. The 75-kg man bungee jumps off the bridge at A with an initial downward speed of 1.5 m>s. Determine the required unstretched length of the elastic cord to which he is attached in order that he stops momentarily just above the surface of the water. The stiffness of the elastic cord is k = 80 N>m. Neglect the size of the man.

A

150 m

SOLUTION

B

Potential Energy: With reference to the datum set at the surface of the water, the gravitational potential energy of the man at positions A and B are A Vg B A = mghA =

75(9.81)(150) = 110362.5 J and A Vg B B = mghB = 75(9.81)(0) = 0. When the man

is at position A, the elastic cord is unstretched (sA = 0), whereas the elastic cord stretches sB = A 150 - l0 B m, where l0 is the unstretched length of the cord.Thus, the elastic potential energy of the elastic cord when the man is at these two positions are 1 1 1 A Ve B A = ksA 2 = 0 and A Ve B B = ksB 2 = (80)(150 - l0)2 = 40(150 - l0)2. 2 2 2 Conservation of Energy:

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1 (75)(1.52) + A 110362.5 + 0 B = 0 + C 0 + 40(150 - l0)2 D 2

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1 1 mvA 2 + B a Vg b + A Ve B A R = mvB 2 + B a Vg b + A Ve B B R 2 2 A B

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l0 = 97.5 m






Ans.

14–93. The 10-kg sphere C is released from rest when u = 0° and the tension in the spring is 100 N. Determine the speed of the sphere at the instant u = 90°. Neglect the mass of rod AB and the size of the sphere.

E

0.4 m

SOLUTION

A

u

Potential Energy: With reference to the datum set in Fig. a, the gravitational potential energy of the sphere at positions (1) and (2) are A Vg B 1 = mgh1 = 10(9.81)(0.45) =

D

44.145 J and A Vg B 2 = mgh2 = 10(9.81)(0) = 0. When the sphere is at position (1), 100 the spring stretches s1 = = 0.2 m. Thus, the unstretched length of the spring is 500

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1 1 m A v B 2 + c A Vg B 1 + A Ve B 1 d = ms A vs B 2 2 + c A Vg B 2 + A Ve B 2 d 2 s s 1 2

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1 (10)(vs)22 + (0 + 40) 2

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(vs)2 = 1.68 m>s






Ans.

C 0.15 m

Conservation of Energy: T1 + V1 = T2 + V2

B

0.3 m

l0 = 30.32 + 0.42 - 0.2 = 0.3 m, and the elastic potential energy of the spring is 1 1 A Ve B 1 = ks12 = (500)(0.22) = 10 J. When the sphere is at position (2), the spring 2 2 stretches s2 = 0.7 - 0.3 = 0.4 m, and the elastic potential energy of the spring is 1 1 A Ve B 2 = ks22 = (500)(0.42) = 40 J. 2 2

0 + (44.145 + 10) =

k ⫽ 500 N/m

14–94. v

The double-spring bumper is used to stop the 1500-lb steel billet in the rolling mill. Determine the maximum displacement of the plate A if the billet strikes the plate with a speed of 8 ft>s. Neglect the mass of the springs, rollers and the plates A and B. Take k1 = 3000 lb>ft, k2 = 45 000 lb>ft.

A

SOLUTION T1 + V1 = T2 + V2 1 1 2 1 1500 a b (8)2 + 0 = 0 + (3000)s2L + (4500)s 2 32.2 2 2 2

(1)

Fs = 3000s1 = 4500s2 ;

(2)

s1 = 1.5s2 Solving Eqs. (1) and (2) yields: s1 = 0.7722 ft

sA = s1 + s2 = 0.7722 + 0.5148 = 1.29 ft

Ans.

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s2 = 0.5148 ft






8 ft/s

B k1

k2

14–95. y

The 2-lb box has a velocity of 5 ft/s when it begins to slide down the smooth inclined surface at A. Determine the point C (x, y) where it strikes the lower incline.

5 ft/s A 15 ft

5

3

4

B

SOLUTION

30 ft

Datum at A:

1

C

TA + VA = TB + VB

2

y x

2 2 1 1 a b (5)2 + 0 = a b v2B - 2(15) 2 32.2 2 32.2

x

vB = 31.48 ft>s + B A:

s = s0 + v0t 4 x = 0 + 31.48 a b t 5

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(1)

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1 2 at 2 c

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3 1 y = 30 - 31.48a b t + (- 32.2)t2 5 2

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Equation of inclined surface:

work

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(2)

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y 1 = ; x 2

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- 16.1t2 - 31.480t + 30 = 0

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30 - 18.888t - 16.1t2 = 12.592t

t = 0.7014 s From Eqs. (1) and (2): 4 x = 31.48 a b (0.7014) = 17.66 = 17.7 ft 5 1 (17.664) = 8.832 = 8.83 ft 2

y =






Ans.

Ans.

*14–96. y

The 2-lb box has a velocity of 5 ft/s when it begins to slide down the smooth inclined surface at A. Determine its speed just before hitting the surface at C and the time to travel from A to C. The coordinates of point C are x = 17.66 ft, and y = 8.832 ft.

5 ft/s A 15 ft

5

3

4

B

SOLUTION

30 ft

Datum at A:

1

C

TA + VA = TC + VC

2

y x

2 1 2 1 a b (5)2 + 0 = a b (vC)2 - 2[15 + (30 - 8.832)] 2 32.2 2 32.2

x

vC = 48.5 ft>s

Ans. 2 3 ba 2a b = a 5 32.2 x¿

+R©Fx¿ = max¿;

or

ax¿ = 19.32 ft>s2

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laws

TA + VA = TB + VB

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1 2 2 1 a b (5)2 + 0 = a b v2B - 2(15) 2 32.2 2 32.2

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vB = 31.48 ft>s

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vB = vA + act by

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31.48 = 5 + 19.32tAB is

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tAB = 1.371 s

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+ B A:

1 3 y = 30 - 31.48 a b t + (- 32.2)t2 5 2 Equation of inclined surface: y 1 = ; x 2

y =

1 x 2

(2)

Thus 30 - 18.888t - 16.1t2 = 12.592t - 16.1t2 - 31.480t + 30 = 0 Solving for the positive root: t = 0.7014 s Total time is t = 1.371 + 0.7014 = 2.07 s © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

14–97. A pan of negligible mass is attached to two identical springs of stiffness k = 250 N>m. If a 10-kg box is dropped from a height of 0.5 m above the pan, determine the maximum vertical displacement d. Initially each spring has a tension of 50 N.

1m

k

250 N/m

1m

k

250 N/m

0.5 m

d

SOLUTION Potential Energy: With reference to the datum set in Fig. a, the gravitational potential energy of the box at positions (1) and (2) are A Vg B 1 = mgh1 = 10(9.81)(0) = 0 and

A Vg B 2 = mgh2 = 10(9.81) C - A 0.5 + d B D = - 98.1 A 0.5 + d B . Initially, the spring 50 = 0.2 m. Thus, the unstretched length of the spring 250 is l0 = 1 - 0.2 = 0.8 m and the initial elastic potential of each spring is 1 A Ve B 1 = (2) ks1 2 = 2(250 > 2)(0.22) = 10 J . When the box is at position (2), the 2

stretches s1 =

spring stretches s2 = a 2d2 + 12 - 0.8 b m. The elastic potential energy of the

in

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or

springs when the box is at this position is 2

1 2

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permitted.

Dissemination

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A Ve B 2 = (2) ks2 2 = 2(250 > 2) c 2d2 + 1 - 0.8 d = 250a d2 - 1.62d2 + 1 + 1.64 b .

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1 1 mv1 2 + B a Vg b + A Ve B 1 R = mv2 2 + B aVg b + A Ve B 2 R 2 2 1 2

integrity

of

and

work

this

assessing

is

0 + A 0 + 10 B = 0 + B - 98.1 A 0.5 + d B + 250 ¢ d2 - 1.6 2d2 + 1 + 1.64 ≤ R






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d = 1.34 m

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Solving the above equation by trial and error,

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250d2 - 98.1d - 400 2d2 + 1 + 350.95 = 0

Ans.

15–1. A 2-lb ball is thrown in the direction shown with an initial speed vA = 18 ft>s. Determine the time needed for it to reach its highest point B and the speed at which it is traveling at B. Use the principle of impulse and momentum for the solution.

B vA A

SOLUTION 1 + c2

m1vy21 + ©

L

F dt = m1vy22

2 118 sin 30°2 - 21t2 = 0 32.2 t = 0.2795 = 0.280 s + B A:

m1vx21 + ©

L

Ans.

Fx dt = m1vx22

laws

or

2 2 (v ) (18 cos 30°) + 0 = 32.2 32.2 B vB = 15.588 = 15.6 ft>s

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Ans.






18 ft/s 30

15–2. A 20-lb block slides down a 30° inclined plane with an initial velocity of 2 ft>s. Determine the velocity of the block in 3 s if the coefficient of kinetic friction between the block and the plane is mk = 0.25.

SOLUTION A +aB

m1vvœ2 + ©

t2

Lt1

Fyœdt = m1vyœ22

0 + N(3) - 20 cos 30°(3) = 0

A +bB

N = 17.32 lb

t2

m1vxœ21 + ©

Lt1

Fxœdt = m1vxœ22

20 20 (2) + 20 sin 30°(3) - 0.25(17.32)(3) = v 32.2 32.2 v = 29.4 ft>s

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Ans.






15–3. A 5-lb block is given an initial velocity of 10 ft>s up a 45° smooth slope. Determine the time for it to travel up the slope before it stops.

SOLUTION A Q+ B

t2

m(vx¿)1 + ©

Lt1

Fx dt = m(vx¿)2

5 (10) + ( -5 sin 45°)t = 0 32.2 t = 0.439 s

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Ans.






*15–4. The 180-lb iron worker is secured by a fall-arrest system consisting of a harness and lanyard AB, which is fixed to the beam. If the lanyard has a slack of 4 ft, determine the average impulsive force developed in the lanyard if he happens to fall 4 feet. Neglect his size in the calculation and assume the impulse takes place in 0.6 seconds.

A

B

SOLUTION T1 + ©U1 - 2 = T2 0 + 180(4) =

1 180 2 ( )v 2 32.2

y = 16.05 ft>s (+ T )

mv1 +

L

F dt = mv2

180 (16.05) + 180(0.6) - F(0.6) = 0 32.2 F = 329.5 lb = 330 lb

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Ans.






15–5. A man hits the 50-g golf ball such that it leaves the tee at an angle of 40° with the horizontal and strikes the ground at the same elevation a distance of 20 m away. Determine the impulse of the club C on the ball. Neglect the impulse caused by the ball’s weight while the club is striking the ball.

v2

C

SOLUTION + ) (:

sx = (s0)x + (v0)x t 20 = 0 + v cos 40°(t)

(+ c)

s = s0 + v0t +

1 2 at 2 c

0 = 0 + v sin 40°(t) -

1 (9.81)t2 2

t = 1.85 s

laws

or

v = 14.115 m>s

L

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F dt = mv2 in

mv1 + ©

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copyright

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World

permitted.

Dissemination

F dt = 10.052114.1152 learning.

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F dt = 0.706 N # s au 40°

the

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40⬚

15–6. A train consists of a 50-Mg engine and three cars, each having a mass of 30 Mg. If it takes 80 s for the train to increase its speed uniformly to 40 km>h, starting from rest, determine the force T developed at the coupling between the engine E and the first car A. The wheels of the engine provide a resultant frictional tractive force F which gives the train forward motion, whereas the car wheels roll freely. Also, determine F acting on the engine wheels.

v A

F

SOLUTION (yx)2 = 40 km>h = 11.11 m>s Entire train: + b a:

m(vx)1 + ©

L

Fx dt = m(vx)2

0 + F(80) = [50 + 3(30)] A 103 B (11.11) F = 19.4 kN

Ans.

teaching

Web)

laws

or

Three cars: in

Fx dt = m(vx)2

Wide Dissemination

0 + T(80) = 3(30) A 103 B (11.11)

T = 12.5 kN

permitted.

Ans. World

L

copyright

m(vx)1 + ©

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+ b a:






E

15–7. A

Crates A and B weigh 100 lb and 50 lb, respectively. If they start from rest, determine their speed when t = 5 s. Also, find the force exerted by crate A on crate B during the motion. The coefficient of kinetic friction between the crates and the ground is mk = 0.25.

P ⫽ 50 lb

SOLUTION Free-Body Diagram: The free-body diagram of crates A and B are shown in Figs. a and b, respectively. The frictional force acting on each crate is (Ff)A = mkNA = 0.25NA and (Ff)B = mkNB = 0.25NB. Principle of Impulse and Momentum: Referring to Fig. a, t2

(+ c )

m(v1)y + ©

Fy dt = m(v2)y

Lt1

100 100 (0) + NA(5) - 100(5) = (0) 32.2 32.2 NA = 100 lb t2

Fx dt = m(v2)x

Lt1

or

m(v1)x + ©

laws

+ ) (:

Wide

copyright

in

teaching

Web)

100 100 (0) + 50(5) - 0.25(100)(5) - FAB (5) = v 32.2 32.2

(1) not

instructors

States

World

permitted.

Dissemination

v = 40.25 - 1.61FAB

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Fy dt = m(v2)y

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50 50 (0) + NB(5) - 50(5) = (0) 32.2 32.2 work

this

NB = 50 lb

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Fx dt = m(v2)x

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50 50 (0) + FAB(5) - 0.25(50)(5) = v 32.2 32.2 v = 3.22 FAB - 40.25

(2)

Solving Eqs. (1) and (2) yields FAB = 16.67 lb = 16.7 lb






v = 13.42 ft>s = 13.4 ft>s

Ans.

B

*15–8. If the jets exert a vertical thrust of T = (500t3>2)N, where t is in seconds, determine the man’s speed when t = 3 s. The total mass of the man and the jet suit is 100 kg. Neglect the loss of mass due to the fuel consumed during the lift which begins from rest on the ground. T

SOLUTION Free-Body Diagram: The thrust T must overcome the weight of the man and jet before they move. Considering the equilibrium of the free-body diagram of the man and jet shown in Fig. a, + c ©Fy = 0;

500t3>2 - 100(9.81) = 0

t = 1.567 s

Principle of Impulse and Momentum: Only the impulse generated by thrust T after t = 1.567 s contributes to the motion. Referring to Fig. a, t2

(+ c )

m(v1)y + ©

Lt1

Fy dt = m(v2)y

3s

L1.567 s

500 t3>2dt - 100(9.81)(3 - 1.567) = 100v or

100(0) +

laws

3s

teaching

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- 1405.55 = 100v 1.567 s

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v = 11.0 m>s

Wide

a 200t5>2 b 2






15–9. Under a constant thrust of T = 40 kN, the 1.5-Mg dragster reaches its maximum speed of 125 m>s in 8 s starting from rest. Determine the average drag resistance FD during this period of time.

FD

SOLUTION Principle of Impulse and Momentum: The final speed of the dragster is v2 = 125 m>s. Referring to the free-body diagram of the dragster shown in Fig. a, t2

+ ) (;

Fx dt = m(v2)x Lt1 1500(0) + 40(103)(8) - (FD)avg (8) = 1500(125) m(v1)x + ©

(FD)avg = 16 562.5 N = 16.6 kN

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Ans.






T ⫽ 40 kN

15–10. P

The 50-kg crate is pulled by the constant force P. If the crate starts from rest and achieves a speed of 10 m>s in 5 s, determine the magnitude of P. The coefficient of kinetic friction between the crate and the ground is mk = 0.2.

30⬚

SOLUTION Impulse and Momentum Diagram: The frictional force acting on the crate is Ff = mkN = 0.2N. Principle of Impulse and Momentum: t2

(+ c )

Fy dt = m(v2)y Lt1 0 + N(5) + P(5) sin 30° - 50(9.81)(5) = 0 m(v1)y + ©

N = 490.5 - 0.5P

(1)

t2

Fx dt = m(v2)x Lt1 50(0) + P(5) cos 30° - 0.2N(5) = 50(10) m(v1)x + ©

laws

or

+ ) (:

4.3301P - N = 500 in

teaching

Web)

(2)

Dissemination

Wide

copyright

Solving Eqs. (1) and (2), yields

of

the

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N = 387.97 N

Ans.

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P = 205 N






15–11. When the 5-kg block is 6 m from the wall, it is sliding at v1 = 14 m>s. If the coefficient of kinetic friction between the block and the horizontal plane is mk = 0.3, determine the impulse of the wall on the block necessary to stop the block. Neglect the friction impulse acting on the block during the collision.

v1

SOLUTION Equation of Motion: The acceleration of the block must be obtained first before one can determine the velocity of the block before it strikes the wall. + c ©Fy = may ;

N - 5(9.81) = 5(0)

N = 49.05 N

+ ©F = ma ; : x x

-0.3(49.05) = - 5a

a = 2.943 m>s2

Kinematics: Applying the equation v2 = v20 + 2ac (s - s0) yields + B A:

v2 = 142 + 2( -2.943)(6 - 0)

v = 12.68 m>s

Principle of Linear Impulse and Momentum: Applying Eq. 15–4, we have or

t2

Web)

laws

Fxdt = m(vx)2 in

5(12.68) - I = 5(0)

Wide

copyright

Dissemination

I = 63.4 N # s

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permitted.

+ B A:

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teaching

m(vx)1 + ©

14 m/s

6m

*15–12. For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is FD = (600t2) N, where t is in seconds. If the van has a speed of 20 km>h when t = 0, determine its speed when t = 5 s.

FD

SOLUTION Principle of Impulse and Momentum: The initial speed of the van is v1 = c 20(103) c

m d h

1h d = 5.556 m>s. Referring to the free-body diagram of the van shown in Fig. a, 3600 s t2

+ ) (:

m(v1)x + ©

Lt1

Fx dt = m(v2)x 5s

2500(5.556) +

2

L0

600t dt = 2500 v2

v2 = 15.6 m>s

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Ans.






15–13. The 2.5-Mg van is traveling with a speed of 100 km>h when the brakes are applied and all four wheels lock. If the speed decreases to 40 km>h in 5 s, determine the coefficient of kinetic friction between the tires and the road.

SOLUTION Free-Body Diagram: The free-body diagram of the van is shown in Fig. a. The frictional force is Ff = mkN since all the wheels of the van are locked and will cause the van to slide. Principle of Impulse and Momentum: The initial and final speeds of the van are m 1h m 1h v1 = c100(103) d c d = 27.78 m>s and v2 = c40(103) d c d = 11.11 m>s. h 3600 s h 3600 s Referring to Fig. a, t2

(+ c )

m(v1)y + ©

Fy dt = m(v2)y Lt1 2500(0) + N(5) - 2500(9.81)(5) = 2500(0) laws

or

N = 24 525 N

Web)

t2

Fx dt = m(v2)x Lt1 2500(27.78) + [ - mk(24525)(5)] = 2500(11.1)

in

teaching

m(v1)x + ©

States

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permitted.

Dissemination

Wide

copyright

+ ) (;

Ans.

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mk = 0.340






15–14. The force acting on a projectile having a mass m as it passes horizontally through the barrel of the cannon is F = C sin (pt>t¿). Determine the projectile’s velocity when t = t¿ . If the projectile reaches the end of the barrel at this instant, determine the length s.

s

SOLUTION + b a:

m(vx)1 + ©

L

Fx dt = m(vx)2

t

0 + - Ca

L0

C sin a

pt b = mv t¿

pt t t¿ b cos a b 2 = mv p t¿ 0 v =

Ct¿ pt a 1 - cos a b b pm t¿

laws

or

When t = t¿ , teaching

Web)

2C t¿ pm

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`Ans. Wide

copyright

v2 =

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ds = v dt

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pt C t¿ b a 1 - cos a b b dt pm t¿

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L0

a

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L0

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ds =

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t¿ pt t¿ C t¿ b c t - sin a b d pm p t¿ 0 is

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s = a

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Ct¿ 2 pm

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s =

15–15. During operation the breaker hammer develops on the concrete surface a force which is indicated in the graph. To achieve this the 2-lb spike S is fired from rest into the surface at 200 ft>s. Determine the speed of the spike just after rebounding.

F (103) lb 112.5 90 67.5 45 22.5

SOLUTION

S

0

t (ms) 0

1+ T 2

mv1 +

L

F dt = mv2

-2 2 (v) (200) + 2(0.0004) - Area = 32.2 32.2 Area =

1 19021103210.42110 - 32 = 18 lb # s 2

Thus, v = 89.8 ft>s

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Ans.






0.1

0.2

0.3

0.4

*15–16. The twitch in a muscle of the arm develops a force which can be measured as a function of time as shown in the graph. If the effective contraction of the muscle lasts for a time t0 , determine the impulse developed by the muscle.

F

F ⫽ F0 ( t )e⫺t/T T

SOLUTION

t0 t0

I =

L

F dt =

t F0 a b e t>Tdt T L0

F0 t0 - 1t>T2 te dt T L0

I = - F0 c Te - t>T a

t0 t + 1b d T 0

I = - F0 c Te - t0>T a

t0 + 1b - T d T t0 bd T

Ans.

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laws

I = TF0 c 1 - e - t0>T a 1 +

or

I =






t

15–17. A hammer head H having a weight of 0.25 lb is moving vertically downward at 40 ft/s when it strikes the head of a nail of negligible mass and drives it into a block of wood. Find the impulse on the nail if it is assumed that the grip at A is loose, the handle has a negligible mass, and the hammer stays in contact with the nail while it comes to rest. Neglect the impulse caused by the weight of the hammer head during contact with the nail.

H v = 40 ft/s A

SOLUTION (+ T)

m(vy)1 + © a

L

Fy dt = m(vy)2

0.25 b (40) F dt = 0 32.2 L F dt = 0.311 lb # s

Ans.

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Dissemination

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laws

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L






15–18. The 40-kg slider block is moving to the right with a speed of 1.5 m>s when it is acted upon by the forces F1 and F2. If these loadings vary in the manner shown on the graph, determine the speed of the block at t = 6 s. Neglect friction and the mass of the pulleys and cords.

F2

F1

F (N)

SOLUTION

40

The impulses acting on the block are equal to the areas under the graph.

30

+ b a:

20

m(vx)1 + ©

L

Fx dt = m(vx)2

0

v2 = 12.0 m>s ( : )

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Ans.





F1

10

40(1.5) + 4[(30)4 + 10(6 - 4)] - [10(2) + 20(4 - 2) + 40(6 - 4)] = 40v2


F2

2

4

6

t (s)

15–19. Determine the velocity of each block 2 s after the blocks are released from rest. Neglect the mass of the pulleys and cord.

SOLUTION

A

Kinematics: The speed of block A and B can be related by using the position coordinate equation. 2sA + sB = l 2yA + yB = 0

(1)

Principle of Linear Impulse and Momentum: Applying Eq. 15–4 to block A, we have Fy dt = m A vy B 2

10 10 b (0) + 2T(2) - 10(2) = - a b (vA) 32.2 32.2

(2) teaching

Web)

-a

Lt1

laws

(+ c)

t2

or

m A vy B 1 + ©

Wide

copyright

in

Applying Eq. 15–4 to block B, we have t2

World

permitted.

Dissemination

Fy dt = m A vy B 2

is

of

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instructors

Lt1

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m A vy B 1 + ©

United

on

learning.

50 50 b(0) + T(2) - 50(2) = - a b(vB) 32.2 32.2

and

use

-a

(3)

for

work

student

the

by

(+ c)

solely

of

protected

the

(including

Solving Eqs. (1),(2) and (3) yields vA = -27.6 ft>s = 27.6 ft>s c work

this

assessing

is

work

vB = 55.2 ft>s T

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T = 7.143 lb






Ans.

10 lb

B

50 lb

*15–20. z

The particle P is acted upon by its weight of 3 lb and forces F1 and F2 , where t is in seconds. If the particle orginally has a velocity of v1 = 53i + 1j + 6k6 ft>s, determine its speed after 2 s.

F1 ⫽ {5i ⫹ 2tj ⫹ tk} lb

SOLUTION 2

mv1 + a

F2 ⫽ {t2i} lb

Fdt = mv2

L0

x

Resolving into scalar components, 2

3 3 132 + 15 + t 22dt = 1v 2 32.2 32.2 x L0 2

3 3 112 + 2tdt = 1vy2 32.2 32.2 L0 2

laws

or

3 3 162 + 1t - 32dt = 1vz2 32.2 32.2 L0 Web)

vz = - 36.933 ft>s

teaching

vy = 43.933 ft>s

in

vx = 138.96 ft>s

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Dissemination

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Ans.

copyright

v = 2(138.96)2 + (43.933)2 + ( -36.933)2 = 150 ft>s






y P

15–21. If it takes 35 s for the 50-Mg tugboat to increase its speed uniformly to 25 km>h, starting from rest, determine the force of the rope on the tugboat.The propeller provides the propulsion force F which gives the tugboat forward motion, whereas the barge moves freely. Also, determine F acting on the tugboat. The barge has a mass of 75 Mg.

F

SOLUTION 1000 25a b = 6.944 m/s 3600 System: + 2 1:

my1 + ©

L

F dt = my2

[0 + 0] + F(35) = (50 + 75)(103)(6.944) F = 24.8 kN

Ans.

laws

or

Barge: teaching

Web)

F dt = mv2

Wide

L

in

mv1 + ©

copyright

+ ) (:

instructors

States

World

permitted.

Dissemination

0 + T(35) = (75)(103)(6.944)

Ans. and

use

United

on

learning.

is

of

the

not

T = 14.881 = 14.9 kN

for

work

student

the

by

Also, using this result for T,

this

F dt = mv2

integrity

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mv1 + ©

the

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Tugboat:

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F = 24.8 kN

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0 + F1352 - 114.88121352 = 15021103216.9442






Ans.

15–22. T (lb)

If the force T exerted on the cable by the motor M is indicated by the graph, determine the speed of the 500-lb crate when t = 4 s, starting from rest. The coefficients of static and kinetic friction are ms = 0.3 and mk = 0.25, respectively.

60 30 t (s) 2

SOLUTION Free-Body Diagram: Here, force 3T must overcome the friction Ff before the crate T - 30 60 - 30 moves. For 0 … t … 2 s, or T = A 15t + 30 B lb. Considering the = t - 0 2 - 0 free-body diagram of the crate shown in Fig. a, where Ff = mk N = 0.3N, + c ©Fy = 0;

N - 500 = 0

N = 500 lb

+ : ©Fx = 0;

3(15t + 30) - 0.3(500) = 0

t = 1.333 s

or

Principle of Impulse and Momentum: Only the impulse of 3T after t = 1.333 s contributes to the motion. The impulse of T is equal to the area under the T vs. t graph. At t = 1.333 s, T = 50 lb. Thus,

Web)

laws

1 3Tdt = 3 c (50 + 60)(2 - 1.333) + 60(4 - 2) d = 470 lb # s 2 in

L

teaching

I =

Dissemination

Wide

copyright

Since the crate moves, Ff = mkN = 0.25(500) = 125 lb. Referring to Fig. a, States

World

Fx dt = m(v2)x

the

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500 500 (0) + 470 - 125(4 - 1.333) = a bv 32.2 32.2

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v = 8.80 ft>s






Ans.

T

M

15–23. The 5-kg block is moving downward at v1 = 2 m>s when it is 8 m from the sandy surface. Determine the impulse of the sand on the block necessary to stop its motion. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand.

v1

8m

SOLUTION Just before impact T1 + ©U1 - 2 = T2 1 1 (5)(2)2 + 8(5)(9.81) = (5)(v2) 2 2 v = 12.687 m>s F dt = mv2

5112.6872 -

teaching

Web)

F dt = 63.4 N # s

in

Ans. Wide

L

F dt = 0

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permitted.

Dissemination

I =

L

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laws

mv1 + ©

copyright

1+ T2






2 m/s

*15–24. The 5-kg block is falling downward at v1 = 2 m>s when it is 8 m from the sandy surface. Determine the average impulsive force acting on the block by the sand if the motion of the block is stopped in 0.9 s once the block strikes the sand. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand.

v1

8m

SOLUTION Just before impact T1 + ©U1 - 2 = T2 1 1 (5)(2)2 + 8(5)(9.81) = (5)(v2) 2 2 v = 12.69 m>s 1+ T2

mv1 + ©

L

F dt = mv2

or

5(12.69) - Fang (0.9) = 0 Fang = 70.5 N

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laws

Ans.






2 m/s

15–25. The 0.1-lb golf ball is struck by the club and then travels along the trajectory shown. Determine the average impulsive force the club imparts on the ball if the club maintains contact with the ball for 0.5 ms.

v 30 500 ft

SOLUTION Kinematics: By considering the x-motion of the golf ball, Fig. a, + b a:

sx = A s0 B + A v0 B x t 500 = 0 + v cos 30° t t =

500 v cos 30°

Subsequently, using the result of t and considering the y-motion of the golf ball, sy = A s0 B y + A v0 B y t +

1 ay t2 2 or

a+cb

in

teaching

Web)

laws

2 1 500 500 ≤ + ( -32.2) ¢ ≤ v cos 30° 2 v cos 30°

Wide

copyright

0 = 0 + v sin 30° ¢

instructors

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permitted.

Dissemination

v = 136.35 ft>s

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not

Principle of Impulse and Momentum: Here, the impulse generated by the weight of the golf ball is very small compared to that generated by the force of the impact. Hence, it can be neglected. By referring to the impulse and momentum diagram shown in Fig. b,

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Fx¿ dt = m A v2 B x¿ work

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m A v1 B x¿ + ©

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0 + Favg (0.5)(10 - 3) =

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Favg = 847 lb






Ans.

15–26. As indicated by the derivation, the principle of impulse and momentum is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which rests on the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x, determine the final speed of the block in 4 s if it has an initial speed of 5 m>s measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x¿ axis that moves at a constant velocity of 2 m>s relative to A.

L

6N

F dt = m v2

10(5) + 6(4) = 10v v = 7.40 m>s

Ans.

Observer B:

L

or

F dt = m v2

teaching

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m v1 + a

laws

+ 2 (:

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in

10(3) + 6(4) = 10v v = 5.40 m>s

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Dissemination

Ans.






B

x¿

5 m/s

Observer A: m v1 + a

x

2 m/s

SOLUTION + 2 1:

A

15–27. The winch delivers a horizontal towing force F to its cable at A which varies as shown in the graph. Determine the speed of the 70-kg bucket when t = 18 s. Originally the bucket is moving upward at v1 = 3 m>s.

A F

F (N) 600 360

v B

12

SOLUTION Principle of Linear Impulse and Momentum: For the time period 12 s … t 6 18 s, 600 - 360 F - 360 = , F = (20t + 120) N. Applying Eq. 15–4 to bucket B, we have t - 12 24 - 12 t2

Fy dt = m A vy B 2

Lt1

or

m A vy B 1 + ©

in

teaching

(20t + 120)dt d - 70(9.81)(18) = 70v2

L12 s

Web)

laws

18 s

70(3) + 2 c 360(12) +

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(+ c)

Ans.

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Dissemination

v2 = 21.8 m>s






24

t (s)

*15–28. The winch delivers a horizontal towing force F to its cable at A which varies as shown in the graph. Determine the speed of the 80-kg bucket when t = 24 s. Originally the bucket is released from rest.

A F

F (N) 600 360

v B

12

SOLUTION Principle of Linear Impulse and Momentum: The total impluse exerted on bucket B

laws

or

can be obtained by evaluating the area under the F–t graph. Thus, t2 1 I = © Fy dt = 2 c 360(12) + (360 + 600)(24 - 12) d = 20160 N # s. Applying 2 Lt1 Eq. 15–4 to the bucket B, we have t2

in

teaching

Web)

Fy dt = m A vy B 2

Lt1

Wide

copyright

m A vy B 1 + ©

permitted.

Dissemination

80(0) + 20160 - 80(9.81)(24) = 80v2

Ans.

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v2 = 16.6m>s






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(+ c)

24

t (s)

15–29. The train consists of a 30-Mg engine E, and cars A, B, and C, which have a mass of 15 Mg, 10 Mg, and 8 Mg, respectively. If the tracks provide a traction force of F = 30 kN on the engine wheels, determine the speed of the train when t = 30 s, starting from rest. Also, find the horizontal coupling force at D between the engine E and car A. Neglect rolling resistance.

C

Principle of Impulse and Momentum: By referring to the free-body diagram of the entire train shown in Fig. a, we can write m A v1 B x + ©

t2

Lt1

Fxdt = m A v2 B x

63 000(0) + 30(103)(30) = 63 000v v = 14.29 m>s

Ans.

Using this result and referring to the free-body diagram of the train’s car shown in Fig. b, t2

or

Fx dt = m A v2 B x

Web)

Lt1

laws

m A v1 B x + ©

teaching

+ b a:

FD = 15 714.29 N = 15.7 kN

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Ans.





permitted.

Dissemination

Wide

copyright

in

33000(0) + FD(30) = 33 000 A 14.29 B


A

E D

SOLUTION

+ b a:

B

F

30 kN

15–30. The crate B and cylinder A have a mass of 200 kg and 75 kg, respectively. If the system is released from rest, determine the speed of the crate and cylinder when t = 3 s. Neglect the mass of the pulleys.

B

SOLUTION Free-Body Diagram: The free-body diagrams of cylinder A and crate B are shown in Figs. b and c. vA and vB must be assumed to be directed downward so that they are consistent with the positive sense of sA and sB shown in Fig. a.

A

Principle of Impulse and Momentum: Referring to Fig. b, t2

m(v1)y + ©

(+ T )

Fy dt = m(v2)y

Lt1

75(0) + 75(9.81)(3) - T(3) = 75vA vA = 29.43 - 0.04T

(1)

From Fig. b,

Web)

laws

Fy dt = m(v2)y

teaching

Lt1

in

m(v1)y + ©

or

t2

(+ T )

Wide

copyright

200(0) + 2500(9.81)(3) - 4T(3) = 200vB

(2)

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Kinematics: Expressing the length of the cable in terms of sA and sB and referring to Fig. a, (3)

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sA + 4sB = l

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Taking the time derivative,

(4)

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and

work

vA + 4vB = 0

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and

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courses

Solving Eqs. (1), (2), and (4) yields

will

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vA = 8.409 m>s = 8.41 m>s T Ans. sale

vB = - 2.102 m>s = 2.10 m>s c T = 525.54 N






not

instructors

States

World

permitted.

Dissemination

vB = 29.43 - 0.06T

15–31. Block A weighs 10 lb and block B weighs 3 lb. If B is moving downward with a velocity 1vB21 = 3 ft>s at t = 0, determine the velocity of A when t = 1 s. Assume that the horizontal plane is smooth. Neglect the mass of the pulleys and cords.

A

SOLUTION sA + 2sB = l (vB)1

vA = -2vB + 2 1;

mv1 + © -

1+ T2

L

F dt = mv2

10 10 (vA)2 (2)(3) - T(1) = 32.2 32.2

mv1 + ©

L

F dt = mv2

in

teaching

Web)

laws

or

(vA)2 3 3 () (3) + 3(1) - 2T(1) = 32.2 32.2 2

Dissemination

Wide

copyright

- 32.2T - 10(vA)2 = 60

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permitted.

- 64.4T + 1.5(vA)2 = - 105.6

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T = 1.40 lb by

(vA22 = -10.5 ft>s = 10.5 ft>s :

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Ans.






3 ft/s

B

*15–32. Block A weighs 10 lb and block B weighs 3 lb. If B is moving downward with a velocity 1vB21 = 3 ft>s at t = 0, determine the velocity of A when t = 1 s. The coefficient of kinetic friction between the horizontal plane and block A is mA = 0.15.

A

SOLUTION sA + 2sB = l (vB)1

vA = -2vB + 2 1;

mv1 + © -

1+ T 2

L

F dt = mv2

10 10 (v ) (2)(3) - T(1) + 0.15(10) = 32.2 32.2 A 2

mv1 + ©

L

F dt = mv2

in

teaching

Web)

laws

or

3 3 (vA)2 (3) - 3(1) - 2T(1) = a b 32.2 32.2 2

Dissemination

Wide

copyright

- 32.2T - 10(vA)2 = 11.70

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permitted.

- 64.4T + 1.51vA22 = - 105.6

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T = 1.50 lb by

(vA)2 = - 6.00 ft>s = 6.00 ft>s :

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Ans.






3 ft/s

B

15–33. The log has a mass of 500 kg and rests on the ground for which the coefficients of static and kinetic friction are ms = 0.5 and mk = 0.4, respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the log when t = 5 s. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the log.

T (N) 1800 T

200 t2

3

t (s) A T

SOLUTION + ©F = 0; : x

F - 0.5(500)(9.81) = 0 F = 2452.5 N

Thus, 2T = F 2(200t 2) = 2452.5 t = 2.476 s to start log moving

L

F dt = m v2

or

m v1 + ©

laws

+ ) (:

teaching

Web)

3 in

200t 2 dt + 211800215 - 32 - 0.41500219.81215 - 2.4762 = 500v2 Wide Dissemination

L2.476

copyright

0 + 2

on

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States

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permitted.

t3 3 400( ) ` + 2247.91 = 500v2 3 2.476

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v2 = 7.65 m>s






Ans.

15–34. The 50-kg block is hoisted up the incline using the cable and motor arrangement shown. The coefficient of kinetic friction between the block and the surface is mk = 0.4. If the block is initially moving up the plane at v0 = 2 m>s, and at this instant (t = 0) the motor develops a tension in the cord of T = (300 + 120 2 t) N, where t is in seconds, determine the velocity of the block when t = 2 s.

v0

2 m/s

30

SOLUTION +a©Fx = 0; (+Q)

NB - 50(9.81)cos 30° = 0

m(vx)1 + ©

L

NB = 424.79 N

Fx dt = m(vx)2

2

A 300 + 120 2t B dt - 0.4(424.79)(2) L0 - 50(9.81)sin 30°(2) = 50v2

50(2) +

v2 = 1.92 m>s

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Dissemination

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in

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or

Ans.






15–35. The bus B has a weight of 15 000 lb and is traveling to the right at 5 ft>s. Meanwhile a 3000-lb car A is traveling at 4 ft>s to the left. If the vehicles crash head-on and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision.

vB

5 ft/s B

A

SOLUTION + ) (:

mA(vA)1 + mB(vB)1 = (mA + mB)v 3000 18 000 15 000 (5) (4) = v 32.2 32.2 32.2 v = 3.5 ft>s :

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Dissemination

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or

Ans.






vA

4 ft/s

*15–36. The 50-kg boy jumps on the 5-kg skateboard with a horizontal velocity of 5 m>s. Determine the distance s the boy reaches up the inclined plane before momentarily coming to rest. Neglect the skateboard’s rolling resistance.

s 30⬚

SOLUTION Free-Body Diagram: The free-body diagram of the boy and skateboard system is shown in Fig. a. Here, Wb,Wsb, and N are nonimpulsive forces. The pair of impulsive forces F resulting from the impact during landing cancel each other out since they are internal to the system. Conservation of Linear Momentum: Since the resultant of the impulsive force along the x axis is zero, the linear momentum of the system is conserved along the x axis. mb(vb)1 + msb(vsb)1 = (mb + msb)v

+ ) (;

50(5) + 5(0) = (50 + 5)v

laws

or

v = 4.545 m>s

teaching

Web)

Conservation of Energy: With reference to the datum set in Fig. b, the in

gravitational potential energy of the boy and skateboard at positions A and B are

learning.

is

of

the

not

instructors

States

World

= 269.775s.

permitted.

Dissemination

Wide

copyright

A Vg B A = (mb + msb)ghA = 0 and A Vg B B = (mb + msb)ghB = (50 + 5)(9.81)(s sin 30°)

the

by

and

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TA + VA = TB + VB

is

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the

(including

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student

1 1 (mb + msb)vA 2 + A Vg B A = (mb + msb)vB 2 + A Vg B B 2 2

part

the

is

This

provided

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of

and

work

this

assessing

1 (50 + 5) A 4.5452 B + 0 = 0 + 269.775s 2 any

courses

Ans.

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and

s = 2.11 m






15–37. 30 km/h

The 2.5-Mg pickup truck is towing the 1.5-Mg car using a cable as shown. If the car is initially at rest and the truck is coasting with a velocity of 30 km>h when the cable is slack, determine the common velocity of the truck and the car just after the cable becomes taut. Also, find the loss of energy.

SOLUTION Free-Body Diagram: The free-body diagram of the truck and car system is shown in Fig. a. Here, Wt,WC, Nt, and NC are nonimpulsive forces. The pair of impulsive forces F generated at the instant the cable becomes taut are internal to the system and thus cancel each other out. Conservation of Linear Momentum: Since the resultant of the impulsive force is zero, the linear momentum of the system is conserved along the x axis. The initial m 1h speed of the truck is A vt B 1 = c 30(103) d c d = 8.333 m>s. h 3600 s mt A vt B 1 + mC A vC B 1 = A mt + mC B v2 laws

or

2500(8.333) + 0 = (2500 + 1500)v2 v2 = 5.208 m>s = 5.21 m>s ;

Ans. Wide

copyright

in

teaching

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+ ) (;

instructors

States

World

permitted.

Dissemination

Kinetic Energy: The initial and final kinetic energy of the system is

United

on

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is

of

the

not

1 1 m (v ) 2 + mC(vC)12 2 t t1 2 by

and

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T1 =

the

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student

the

1 (2500)(8.3332) + 0 2 solely

of

protected

=

integrity

of

and

work

this

assessing

is

work

= 86 805.56 J

any will

1 (2500 + 1500)(5.2082) 2

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=

their

destroy

of

and

T2 = (mt + mC)v22

courses

part

the

is

This

provided

and

= 54 253.47 Thus, the loss of energy during the impact is ¢E = T1 - T2 = 86 805.56 - 54 253.47 = 32.55(103) J = 32.6 kJ






Ans.

15–38. A railroad car having a mass of 15 Mg is coasting at 1.5 m>s on a horizontal track. At the same time another car having a mass of 12 Mg is coasting at 0.75 m>s in the opposite direction. If the cars meet and couple together, determine the speed of both cars just after the coupling. Find the difference between the total kinetic energy before and after coupling has occurred, and explain qualitatively what happened to this energy.

SOLUTION + ) (:

©mv1 = ©mv2 15 000(1.5) - 12 000(0.75) = 27 000(v2) v2 = 0.5 m>s

Ans.

1 1 (15 000)(1.5)2 + (12 000)(0.75)2 = 20.25 kJ 2 2

T2 =

1 (27 000)(0.5)2 = 3.375 kJ 2 or

T1 =

teaching

Web)

laws

¢T = T1 - T2 = 20.25 - 3.375 = 16.9 kJ

Dissemination

Wide

copyright

in

Ans.

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This energy is dissipated as noise, shock, and heat during the coupling.






15–39. The car A has a weight of 4500 lb and is traveling to the right at 3 ft>s. Meanwhile a 3000-lb car B is traveling at 6 ft>s to the left. If the cars crash head-on and become entangled, determine their common velocity just after the collision. Assume that the brakes are not applied during collision.

vA

3 ft/s

A

SOLUTION + ) (:

mA (vA)1 + mB(vB)1 = (mA + mB)v 2 3000 7500 4500 (3) (6) = v 32.2 32.2 32.2 2 v 2 = -0.600 ft>s = 0.600 ft>s ;

destroy

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Dissemination

Wide

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Ans.

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vB B

6 ft/s

*15–40. The 200-g projectile is fired with a velocity of 900 m>s towards the center of the 15-kg wooden block, which rests on a rough surface. If the projectile penetrates and emerges from the block with a velocity of 300 m>s, determine the velocity of the block just after the projectile emerges. How long does the block slide on the rough surface, after the projectile emerges, before it comes to rest again? The coefficient of kinetic friction between the surface and the block is mk = 0.2.

900 m/s

Before

300 m/s

SOLUTION Free-Body Diagram: The free-body diagram of the projectile and block system is shown in Fig. a. Here, WB, WP, N, and Ff are nonimpulsive forces. The pair of impulsive forces F resulting from the impact cancel each other out since they are internal to the system.

After

Conservation of Linear Momentum: Since the resultant of the impulsive force along the x axis is zero, the linear momentum of the system is conserved along the x axis. + B A:

mP A vP B 1 + mB A vB B 1 = mP A vP B 2 + mB A vB B 2 laws

or

0.2(900) + 15(0) = 0.2(300) + 15(vB)2 teaching

Web)

Ans.

(vB)2 = 8 m>s :

instructors

States

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permitted.

Dissemination

Wide

copyright

in

Principle of Impulse and Momentum: Using the result of A vB B 2, and referring to the free-body diagram of the block shown in Fig. b, is

of

the

not

t2 United

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Fy dt = m(v2)y

and

use

Lt1

by

m A v1 B y + ©

the

A+cB

the

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15(0) + N(t) - 15(9.81)(t) = 15(0)

assessing

is

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N = 147.15 N work

this

t2 provided

integrity

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m A v1 B x + ©

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Ans.

15–41. The block has a mass of 50 kg and rests on the surface of the cart having a mass of 75 kg. If the spring which is attached to the cart and not the block is compressed 0.2 m and the system is released from rest, determine the speed of the block relative to the ground after the spring becomes undeformed. Neglect the mass of the cart’s wheels and the spring in the calculation. Also neglect friction. Take k = 300 N>m.

B C

SOLUTION T1 + V1 = T2 + V2 (0 + 0) +

1 1 1 (300)(0.2)2 = (50)(vb)2 + (75)(vc)2 2 2 2

12 = 50 v2b + 75 v2c + ) (:

©mv1 = ©mv2 0 + 0 = 50 vb - 75 vc vb = 1.5vc laws

or

vc = 0.253 m>s ; vb = 0.379 m s :

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Ans.






15–42. The block has a mass of 50 kg and rests on the surface of the cart having a mass of 75 kg. If the spring which is attached to the cart and not the block is compressed 0.2 m and the system is released from rest, determine the speed of the block with respect to the cart after the spring becomes undeformed. Neglect the mass of the wheels and the spring in the calculation. Also neglect friction. Take k = 300 N>m.

B C

SOLUTION T1 + V1 = T2 + V2 (0 + 0) +

1 1 1 (300)(0.2)2 = (50)(vb)2 + (75)(vc)2 2 2 2

12 = = 50 v2b + 75 v2c + ) (:

©mv1 = ©mv2 0 + 0 = 50 vb - 75 vc vb = 1.5vc laws

or

vc = 0.253 m>s ; in

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vb = 0.379 m>s : Dissemination

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copyright

vb = vc + vb>c

World

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vb c = 0.632 m s :

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15–43. 20 km/h

The three freight cars A, B, and C have masses of 10 Mg, 5 Mg, and 20 Mg, respectively. They are traveling along the track with the velocities shown. Car A collides with car B first, followed by car C. If the three cars couple together after collision, determine the common velocity of the cars after the two collisions have taken place.

5 km/h

A

B

SOLUTION Free-Body Diagram: The free-body diagram of the system of cars A and B when they collide is shown in Fig. a. The pair of impulsive forces F1 generated during the collision cancel each other since they are internal to the system. The free-body diagram of the coupled system composed of cars A and B and car C when they collide is shown in Fig. b. Again, the internal pair of impulsive forces F2 generated during the collision cancel each other. Conservation of Linear Momentum: When A collides with B, and then the coupled cars A and B collide with car C, the resultant impulsive force along the x axis is zero. Thus, the linear momentum of the system is conserved along the x axis. The initial speed of the cars A, B, and C are

or in

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m 1h da b = 1.389 m>s, h 3600 s

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A vB B 1 = c5(103)

m 1h da b = 5.556 m>s h 3600 s

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A vA B 1 = c20(103)

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mA(vA)1 + mB(vB)1 = (mA + mB)v2

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10000(5.556) + 5000(1.389) = (10000 + 5000)vAB

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vAB = 4.167 m>s :

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Using the result of vAB and considering the second case, (mA + mB)vAB + mC(vC)1 = (mA + mB + mC)vABC (10000 + 5000)(4.167) + [ -20000(6.944)] = (10000 + 5000 + 20000)vABC vABC = - 2.183 m>s = 2.18 m>s ;






Ans.

25 km/h C

*15–44. Two men A and B, each having a weight of 160 lb, stand on the 200-lb cart. Each runs with a speed of 3 ft>s measured relative to the cart. Determine the final speed of the cart if (a) A runs and jumps off, then B runs and jumps off the same end, and (b) both run at the same time and jump off at the same time. Neglect the mass of the wheels and assume the jumps are made horizontally.

A

SOLUTION (a) A jumps first. + ) 0 + 0 = m v - (m + m )vœ (; A A C B C 0 =

However, vA = -vCœ + 3

160 360 œ ( - vCœ + 3) vC 32.2 32.2 vCœ = 0.9231 ft>s :

And then B jumps 0 + (mC + mB) vCœ = mB vB - mCvC

However, vB = - vC + 3

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laws

or

160 200 360 ( - 0.9231) = ( -vC + 3) vC 32.2 32.2 32.2 in

vC = 2.26 ft>s :

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Ans.

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+ B 0 + 0 = (m + m )v - m v A; A B C C

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160 160 200 + b ( - vC + 3) v 32.2 32.2 32.2 C

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vC = 1.85 ft>s :






Ans.

B

15–45. The block of mass m is traveling at v1 in the direction u1 shown at the top of the smooth slope. Determine its speed v2 and its direction u2 when it reaches the bottom.

z v1 y

x

u1

h

SOLUTION There are no impulses in the v direction:

v2

mv1 sin u1 = mv2 sin u2

u2

T1 + V1 = T2 + V2 1 1 mv12 + mgh = mv22 + 0 2 2 v2 = 3v21 + 2gh v1 sin u1 or

3v21 + 2gh in

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sin u2 =

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3v21 + 2gh

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u2 = sin - 1 £






15–46. The barge B weighs 30 000 lb and supports an automobile weighing 3000 lb. If the barge is not tied to the pier P and someone drives the automobile to the other side of the barge for unloading, determine how far the barge moves away from the pier. Neglect the resistance of the water.

200 ft

P B

SOLUTION Relative Velocity: The relative velocity of the car with respect to the barge is y c/b. Thus, the velocity of the car is + B A:

vc = -vb + vc>b

(1)

Conservation of Linear Momentum: If we consider the car and the barge as a system, then the impulsive force caused by the traction of the tires is internal to the system. Therefore, it will cancel out. As the result, the linear momentum is conserved along the x axis. 0 = mc vc + mb vb 3000 30 000 b vc - a b vb 32.2 32.2

or

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0 + 0 = a

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+ B A:

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Substituting Eq. (1) into (2) yields

(3)

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11vb - vc>b = 0

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11sb - sc>b = 0

(4)

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+ B A:

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Integrating Eq. (3) becomes

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Here, s c/b = 200 ft. Then, from Eq. (4) work

sb = 18.2 ft

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11sb - 200 = 0






Ans.

15–47. 20 km/h

The 30-Mg freight car A and 15-Mg freight car B are moving towards each other with the velocities shown. Determine the maximum compression of the spring mounted on car A. Neglect rolling resistance.

SOLUTION Conservation of Linear Momentum: Referring to the free-body diagram of the freight cars A and B shown in Fig. a, notice that the linear momentum of the system is conserved along the x axis. The initial speed of freight cars A and B are m 1h m 1h (vA)1 = c20(103) d a b = 5.556 m>s and (vB)1 = c10(103) d a b h 3600 s h 3600 s = 2.778 m>s. At this instant, the spring is compressed to its maximum, and no relative motion occurs between freight cars A and B and they move with a common speed. + ) (:

mA(vA)1 + mB(vB)1 = (mA + mB)v2 30(103)(5.556) + c -15(103)(2.778) d = c 30(103) + 15(103) dv2

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v2 = 2.778 m>s :

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©T1 + ©V1 = ©T2 + ©V2

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1 1 (30) A 103 B (5.5562) + (15) A 103 B A 2.7782 B + 0 2 2

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1 1 1 c mA(vA)12 + mB(vB)12 d + (Ve)1 = (mA + mB)v22 + (Ve)2 2 2 2

any

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1 c30 A 103 B + 15 A 103 B d A 2.7782 B + 1.5 A 106 B smax2 2

smax = 0.4811 m = 481 mm






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Conservation of Energy: The initial and final elastic potential energy of the spring 1 1 1 is (Ve)1 = ks12 = 0 and (Ve)2 = ks22 = (3)(106)smax2 = 1.5(106)smax2. 2 2 2

10 km/h k ⫽ 3 MN/m

A

B

*15–48. The barge weighs 45 000 lb and supports two automobiles A and B, which weigh 4000 lb and 3000 lb, respectively. If the automobiles start from rest and drive towards each other, accelerating at aA = 4 ft>s2 and aB = 8 ft>s2 until they reach a constant speed of 6 ft/s relative to the barge, determine the speed of the barge just before the automobiles collide. How much time does this take? Originally the barge is at rest. Neglect water resistance.

30 ft A

B C

SOLUTION + B A;

vA = vC + vA>C = vC - 6

+ B A;

vB = vC + vB>C = vC + 6

+ B A;

©mv1 = ©mv2 0 = mA(vC - 6) + mB(vC + 6) + mCvC 0 = a

3000 45 000 4000 b(vC - 6) + a b (vC + 6) + a bvC 32.2 32.2 32.2

vC = 0.1154 ft>s = 0.115 ft>s

laws

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Ans.

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For A: v = v0 + act

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6 = 0 + 4tA

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tA = 1.5 s

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1 (4)(1.5)2 = 4.5 ft 2

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s = 0 + 0 +

1 2 at 2 c

for

s = s0 + v0t +

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v = v0 + ac t

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6 = 0 + 8tB tB = 0.75 s + B A;

s = s0 + v0 t + s = 0 + 0 +

1 2 at 2 c

1 (8)(0.75)2 = 2.25 ft 2

For the remaining (1.5 - 0.75) s = 0.75 s s = vt = 6(0.75) = 4.5 ft Thus, s = 30 - 4.5 - 4.5 - 2.25 = 18.75 ft t¿ =

18.75>2 s>2 = = 1.5625 v 6

t = 1.5 + 1.5625 = 3.06 s






Ans.

15–49. The man M weighs 150 lb and jumps onto the boat B which has a weight of 200 lb. If he has a horizontal component of velocity relative to the boat of 3 ft>s, just before he enters the boat, and the boat is traveling vB = 2 ft>s away from the pier when he makes the jump, determine the resulting velocity of the man and boat.

M

B

SOLUTION + ) (:

vM = vB + vM>B vM = 2 + 3 vM = 5 ft>s

+ ) (:

©m v1 = ©m v2 200 350 150 152 + 122 = 1vB22 32.2 32.2 32.2 1vB22 = 3.29 ft>s

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Ans.






15–50. The man M weighs 150 lb and jumps onto the boat B which is originally at rest. If he has a horizontal component of velocity of 3 ft>s just before he enters the boat, determine the weight of the boat if it has a velocity of 2 ft>s once the man enters it.

M

B

SOLUTION + ) (:

vM = vB + vM/B vM = 0 + 3 vM = 3 ft>s

+ ) (:

©m1v12 = ©m1v22 1WB + 1502 WB 150 132 + 102 = 122 32.2 32.2 32.2 WB = 75 lb

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Ans.






15–51. 1.5 m/s

The 20-kg package has a speed of 1.5 m>s when it is delivered to the smooth ramp. After sliding down the ramp it lands onto a 10-kg cart as shown. Determine the speed of the cart and package after the package stops sliding on the cart.

2m

SOLUTION Conservation of Energy: With reference to the datum set in Fig. a, the gravitational potential energy of the package at positions (1) and (2) are A Vg B 1 = mgh1 = 20(9.81)(2) = 392.4 J and A Vg B 2 = mgh2 = 0. T1 + V1 = T2 + V2 1 1 m(vP)12 + (Vg)1 = m(vP)22 + (Vg)2 2 2

laws

or

1 1 (20)(1.52) + 392.4 = (20)(vP)22 + 0 2 2

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(vP)2 2 = 6.441 m>s ;

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mP(vP)2 + mC(vC)1 = (mP + mC)v

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+ ) (;

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20(6.441) + 0 = (20 + 10)v

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v = 4.29 m>s ;






Ans.

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Dissemination

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copyright

Conservation of Linear Momentum: Referring to the free-body diagram of the package and cart system shown in Fig. b, the linear momentum is conserved along the x axis since no impulsive force acts along it. The package stops sliding on the cart when they move with a common speed. At this instant,

*15–52. The free-rolling ramp has a mass of 40 kg. A 10-kg crate is released from rest at A and slides down 3.5 m to point B. If the surface of the ramp is smooth, determine the ramp’s speed when the crate reaches B. Also, what is the velocity of the crate?

3.5 m

A

30

SOLUTION Conservation of Energy: The datum is set at lowest point B. When the crate is at point A, it is 3.5 sin 30° = 1.75 m above the datum. Its gravitational potential energy is 1019.81211.752 = 171.675 N # m. Applying Eq. 14–21, we have T1 + V1 = T2 + V2 0 + 171.675 =

1 1 1102v2C + 1402v2R 2 2

171.675 = 5 v2C + 20 v2R

(1)

laws

or

Relative Velocity: The velocity of the crate is given by

in

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vC = vR + vC>R

Dissemination

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copyright

= -vRi + 1vC>R cos 30°i - vC>R sin 30°j2

permitted.

(2) learning.

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= 10.8660 vC>R - vR2i - 0.5 vC>Rj

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The magnitude of vC is

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vC = 2(0.8660 vC>R - vR22 + 1-0.5 vC>R22

(3)

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= 2v2C>R + v2R - 1.732 vR vC>R

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Conservation of Linear Momentum: If we consider the crate and the ramp as a system, from the FBD, one realizes that the normal reaction NC (impulsive force) is internal to the system and will cancel each other. As the result, the linear momentum is conserved along the x axis. sale

will

their

0 = mC 1vC2x + mR vR + ) (:

0 = 1010.8660 vC>R - vR2 + 401-vR2 0 = 8.660 vC>R - 50 vR

(4)

Solving Eqs. (1), (3), and (4) yields vR = 1.101 m>s = 1.10 m>s vC = 5.43 m>s

Ans.

vC>R = 6.356 m>s From Eq. (2) vC = 30.866016.3562 - 1.1014i - 0.516.3562j = 54.403i - 3.178j6 m>s Thus, the directional angle f of vC is f = tan - 1






3.178 = 35.8° 4.403

cf

Ans.

B

15–53. 20 ft

The 80-lb boy and 60-lb girl walk towards each other with a constant speed on the 300-lb cart. If their velocities, measured relative to the cart, are 3 ft>s to the right and 2 ft>s to the left, respectively, determine the velocities of the boy and girl during the motion. Also, find the distance the cart has traveled at the instant the boy and girl meet.

A

SOLUTION Conservation of Linear Momentum: From the free-body diagram of the boy, girl, and cart shown in Fig. a, the pairs of impulsive forces F1 and F2 generated during the walk cancel each other since they are internal to the system. Thus, the resultant of the impulsive forces along the x axis is zero, and the linear momentum of the system is conserved along the x axis. + ) (:

©mv1 = ©mv2 0 + 0 + 0 =

60 300 80 v (v ) v 32.2 b 32.2 g 32.2 c (1)

80vb - 60vg - 300vc = 0

laws

or

Kinematics: Applying the relative velocity equation and considering the motion of the boy,

vb = - vc + 3

(2)

permitted.

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vb = vc + vb>c

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vg = vc + vg>c

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-vg = - vc - 2

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vg = vc + 2

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Solving Eqs. (1), (2), and (3), yields

Ans.

and

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vb = 2.727 ft>s = 2.73 ft>s :

Ans.

vc = 0.2727 ft>s ;

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vg = 2.273 ft>s = 2.27 ft>s ;

The velocity of the girl relative to the boy can be determined from vg = vb + vg>b + ) (:

- 2.273 = 2.727 + vg>b vg>b = - 5 ft>s = 5 ft>s ;

Here, sg>b = 20 ft and vg>b = 5 ft>s is constant. Thus, + ) (;

sg>b = (sg>b)0 + vg>b t 20 = 0 + 5t t = 4s

Thus, the distance the cart travels is given by + ) (;

sc = (sc)0 + vc t = 0 + 0.2727(4)






= 1.09 ft

Ans.

A

15–54. 20 ft

The 80-lb boy and 60-lb girl walk towards each other with constant speed on the 300-lb cart. If their velocities measured relative to the cart are 3 ft>s to the right and 2 ft>s to the left, respectively, determine the velocity of the cart while they are walking.

A

SOLUTION Conservation of Linear Momentum: From the free-body diagram of the body, girl, and cart shown in Fig. a, the pairs of impulsive forces F1 and F2 generated during the walk cancel each other since they are internal to the system. Thus, the resultant of the impulsive forces along the x axis is zero, and the linear momentum of the system is conserved along the x axis. + ) (:

©mv1 = ©mv2 0 + 0 + 0 =

60 300 80 v (v ) v 32.2 b 32.2 g 32.2 c (1)

80vb - 60vg - 300vc = 0

laws

or

Kinematics: Applying the relative velocity equation and considering the motion of the boy,

in

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vb = vc + vb>c

permitted.

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vb = - vc + 3

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vg = vc + vg>c

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-vg = - vc - 2

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vg = vc + 2

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vb = 2.727 ft>s :

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this

Solving Eqs. (1), (2), and (3), yields

vc = 0.2727 ft>s = 0.273 ft>s ;






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vg = 2.273 ft>s ;

Ans.

A

15–55. A tugboat T having a mass of 19 Mg is tied to a barge B having a mass of 75 Mg. If the rope is “elastic” such that it has a stiffness k = 600 kN>m, determine the maximum stretch in the rope during the initial towing. Originally both the tugboat and barge are moving in the same direction with speeds 1vT21 = 15 km>h and 1vB21 = 10 km>h, respectively. Neglect the resistance of the water.

(vB)1 B

SOLUTION (vT)1 = 15 km>h = 4.167 m>s (vB)1 = 10 km>h = 2.778 m>s When the rope is stretched to its maximum, both the tug and barge have a common velocity. Hence, + ) (:

©mv1 = ©mv2 19 000(4.167) + 75 000(2.778) = (19 000 + 75 000)v2 v2 = 3.059 m>s

use

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Dissemination

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1 (19 000 + 75 000)(3.059)2 = 439.661 kJ 2

teaching

T2 =

in

1 1 (19 000)(4.167)2 + (75 000)(2.778)2 = 454.282 kJ 2 2

copyright

T1 =

Web)

laws

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T1 + V1 = T2 + V2

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by

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Hence,

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the

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1 (600)(103)x 2 2

454.282(103) + 0 = 439.661(103) +

assessing

x = 0.221 m

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(vT)1 T

*15–56. Two boxes A and B, each having a weight of 160 lb, sit on the 500-lb conveyor which is free to roll on the ground. If the belt starts from rest and begins to run with a speed of 3 ft>s, determine the final speed of the conveyor if (a) the boxes are not stacked and A falls off then B falls off, and (b) A is stacked on top of B and both fall off together.

A

B

SOLUTION a) Let vb be the velocity of A and B. + b a:

©mv1 = ©mv2 0 = a

+ b a:

320 500 b (vb) - a b(vc) 32.2 32.2

vb = vc + vb>c vb = -vc + 3

Thus, vb = 1.83 ft>s :

vc = 1.17 ft>s ;

in

teaching

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laws

or

When a box falls off, it exerts no impulse on the conveyor, and so does not alter the momentum of the conveyor. Thus,

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a) vc = 1.17 ft>s ; b) vc = 1.17 ft>s ;

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Ans.






15–57. The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the 10-g bullet is traveling at 300 m>s when it becomes embedded in the 10-kg block, determine the distance the block will slide up along the plane before momentarily stopping.

300 m/s

A

30

SOLUTION Conservation of Linear Momentum: If we consider the block and the bullet as a system, then from the FBD, the impulsive force F caused by the impact is internal to the system. Therefore, it will cancel out. Also, the weight of the bullet and the block are nonimpulsive forces. As the result, linear momentum is conserved along the x œ axis. mb(vb)x¿ = (mb + mB) vx œ 0.01(300 cos 30°) = (0.01 + 10) v v = 0.2595 m>s

Wide

copyright

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Conservation of Energy: The datum is set at the blocks initial position. When the block and the embedded bullet is at their highest point they are h above the datum. Their gravitational potential energy is (10 + 0.01)(9.81)h = 98.1981h. Applying Eq. 14–21, we have

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T1 + V1 = T2 + V2

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1 (10 + 0.01) A 0.25952 B = 0 + 98.1981h 2 by

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h = 0.003433 m = 3.43 mm

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d = 3.43 > sin 30° = 6.87 mm






Ans.

15–58. A ball having a mass of 200 g is released from rest at a height of 400 mm above a very large fixed metal surface. If the ball rebounds to a height of 325 mm above the surface, determine the coefficient of restitution between the ball and the surface.

SOLUTION Before impact T1 + V1 = T2 + V2 0 + 0.2(9.81)(0.4) =

1 (0.2)v21 + 0 2

v1 = 2.801 m>s After the impact

or

1 (0.2)v22 = 0 + 0.2(9.81)(0.325) 2

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v2 = 2.525 m>s

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(vA)2 - (vB)2 (vB)1 - (vA)1 0 - ( - 2.525) = 2.801 - 0

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15–59. The 5-Mg truck and 2-Mg car are traveling with the freerolling velocities shown just before they collide. After the collision, the car moves with a velocity of 15 km>h to the right relative to the truck. Determine the coefficient of restitution between the truck and car and the loss of energy due to the collision.

30 km/h 10 km/h

SOLUTION Conservation of Linear Momentum: The linear momentum of the system is conserved along the x axis (line of impact). 1h m da b = 8.333 m>s h 3600 s

The initial speeds of the truck and car are (vt)1 = c30 A 103 B and (vc)1 = c10 A 103 B

m 1h da b = 2.778 m>s. h 3600 s

By referring to Fig. a, mt A vt B 1 + mc A vc B 1 = mt A vt B 2 + mc A vc B 2 or

+ b a:

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5000(8.333) + 2000(2.778) = 5000 A vt B 2 + 2000 A vc B 2

(1)

1h m da b = 4.167 m>s : . h 3600 s of

the

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(vc)2 = (vt)2 + 4.167

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(vc)2 = (vt)2 + (vc>t)2

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(vc)2 - (vt)2 = 4.167






(vc)2 - (vt)2 8.333 - 2.778

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5 A vt B 2 + 2 A vc B 2 = 47.22

15–59. continued Substituting Eq. (2) into Eq. (3), e =

4.167 = 0.75 8.333 - 2.778

Ans.

Solving Eqs. (1) and (2) yields (vt)2 = 5.556 m>s (vc)2 = 9.722 m>s Kinetic Energy: The kinetic energy of the system just before and just after the collision are T1 = =

1 1 mt(vt)1 2 + mc(vc)1 2 2 2 1 1 (5000)(8.3332) + (2000)(2.7782) 2 2

= 181.33 A 103 B J

or

1 1 m (v ) 2 + mc(vc)2 2 2 t t2 2 laws

T2 =

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1 1 (5000)(5.5562) + (2000)(9.7222) 2 2

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= 171.68 A 103 B J

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= 9.65 kJ






*15–60. Disk A has a mass of 2 kg and is sliding forward on the smooth surface with a velocity 1vA21 = 5 m/s when it strikes the 4-kg disk B, which is sliding towards A at 1vB21 = 2 m/s, with direct central impact. If the coefficient of restitution between the disks is e = 0.4, compute the velocities of A and B just after collision.

SOLUTION Conservation of Momentum : mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2 + B A:

2(5) + 4(- 2) = 2(vA)2 + 4(vB)2

(1)

Coefficient of Restitution : (vB)2 - (vA)2 (vA)1 - (vB)1

0.4 =

(vB)2 - (vA)2 5 - ( - 2)

(2) laws

or

+ B A:

e =

(vB)2 = 1.27m s :

Ans.

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(vA)2 = - 1.53 m s = 1.53 m s ;

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(vA)1 = 5 m/s

(vB)1 = 2 m/s

A

B

15–61. Block A has a mass of 3 kg and is sliding on a rough horizontal surface with a velocity 1vA21 = 2 m>s when it makes a direct collision with block B, which has a mass of 2 kg and is originally at rest. If the collision is perfectly elastic 1e = 12, determine the velocity of each block just after collision and the distance between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is mk = 0.3.

(vA)1 A

SOLUTION + ) (:

a mv1 = a mv2 3122 + 0 = 31vA22 + 21vB22

+ ) (:

e = 1 =

1vB22 - 1vA222 1vA21 - 1vB21 1vB22 - 1vA22 2 - 0

Ans.

1vB22 = 2.40 m>s :

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1vA22 = 0.400 m>s :

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T1 + a U1 - 2 = T2

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T1 + a U1 - 2 = T2

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Block B:

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1 12212.4022 - 219.81210.32dB = 0 2 dB = 0.9786 m d = dB - dA = 0.951 m






Ans.

B

15–62. If two disks A and B have the same mass and are subjected to direct central impact such that the collision is perfectly elastic 1e = 12, prove that the kinetic energy before collision equals the kinetic energy after collision. The surface upon which they slide is smooth.

SOLUTION + B A:

©m v1 = ©m v2 mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2 mA C (vA)1 - (vA)2 D = mB C (vB)2 - (vB)1 D

+ B A:

e =

(1)

(vB)2 - (vA)2 = 1 (vA)1 - (vB)1

(vB)2 - (vA)2 = (vA)1 - (vB)1

(2)

Combining Eqs. (1) and (2):

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laws

or

mA C (vA)1 - (vA)2 D C (vA)1 + (vA)2 D = mB C (vB)2 - (vB)1 D C (vB)2 + (vB)1 D

Dissemination

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1 Expand and multiply by : 2 States

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1 1 1 1 m (v )21 + mB (vB)21 = mA(vA)22 + mB (vB)22 2 A A 2 2 2

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Q.E.D.






15–63. Each ball has a mass m and the coefficient of restitution between the balls is e. If they are moving towards one another with a velocity v, determine their speeds after collision. Also, determine their common velocity when they reach the state of maximum deformation. Neglect the size of each ball.

SOLUTION + B A:

©m v1 = ©m v2 mv - mv = mvA + mvB vA = - vB

+ B A:

e =

(vB)2 - (vA)2 vB - vA = (vA)1 - (vB)1 v - (-v)

Ans.

vA = - ve = ve ;

Ans. laws

vB = ve :

or

2ve = 2vB

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©m v1 = ©m v2

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At maximum deformation vA = vB = v¿ .

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mv - mv = (2m) v¿

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Ans.

v

v

A

B

*15–64. The three balls each have a mass m. If A has a speed v just before a direct collision with B, determine the speed of C after collision. The coefficient of restitution between each pair of balls e. Neglect the size of each ball.

v B A

SOLUTION Conservation of Momentum: When ball A strikes ball B, we have mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2 + B A:

my + 0 = m(vA)2 + m(vB)2

(1)

Coefficient of Restitution: (vB)2 - (vA)2 (vA)1 - (vB)1

e =

(vB)2 - (vA)2 v - 0

(2) laws

or

+ B A:

e =

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Solving Eqs. (1) and (2) yields y(1 + e) 2

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(yB)2 =

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y(1 - e) 2

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Conservation of Momentum:When ball B strikes ball C, we have

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mB (vB)2 + mC (vC)1 = mB (vB)3 + mC (vC)2

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v(1 + e) d + 0 = m(vB)3 + m(vC)2 2

(3)

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(vC)2 - (vB)3 v(1 + e) - 0 2

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Coefficient of Restitution:

(4)







(vC)2 =

v(1 + e)2 4

(vB)3 =

v(1 - e2) 4

Ans.

C

15–65. A 1-lb ball A is traveling horizontally at 20 ft>s when it strikes a 10-lb block B that is at rest. If the coefficient of restitution between A and B is e = 0.6, and the coefficient of kinetic friction between the plane and the block is mk = 0.4, determine the time for the block B to stop sliding.

SOLUTION + b a:

©m1 v1 = ©m2 v2 a

1 10 1 b (20) + 0 = a b(vA)2 + a b(vB)2 32.2 32.2 32.2

(vA)2 + 10(vB)2 = 20 + b a:

e =

(vB)2 - (vA)2 (vA)1 - (vB)1

0.6 =

(vB)2 - (vA)2 20 - 0

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laws

or

(vB)2 - (vA)2 = 12

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(vB)2 = 2.909 ft>s :

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(vA)2 = -9.091 ft>s = 9.091 ft>s ;

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10 b(2.909) - 4t = 0 32.2

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t = 0.226 s






15–66. If the girl throws the ball with a horizontal velocity of 8 ft>s, determine the distance d so that the ball bounces once on the smooth surface and then lands in the cup at C. Take e = 0.8.

A

vA

3 ft C B d

SOLUTION (+ T )

v2 = v20 + 2ac1s - s02 (v122y = 0 + 2132.22132 1v12y = 13.90 T

(+ T)

s = s0 + v0 t + 3 = 0 + 0 +

1 2 ac t 2

1 132.221tAB22 2

or

tAB = 0.43167 s laws

1v22y

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1v22y

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11.1197 = -11.1197 + 32.21tBC2

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Total time is tAC = 1.1224 s

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tBC = 0.6907 s

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d = vA1tAC2

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Since the x component of momentum is conserved

d = 811.12242 d = 8.98 ft






Ans.

15–67. The three balls each weigh 0.5 lb and have a coefficient of restitution of e = 0.85. If ball A is released from rest and strikes ball B and then ball B strikes ball C, determine the velocity of each ball after the second collision has occurred. The balls slide without friction.

A r

3 ft B

SOLUTION Ball A: Datum at lowest point. T1 + V1 = T2 + V2 0 + 10.52132 =

1 0.5 1 21vA221 + 0 2 32.2

1vA21 = 13.90 ft>s Balls A and B: + B A:

laws in

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0.5 0.5 0.5 2113.902 + 0 = 1 21vA22 + 1 21vB22 32.2 32.2 32.2

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1vB22 - 1vA222

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1vB22 - 1vA22

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13.90 - 0

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1vA22 = 1.04 ft>s

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1vB22 = 12.86 ft>s

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Balls B and C:

will

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1 + B A:

sale

©mv 2 = ©mv 3

0.5 0.5 0.5 2112.862 + 0 = 1 21v 2 + 1 21v 2 32.2 32.2 B 3 32.2 C 3

e = 0.85 =

1vC23 - 1vB23 1vB22 - 1vC22 1vC23 - 1vB23 12.86 - 0

Solving:






1vB23 = 0.964 ft>s

Ans.

1vC23 = 11.9 ft>s

Ans.

C

*15–68. The girl throws the ball with a horizontal velocity of v1 = 8 ft>s. If the coefficient of restitution between the ball and the ground is e = 0.8, determine (a) the velocity of the ball just after it rebounds from the ground and (b) the maximum height to which the ball rises after the first bounce.

v1 3 ft

h

SOLUTION Kinematics: By considering the vertical motion of the falling ball, we have ( + T)

(v1)2y = (v0)2y + 2ac C sy - (s0)y D (v1)2y = 02 + 2(32.2)(3 - 0) (v1)y = 13.90 ft>s

Coefficient of Restitution (y):

(v1)y - A vg B 1 0 - (v2)y

or

0.8 =

-13.90 - 0

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(+ c)

A vg B 2 - (v2)y

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Conservation of “x” Momentum: The momentum is conserved along the x axis.

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(v2)y = 11.12 ft>s

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(vx)2 = 8 ft>s :

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m(vx)1 = m(vx)2 ;

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The magnitude and the direction of the rebounding velocity for the ball is Ans.

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11.12 b = 54.3° 8

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v2 = 2(vx)22 + A vy B 22 = 282 + 11.122 = 13.7 ft>s u = tan - 1 a

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(v)2y = (v2)2y + 2ac C sy - (s2)y D

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Kinematics: By considering the vertical motion of tfie ball after it rebounds from the ground, we have (+ c)

0 = 11.122 + 2( - 32.2)(h - 0) h = 1.92 ft






8 ft/s

Ans.

15–69. A 300-g ball is kicked with a velocity of vA = 25 m>s at point A as shown. If the coefficient of restitution between the ball and the field is e = 0.4, determine the magnitude and direction u of the velocity of the rebounding ball at B.

vA

v¿B

30

u B

A

SOLUTION Kinematics: The parabolic trajectory of the football is shown in Fig. a. Due to the symmetrical properties of the trajectory, vB = vA = 25 m>s and f = 30°. Conservation of Linear Momentum: Since no impulsive force acts on the football along the x axis, the linear momentum of the football is conserved along the x axis. + b a;

m A vB B x = m A v Bœ B x 0.3 A 25 cos 30° B = 0.3 Av Bœ B x

A v Bœ B x = 21.65 m>s ; laws

or

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0.4 =

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A vB B y - 0 - A v Bœ B y

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A v Bœ B y = 5 m>s c

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v Bœ = 2 A v Bœ B x + A vBœ B y = 221.652 + 52 = 22.2 m>s






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A v Bœ B y

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u = tan - 1 C

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Ans.

25 m>s

15–70. Two smooth spheres A and B each have a mass m. If A is given a velocity of v0, while sphere B is at rest, determine the velocity of B just after it strikes the wall. The coefficient of restitution for any collision is e.

v0

A

SOLUTION Impact: The first impact occurs when sphere A strikes sphere B. When this occurs, the linear momentum of the system is conserved along the x axis (line of impact). Referring to Fig. a, + ) (:

mAvA + mBvB = mA(vA)1 + mB(vB)1 mv0 + 0 = m(vA)1 + m(vB)1 (vA)1 + (vB)1 = v0 (vB)1 - (vA)1 vA - vB

e =

(vB)1 - (vA)1 v0 - 0

or

e =

laws

+ ) (:

(1)

(vB)1 - (vA)1 = ev0 in

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(2)

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1 - e b v0 : 2

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1 + e b v0 : 2

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(vB)1 = a

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The second impact occurs when sphere B strikes the wall, Fig. b. Since the wall does not move during the impact, the coefficient of restitution can be written as work

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the

(including

0 - C - (vB)2 D (vB)1 - 0

this

assessing

e =

is

+ ) (:

integrity

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the

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will

e(1 + e) v0 2

sale

(vB)2 =

their

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1 + e d v0 - 0 2

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0 + (vB)2 and

e =

Ans.

B

15–71. It was observed that a tennis ball when served horizontally 7.5 ft above the ground strikes the smooth ground at B 20 ft away. Determine the initial velocity vA of the ball and the velocity vB (and u) of the ball just after it strikes the court at B. Take e = 0.7.

vA

A

7.5 ft

vB u 20 ft

SOLUTION + ) (:

s = s0 + v0t 20 = 0 + vA t

(+ T)

s = s0 + v0 t +

1 2 a t 2 c

7.5 = 0 + 0 +

1 (32.2)t 2 2

t = 0.682524 vA = 29.303 = 29.3 ft>s

or

Ans.

teaching

Web)

laws

vB x 1 = 29.303 ft>s

Wide

copyright

in

(+ T) v = v0 + ac t

the

not

mv1 = mv 2 use

United

on

learning.

is

of

+ ) (:

instructors

States

World

permitted.

Dissemination

vBy 1 = 0 + 32.2(0.68252) = 21.977 ft>s

work

student

the

by

and

vB2x = vB 1x = 29.303 ft>s :

the

(including

for

vBy 2

work

solely

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protected

vBy 1

21.977

vBy 2 = 15.384 ft>s c

integrity

of

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work

,

provided

vBy 2

part

the

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This

0.7 =

this

assessing

is

e =

Ans.






will

15.384 = 27.7° 29.303

au

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u = tan - 1

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any

courses

vB 2 = 2(29.303)2 + (15.384)2 = 33.1 ft>s

Ans.

B

*15–72. The tennis ball is struck with a horizontal velocity vA , strikes the smooth ground at B, and bounces upward at u = 30°. Determine the initial velocity vA , the final velocity vB , and the coefficient of restitution between the ball and the ground.

vA

A

7.5 ft

vB u 20 ft

SOLUTION (+ T)

v2 = v20 + 2 ac (s - s0) (vBy)21 = 0 + 2(32.2)(7.5 - 0) vBy1 = 21.9773 m>s

(+ T)

v = v0 + ac t 21.9773 = 0 + 32.2 t t = 0.68252 s

( + T)

s = s0 + v0 t

Web)

laws

or

20 = 0 + vA (0.68252) in

Ans. Wide

copyright

mv1 = mv 2 States

World

permitted.

Dissemination

+ ) (:

teaching

vA = 29.303 = 29.3 ft>s

on

learning.

is

of

the

not

instructors

vB x 2 = vB x 1 = vA = 29.303

Ans.

the

by

and

use

United

vB2 = 29.303>cos 30° = 33.8 ft>s

the

(including

for

work

student

vBy 2 = 29.303 tan 30° = 16.918 ft>s

work

solely

of

protected

16.918 = 0.770 21.9773

this

=

assessing

vBy 1

is

vBy 2

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e =






Ans.

B

15–73. The 1 lb ball is dropped from rest and falls a distance of 4 ft before striking the smooth plane at A. If e = 0.8, determine the distance d to where it again strikes the plane at B.

4 ft

A

d

SOLUTION T1 + V1 = T2 + V2

B 3

5 4

0 + 0 =

1 1m21vA221 - m132.22142 2

1vA21 = 22132.22142 = 16.05 ft>s R+ 1vA22x =

3 116.052 = 9.63 ft>s 5

4 Q+ 1vA22y = 0.8 a b116.052 = 10.27 ft>s 5

Web)

laws

or

1vA22 = 2(9.6322 + 110.2722 = 14.08 ft>s in

teaching

10.27 b = 46.85° 9.63

Dissemination

Wide

copyright

u = tan - 1 a

on

learning.

is

of

the

not

instructors

States

World

permitted.

3 f = 46.85° - tan - 1 a b = 9.977° 4

work

student

the

by

and

use

United

+ ) s = s + v t (: 0 0

assessing

is

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protected

the

(including

for

4 da b = 0 + 14.08 cos 9.977° 1t2 5

provided

integrity

of

and

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this

1 2 at 2 c

part

the

is

This

( + T) s = s0 + v0 t +

t = 0.798 s d = 13.8 ft






sale

will

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destroy

of

and

any

courses

1 3 da b = 0 - 14.08 sin 9.977° 1t2 + 132.22t2 5 2

Ans.

15–74. The 1 lb ball is dropped from rest and falls a distance of 4 ft before striking the smooth plane at A. If it rebounds and in t = 0.5 s again strikes the plane at B, determine the coefficient of restitution e between the ball and the plane. Also, what is the distance d?

4 ft

A

d


B 3

5 4

0 + 0 =

1 1m21vA221 - m132.22142 2

1vA21 = 22132.22142 = 16.05 ft>s 3 116.052 = 9.63 ft>s 5

+R

1vA22x¿ =

Q+

4 1vA22y¿ = ea b 116.052 = 12.84e ft>s 5 or

s = s0 + v0 t

teaching

Web)

laws

+ ) (:

permitted.

Dissemination

Wide

copyright

in

4 1d2 = 0 + vA2x10.52 5

the

not

instructors

States

World

1 2 a t 2 c

is and

use

United

on

learning.

s = s0 + v0 t +

of

(+ T)

the

(including

for

work

student

the

by

3 1 1d2 = 0 - vA2y 10.52 + 132.2210.522 5 2 3 4 4 0.5 c9.63 a b + 12.84e a b d = d 5 5 5

(+ c)

4 3 3 0.5 c -9.63 a b + 12.84e a b d = 4.025 - d 5 5 5

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e = 0.502

Ans.

d = 7.23 ft

Ans.

15–75. The 1-kg ball is dropped from rest at point A, 2 m above the smooth plane. If the coefficient of restitution between the ball and the plane is e = 0.6, determine the distance d where the ball again strikes the plane.

A

2m B

SOLUTION

30⬚

Conservation of Energy: By considering the ball’s fall from position (1) to position (2) as shown in Fig. a,

d

TA + VA = TB + VB 1 1 2 2 m v + (Vg)A = mBvB + (Vg)B 2 A A 2 1 2 0 + 1(9.81)(2) = (1)vB + 0 2 vB = 6.264 m>s T

laws

or

Conservation of Linear Momentum: Since no impulsive force acts on the ball along the inclined plane (x¿ axis) during the impact, linear momentum of the ball is conserved along the x¿ axis. Referring to Fig. b, in

teaching

Web)

mB(vB)x¿ = mB(v¿B)x¿ Dissemination

Wide

copyright

1(6.264) sin 30° = 1(v¿B) cos u v¿B cos u = 3.1321

the

by

and

use

United

on

learning.

Coefficient of Restitution: Since the inclined plane does not move during the impact,

is

of

the

not

instructors

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World

permitted.

(1)

the

(including

for

work

student

0 - (v¿B)y¿ (v¿B)y¿ - 0

assessing

is

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protected

e =

provided

integrity

of

and

work

this

0 - v¿B sin u - 6.264 cos 30° - 0

part

the

is

This

0.6 =

(2)

their

destroy

of

and

any

courses

v¿B sin u = 3.2550

u = 46.10°

sale

will


v¿B = 4.517 m>s

Kinematics: By considering the x and y motion of the ball after the impact, Fig. c, + ) (:

sx = (s0)x + (v¿B)x t d cos 30° = 0 + 4.517 cos 16.10°t t = 0.1995d

(+ c )

sy = (s0)y + (v¿B)y t +

(3)

1 2 at 2 y

- d sin 30° = 0 + 4.517 sin 16.10°t + 4.905t2 - 1.2528t - 0.5d = 0

1 ( -9.81)t2 2 (4)

Solving Eqs. (3) and (4) yields d = 3.84 m t = 0.7663 s © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

C

*15–76. A ball of mass m is dropped vertically from a height h0 above the ground. If it rebounds to a height of h1, determine the coefficient of restitution between the ball and the ground.

h0 h1

SOLUTION Conservation of Energy: First, consider the ball’s fall from position A to position B. Referring to Fig. a, TA + VA = TB + VB 1 1 2 2 mvA + (Vg)A = mvB + (Vg)B 2 2 1 2 0 + mg(h0) = m(vB)1 + 0 2 Subsequently, the ball’s return from position B to position C will be considered. TB + VB = TC + VC

laws

or

1 1 2 2 mvB + (Vg)B = mvC + (Vg)C 2 2

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

1 2 m(vB)2 + 0 = 0 + mgh1 2 (vB)2 = 22gh1 c

and

use

United

on

learning.

is

of

the

Coefficient of Restitution: Since the ground does not move, the

by

(vB)2 (vB)1

work

student

(including solely

of

protected

the

e = -

for

(+ c )

is

work

22gh1

assessing

this

integrity

of

and

part

the

is

any

courses

destroy

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and

will

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Ans.

work

- 22gh0

h1 A h0

provided

=

This

e = -

15–77. The cue ball A is given an initial velocity (vA)1 = 5 m>s. If it makes a direct collision with ball B (e = 0.8), determine the velocity of B and the angle u just after it rebounds from the cushion at C (e¿ = 0.6). Each ball has a mass of 0.4 kg. Neglect the size of each ball.

(vA)1 ⫽ 5 m/s A B 30⬚ C u

SOLUTION Conservation of Momentum: When ball A strikes ball B, we have mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2 0.4(5) + 0 = 0.4(vA)2 + 0.4(vB)2

(1)

Coefficient of Restitution: e =

0.8 =

(vB)2 - (vA)2 5 - 0

(2) laws

or

+) (;

(vB)2 - (vA)2 (vA)1 - (vB)1

Wide

copyright

in

teaching

Web)

Solving Eqs. (1) and (2) yields (vB)2 = 4.50 m>s

not

instructors

States

World

permitted.

Dissemination

(vA)2 = 0.500 m>s

the

by

and

use

United

on

learning.

is

of

the

Conservation of “y” Momentum: When ball B strikes the cushion at C, we have

of

protected

the

(including

for

work

student

mB(vBy)2 = mB(vBy)3

work

solely

0.4(4.50 sin 30°) = 0.4(vB)3 sin u

this

assessing

is

(+ T )

(3)

part

the

is

This

provided

integrity

of

and

work

(vB)3 sin u = 2.25 and

any

courses

Coefficient of Restitution (x):

+ ) (;

(vBx)2 - (vC)1

0.6 =

0 - [- (vB)3 cos u] 4.50 cos 30° - 0

will

their

destroy

of

(vC)2 - (vBx)3 sale

e =

(4)

Solving Eqs. (1) and (2) yields (vB)3 = 3.24 m>s






u = 43.9°

Ans.

15–78. Using a slingshot, the boy fires the 0.2-lb marble at the concrete wall, striking it at B. If the coefficient of restitution between the marble and the wall is e = 0.5, determine the speed of the marble after it rebounds from the wall.

vA

45

A

5 ft

SOLUTION A sB B x = A sA B x + A vA B x t 100 = 0 + 75 cos 45° t t = 1.886 s and 1 a t2 2 y 1 ( -32.2)(1.8862) 2

teaching

Web)

A sB B y = 0 + 75 sin 45°(1.886) +

or

A sB B y = A sA B y + A vA B y t +

laws

a+cb

Dissemination

Wide

copyright

in

= 42.76 ft

not

instructors

States

World

permitted.

and United

on

learning.

is

of

the

A vB B y = A vA B y + ay t

and

use

a+cb

for

work

student

the

by

A vB B y = 75 sin 45° + ( -32.2)(1.886) = -7.684 ft>s = 7.684 ft>s T

is

work

solely

of

protected

the

(including

Since A vB B x = A vA B x = 75 cos 45° = 53.03 ft>s, the magnitude of vB is

part

the

is

any

destroy

of

and

7.684 ≤ = 8.244° 53.03 will

A vB B x

S = tan - 1 ¢ their

A vB B y

sale

u = tan - 1 C

courses

and the direction angle of vB is

This

provided

integrity

of

and

work

this

assessing

vB = 2 A vB B x 2 + A vB B y 2 = 253.032 + 7.6842 = 53.59 ft>s

Conservation of Linear Momentum: Since no impulsive force acts on the marble along the inclined surface of the concrete wall (x¿ axis) during the impact, the linear momentum of the marble is conserved along the x¿ axis. Referring to Fig. b,

A +Q B

mB A v Bœ B x¿ = mB A v Bœ B x¿ 0.2 0.2 A 53.59 sin 21.756° B = A v Bœ cos f B 32.2 32.2 v Bœ cos f = 19.862






C

100 ft

Kinematics: By considering the x and y motion of the marble from A to B, Fig. a, + b a:

B

75 ft>s

(1)

60

15–78. continued

Coefficient of Restitution: Since the concrete wall does not move during the impact, the coefficient of restitution can be written as

A +aB

e =

0 - A v Bœ B y¿

A v Bœ B y¿ - 0

0.5 =

-v Bœ sin f - 53.59 cos 21.756°

v Bœ sin f = 24.885

(2)

Solving Eqs. (1) and (2) yields v Bœ = 31.8 ft>s

sale

will

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destroy

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and

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courses

part

the

is

This

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is

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of

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the

(including

for

work

student

the

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use

United

on

learning.

is

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the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






15–79. The sphere of mass m falls and strikes the triangular block with a vertical velocity v. If the block rests on a smooth surface and has a mass 3 m, determine its velocity just after the collision. The coefficient of restitution is e.

SOLUTION Conservation of “x¿ ” Momentum: m (v)1 = m (v)2 ( R +)

45°

m( v sin 45°) = m(vsx¿)2 (vsx¿)2 =

22 v 2

Coefficient of Restitution (y¿ ):

or laws

vb cos 45° - [- A vsy¿ B 2]

Web)

v cos 45° - 0

teaching

e =

A vsy¿ B 1 - (vb)1

Wide

copyright

(+ b )

(vb)2 - A vs y¿ B 2

in

e =

World

permitted.

Dissemination

22 (ev - vb) 2

is

of

the

not

instructors

States

A vsy¿ B 2 =

by

and

use

United

on

learning.

Conservation of “ x ” Momentum:

work

student

the

(including

for

0 + 0 = 3mvb - m A vsy¿ B 2 cos 45° - m(vsx¿)2 cos 45° this

integrity

of

and

work

22 22 22 22 (ev - vb) v = 0 2 2 2 2 any

destroy

of

will sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

and

1 + e v 7

their

vb =

courses

part

the

is

This

provided

3vb -

assessing

is

work

solely

of

protected

+ B A;

the

0 = ms (vs)x + mb vb

*15–80. Block A, having a mass m, is released from rest, falls a distance h and strikes the plate B having a mass 2m. If the coefficient of restitution between A and B is e, determine the velocity of the plate just after collision. The spring has a stiffness k.

A h B k

SOLUTION Just before impact, the velocity of A is T1 + V1 = T2 + V2 0 + 0 =

1 mv 2A - mgh 2

vA = 22gh (+ T )

e =

(vB)2 - (vA)2 22gh

e 22gh = (vB)2 - (vA)2

or

(1)

©mv1 = ©mv 2

teaching

Web)

laws

(+ T )

(2)

Dissemination

Wide

copyright

in

m(vA) + 0 = m(vA)2 + 2m(vB)2

of

is

2gh(1 + e)

on

learning. United

and work

student

the

the

(including

for

work

solely

of

protected

this

assessing

is

integrity

of

and

work

provided

any

courses

part

the

is

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destroy

of

and

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their






Ans.

use

1 3

by

(vB)2 =

the

not

instructors

States

World

permitted.

Solving Eqs. (1) and (2) for (vB)2 yields;

15–81. The girl throws the 0.5-kg ball toward the wall with an initial velocity vA = 10 m>s. Determine (a) the velocity at which it strikes the wall at B, (b) the velocity at which it rebounds from the wall if the coefficient of restitution e = 0.5, and (c) the distance s from the wall to where it strikes the ground at C.

B

vA

A

10 m/s

30 1.5 m C

SOLUTION

3m

Kinematics: By considering the horizontal motion of the ball before the impact, we have + B A:

s

sx = (s0)x + vx t 3 = 0 + 10 cos 30°t

t = 0.3464 s

By considering the vertical motion of the ball before the impact, we have (+ c)

vy = (v0)y + (ac)y t = 10 sin 30° + ( - 9.81)(0.3464)

laws

or

= 1.602 m>s

copyright

in

teaching

Web)

The vertical position of point B above the ground is given by

permitted.

Dissemination

Wide

1 (a ) t2 2 cy

World

sy = (s0)y + (v0)y t +

on

learning.

work

student

the

by

and

use

United

(sB)y = 1.5 + 10 sin 30°(0.3464) +

is

of

1 ( -9.81) A 0.34642 B = 2.643 m 2

the

not

instructors

States

(+ c)

Ans.

this

assessing

is

work

solely

(vb)1 = 2(10 cos 30°)2 + 1.6022 = 8.807 m>s = 8.81 m>s

of

protected

the

(including

for

Thus, the magnitude of the velocity and its directional angle are

integrity

of

and

work

1.602 = 10.48° = 10.5° 10 cos 30° provided

Ans.

and

any

courses

part

the

is

This

u = tan - 1

sale

will

their

destroy

of

Conservation of “y” Momentum: When the ball strikes the wall with a speed of (vb)1 = 8.807 m>s, it rebounds with a speed of (vb)2. mb A vby B 1 = mb A vby B 2 + B A;

mb (1.602) = mb C (vb)2 sin f D (vb)2 sin f = 1.602

(1)

Coefficient of Restitution (x): e = + B A:






0.5 =

(vw)2 - A vbx B 2

A vbx B 1 - (vw)1 0 - C -(vb)2 cos f D 10 cos 30° - 0

(2)

15–81. continued Solving Eqs. (1) and (2) yields f = 20.30° = 20.3°

(vb)2 = 4.617 m>s = 4.62 m>s

Ans.

Kinematics: By considering the vertical motion of the ball after the impact, we have (+ c)

sy = (s0)y + (v0)y t +

1 (a ) t2 2 cy 1 ( - 9.81)t21 2

- 2.643 = 0 + 4.617 sin 20.30°t1 + t1 = 0.9153 s

By considering the horizontal motion of the ball after the impact, we have + B A;

sx = (s0)x + vx t s = 0 + 4.617 cos 20.30°(0.9153) = 3.96 m

sale

will

their

destroy

of

and

any

courses

part

the

is

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provided

integrity

of

and

work

this

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is

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of

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the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






15–82. The 20-lb box slides on the surface for which mk = 0.3. The box has a velocity v = 15 ft>s when it is 2 ft from the plate. If it strikes the smooth plate, which has a weight of 10 lb and is held in position by an unstretched spring of stiffness k = 400 lb>ft, determine the maximum compression imparted to the spring. Take e = 0.8 between the box and the plate. Assume that the plate slides smoothly.

v

k

2 ft

SOLUTION T1 + a U1 - 2 = T2 1 20 1 20 a b (15)2 - (0.3)(20)(2) = a b (v2)2 2 32.2 2 32.2 v2 = 13.65 ft>s + ) (:

a mv1 = a mv2

20 20 10 b (13.65) = a b vA + vB 32.2 32.2 32.2

laws

teaching

Web)

vP - vA 13.65

Wide

permitted.

Dissemination

0.8 =

or

(vB)2 - (vA)2 (vA)1 - (vB)1 in

e =

copyright

a

is

of

the

not

instructors

States

World

Solving,

by

and

use

United

on

learning.

vP = 16.38 ft>s, vA = 5.46 ft>s

(including

for

work

student

the

T1 + V1 = T2 + V2

this

assessing

is

work

solely

of

protected

the

1 1 10 a b (16.38)2 + 0 = 0 + (400)(s)2 2 32.2 2

sale

will

their

destroy

of

and

any

courses

part

the

is

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provided

integrity

of

and

work

s = 0.456 ft






15 ft/s

Ans.

15–83. Before a cranberry can make it to your dinner plate, it must pass a bouncing test which rates its quality. If cranberries having an e Ú 0.8 are to be accepted, determine the dimensions d and h for the barrier so that when a cranberry falls from rest at A it strikes the plate at B and bounces over the barrier at C.

A

SOLUTION

T1 + V1 = T2 + V2 0 + 3.5W =

1 W a b (vc)21 + 0 2 32.2

(vc)1 = 15.01 ft>s

or

Conservation of “x¿ ” Momentum: When the cranberry strikes the plate with a speed of (vc)1 = 15.01 ft>s, it rebounds with a speed of (vc)2.

teaching

Web)

laws

mc A vcx¿ B 1 = mc A vcx¿ B 2

permitted.

Dissemination

Wide

copyright

in

3 mc (15.01) a b = mc C (vc)2 cos f D 5

learning.

is

of

the

instructors

States

(1)

United

and

use by

(including

for

work

student

the

(vP)2 - A vcy¿ B 2

the

A vcy¿ B 1 - (vP)1 is

work

solely

of

protected

e =

on

Coefficient of Restitution (y¿ ):

assessing

0 - (vc)2 sin f

integrity

of

and

work

this

(2)

provided

4 -15.01 a b - 0 5 and

any

courses

part

the

is

This

0.8 =

will

their

destroy

of


(vc)2 = 13.17 ft>s sale

f = 46.85°

Kinematics: By considering the vertical motion of the cranberry after the impact, we have vy = (v0)y + ac t

0 = 13.17 sin 9.978° + ( - 32.2) t (+ c)

sy = (s0)y + (v0)y t +

t = 0.07087 s

1 (a ) t2 2 cy

= 0 + 13.17 sin 9.978° (0.07087) + = 0.080864 ft






1 (- 32.2) A 0.070872 B 2

not

World

(vc)2 cos f = 9.008

(+ c)

B

d

Conservation of Energy: The datum is set at point B. When the cranberry falls from a height of 3.5 ft above the datum, its initial gravitational potential energy is W(3.5) = 3.5 W. Applying Eq. 14–21, we have

( a +)

3.5 ft 4

h

(+ b )

5

C

3

15–83. continued By considering the horizontal motion of the cranberry after the impact, we have + B A;

sx = (s0)x + vx t 4 d = 0 + 13.17 cos 9.978° (0.07087) 5 d = 1.149 ft = 1.15 ft

Ans.

Thus, 3 3 d = 0.080864 + (1.149) = 0.770 ft 5 5

Ans.

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laws

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h = sy +






*15–84. A ball is thrown onto a rough floor at an angle u. If it rebounds at an angle f and the coefficient of kinetic friction is m, determine the coefficient of restitution e. Neglect the size of the ball. Hint: Show that during impact, the average impulses in the x and y directions are related by Ix = mIy . Since the time of impact is the same, Fx ¢t = mFy ¢t or Fx = mFy .

y

u

x

SOLUTION 0 - 3 - v2 sin f4

( + T)

e =

+ ) (:

m1vx21 +

v1 sin u - 0

e =

v2 sin f v1 sin u

(1)

t2

Fx dx = m1vx22

Lt1

mv1 cos u - Fx ¢t = mv2 cos f

m1vy21 +

(2)

t2

Lt1

Fy dx = m1vy22 laws

(+ T)

mv1 cos u - mv2 cos f ¢t

or

Fx =

in

teaching

Web)

mv1 sin u - Fy ¢t = - mv2 sin f mv1 sin u + mv2 sin f ¢t

(3)

instructors

States

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permitted.

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Fy =

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Since Fx = mFy, from Eqs. (2) and (3)

the

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m1mv1 sin u + mv2 sin f) mv1 cos u - mv2 cos f = ¢t ¢t

(4)

part

the

is

Substituting Eq. (4) into (1) yields:

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cos u - m sin u v2 = v1 m sin f + cos f






destroy

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sin f cos u - m sin u a b sin u m sin f + cos f their

e =

f

Ans.

15–85. A ball is thrown onto a rough floor at an angle of u = 45°. If it rebounds at the same angle f = 45°, determine the coefficient of kinetic friction between the floor and the ball. The coefficient of restitution is e = 0.6. Hint: Show that during impact, the average impulses in the x and y directions are related by Ix = mIy . Since the time of impact is the same, Fx ¢t = mFy ¢t or Fx = mFy .

y

u

x

SOLUTION (+ T ) + B A:

e =

0 - [-v2 sin f] v1 sin u - 0

v2 sin f v1 sin u

e =

(1)

t2

m(vx)1 +

Fx dx = m(vx)2

Lt1

mv1 cos u - Fx ¢t = mv2 cos f

m A vy B 1 +

(2)

t2

Lt1

Fydx = m A yy B 2 laws

(+ c )

mv1 cos u - mv2 cos f ¢t

or

Fx =

in

teaching

Web)

mv1 sin u - Fy ¢t = -mv2 sin f mv1 sin u + mv2 sin f ¢t

Wide

copyright

permitted.

Dissemination

(3)

instructors

States

World

Fy =

the

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by

m(mv1 sin u + mv2 sin f) mv1 cos u - mv2 cos f = ¢t ¢t

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Since Fx = mFy, from Eqs. (2) and (3)

(4)

part

the

is

Substituting Eq. (4) into (1) yields:

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cos u - m sin u v2 = v1 m sin f + cos f

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sin f cos u - m sin u a b sin u m sin f + cos f sale

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e =






0.6 =

sin 45° cos 45° - m sin 45° a b sin 45° m sin 45° + cos 45°

0.6 =

1-m 1+m

m = 0.25

f

Ans.

15–86. The “stone” A used in the sport of curling slides over the ice track and strikes another “stone” B as shown. If each “stone” is smooth and has a weight of 47 lb, and the coefficient of restitution between the “stones” is e = 0.8, determine their speeds just after collision. Initially A has a velocity of 8 ft>s and B is at rest. Neglect friction.

3 ft y B A

SOLUTION

(vA)1

Line of impact (x-axis): ©mv1 = ©mv2 47 47 47 (8) cos 30° = (v ) + (v ) 32.2 32.2 B 2x 32.2 A 2x

( + a)

0 +

( + a)

e = 0.8 =

(vB)2x - (vA)2x 8 cos 30° - 0

Solving:

or

(vA)2x = 0.6928 ft>s

in

teaching

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laws

(vB)2x = 6.235 ft>s

permitted.

Dissemination

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copyright

Plane of impact (y-axis):

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Stone A:

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mv1 = mv2

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47 47 (8) sin 30° = (v ) 32.2 32.2 A 2y

(Q+)

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(vA)2y = 4

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Stone B:

47 (vB)2y 32.2

will

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mv1 = mv2

(vB)2y = 0






(vA)2 = 2(0.6928)2 + (4)2 = 4.06 ft>s

Ans.

(vB)2 = 2(0)2 + (6.235)2 = 6.235 = 6.24 ft>s

Ans.

30 8 ft/s x

15–87. Two smooth disks A and B each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide, determine their final velocities just after collision. The coefficient of restitution is e = 0.75.

y (vA)1

x

B

SOLUTION + ) (:

(vB)1

©mv1 = ©mv2 3 0.5(4)( ) - 0.5(6) = 0.5(yB)2x + 0.5(yA)2x 5

+ ) (:

e =

(vA)2 - (vB)2 (vB)1 - (vA)1

0.75 =

(vA)2x - (vB)2x 4 A 35 B - ( - 6)

(vA)2x = 1.35 m>s : laws

or

(vB)2x = 4.95 m>s ; in

teaching

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mv1 = mv2

Wide

copyright

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4 0.5( )(4) = 0.5(vB)2y 5

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(vB)2y = 3.20 m>s c by

vA = 1.35 m>s :

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the

Ans.

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vB = 2(4.59)2 + (3.20)2 = 5.89 m>s

assessing

3.20 = 32.9˚ ub 4.95

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Ans.

work

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This

provided

u = tan-1

6 m/s

5 4 3

4 m/s

A

*15–88. Two smooth disks A and B each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide, determine the coefficient of restitution between the disks if after collision B travels along a line, 30° counterclockwise from the y axis.

y (vA)1

x

B

SOLUTION

(vB)1

©mv1 = ©mv2 + ) (:

3 0.5(4)( ) - 0.5(6) = - 0.5(vB)2x + 0.5(vA)2x 5 - 3.60 = -(vB)2x + (vA)2x

(+ c )

4 0.5(4)( ) = 0.5(vB)2y 5 (vB)2y = 3.20 m>s c

laws

or

(vB)2x = 3.20 tan 30° = 1.8475 m>s ;

in

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(vA)2x = - 1.752 m>s = 1.752 m>s ; (vA)2 - (vB)2 (vB)1 - (vA)1

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= 0.0113

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- 1.752- ( -1.8475)

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6 m/s

5 4 3

4 m/s

A

15–89. Two smooth disks A and B have the initial velocities shown just before they collide at O. If they have masses mA = 8 kg and mB = 6 kg, determine their speeds just after impact. The coefficient of restitution is e = 0.5.

y 13 12 5

A

vA

O

SOLUTION

vB

+b©mv1 = ©mv2 -6(3 cos 67.38°) + 8(7 cos 67.38°) = 6(vB)x¿ + 8(vA)xœ (+ b) 0.5 =

e =

(vB)2 - (vA)2 (vA)1 - (vB )1

(vB)xœ - (vA)xœ 7 cos 67.38° + 3 cos 67.38°

Solving,

laws

or

(vB)xœ = 2.14 m>s

in

teaching

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(vA)xœ = 0.220 m>s

Dissemination

Wide

copyright

(vB)yœ = 3 sin 67.38° = 2.769 m>s

not

instructors

States

World

permitted.

(vA)y¿ = -7 sin 67.38° = -6.462 m>s

of

the

vB = 2(2.14)2 + (2.769)2 = 3.50 m>s

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vA = 2(0.220)2 + (6.462)2 = 6.47 m>s






3 m/s

7 m/s x

B

15–90. If disk A is sliding along the tangent to disk B and strikes B with a velocity v, determine the velocity of B after the collision and compute the loss of kinetic energy during the collision. Neglect friction. Disk B is originally at rest. The coefficient of restitution is e, and each disk has the same size and mass m.

B v

A

SOLUTION Impact: This problem involves oblique impact where the line of impact lies along x¿ axis (line jointing the mass center of the two impact bodies). From the geometry r u = sin - 1 a b = 30°. The x¿ and y¿ components of velocity for disk A just before 2r impact are

A vAx¿ B 1 = - v cos 30° = - 0.8660v

A vAy¿ B 1 = - v sin 30° = - 0.5 v

Conservation of “x¿ ” Momentum:

or

mA A vAx¿ B 1 + mB A vBx¿ B 1 = mA A vAx¿ B 2 + mB A vBx¿ B 2 m( -0.8660v) + 0 = m A vAx B 2 + m A vBx¿ B 2

laws

(1) in

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permitted.

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on

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is

of

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not

A vAx¿ B 1 - A vBx¿ B 1 A vBx¿ B 2 - A vAx¿ B 2

(2)

by

- 0.8660v - 0

work

student

the

(including

for

e =

A vBx¿ B 2 - A vAx¿ B 2

the

e =

copyright

Coefficient of Restitution (x¿ ):

assessing

is

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solely

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23 (e - 1) v 4 integrity

of

and

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this

A vAx¿ B 2 = provided

23 (1 + e) v 4

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the

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A vBx¿ B 2 = -

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Conservation of “y¿ ” Momentum: The momentum is conserved along y¿ axis for both disks A and B. (+Q) (+Q)

mB A vBy¿ B 1 = mB A vBy¿ B 2; mA A vAy¿ B 1 = mA A vAy¿ B 2;

A vBy¿ B 2 = 0 A vAy¿ B 2 = - 0.5 v

Thus, the magnitude the velocity for disk B just after the impact is (vB)2 = 2(vBx¿)22 + (vBy¿)22 =

D

a-

2 23 23 (1 + e) vb + 0 = (1 + e) v 4 4

and directed toward negative x¿ axis.






Ans. Ans.

15–90. continued The magnitude of the velocity for disk A just after the impact is (vA)2 = 2( vAx¿)22 + (vAy¿)22 =

2 23 (e - 1) v d + ( -0.5 v)2 D 4

=

v2 (3e2 - 6e + 7) A 16

c

Loss of Kinetic Energy: Kinetic energy of the system before the impact is Uk =

1 mv2 2

Kinetic energy of the system after the impact is Uk œ =

mv2 A 6e2 + 10 B 32 or

=

2 2 2 23 1 1 m B v (3e2 - 6e + 7) R + m B (1 + e) v R 2 A 16 2 4

Web)

laws

Thus, the kinetic energy loss is in

teaching

1 m v2 mv2 (6e2 + 10) 2 32

Dissemination

Wide

copyright

¢Uk = Uk - Uk œ =

States

World

permitted.

3mv2 (1 - e2) 16

the

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15–91. Two disks A and B weigh 2 lb and 5 lb, respectively. If they are sliding on the smooth horizontal plane with the velocities shown, determine their velocities just after impact. The coefficient of restitution between the disks is e = 0.6.

y

SOLUTION

30

mA A vA B n + mB A vB B n = mA A vAœ B n + mB A vBœ B n -

5 2 œ 2 5 œ (5) cos 45° + (10) cos 30° = vA cos uA + vB cos uB 32.2 32.2 32.2 32.2

œ 2vA cos uA + 5vBœ cos uB = 36.23

(1)

Also, we notice that the linear momentum a of disks A and B are conserved along the t axis (tangent to the plane of impact). Thus, a+Tb

œ mA A vA B t = mA A vA Bt

Web)

laws

or

2 2 œ sin uA (5) sin 45° = vA 32.2 32.2

(2) Wide

copyright

in

teaching

œ vA sin uA = 3.5355

instructors

States

World

permitted.

Dissemination

and the

not

mB A vB B t = mB A v Bœ B t use

United

on

learning.

is

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a+Tb

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for

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the

by

and

2 2 (10) sin 30° = = v Bœ sin uB 32.2 32.2

(3)

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solely

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the

v Bœ sin uB = 5

part

the

is

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provided

integrity

of

and

work

this

Coefficient of Restitution: The coefficient of restitution equation written along the n axis (line of impact) gives any

courses

A vAœ B n - A vBœ B n

and

A vB B n - A vA B n

destroy

of

0.6 = œ vA

œ vA cos uA - v Bœ cos uB

10 cos 30° -

cos uA -

v Bœ

sale

will

e =

their

+ b a:

A -5 cos 45° B

cos uB = 7.317

(4)

Solving Eqs. (1), (2), (3), and (4), yields






œ = 11.0 ft>s vA

uA = 18.8°

Ans.

v Bœ = 5.88 ft>s

uB = 58.3°

Ans.

B

A

45 x

Conservation of Linear Momentum:By referring to the impulse and momentum of the system of disks shown in Fig. a, notice that the linear momentum of the system is conserved along the n axis (line of impact). Thus, + b a:

5 ft/s

10 ft/s

*15–92. Two smooth coins A and B, each having the same mass, slide on a smooth surface with the motion shown. Determine the speed of each coin after collision if they move off along the blue paths. Hint: Since the line of impact has not been defined, apply the conservation of momentum along the x and y axes, respectively.

y

(vA)2

(vA)1 A

0.5 ft/s

45 5

A

3

4

30

SOLUTION

30

©mv1 = mv2 + ) (:

(vB)2

4 -m(0.8) sin 30° - m(0.5)( ) = -m(vA)2 sin 45° - m(vB)2 cos 30° 5

3 m(0.8) cos 30° - m(0.5)( ) = m(vA)2 cos 45° - m(vB)2 sin 30° 5 -0.3928 = - 0.707(vA)2 + 0.5(vB)2

laws

or

Solving, Ans.

(vA)2 = 0.766 ft>s

Ans.

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(vB)2 = 0.298 ft>s






(vB)1

B

0.8 = 0.707(vA)2 + 0.866(vB)2 (+ c)

x

O

B

0.8 ft/s

15–93. Disks A and B have a mass of 15 kg and 10 kg, respectively. If they are sliding on a smooth horizontal plane with the velocities shown, determine their speeds just after impact. The coefficient of restitution between them is e = 0.8.

y 5

4

Line of impact

3

A

10 m/s x

SOLUTION

8 m/s

Conservation of Linear Momentum: By referring to the impulse and momentum of the system of disks shown in Fig. a, notice that the linear momentum of the system is conserved along the n axis (line of impact). Thus, ¿ + Q mA A vA B n + mB A vB B n = mA A vA B n + mB A vB¿ B n

3 3 ¿ 15(10)a b - 10(8)a b = 15vA cos fA + 10vB¿ cos fB 5 5 ¿ 15vA cos fA + 10vB¿ cos fB = 42

(1)

or

Also, we notice that the linear momentum of disks A and B are conserved along the t axis (tangent to? plane of impact). Thus,

teaching

Web)

laws

¿ + a mA A vA B t = mA A vA Bt

Dissemination

Wide

copyright

in

4 ¿ 15(10)a b = 15vA sin fA 5

permitted.

¿ vA sin fA = 8

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+a mB A vB B t = mB A vB¿ B t

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4 10(8)a b = 10 vB¿ sin fB 5

(3)

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this

vB¿ sin fB = 6.4

+Q e =

0.8 =

sale

¿ (vB¿ )n - (vA )n (vA)n - (vB)n

will

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Coefficient of Restitution: The coefficient of restitution equation written along the n axis (line of impact) gives

¿ cos fA vB¿ cos fB - vA

3 3 10 a b - c -8a b d 5 5

¿ vB¿ cos fB - vA cos fA = 8.64

(4)

Solving Eqs. (1), (2), (3), and (4), yeilds ¿ = 8.19 m>s vA

Ans.

fA = 102.52° vB¿ = 9.38 m>s fB = 42.99°






Ans.

B

15–94. z

Determine the angular momentum H O of the particle about point O. 1.5 kg A 6 m/s 4m

SOLUTION vA = 6 ¢

rOB = { - 7j} m

4m

2i - 4j - 4k 2(22 + ( -4)2 + (- 4)2

j -7 1.5( -4)

x

k 3 = {42i + 21k} kg # m2>s 0 1.5(- 4)

Ans.

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in

teaching

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laws

or

i 0 1.5(2)






3m

≤ = {2i - 4j - 4k} m>s

HO = rOB * mvA HO = 3

2m

B O

y

15–95. Determine the angular momentum H O of the particle about point O.

z A

10 lb

14 ft/s

P

6 ft O

5 ft

SOLUTION rOB = {8i + 9j}ft

HO = 4

10 a b(12) 32.2

vA = 14 ¢

12i + 4j - 6k 2122 + 42 + (- 6)2

j 9 10 a b(4) 32.2

y 8 ft

≤ = {12i + 4j - 6k} ft>s

9 ft

x

B

k 0

4

10 a b ( -6) 32.2 ft2>s

Ans.

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in

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Web)

laws

or

= {-16.8i + 14.9j - 23.6k} slug






4 ft

2 ft

HO = rOB * mvA i 8

5 ft 3 ft

*15–96. Determine the angular momentum H P of the particle about point P.

z A

10 lb

14 ft/s

P

6 ft O

5 ft

5 ft 4 ft

3 ft

SOLUTION

2 ft

y

rPB = {5i + 11j - 5k} ft vA = 14 a

8 ft

12i + 4j - 6k 212)2 + (4)2 + ( - 6)2 i 5

H P = rPB * mvA = 4 a

9 ft

x

B

b = {12i + 4j - 6k} ft>s

10 b (12) 32.2

j 11 10 a b (4) 32.2

k -5 4 10 a b (- 6) 32.2

= {-14.3i - 9.32j - 34.8k} slug # ft2>s

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15–97. z

Determine the total angular momentum H O for the system of three particles about point O. All the particles are moving in the x–y plane.

3 kg 6 m/s C O 900 mm

SOLUTION

4 m/s A 1.5 kg

HO = ©r * mv i = 3 0.9 0

j 0 - 1.5(4)

x

i k 0 3 + 3 0.6 -2.5(2) 0

i k 0 3 + 3 -0.8 0 0

j 0.7 0

j -0.2 3(-6)

k 03 0

= {12.5k} kg # m2>s

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700 mm

200 mm

800 mm y

B

2 m/s 600 mm 2.5 kg

15–98. Determine the angular momentum H O of each of the two particles about point O. Use a scalar solution.

y

5m

P

15 m/s 5

A

3

4m

4

2 kg

1.5 m

O

x

2m

SOLUTION

4m

4 3 a +(HA)O = -2(15)a b (1.5) - 2(15)a b (2) 5 5

30⬚

1.5 kg 10 m/s

= -72.0 kg # m2>s = 72.0 kg # m2>s b

Ans.

a + (HB)O = -1.5(10)(cos 30°)(4) -1.5(10)(sin 30°)(1) = -59.5 kg # m2>s = 59.5 kg # m2>s b

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B

1m

15–99. Determine the angular momentum H P of each of the two particles about point P. Use a scalar solution.

y

5m

P

15 m/s 5

A

3

4m

4

2 kg

1.5 m

O

x

2m

SOLUTION

4m

4 3 a +(HA)P = 2(15) a b (2.5) - 2(15) a b (7) 5 5

30⬚

1.5 kg 10 m/s

= -66.0 kg # m2>s = 66.0 kg # m2>s b

Ans.

a + (HB)P = - 1.5(10)(cos 30°)(8) + 1.5(10)(sin 30°)(4) = -73.9 kg # m2>s b

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B

1m

*15–100. The small cylinder C has a mass of 10 kg and is attached to the end of a rod whose mass may be neglected. If the frame is subjected to a couple M = 18t2 + 52 N # m, where t is in seconds, and the cylinder is subjected to a force of 60 N, which is always directed as shown, determine the speed of the cylinder when t = 2 s. The cylinder has a speed v0 = 2 m>s when t = 0.

z 60 N

5 4

0.75 m

(Hz)1 + ©

L

Mz dt = (Hz)2 2

8 69 + [ t3 + 5t]20 = 7.5v 3 v = 13.4 m>s

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C M

3 (8t 2 + 5)dt = 10v(0.75) (10)(2)(0.75) + 60(2)( )(0.75) + 5 L0


v

x

SOLUTION

3

(8t2

y 5) N m

15–101. The 10-lb block rests on a surface for which mk = 0.5. It is acted upon by a radial force of 2 lb and a horizontal force of 7 lb, always directed at 30° from the tangent to the path as shown. If the block is initially moving in a circular path with a speed v1 = 2 ft>s at the instant the forces are applied, determine the time required before the tension in cord AB becomes 20 lb. Neglect the size of the block for the calculation.

A 4 ft B 7 lb

SOLUTION ©Fn = man ; 20 - 7 sin 30° - 2 =

10 v 2 ( ) 32.2 4

v = 13.67 ft>s (HA) t + ©

MA dt = (HA)2

teaching

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laws

or

10 10 )(2)(4) + (7 cos 30°)(4)(t) - 0.5(10)(4) t = (13.67)(4) 32.2 32.2 in

(

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t = 3.41 s






30

2 lb

15–102. The 10-lb block is originally at rest on the smooth surface. It is acted upon by a radial force of 2 lb and a horizontal force of 7 lb, always directed at 30° from the tangent to the path as shown. Determine the time required to break the cord, which requires a tension T = 30 lb. What is the speed of the block when this occurs? Neglect the size of the block for the calculation.

A 4 ft B 7 lb

SOLUTION ©Fn = man ; 30 - 7 sin 30° - 2 =

10 v2 ( ) 32.2 4

v = 17.764 ft>s

L

MA dt = (HA)2

0 + (7 cos 30°)(4)(t) =

10 (17.764)(4) 32.2

t = 0.910 s

Ans.

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(HA)1 + ©






30

2 lb

15–103. A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed 1vB21 = 6 ft>s. If the attached cord is pulled down through the hole with a constant speed vr = 2 ft>s, determine the ball’s speed at the instant r2 = 2 ft. How much work has to be done to pull down the cord? Neglect friction and the size of the ball.

B

r1 = 3 ft

(v B )1 = 6 ft>s

SOLUTION H1 = H2

v r = 2 ft>s

4 4 v (2) (6)(3) = 32.2 32.2 u vu = 9 ft>s v2 = 292 + 22 = 9.22 ft>s

Ans.

T1 + ©U1 - 2 = T2 1 4 1 4 ( )(6)2 + ©U1 - 2 = ( )(9.22)2 2 32.2 2 32.2 or

©U1 - 2 = 3.04 ft # lb

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*15–104. A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed 1vB21 = 6 ft>s. If the attached cord is pulled down through the hole with a constant speed vr = 2 ft>s, determine how much time is required for the ball to reach a speed of 12 ft/s. How far r2 is the ball from the hole when this occurs? Neglect friction and the size of the ball.

B

r1 = 3 ft

(v B )1 = 6 ft>s

SOLUTION v = 2(vu)2 + (2)2

v r = 2 ft>s

12 = 2(vu)2 + (2)2 vu = 11.832 ft>s H1 = H2 4 4 (6)(3) = (11.832)(r2) 32.2 32.2 r2 = 1.5213 = 1.52 ft

Ans.

laws

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¢r = v rt

Ans.

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t = 0.739 s

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(3 - 1.5213) = 2t






15–105. The four 5-lb spheres are rigidly attached to the crossbar frame having a negligible weight. If a couple moment M = (0.5t + 0.8) lb # ft, where t is in seconds, is applied asshown, determine the speed of each of the spheres in 4 seconds starting from rest. Neglect the size of the spheres.

M

SOLUTION (Hz)1 + ©

L

Mz dt = (Hz)2

4

0 +

L0

0.6 ft

(0.5 t + 0.8) dt = 4 B a

5 b (0.6 v 2) R 32.2

7.2 = 0.37267 v2 v2 = 19.3 ft>s

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(0.5t

0.8) lb · ft

15–106. A small particle having a mass m is placed inside the semicircular tube. The particle is placed at the position shown and released. Apply the principle of angular # momentum about point O 1©MO = HO2, and show that the motion $ of the particle is governed by the differential equation u + 1g>R2 sin u = 0.

O

u

SOLUTION c + ©MO =

dHO ; dt

-Rmg sin u =

d (mvR) dt dv d 2s = - 2 dt dt

g sin u = But, s = R u $ Thus, g sin u = -Ru $ g or, u + a b sin u = 0 R

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Q.E.D.






R

15–107. The ball B has a weight of 5 lb and is originally rotating in a circle. As shown, the cord AB has a length of 3 ft and passes through the hole A, which is 2 ft above the plane of motion. If 1.5 ft of cord is pulled through the hole, determine the speed of the ball when it moves in a circular path at C.

1.5 ft C 2 ft 3 ft

SOLUTION

B vB

Equation of Motion: When the ball is travelling around the first circular path, 2 u = sin - 1 = 41.81° and r1 = 3 cos 41.81° = 2.236. Applying Eq. 13–8, we have 3 ©Fb = 0; ©Fn = man;

2 T1 a b -5 = 0 3

T1 = 7.50 lb

7.50 cos 41.81° =

v2t 5 a b 32.2 2.236

v1 = 8.972 ft>s

laws

or

When the ball is traveling around the second circular path, r2 = 1.5 cos f. Applying Eq. 13–8, we have Web)

T2 sin f - 5 = 0

in

teaching

(1)

copyright

©Fb = 0;

permitted.

(2) World

Dissemination

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v22 5 a b 32.2 1.5 cos f

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T2 cos f =

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©Fn = man;

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learning.

Conservation of Angular Momentum: Since no force acts on the ball along the tangent of the circular, path the angular momentum is conserved about z axis. Applying Eq. 15–23, we have

provided

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r1mv1 = r2mv2

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the

(Ho)1 = (Ho)2

5 5 b (8.972) = 1.5 cos f a b v2 32.2 32.2

the

(3)

will sale

Solving Eqs. (1),(2) and (3) yields

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2.236 a

f = 13.8678°

T2 = 20.85 lb

v2 = 13.8 ft>s






T

A

Ans.

vC

*15–108. A child having a mass of 50 kg holds her legs up as shown as she swings downward from rest at u1 = 30°. Her center of mass is located at point G1 . When she is at the bottom position u = 0°, she suddenly lets her legs come down, shifting her center of mass to position G2 . Determine her speed in the upswing due to this sudden movement and the angle u2 to which she swings before momentarily coming to rest. Treat the child’s body as a particle.

3m

G2

First before u = 30°; T1 + V1 = T2 + V2 1 (50)(v1)2 + 0 2

v1 = 2.713 m>s H1 = H2 50(2.713)(2.80) = 50(v2)(3) v2 = 2.532 = 2.53 m>s

laws

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Ans.

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Just after u = 0°;

States

World

permitted.

Dissemination

T2 + V2 = T3 + V3

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1 (50)(2.532)2 + 0 = 0 + 50(9.81)(3)(1 - cos u2) 2

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the

0.1089 = 1 - cos u2

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the

u2 = 27.0°





Ans.

30°

G1

G2

0 + 2.80(1 - cos 30°)(50)(9.81) =

2.80 m u1

SOLUTION


u2

15–109. The 150-lb car of an amusement park ride is connected to a rotating telescopic boom. When r = 15 ft, the car is moving on a horizontal circular path with a speed of 30 ft>s. If the boom is shortened at a rate of 3 ft>s, determine the speed of the car when r = 10 ft. Also, find the work done by the axial force F along the boom. Neglect the size of the car and the mass of the boom. F

SOLUTION Conservation of Angular Momentum: By referring to Fig. a, we notice that the angular momentum of the car is conserved about an axis perpendicular to the page passing through point O, since no angular impulse acts on the car about this axis. Thus,

A HO B 1 = A HO B 2 r1mv1 = r2m A v2 B u

Av2Bu =

15(30) r1 v1 = 45 ft>s = r2 10

or

Thus, the magnitude of v2 is laws

v 2 = 4A v 2 B r 2 - A v2 B u 2 = 232 + 452 = 45.10 ft>s = 45.1 ft>s

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Ans.

Dissemination

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copyright

Principle of Work and Energy: Using the result of v2,

of

the

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permitted.

T1 + ©U1 - 2 = T2

(including

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protected

1 150 1 150 a b (302) + UF = a b (45.102) 2 32.2 2 32.2

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1 1 mv1 2 + UF = mv 2 2 2 2

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is

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UF = 2641 ft # lb

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Ans.






r

15–110. An amusement park ride consists of a car which is attached to the cable OA. The car rotates in a horizontal circular path and is brought to a speed v1 = 4 ft>s when r = 12 ft. The cable is then pulled in at the constant rate of 0.5 ft>s. Determine the speed of the car in 3 s.

O

A r

SOLUTION Conservation of Angular Momentum: Cable OA is shorten by ¢r = 0.5(3) = 1.50 ft. Thus, at this instant r2 = 12 - 1.50 = 10.5 ft. Since no force acts on the car along the tangent of the moving path, the angular momentum is conserved about point O. Applying Eq. 15–23, we have (HO)1 = (HO)2 r1 mv1 = r2 mv¿ 12(m)(4) = 10.5(m) v¿ v¿ = 4.571 ft>s

laws

or

The speed of car after 3 s is v2 = 20.52 + 4.5712 = 4.60 ft>s

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Ans.






15–111. The 800-lb roller-coaster car starts from rest on the track having the shape of a cylindrical helix. If the helix descends 8 ft for every one revolution, determine the speed of the car in t = 4 s. Also, how far has the car descended in this time? Neglect friction and the size of the car.

r = 8 ft 8 ft

SOLUTION u = tan-1(

8 ) = 9.043° 2p(8)

©Fy = 0;

N - 800 cos 9.043° = 0 N = 790.1 lb

H1 +

L

M dt = H2

4

0 +

L0

8(790.1 sin 9.043°) dt =

800 (8)vt 32.2 laws

or

vt = 20.0 ft>s

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Web)

20 = 20.2 ft>s cos 9.043°

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Ans. Dissemination

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v =

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T1 + ©U1 - 2 = T2

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1 800 ( )(20.2)2 2 32.2

the

0 + 800h =

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h = 6.36 ft






Ans.

*15–112. A small block having a mass of 0.1 kg is given a horizontal velocity of v1 = 0.4 m>s when r1 = 500 mm. It slides along the smooth conical surface. Determine the distance h it must descend for it to reach a speed of v2 = 2 m>s. Also, what is the angle of descent u, that is, the angle measured from the horizontal to the tangent of the path?

r1 = 500 mm υ 1 = 0.4 m>s

h r2

θ v2


30°

1 1 (0.1)(0.4)2 + 0.1(9.81)(h) = (0.1)(2)2 2 2 h = 0.1957 m = 196 mm

Ans.

From similar triangles r2 =

(0.8660 - 0.1957) (0.5) = 0.3870 m 0.8660

laws

or

(H0)1 = (H0)2

in

teaching

Web)

0.5(0.1)(0.4) = 0.3870(0.1)(v 2 ¿)

Dissemination

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copyright

v 2 ¿ = 0.5168 m>s

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v 2 = cos u = v 2 ¿

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the

2 cos u = 0.5168

Ans.

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u = 75.0°






15–113. An earth satellite of mass 700 kg is launched into a freeflight trajectory about the earth with an initial speed of vA = 10 km>s when the distance from the center of the earth is rA = 15 Mm. If the launch angle at this position is fA = 70°, determine the speed vB of the satellite and its closest distance rB from the center of the earth. The earth has a mass Me = 5.976110242 kg. Hint: Under these conditions, the satellite is subjected only to the earth’s gravitational force, F = GMems>r2, Eq. 13–1. For part of the solution, use the conservation of energy.

vB

rA vA

SOLUTION (HO)1 = (HO)2 ms (vA sin fA)rA = ms (vB)rB 700[10(103) sin 70°](15)(106) = 700(vB)(rB)

(1)

TA + VA = TB + VB

laws

or

GMe ms GMems 1 1 m (v )2 = ms (vB)2 rA rB 2 s A 2

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copyright

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66.73(10-12)(5.976)(1024)(700) 1 1 = (700)(vB)2 (700)[10(103)]2 6 2 2 [15(10 )] Dissemination

66.73(10-12)(5.976)(1024)(700) rB

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vB = 10.2 km>s

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rB = 13.8 Mm






permitted.

(2)

States

-

rB

fA

Ans.

15–114. The fire boat discharges two streams of seawater, each at a flow of 0.25 m3>s and with a nozzle velocity of 50 m> s. Determine the tension developed in the anchor chain, needed to secure the boat. The density of seawater is rsw = 1020 kg>m3.

30⬚ 45⬚

60⬚

SOLUTION Steady Flow Equation: Here, the mass flow rate of the sea water at nozzles A and dmA dmB B are = = rsw Q = 1020(0.25) = 225 kg>s. Since the sea water is coldt dt lected from the larger reservoir (the sea), the velocity of the sea water entering the control volume can be considered zero. By referring to the free-body diagram of the control volume (the boat), + ©F = dmA A v B + dmB A v B ; ; x A x B x dt dt T cos 60° = 225(50 cos 30°) + 225(50 cos 45°) T = 40 114.87 N = 40.1 kN

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Ans.






15–115. A jet of water having a cross-sectional area of 4 in2 strikes the fixed blade with a speed of 25 ft>s. Determine the horizontal and vertical components of force which the blade exerts on the water, gw = 62.4 lb>ft3.

130 25 ft/s

SOLUTION Q = Av = a

4 b(25) = 0.6944 ft3>s 144

62.4 dm = rQ = a b (0.6944) = 1.3458 slug>s dt 32.2 vAx = 25 ft>s

vAy = 0

vBx = -25 cos 50°ft>s

vBy = 25 sin 50°ft>s

+ ©F = dm (v - v ); : x A dt Bx

- Fx = 1.3458[- 25 cos 50° - 25] or

Fx = 55.3 lb

teaching

Web)

laws

Ans. in

Fy = 1.3458(25 sin 50° - 0)

Wide Dissemination

dm A vBy - vAy); dt

copyright

+ c ©Fy =

permitted.

Ans.

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Fy = 25.8 lb






*15–116. The 200-kg boat is powered by the fan which develops a slipstream having a diameter of 0.75 m. If the fan ejects air with a speed of 14 m/s, measured relative to the boat, determine the initial acceleration of the boat if it is initially at rest.Assume that air has a constant density of ra = 1.22 kg/m3 and that the entering air is essentially at rest. Neglect the drag resistance of the water.

0.75 m

SOLUTION Equations of Steady Flow: Initially, the boat is at rest hence vB = va>b p dm = 14 m>s. Then, Q = vBA = 14 c A 0.752 B d = 6.185 m3>s and = raQ 4 dt = 1.22(6.185) = 7.546 kg>s. Applying Eq. 15–26, we have ©Fx =

dm (v - vAx); dt Bx

-F = 7.546( - 14 - 0)

F = 105.64 N

Equation of Motion : 105.64 = 200a

a = 0.528 m>s2

Ans.

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or

+ ©F = ma ; : x x






15–117. The chute is used to divert the flow of water, Q = 0.6 m3>s. If the water has a cross-sectional area of 0.05 m2, determine the force components at the pin D and roller C necessary for equilibrium. Neglect the weight of the chute and weight of the water on the chute, rw = 1 Mg>m3 .

A

0.12 m

C

SOLUTION

2m

Equations of Steady Flow: Here, the flow rate Q = 0.6 m2>s. Then, Q 0.6 dm = = 12.0 m>s. Also, = rw Q = 1000 (0.6) = 600 kg>s. Applying y = A 0.05 dt Eqs. 15–26 and 15–28, we have

D

1.5 m

dm (dDB yB - dDA yA); dt - Cx (2) = 600 [0 - 1.38(12.0)]

a + ©MA =

Cx = 4968 N = 4.97 kN

Dx = 2232N = 2.23 kN

Ans. laws

or

+ ©F = dm A y - y ); : x Bx Ax dt Dx + 4968 = 600 (12.0 - 0)

Ans.

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Dissemination

copyright

Dy = 7200 N = 7.20 kN

Wide

dm A youty - yiny B ; dt Dy = 600[0 - ( -12.0)]

+ c ©Fy = ©






B

15–118. The buckets on the Pelton wheel are subjected to a 2-indiameter jet of water, which has a velocity of 150 ft>s. If each bucket is traveling at 95 ft>s when the water strikes it, determine the power developed by the bucket. gw = 62.4 lb>ft3.

20 150 ft/s

SOLUTION vA = 150 - 95 = 55 ft>s : + )(v ) = -55 cos 20° + 95 = 43.317 ft>s : (: B x + ©F = dm (v - v ) ; x Ax dt Bx 1 62.4 b(p)a b 2[(55)(55 cos 20° - ( - 55)] Fx = a 32.2 12 Fx = 248.07 lb

or

P = 248.07(95) = 23,567 ft # lb>s laws

P = 42.8 hp

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Ans.






20

95 ft/s

15–119. The blade divides the jet of water having a diameter of 3 in. If one-fourth of the water flows downward while the other three-fourths flows upwards, and the total flow is Q = 0.5 ft3>s, determine the horizontal and vertical components of force exerted on the blade by the jet, gw = 62.4 lb>ft3.

3 in.

SOLUTION Equations of Steady Flow: Here, the flow rate Q = 0.5 ft2>s. Then, Q 0.5 dm 62.4 y = = = rw Q = (0.5) = 0.9689 slug>s. = 10.19 ft>s. Also, p 3 2 A dt 32.2 A B 4

12

Applying Eq. 15–25 we have ©Fx = ©

dm A youts - yins B ; - Fx = 0 - 0.9689 (10.19) dt

©Fy = ©

1 dm 3 A youty - yiny B ; Fy = (0.9689)(10.19) + (0.9689)(-10.19) dt 4 4

Fx = 9.87 lb

Ans.

Fy = 4.93 lb

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*15–120. The fan draws air through a vent with a speed of 12 ft>s. If the cross-sectional area of the vent is 2 ft2, determine the horizontal thrust on the blade. The specific weight of the air is ga = 0.076 lb>ft3. v

SOLUTION dm = rvA dt =

0.076 (12)(2) 32.2

= 0.05665 slug>s ©F =

dm (v - vA) dt B

T = 0.05665(12 - 0) = 0.680 lb

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12 ft/s

15–121. The gauge pressure of water at C is 40 lb>in2 . If water flows out of the pipe at A and B with velocities vA = 12 ft>s and vB = 25 ft>s, determine the horizontal and vertical components of force exerted on the elbow necessary tohold the pipe assembly in equilibrium. Neglect the weight of water within the pipe and the weight of the pipe. The pipe has a diameter of 0.75 in. at C, and at A and B the diameter is 0.5 in. gw = 62.4 lb>ft3.

vA

3

B A

dmA 62.4 0.25 2 = (12)(p)a b = 0.03171 slug>s dt 32.2 12

vC C

dmB 0.25 2 62.4 (25)(p) a = b = 0.06606 slug>s dt 32.2 12 dm C = 0.03171 + 0.06606 = 0.09777 slug>s dt vC AC = vA AA + vB AB

in

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laws

or

0.375 2 0.25 2 0.25 2 b = 12(p)a b + 25(p)a b 12 12 12

Dissemination

Wide

copyright

vC = 16.44 ft>s

(including

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the

by

3 40(p)(0.375)2 - Fx = 0 - 0.03171(12) a b - 0.09777(16.44) 5

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permitted.

+ ©F = dmB v + dmA v - dm C v : x dt B s dt As dt Cs

Ans.

assessing

is

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the

Fx = 19.5 lb

provided

integrity

of

and

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this

dmB dmA dm C v + v v dt By dt Ay dt Cy

any

destroy

of

and

Fy = 1.9559 = 1.96 lb






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4 Fy = 0.06606(25) + 0.03171 a b(12) - 0 5

courses

part

the

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This

+ c ©Fy =

vB

5

4

SOLUTION

vC(p)a

12 ft/s

Ans.

25 ft/s

15–122. A speedboat is powered by the jet drive shown. Seawater is drawn into the pump housing at the rate of 20 ft3>s through a 6-in.-diameter intake A. An impeller accelerates the water flow and forces it out horizontally through a 4-in.- diameter nozzle B. Determine the horizontal and vertical components of thrust exerted on the speedboat. The specific weight of seawater is gsw = 64.3 lb>ft3.

B

SOLUTION

45

Steady Flow Equation: The speed of the sea water at the hull bottom A and B are Q Q 20 20 vA = = = 101.86 ft>s and vB = = = 229.18 ft>s and AA AB p 6 2 p 4 2 a b a b 4 12 4 12 the mass flow rate at the hull bottom A and nozle B are the same, i.e., dmA dmB 64.3 dm = = = rsw Q = a b(20) = 39.94 slug>s. By referring to the dt dt dt 32.2 free-body diagram of the control volume shown in Fig. a,

C A vB B x - A vA B x D ;

Fx = 39.94 A 229.18 - 101.86 cos 45° B laws

or

+ b ©F = dm a: x dt

= 6276.55 lb = 6.28 kip in

teaching

Web)

Ans.

C A vB B y - A vA B y D ;

Fy = 39.94 A 101.86 sin 45° - 0 B

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Dissemination

dm dt

copyright

a + c b ©Fy =

instructors

= 2876.53 lb = 2.28 kip

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Ans.






A

15–123. A plow located on the front of a locomotive scoops up snow at the rate of 10 ft3>s and stores it in the train. If the locomotive is traveling at a constant speed of 12 ft>s, determine the resistance to motion caused by the shoveling. The specific weight of snow is gs = 6 lb>ft3.

SOLUTION ©Fx = m

dm t dv + vD>t dt dt

F = 0 + (12 - 0) a

10(6) b 32.2

F = 22.4 lb

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*15–124. The boat has a mass of 180 kg and is traveling forward on a river with a constant velocity of 70 km>h, measured relative to the river. The river is flowing in the opposite direction at 5 km>h. If a tube is placed in the water, as shown, and it collects 40 kg of water in the boat in 80 s, determine the horizontal thrust T on the tube that is required to overcome the resistance due to the water collection and yet maintain the constant speed of the boat. rw = 1 Mg>m3.

T

SOLUTION 40 dm = = 0.5 kg>s dt 80 vD>t = (70)a

dm i dv + vD>i dt dt

T = 0 + 19.444(0.5) = 9.72 N

Ans.

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©Fs = m

1000 b = 19.444 m>s 3600






vR

5 km/h

15–125. Water is discharged from a nozzle with a velocity of 12 m>s and strikes the blade mounted on the 20-kg cart. Determine the tension developed in the cord, needed to hold the cart stationary, and the normal reaction of the wheels on the cart. The nozzle has a diameter of 50 mm and the density of water is rw = 1000 kg>mg3.

45⬚ A

SOLUTION Steady Flow Equation: Here, the mass flow rate at sections A and B of the control p dm = rWQ = rWAv = 1000 c (0.052) d (12) = 7.5p kg>s volume is dt 4 Referring to the free-body diagram of the control volume shown in Fig. a, dm + : ©Fx = dt [(vB)x - (vA)x];

- Fx = 7.5p(12 cos 45° - 12) Fx = 82.81 N

dm + c ©Fy = [(vB)y - (vA)y]; dt

Fy = 7.5p(12 sin 45° - 0)

or

Fy = 199.93 N

in

teaching

Web)

laws

Equilibrium: Using the results of Fx and Fy and referring to the free-body diagram of the cart shown in Fig. b,

+ c ©Fy = 0;

N - 20(9.81) - 199.93 = 0

N = 396 N

Ans.

Wide

T = 82.8 N

Dissemination

82.81 - T = 0

copyright

+ ©F = 0; : x

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Ans.






B

15–126. Water is flowing from the 150-mm-diameter fire hydrant with a velocity vB = 15 m>s. Determine the horizontal and vertical components of force and the moment developed at the base joint A, if the static (gauge) pressure at A is 50 kPa. The diameter of the fire hydrant at A is 200 mm. rw = 1 Mg>m3.

vB

15 m/s

B 500 mm

SOLUTION dm = rvAA B = 1000(15)(p)(0.075)2 dt

A

dm = 265.07 kg>s dt vA = (

dm 1 265.07 ) = dt rA A 1000(p)(0.1)2

vA = 8.4375 m>s

laws

or

+ ©F = dm (v - v ) ; x Ax dt Bx A x = 265.07(15 - 0) = 3.98 kN

in

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Web)

Ans.

dm (vBy - vAy) dt instructors

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+ c ©Fy =

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-A y + 50(103)(p)(0.1)2 = 265.07(0- 8.4375)

Ans.

work

student

the

by

and

use

Ay = 3.81 kN

the

(including

for

dm (dAB vB - dAA vA) dt

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a + ©MA =

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This

M = 1.99 kN # m

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this

M = 265.07(0.5(15) - 0)






Ans.

15–127. A coil of heavy open chain is used to reduce the stopping distance of a sled that has a mass M and travels at a speed of v0. Determine the required mass per unit length of the chain needed to slow down the sled to (1>2)v0 within a distance x = s if the sled is hooked to the chain at x = 0. Neglect friction between the chain and the ground.

SOLUTION Observing the free-body diagram of the system shown in Fig. a, notice that the pair of forces F, which develop due to the change in momentum of the chain, cancel each other since they are internal to the system. Here, vD>s = v since the chain on the dms ground is at rest. The rate at which the system gains mass is = m¿v and the dt mass of the system is m = m¿x + M. Referring to Fig. a, dv + v A m¿v B dt

0 = A m¿x + M B dv + m¿v2 dt

(1) in

teaching

Web)

laws

0 = A m¿x + M)

or

+ b ©F = m dv + v dms ; a: s D>s dt dt

dx dx = v or dt = , dt v instructors

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Dissemination

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Since

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dv + m¿v2 = 0 dx

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by

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A m¿x + M B v

(2)

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the

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work

student

dv m¿ = -¢ ≤ dx v m¿x + M

this

assessing

1 v at x = s, 2 0

provided

integrity

of

and

work

Integrating using the limit v = v0 at x = 0 and v =

v0

= -1n A m¿x + M B

the

is

s

part any

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1 2 v0

will

1n v

dv m¿ = b dx a v L0 m¿x + M and

Lv0

s

This

v0

sale

1 2

0

1 M = 2 m¿s + M m¿ =






M s

Ans.

*15–128. The car is used to scoop up water that is lying in a trough at the tracks. Determine the force needed to pull the car forward at constant velocity v for each of the three cases. The scoop has a cross-sectional area A and the density of water is rw .

v

v

F1

F2

(a)

(b) v

SOLUTION F3

The system consists of the car and the scoop. In all cases ©Ft = m

dme dv - vD>e dt dt

(c)

F = 0 - v(r)(A) v F = v2 r A

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15–129. 250 mm

The water flow enters below the hydrant at C at the rate of 0.75 m3>s. It is then divided equally between the two outlets at A and B. If the gauge pressure at C is 300 kPa, determine the horizontal and vertical force reactions and the moment reaction on the fixed support at C. The diameter of the two outlets at A and B is 75 mm, and the diameter of the inlet pipe at C is 150 mm. The density of water is rw = 1000 kg>m3. Neglect the mass of the contained water and the hydrant.

30⬚ A B 650 mm 600 mm

SOLUTION Free-Body Diagram: The free-body diagram of the control volume is shown in Fig. a. The force exerted on section A due to the water pressure is FC = pCAC = p 300(103)c A 0.152 B d = 5301.44 N. The mass flow rate at sections A, B, and C, are 4 Q dmC dmA dmB 0.75 = = rW a b = 1000a b = 375 kg>s = rWQ = and dt dt 2 2 dt 1000(0.75) = 750 kg>s.

C

Web)

laws

Q 0.75 = = 42.44 m>s. p AC (0.152) 4 in

vC =

teaching

Q>2 0.75>2 = = 84.88 m>s p AA (0.0752) 4

Wide

copyright

vA = vB =

or

The speed of the water at sections A, B, and C are

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is

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for

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student

the

by

Cx = - 375(84.88 cos 30°) + 375(84.88) - 0

Ans.

and

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assessing

is

work

solely

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protected

the

(including

Cx = 4264.54 N = 4.26 kN dmA dmB dmC (vA)y + (vB)y (vC)y; + c ©Fy = dt dt dt

part

the

is

This

provided

integrity

of

- Cy + 5301.44 = 375(84.88 sin 30°) + 0 - 750(42.44) any

courses

Ans. destroy

of

and

Cy = 21 216.93 N = 2.12 kN

sale

will

their

Writing the steady flow equation about point C, dmA dmB dmC dvA + dvB dvC; + ©MC = dt dt dt

-MC = 375(0.65)(84.88 cos 30°) - 375(0.25)(84.88 sin 30°) + [- 375(0.6)(84.88)] - 0 MC = 5159.28 N # m = 5.16 kN # m






Ans.

not

instructors

States

World

permitted.

Dissemination

Steady Flow Equation: Writing the force steady flow equations along the x and y axes, + ©F = dmA (v ) + dmB (v ) - dm C (v ) ; : x A x B x C x dt dt dt

15–130. The mini hovercraft is designed so that air is drawn in at a constant rate of 20 m3>s by the-fan blade and channeled to provide a vertical thrust F, just sufficient to lift the hovercraft off the water, while the remaining air is channeled to produce a horizontal thrust T on the hovercraft. If the air is discharged horizontally at 200 m>s and vertically at 800 m>s, determine the thrust T produced. The hovercraft and its passenger have a total mass of 1.5 Mg, and the density of the air is ra = 1.20 kg>m3.

T

F

SOLUTION Steady Flow Equation: The free-body diagram of the control volume A is shown in dmA Fig. a. The mass flow rate through the control volume is = raQA = 1.20QA. dt Since the air intakes from a large reservoir (atmosphere), the velocity of the air entering the control volume can be considered zero; i.e., (vA)in = 0. Also, the force acting on the control volume is required to equal the weight of the hovercraft. Thus, F = 1500(9.81) N. dmA c (vA)out - (vA)in d ; 1500(9.81) = 1.20QA (800 - 0) dt laws

or

QA = 15.328 m3>s

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the

(including

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work

student

the

by

+ )©Fx = dmB c(vB)out - (vB)in d ; T = 5.606(200 - 0) = 1121.25 N = 1.12 kN (; dt Ans.






permitted.

Dissemination

Wide

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The flow rate through the control volume B, Fig. b, is QB = 20 - QA = 20 - 15.328 = 4.672 m3>s. Thus, the mass flow rate of this control volume is dmB = raQB = 1.20(4.672) = 5.606 kg>s. Again, the intake velocity of the control dt volume B is equal to zero; i.e., (vB)in = 0. Referring to the free-body diagram of this control volume, Fig. b,

Web)

( + T )©Fy =

15–131. B

Sand is discharged from the silo at A at a rate of 50 kg>s with a vertical velocity of 10 m>s onto the conveyor belt, which is moving with a constant velocity of 1.5 m>s. If the conveyor system and the sand on it have a total mass of 750 kg and center of mass at point G, determine the horizontal and vertical components of reaction at the pin support B roller support A. Neglect the thickness of the conveyor.

1.5 m/s

G

30⬚

A

SOLUTION

10 m/s 4m

Steady Flow Equation: The moment steady flow equation will be written about point B to eliminate Bx and By. Referring to the free-body diagram of the control volume shown in Fig. a, + ©MB =

dm (dvB - dvA); dt

750(9.81)(4) - Ay(8) = 50[0 - 8(5)] Ay = 4178.5 N = 4.18 kN

Ans.

Writing the force steady flow equation along the x and y axes, -Bx = 50(1.5 cos 30° - 0) Bx = | -64.95 N| = 65.0 N :

Ans. laws

or

+ ©F = dm [(v ) - (v ) ]; : x B x A x dt

teaching

Web)

By + 4178.5 - 750(9.81)

in

dm [(vB)y - (vA)y]; dt

Wide

copyright

+ c ©Fy =

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Dissemination

= 50[1.5 sin 30° - (-10)]

Ans.

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the

By = 3716.25 N = 3.72 kN c






4m

*15–132. The fan blows air at 6000 ft3>min. If the fan has a weight of 30 lb and a center of gravity at G, determine the smallest diameter d of its base so that it will not tip over. The specific weight of air is g = 0.076 lb>ft3.

1.5 ft

G

0.5 ft

SOLUTION

4 ft

1 min 6000 ft3 b * a b = 100 ft3>s. Then, Equations of Steady Flow: Here Q = a min 60 s Q dm 100 0.076 y = = p = ra Q = (100) = 0.2360 slug>s. = 56.59 ft>s. Also, 2 A dt 32.2 (1.5 ) 4 Applying Eq. 15–26 we have a + ©MO =

dm d (dOB yB - dOA yA B ; 30 a 0.5 + b = 0.2360 [4(56.59) - 0] dt 2

d

d = 2.56 ft

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Ans.






15–133. The bend is connected to the pipe at flanges A and B as shown. If the diameter of the pipe is 1 ft and it carries a discharge of 50 ft3>s, determine the horizontal and vertical components of force reaction and the moment reaction exerted at the fixed base D of the support. The total weight of the bend and the water within it is 500 lb, with a mass center at point G. The gauge pressure of the water at the flanges at A and B are 15 psi and 12 psi, respectively. Assume that no force is transferred to the flanges at A and B. The specific weight of water is gw = 62.4 lb>ft3.

G A

45 4 ft D

SOLUTION

teaching

Web)

laws

or

Free-Body Diagram: The free-body of the control volume is shown in Fig. a. The force exerted on sections A and B due to the water pressure is p p FA = PA AA = 15c A 122 B d = 1696.46 lb and FB = PB AB = 12c A 122 B d 4 4 = 1357.17 lb. The speed of the water at, sections A and B are Q 50 = = 63.66 ft>s. Also, the mass flow rate at these two sections vA = vB = p 2 A A1 B 4 dm A dm B 62.4 are = = rW Q = a b (50) = 96.894 slug>s. dt dt 32.2

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in

Steady Flow Equation: The moment steady flow equation will be wr titen about point D to eliminate Dx and Dy.

the

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World

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dm B dm A dvB dvA ; dt dt use

United

on

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of

a + ©MD =

for

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the

by

and

MD + 1357.17 cos 45°(4) - 500 A 1.5 cos 45° B - 1696.46(4)

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the

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= -96.894(4) A 63.66 cos 45° B - [-96.894(4)(63.66)] MD = 10 704.35 lb # ft = 10.7 kip # ft

B

1.5 ft

15–133. continued Writing the force steady flow equation along the x and y axes,

A + c B ©Fy =

dm c A vB B y - A vA B y d ; dt

Dy - 500 - 1357.17 sin 45° = 96.894(63.66 sin 45° - 0) Dy = 5821.44 lb = 5.82 kip + b ©F = dm a: x dt

Ans.

C A vB B x - A vA B x D ;

1696.46 - 1357.17 cos 45° - Dx = 96.894[63.66 cos 45° - 63.66 D Dx = 2543.51 lb = 2.54 kip

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Ans.






15–134. Each of the two stages A and B of the rocket has a mass of 2 Mg when their fuel tanks are empty. They each carry 500 kg of fuel and are capable of consuming it at a rate of 50 kg>sand eject it with a constant velocity of 2500 m> s, measured with respect to the rocket. The rocket is launched vertically from rest by first igniting stage B. Then stage A is ignited immediately after all the fuel in B is consumed and A has separated from B. Determine the maximum velocity of stage A. Neglect drag resistance and the variation of the rocket’s weight with altitude.

A

B

SOLUTION The mass of the rocket at any instant t is m = (M + m0) - qt. Thus, its weight at the same instant is W = mg = [(M + m0) - qt]g. + c ©Fs = m

dme dv dv ; -[(M + m0) - qt]g = [(M + m0) - qt] - vD>e - vD>eq dt dt dt

vD>eq dv = - g dt (M + m0) - qt M = 4000 kg,

q = 50 kg>s ,

and or

m0 = 1000 kg ,

laws

During the first stage, vD>e = 2500 m>s. Thus,

Dissemination

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2500(50) dv = - 9.81 dt (4000 + 1000) - 50t

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2500 dv = a - 9.81 b m>s2 dt 100 - t

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The time that it take to complete the first stage is equal to the time for all the fuel 500 in the rocket to be consumed, i.e., t = = 10 s. Integrating, 50 this

assessing

2500 - 9.81 b dt 100 - t

integrity

of

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L0

a

the

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L0

10 s

dv =

provided

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10 s

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v1 = [- 2500 ln(100 - t) - 9.81t] 2 sale

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= 165.30 m>s

During the second stage of launching, M = 2000 kg, m0 = 500 kg, q = 50 kg>s, and vD>e = 2500 m>s. Thus, Eq. (1) becomes 2500(50) dv = - 9.81 dt (2000 + 500) - 50t 2500 dv = a - 9.81 b m>s2 dt 50 - t The maximum velocity of rocket A occurs when it has consumed all the fuel. Thus, 500 the time taken is given by t = = 10 s. Integrating with the initial condition 50 v = v1 = 165.30 m>s when t = 0 s, vmax

10 s

dv =

L165.30 m>s

L0

a

2500 - 9.81b dt 50 - t

vmax - 165.30 = [ - 2500 ln(50 - t) - 9.81t] 2

10 s 0






vmax = 625 m>s

Ans.

15–135. A power lawn mower hovers very close over the ground. This is done by drawing air in at a speed of 6 m>s through an intake unit A, which has a cross-sectional area of A A = 0.25 m2, and then discharging it at the ground, B, where the cross-sectional area is A B = 0.35 m2. If air at A is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has a mass of 15 kg with center of mass at G. Assume that air has a constant density of ra = 1.22 kg>m3.

vA A G B

SOLUTION dm = r A A vA = 1.22(0.25)(6) = 1.83 kg>s dt + c ©Fy =

dm ((vB)y - (vA)y) dt

P = (0.35) - 15(9.81) = 1.83(0 - ( -6)) P = 452 Pa

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Ans.






*15–136. The 12-ft-long open-link chain has 2 ft of its end suspended from the hole as shown. If a force of P = 10 lb is applied to its end and the chain is released from rest, determine the velocity of the chain at the instant the entire chain has been extended. The chain has a weight of 2 lb> ft.

2 ft

SOLUTION P ⫽ 10 lb

From the free-body diagram of the system shown in Fig. a, F cancels since it is internal to the system. Here, vD>s = v since the chain on the horizontal surface is at dms 2 rest. The rate at which the chain gains mass is = a vb slug>s, and the mass dt 32.2 2y of the chain is m = a b slug. 32.2 Referring to Fig. a,

+ T ©Fs = m

dms dv ; + vD>s dt dt

10 + 2y = a

2y dv 2 b + va vb 32.2 dt 32.2 dv + 2v2 = 322 + 64.4y dt laws

or

2y

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dv + 2v2 = 322 + 64.4y dy United

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dv + 2yv2 b dy = A 322y + 64.4y2 B dy dy

dy

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= 2vy2

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d A v2y2 B = A 322y + 64.4y2 B dy sale

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Integrating,

v2y2 = 161y2 + 21.467y3 + C Substituting the initial condition v = 0 at y = 2 ft, 0 = 161 A 22 B + 21.467 A 23 B + C

C = - 815.73

ft4 s2

Thus, v2y2 = 161y2 + 21.467y3 - 815.73 At the instant the entire chain is in motion y = 12 ft. v2 A 122 B = 161 A 122 B + 21.467 A 123 B - 815.73 v = 20.3 ft>s






Ans.

15–137. P

A chain of mass m0 per unit length is loosely coiled on the floor. If one end is subjected to a constant force P when y = 0, determine the velocity of the chain as a function of y. y

SOLUTION From the free-body diagram of the system shown in Fig. a, F cancels since it is internal to the system. Here, vD>s = v since the chain on the horizontal surface is at dms rest. The rate at which the chain gains mass is = m0v, and the mass of the dt system is m = m0y. Referring to Fig. a, P - m0gy = m0y

dv + v(m0v) dt

P - m0gy = m0y

dv + m0v2 dt

dv P + v2 = - gy m0 dt in

teaching

Web)

y

or

dms dv ; + vD>s dt dt

laws

+ c ©Fs = m

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dy dy , = v or dt = v dt instructors

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permitted.

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dv P + v2 = - gy m0 dy

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dv 2P + 2v2y b dy = a y - 2gy2 bdy m0 dy

(1)

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and provided

dv + 2yv2, then Eq. (1) can be written as dy part

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= 2vy2

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2P y - 2gy2 b dy m0

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d A v2y2 B = a

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d A v2y2 B

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assessing

is

a2vy2

Integrating,

L

d A v2y2 B =

v2y2 =

2P a y - 2gy2 b dy L m0

P 2 2 y - gy3 + C m0 3

Substituting v = 0 at y = 0, 0 = 0 - 0 + C

C = 0

Thus, v2y2 =

v = © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





P 2 2 y - gy3 m0 3

2 P - gy 3 B m0

Ans.

15–138. The second stage of a two-stage rocket weighs 2000 lb (empty) and is launched from the first stage with a velocity of 3000 mi>h. The fuel in the second stage weighs 1000 lb. If it is consumed at the rate of 50 lb>s and ejected with a relative velocity of 8000 ft>s, determine the acceleration of the second stage just after the engine is fired. What is the rocket’s acceleration just before all the fuel is consumed? Neglect the effect of gravitation.

SOLUTION Initially, ©Fs = m 0 =

dme dv - v D>e a b dt dt

50 3000 a - 8000 a b 32.2 32.2

a = 133 ft>s2

Ans.

Finally, or

50 2000 a - 8000 a b 32.2 32.2

teaching

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laws

0 =

a = 200 ft>s2

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15–139. The missile weighs 40 000 lb. The constant thrust provided by the turbojet engine is T = 15 000 lb. Additional thrust is provided by two rocket boosters B. The propellant in each booster is burned at a constant rate of 150 lb/s, with a relative exhaust velocity of 3000 ft>s. If the mass of the propellant lost by the turbojet engine can be neglected, determine the velocity of the missile after the 4-s burn time of the boosters.The initial velocity of the missile is 300 mi/h.

T B

SOLUTION + ©F = m dv - v dme : s D>e dt dt At a time t, m = m0 - ct, where c =

T + cvD>e m0 - ct

c

b1n a

m0 b + y0 m0 - ct

or

T + cvD>e

b dt

(1) in

v = a

L0

a

Web)

Lv0

t

dv =

laws

v

dv - vD>e c dt

teaching

T = (m0 - ct)

dme . dt

40 000 150 = 1242.24 slug, c = 2 a b = 9.3168 slug>s, vD>e = 3000 ft>s, 32.2 32.2 the on

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vmax = 580 ft s

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and

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vmax = a

assessing

is

work

solely

15 000 + 9.3168(3000) 1242.24 b 1n a b + 440 9.3168 1242.24 - 9.3168(4)

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(including

Substitute the numerical values into Eq. (1):






Ans.

permitted.

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not

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300(5280) = 440 ft>s. 3600

the

t = 4 s, v0 =

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Here, m0 =

*15–140. The 10-Mg helicopter carries a bucket containing 500 kg of water, which is used to fight fires. If it hovers over the land in a fixed position and then releases 50 kg>s of water at 10 m>s, measured relative to the helicopter, determine the initial upward acceleration the helicopter experiences as the water is being released.

a

SOLUTION + c ©Ft = m

dm e dy - yD>e dt dt

Initially, the bucket is full of water, hence m = 10(103) + 0.5(103) = 10.5(103) kg 0 = 10.5(103)a - (10)(50) a = 0.0476 m>s2

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Ans.






15–141. The earthmover initially carries 10 m3 of sand having a density of 1520 kg>m3. The sand is unloaded horizontally through a 2.5 m 2 dumping port P at a rate of 900 kg>s measured relative to the port. Determine the resultant tractive force F at its front wheels if the acceleration of the earthmover is 0.1 m>s2 when half the sand is dumped. When empty, the earthmover has a mass of 30 Mg. Neglect any resistance to forward motion and the mass of the wheels. The rear wheels are free to roll.

P F

SOLUTION When half the sand remains, m = 30 000 +

1 (10)(1520) = 37 600 kg 2

dm = 900 kg>s = r vD>e A dt 900 = 1520(vD>e)(2.5)

laws

or

vD>e = 0.237 m>s

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dv = 0.1 dt

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a =

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+ ©F = m dv - dm v ; s dt dt

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F = 37 600(0.1) - 900(0.237)

Ans.

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F = 3.55 kN






15–142. The 12-Mg jet airplane has a constant speed of 950 km>h when it is flying along a horizontal straight line. Air enters the intake scoops S at the rate of 50 m3>s. If the engine burns fuel at the rate of 0.4 kg>s and the gas (air and fuel) is exhausted relative to the plane with a speed of 450 m>s, determine the resultant drag force exerted on the plane by air resistance. Assume that air has a constant density of 1.22 kg>m3. Hint: Since mass both enters and exits the plane, Eqs. 15–28 and 15–29 must be combined to yield dme dmi dv ©Fs = m - vD>e + vD>i . dt dt dt

v

950 km/h

S

SOLUTION ©Fs = m

dm e dm i dv (vD>E) + (v ) dt dt dt D>i

v = 950 km>h = 0.2639 km>s,

(1)

dv = 0 dt

or

vD>E = 0.45 km>s teaching

Web)

laws

vD>t = 0.2639 km>s

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permitted.

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in

dm t = 50(1.22) = 61.0 kg>s dt

the

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dm e = 0.4 + 61.0 = 61.4 kg>s dt

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work

student

Forces T and R are incorporated into Eq. (1) as the last two terms in the equation.

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the

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+ ) - F = 0 - (0.45)(61.4) + (0.2639)(61) (; D

assessing

FD = 11.5 kN

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Ans.






15–143. The jet is traveling at a speed of 500 mi>h, 30° with the horizontal. If the fuel is being spent at 3 lb>s,and the engine takes in air at 400 lb>s, whereas the exhaust gas (air and fuel) has a relative speed of 32 800 ft>s, determine the acceleration of the plane at this instant. The drag resistance of the air is FD = 10.7v22 lb, where the speed is measured in ft>s. The jet has a weight of 15 000 lb. Hint: Since mass both enters and exits the plane, Eqs. 15-28 and 15-29 must be combined to yield ©Fx = m

500 mi/ h

30

dme dmi dv . - vD>e + vD>i dt dt dt

SOLUTION dmi 400 = = 12.42 slug>s dt 32.2 dme 403 = = 12.52 slug>s dt 32.2 v = vD>i = 500 mi>h = 733.3 ft>s

teaching

Web)

laws

or

dme dmi dv - vD>e + vD>i dt dt dt in

a + ©Fs = m

15 000 dv - 32 800(12.52) + 733.3(12.42) 32.2 dt instructors

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permitted.

Dissemination

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-(15 000) sin 30° - 0.7(733.3)2 =

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dv = 37.5 ft>s2 dt

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*15–144. A four-engine commercial jumbo jet is cruising at a constant speed of 800 km>h in level flight when all four engines are in operation. Each of the engines is capable of discharging combustion gases with a velocity of 775 m>s relative to the plane. If during a test two of the engines, one on each side of the plane, are shut off, determine the new cruising speed of the jet. Assume that air resistance (drag) is proportional to the square of the speed, that is, FD = cv2, where c is a constant to be determined. Neglect the loss of mass due to fuel consumption.

SOLUTION Steady Flow Equation: Since the air is collected from a large source (the atmosphere), its entrance speed into the engine is negligible. The exit speed of the air from the engine is + b a:

ve + vp + ve>p

ve = - 222.22 + 775 = 552.78 m>s : in

teaching

Web)

+ b a:

laws

or

When the four engines are in operation, the airplane has a constant speed of m 1h b = 222.22 m>s. Thus, vp = c 800(103) d a h 3600 s

Dissemination

Wide

copyright

Referring to the free-body diagram of the airplane shown in Fig. a,

the

not

instructors

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World

permitted.

dm (552.78 - 0) dt

C(222.222) = 4

dm dt

work

student

the

(including

for

the

by

C = 0.044775

and

use

United

on

learning.

is

of

+ ©F = dm C A v B - A v B D ; : x B x A x dt

integrity

of

and

work

this

ve = - vp + 775

the

is

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+ b a:

assessing

is

work

solely

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protected

When only two engines are in operation, the exit speed of the air is

and

any

courses

part

Using the result for C,

dm dm ≤ A vp 2 B = 2 C - vp + 775 B - 0 D dt dt will

destroy

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their

C A vB B x - A vA B x D ; ¢ 0.044775 sale

+ ©F = dm : x dt

0.044775vp 2 + 2vp - 1550 = 0 Solving for the positive root, vp = 165.06 m s = 594 km h






Ans.

15–145. v

The car has a mass m0 and is used to tow the smooth chain having a total length l and a mass per unit of length m¿. If the chain is originally piled up, determine the tractive force F that must be supplied by the rear wheels of the car, necessary to maintain a constant speed v while the chain is being drawn out. F

SOLUTION + ©F = m dv + v dmi : s D>i dt dt At a time t, m = m0 + ct, where c =

Here, vD>i = v,

dmi m¿dx = = m¿v. dt dt

dv = 0. dt F = (m0 - m¿v )(0) + v(m¿v) = m¿v2

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Ans.






15–146. A rocket has an empty weight of 500 lb and carries 300 lb of fuel. If the fuel is burned at the rate of 1.5 lb / s and ejected with a velocity of 4400 ft/ s relative to the rocket, determine the maximum speed attained by the rocket starting from rest. Neglect the effect of gravitation on the rocket.

SOLUTION dme dv - vD>e dt dt

At a time t, m = m0 - ct, where c = 0 = (m0 - ct) t

dv =

L0

L0

cvD>e m0 - ct

≤ dt

m0 ≤ m0 - ct

(1)

in

teaching

Web)

v = vD> e ln ¢

¢

dv - vD>e c dt

or

v

dme . In space the weight of the rocket is zero. dt

laws

+ c ©Fs =

the

= 0.4658 slug>s, vD>e = 4400 ft>s.

not

15 32.2

is

and

use

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= 24.8447 slug, c =

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500 + 300 32.2

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Here, m0 =

instructors

States

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permitted.

Dissemination

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The maximum speed occurs when all the fuel is consumed, that is, when t = 300 15 = 20 s.

for

work

student

the

by

Substitute the numerical into Eq. (1):

work

solely

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protected

the

(including

24.8447 b 24.8447 - (0.4658(20))

assessing

is

vmax = 4400 1na

integrity

of

and

work

provided

any

courses

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Ans.

this

vmax = 2068 ft>s

15–147. Determine the magnitude of force F as a function of time, which must be applied to the end of the cord at A to raise the hook H with a constant speed v = 0.4 m>s. Initially the chain is at rest on the ground. Neglect the mass of the cord and the hook. The chain has a mass of 2 kg>m.

A

H v

0.4 m/s

SOLUTION dv = 0, dt

y = vt

mi = my = mvt dmi = mv dt + c ©Fs = m

dm i dv + vD>i ( ) dt dt

F - mgvt = 0 + v(mv) laws

or

F = m(gvt + v2)

Wide

copyright

in

teaching

Web)

= 2[9.81(0.4)t + (0.4)2]

Ans.

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learning.

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of

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States

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permitted.

Dissemination

F = (7.85t + 0.320) N






*15–148. The truck has a mass of 50 Mg when empty.When it is unloading 5 m3 of sand at a constant rate of 0.8 m3>s, the sand flows out the back at a speed of 7 m> s, measured relative to the truck, in the direction shown. If the truck is free to roll, determine its initial acceleration just as the load begins to empty. Neglect the mass of the wheels and any frictional resistance to motion. The density of sand is rs = 1520 kg>m3.

a

45⬚ 7 m/s

SOLUTION A System That Loses Mass: Initially, the total mass of the truck is dme and m = 50(103) + 5(1520) = 57.6(103) kg = 0.8(1520) = 1216 kg>s . dt Applying Eq. 15–29, we have + ©F = m dv - v dme ; : s D>e dt dt

0 = 57.6(103)a - (0.8 cos 45°)(1216) a = 0.104 m>s2

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Wide

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laws

or

Ans.






15–149. The chain has a total length L 6 d and a mass per unit length of m¿ . If a portion h of the chain is suspended over the table and released, determine the velocity of its end A as a function of its position y. Neglect friction.

h y d A

SOLUTION ©Fs = m

dm e dv + vD>e dt dt

m¿gy = m¿y

dv + v(m¿v) dt

m¿gy = m¿ ay

Since dt =

dy , we have v or

dv + v2 dy in

teaching

Web)

laws

gy = vy

dv + v2 b dt

Dissemination

Wide

copyright

Multiply by 2y and integrate:

the

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instructors

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permitted.

dv + 2yv2 b dy dy

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2gy2 dy =

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2 3 3 g y + C = v2y2 3






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y3 - h3 2 ga b B3 y2

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v =

2 y3 - h3 ga b 3 y2

sale

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2 when v = 0, y = h, so that C = - gh3 3

Ans.

16–1. The angular velocity of the disk is defined by v = 15t2 + 22 rad>s, where t is in seconds. Determine the magnitudes of the velocity and acceleration of point A on the disk when t = 0.5 s.

A

0.8 m

SOLUTION v = (5 t2 + 2) rad>s a =

dv = 10 t dt

t = 0.5 s v = 3.25 rad>s a = 5 rad>s2 vA = vr = 3.25(0.8) = 2.60 m>s

Ans. laws

or

a z = ar = 5(0.8) = 4 m>s2 in

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a n = v2r = (3.25)2(0.8) = 8.45 m>s2

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Ans.

copyright

a A = 2(4)2 + (8.45)2 = 9.35 m>s2






16–2. A flywheel has its angular speed increased uniformly from 15 rad>s to 60 rad>s in 80 s. If the diameter of the wheel is 2 ft, determine the magnitudes of the normal and tangential components of acceleration of a point on the rim of the wheel when t = 80 s, and the total distance the point travels during the time period.

SOLUTION v = v0 + ac t 60 = 15 + ac(80) ac = 0.5625 rad/s2 a t = ar = (0.5625)(1) = 0.562 ft/s2

Ans.

a n = v2r = (60)2(1) = 3600 ft/s2

Ans.

v2 = v20 + 2ac (u - u0) (60)2 = (15)2 + 2(0.5625)(u-0) laws

or

u = 3000 rad

teaching

Web)

s = ur = 3000(1) = 3000 ft

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Ans.






16–3. The disk is originally rotating at v0 = 8 rad>s. If it is subjected to a constant angular acceleration of a = 6 rad>s2, determine the magnitudes of the velocity and the n and t components of acceleration of point A at the instant t = 0.5 s.

V0 B 1.5 ft 2 ft

SOLUTION v = v0 + ac t v = 8 + 6(0.5) = 11 rad>s v = rv;

vA = 2(11) = 22 ft>s

Ans.

at = ra;

(aA)t = 2(6) = 12.0 ft>s2

Ans.

(aA)n = (11)2(2) = 242 ft>s2

Ans.

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laws

or

an = v2r;






8 rad/s

A

*16–4. The disk is originally rotating at v0 = 8 rad>s. If it is subjected to a constant angular acceleration of a = 6 rad>s2, determine the magnitudes of the velocity and the n and t components of acceleration of point B just after the wheel undergoes 2 revolutions.

V0 B 1.5 ft 2 ft

SOLUTION v2 = v20 + 2ac (u - u0) v2 = (8)2 + 2(6)[2(2p) - 0] v = 14.66 rad>s Ans.

(aB)t = ar = 6(1.5) = 9.00 ft>s2

Ans.

(aB)n = v2r = (14.66)2(1.5) = 322 ft>s2

Ans.

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laws

or

vB = vr = 14.66(1.5) = 22.0 ft>s






8 rad/s

A

16–5. Initially the motor on the circular saw turns its drive shaft at v = 120t2>32 rad>s, where t is in seconds. If the radii of gears A and B are 0.25 in. and 1 in., respectively, determine the magnitudes of the velocity and acceleration of a tooth C on the saw blade after the drive shaft rotates u = 5 rad starting from rest.

SOLUTION v = 20 t2>3 dv 40 -1>3 = t dt 3

a =

du = v dt t

u

L0

du =

L0

20 t2>3 dt

3 u = 20a b t5>3 5

teaching

Web)

laws

or

When u = 5 rad

Wide

copyright

in

t = 0.59139 s

States

World

permitted.

Dissemination

a = 15.885 rad>s2

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v = 14.091 rad>s

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v A r A = vB r B

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14.091(0.25) = vB(1) assessing

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vB = 3.523 rad>s

Ans.

part

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courses

aA rA = aB rB

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this

vC = vB r = 3.523(2.5) = 8.81 in.>s

aB = 3.9712 rad>s2

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15.885(0.25) = aB(1)

(aC)t = aB r = 3.9712(2.5) = 9.928 in.>s2 (aC)n = v2B r = (3.523)2 (2.5) = 31.025 in.>s2 aC = 2(9.928)2 + (31.025)2 = 32.6 in.>s2






Ans.

16–6. A wheel has an initial clockwise angular velocity of 10 rad>s and a constant angular acceleration of 3 rad>s2. Determine the number of revolutions it must undergo to acquire a clockwise angular velocity of 15 rad>s. What time is required?

SOLUTION v2 = v20 + 2ac(u - u0) (15)2 = (10)2 + 2(3)(u- 0) 1 ≤ = 3.32 rev. 2p

u = 20.83 rad = 20.83 ¢

Ans.

v = v0 + ac t 15 = 10 + 3t t = 1.67 s

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Ans.






16–7. If gear A rotates with a constant angular acceleration of aA = 90 rad>s2, starting from rest, determine the time required for gear D to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear D to attain this angular velocity. Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

D

F

SOLUTION aB rB = aA rA rA 15 aB = a baA = a b (90) = 27 rad>s2 rB 50 Since gears C and B share the same shaft, aC = aB = 27 rad>s2. Also, gear D is in mesh with gear C. Thus,

or

aD rD = aC rC rC 25 aD = a b aC = a b (27) = 9 rad>s2 rD 75 600 rev 2p rad 1 min ba ba b = min 1 rev 60 s 20p rad>s. Applying the constant acceleration equation, Dissemination

Wide

copyright

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laws

The final angular velocity of gear D is vD = a

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vD = (vD)0 + aD t

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learning.

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20p = 0 + 9t

Ans. (including

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t = 6.98 s

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the

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1 rev b 2p rad

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This

(20p)2 = 02 + 2(9)(uD - 0)

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of

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this

assessing

vD2 = (vD)0 2 + 2aD [uD - (uD)0]

uD = (219.32 rad) a

sale

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= 34.9 rev






B C

Gear B is in mesh with gear A. Thus,

Ans.

A

*16–8. If gear A rotates with an angular velocity of vA = (uA + 1) rad>s, where uA is the angular displacement of gear A, measured in radians, determine the angular acceleration of gear D when uA = 3 rad, starting from rest. Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

D

F

SOLUTION aA duA = vA dvA aA duA = (uA + 1) d(uA + 1) aA duA = (uA + 1) duA aA = (uA + 1) At uA = 3 rad,

laws

or

aA = 3 + 1 = 4 rad>s2

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Motion of Gear D: Gear A is in mesh with gear B. Thus,

of

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Dissemination

Wide

copyright

aB rB = aA rA rA 15 aB = a baA = a b (4) = 1.20 rad>s2 rB 50

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aD rD = aC rC rC 25 aD = a b aC = a b (1.20) = 0.4 rad>s2 rD 75

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Since gears C and B share the same shaft aC = aB = 1.20 rad>s2. Also, gear D is in mesh with gear C. Thus,






B C

Motion of Gear A:

Ans.

A

16–9. The vertical-axis windmill consists of two blades that have a parabolic shape. If the blades are originally at rest and begin to turn with a constant angular acceleration of ac = 0.5 rad>s2, determine the magnitude of the velocity and acceleration of points A and B on the blade after the blade has rotated through two revolutions.

ac

0.5 rad/s2

B

SOLUTION

10 ft

Angular Motion: The angular velocity of the blade after the blade has rotated 2(2p) = 4p rad can be obtained by applying Eq. 16–7.

20 ft

v2 = v20 + 2ac (u - u0) v2 = 02 + 2(0.5)(4p - 0) v = 3.545 rad>s Motion of A and B: The magnitude of the velocity of point A and B on the blade can be determined using Eq. 16–8. Ans.

vB = vrB = 3.545(10) = 35.4 ft>s

Ans. in

teaching

Web)

laws

or

vA = vrA = 3.545(20) = 70.9 ft>s

States

World

permitted.

Dissemination

Wide

copyright

The tangential and normal components of the acceleration of point A and B can be determined using Eqs. 16–11 and 16–12 respectively.

on

learning.

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(at)A = arA = 0.5(20) = 10.0 ft>s2

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(an)A = v2 rA = A 3.5452 B (20) = 251.33 ft>s2

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(at)B = arB = 0.5(10) = 5.00 ft>s2

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(an)B = v2 rB = A 3.5452 B (10) = 125.66 ft>s2

part

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is

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The magnitude of the acceleration of points A and B are

Ans.

and

any

courses

(a)A = 2(at)2A + (an)2A = 210.02 + 251.332 = 252 ft>s2 5.002 + 125.662 = 126 ft s2 destroy

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(at)2B + (an)2B =

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(a)B = =






Ans.

A

16–10. The vertical-axis windmill consists of two blades that have a parabolic shape. If the blades are originally at rest and begin to turn with a constant angular acceleration of ac = 0.5 rad>s2, determine the magnitude of the velocity and acceleration of points A and B on the blade when t=4 s.

ac

0.5 rad/s2

B

SOLUTION

10 ft

Angular Motion: The angular velocity of the blade at t = 4 s can be obtained by applying Eq. 16–5.

20 ft

v = v0 + ac t = 0 + 0.5(4) = 2.00 rad>s Motion of A and B: The magnitude of the velocity of points A and B on the blade can be determined using Eq. 16–8. vA = vrA = 2.00(20) = 40.0 ft>s

Ans.

vB = vrB = 2.00(10) = 20.0 ft>s

Ans.

teaching

Web)

laws

or

The tangential and normal components of the acceleration of points A and B can be determined using Eqs. 16–11 and 16–12 respectively.

Wide

copyright

in

(at)A = arA = 0.5(20) = 10.0 ft>s2

instructors

States

World

permitted.

Dissemination

(an)A = v2 rA = A 2.002 B (20) = 80.0 ft>s2

on

learning.

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(at)B = arB = 0.5(10) = 5.00 ft>s2

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(an)B = v2 rB = A 2.002 B (10) = 40.0 ft>s2

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The magnitude of the acceleration of points A and B are

Ans.

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solely

(a)A = 2(at)2A + (an)2A = 210.02 + 80.02 = 80.6 ft>s2

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(a)B = 2(at)2B + (an)2B = 25.002 + 40.02 = 40.3 ft>s2






Ans.

A

16–11. If the angular velocity of the drum is increased uniformly from 6 rad>s when t = 0 to 12 rad>s when t = 5 s, determine the magnitudes of the velocity and acceleration of points A and B on the belt when t = 1 s. At this instant the points are located as shown.

45° B

SOLUTION v = v0 + act

A

a = 1.2 rad>s2

12 = 6 + a(5) At t = 1 s,

v = 6 + 1.2(1) = 7.2 rad>s vA = vB = v r = 7.2 ¢

Ans.

4 ≤ = 0.4 ft>s2 12

Ans. laws

or

aA = ar = 1.2 ¢

4 ≤ = 2.4 ft>s 12

in

teaching

Web)

4 ≤ = 0.4 ft>s2 12

Wide

copyright

(aB)t = ar = 1.2 ¢

World

permitted.

Dissemination

= 17.28 ft>s2

is

of

the

not

instructors

4 12

States

(aB)n = v2r = (7.2)2

Ans.

on

learning.

0.42 + 17.282 = 17.3 ft s2

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(aB)2t + (aB)2n =

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aB =






4 in.

*16–12. A motor gives gear A an angular acceleration of aA = 10.25u3 + 0.52 rad>s2, where u is in radians. If this gear is initially turning at 1vA20 = 20 rad>s, determine the angular velocity of gear B after A undergoes an angular displacement of 10 rev.

(vA)0 ⫽ 20 rad/s B 0.05 m

A

AA

SOLUTION aA = 0.25 u3 + 0.5 a du = v dv vA

20p

L0

(0.25 u3 + 0.5)duA = 20p

(0.0625 u4 + 0.5 u)冷 0

=

vA dvA

L20

vA 1 (vA)2冷 20 2

vA = 1395.94 rad>s vA rA = vB rB laws

or

1395.94(0.05) = vB (0.15) vB = 465 rad>s

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Dissemination

Wide

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Ans.






0.15 m

16–13. A motor gives gear A an angular acceleration of aA = 14t32 rad>s2, where t is in seconds. If this gear is initially turning at 1vA20 = 20 rad>s, determine the angular velocity of gear B when t = 2 s.

(vA)0 ⫽ 20 rad/s B 0.05 m

A 0.15 m AA

SOLUTION a A = 4 t3 dv = a dt t

vA

L20

dvA =

L0

t

aA dt =

4 t3dt

L0

vA = t4 + 20 When t = 2 s, vA = 36 rad>s laws

or

vA rA = vB rB in

teaching

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36(0.05) = vB (0.15)

Wide

copyright

vB = 12 rad>s

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Dissemination

Ans.






16–14. The disk starts from rest and is given an angular acceleration a = (2t 2) rad>s2, where t is in seconds. Determine the angular velocity of the disk and its angular displacement when t = 4 s.

P 0.4 m

SOLUTION dv = 2 t2 dt

a =

t

v

2

dv =

L0

L0

v =

2 32t t 3 0

v =

23 t 3

2 t dt

When t = 4 s, 2 3 (4) = 42.7 rad>s 3

in

teaching

Web)

2 3 t dt L0 3

permitted.

Dissemination

Wide

copyright

1 4 t 6 States

World

u =

du =

laws

t

u

L0

Ans. or

v =

learning.

is

of

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When t = 4 s, by

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United

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1 4 (4) = 42.7 rad 6

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u =






Ans.

16–15. The disk starts from rest and is given an angular acceleration a = (5t1>2) rad>s2, where t is in seconds. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when t = 2 s.

P 0.4 m

SOLUTION Motion of the Disk: Here, when t = 0, v = 0. dv = adt t

v

L0 v2

dv =

v

= 0

v = e

L0

1

5t2dt

10 3 2 t t2 3 0 10 3 t2 f rad>s 3

teaching

Web)

laws

10 3 A 2 2 B = 9.428 rad>s 3

in

v =

or

When t = 2 s,

Dissemination

Wide

copyright

When t = 2 s,

permitted.

a = 5 A 2 2 B = 7.071 rad>s2 United

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Motion of point P: The tangential and normal components of the acceleration of point P when t = 2 s are Ans.

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the

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at = ar = 7.071(0.4) = 2.83 m>s2

Ans.

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This

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this

assessing

a n = v2r = 9.4282(0.4) = 35.6 m>s2






*16–16. The disk starts at v0 = 1 rad>s when u = 0, and is given an angular acceleration a = (0.3u) rad>s2, where u is in radians. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when u = 1 rev.

P 0.4 m

SOLUTION a = 0.3u v

L1

vdv =

u

L0

0.3udu

u 1 22v v = 0.15u2 2 2 1 0

v2 - 0.5 = 0.15u2 2

laws

or

v = 20.3u2 + 1

in

teaching

Web)

At u = 1 rev = 2p rad Dissemination

Wide

copyright

v = 20.3(2p)2 + 1

learning.

is

of

the

not

instructors

States

World

permitted.

v = 3.584 rad>s

Ans.

the

by

and

use

United

on

a t = ar = 0.3(2p) rad>s 2 (0.4 m) = 0.7540 m>s2

protected

the

(including

for

work

student

a n = v2r = (3.584 rad>s)2(0.4 m) = 5.137 m>s2

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of

and

any

courses

part

the

is

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provided

integrity

of

and

work

this

assessing

is

work

solely

of

a p = 2(0.7540)2 + (5.137)2 = 5.19 m>s2






Ans.

16–17. Starting at (vA)0 = 3 rad>s, when u = 0, s = 0, pulley A is given an angular acceleration a = (0.6u) rad>s2, where u is in radians. Determine the speed of block B when it has risen s = 0.5 m. The pulley has an inner hub D which is fixed to C and turns with it.

150 mm C

D A 50 mm

75 mm

SOLUTION aa = 0.6uA uC =

B

0.5 = 6.667 rad 0.075 s

uA(0.05) = (6.667)(0.15) uA = 20 rad adu = vdv vA

0.6uAduA =

L3

vAdvA laws

L0

or

20

teaching

Web)

1 2 vA v 2 2 A 3

Wide States

World

permitted.

Dissemination

0

=

in

20

copyright

0.3u2A 2

United

on

learning.

is

of

the

not

instructors

1 2 v - 4.5 2 A by

and

use

120 =

the

(including

for

work

student

the

vA = 15.780 rad>s

assessing

is

work

solely

of

protected

15.780(0.05) = vC (0.15)

the

is

This

provided

integrity

of

and

work

this

vC = 5.260 rad>s

part

vB = 5.260(0.075) = 0.394 m>s

sale

will

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destroy

of

and

any

courses

Ans.






16–18. Starting from rest when s = 0, pulley A is given a constant angular acceleration ac = 6 rad>s2. Determine the speed of block B when it has risen s = 6 m. The pulley has an inner hub D which is fixed to C and turns with it.

150 mm C

D A 50 mm

75 mm

SOLUTION aA r A = aC r C 6(50) = aC(150)

B

aC = 2 rad>s2

s

aB = aC rB = 2(0.075) = 0.15 m>s2 (+ c)

v2 = v20 + 2 ac (s - s0) v2 = 0 + 2(0.15)(6 - 0) v = 1.34 m>s

sale

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is

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provided

integrity

of

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solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






16–19. The vacuum cleaner’s armature shaft S rotates with an angular acceleration of a = 4v3>4 rad>s2, where v is in rad>s. Determine the brush’s angular velocity when t = 4 s, starting from v0 = 1 rad>s, at u = 0. The radii of the shaft and the brush are 0.25 in. and 1 in., respectively. Neglect the thickness of the drive belt.

SOLUTION Motion of the Shaft: The angular velocity of the shaft can be determined from dt =

L t

L0 2

t

dt =

A

dvS L aS vs

dvS

L1 4vS 3>4 2

vs

t 0 = vS 1>4 1 t = vS 1>4 – 1

laws

or

vS = (t+1) 4

in

teaching

Web)

When t = 4 s

States

World

permitted.

Dissemination

Wide

copyright

vs = 54 = 625 rad>s

and

use

United

on

learning.

is

of

the

for

work

student

the

by

vB rB = vs rs

work

solely

of

protected

the

(including

rs 0.25 b(625) = 156 rad>s ≤v = a rB s 1

sale

will

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destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

vB = ¢






Ans.

not

instructors

Motion of the Beater Brush: Since the brush is connected to the shaft by a non-slip belt, then

S

A

S

*16–20. The operation of “reverse” for a three-speed automotive transmission is illustrated schematically in the figure. If the crank shaft G is turning with an angular speed of 60 rad>s, determine the angular speed of the drive shaft H. Each of the gears rotates about a fixed axis. Note that gears A and B, C and D, E and F are in mesh. The radii of each of these gears are reported in the figure.

A

vH H

G

F

C B D

E

rA ⫽ 90 mm rB ⫽ rC ⫽ 30 mm rD ⫽ 50 mm rE ⫽ 70 mm rF ⫽ 60 mm

SOLUTION 60(90) = vBC (30) vBC = 180 rad>s 180(30) = 50(vDE) vDE = 108 rad>s 108(70) = (60) (vH) vH = 126 rad>s

will

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destroy

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of

and

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is

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solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






vG ⫽ 60 rad/s

16–21. V0 ⫽ 6 rad/s

A motor gives disk A an angular acceleration of aA = 10.6t2 + 0.752 rad>s2, where t is in seconds. If the initial angular velocity of the disk is v0 = 6 rad>s, determine the magnitudes of the velocity and acceleration of block B when t = 2 s.

A

0.15 m

SOLUTION dv = a dt v

L6

2

dv =

L0

(0.6 t2 + 0.75) dt

B

v - 6 = (0.2 t3 + 0.75 t)|20 v = 9.10 rad>s Ans.

aB = at = ar = [0.6(2)2 + 0.75](0.15) = 0.472 m>s2

Ans.

sale

will

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of

and

any

courses

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is

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provided

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of

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is

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solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

vB = vr = 9.10(0.15) = 1.37 m>s






16–22. For a short time the motor turns gear A with an angular acceleration of aA = (30t1>2) rad>s2, where t is in seconds. Determine the angular velocity of gear D when t = 5 s, starting from rest. Gear A is initially at rest. The radii of gears A, B, C, and D are rA = 25 mm, rB = 100 mm, rC = 40 mm, and rD = 100 mm, respectively.

A C

SOLUTION Motion of the Gear A: The angular velocity of gear A can be determined from dvA =

L

adt

L

t

vA

dvA =

L0 vA

vA 0

30t1>2dt

L0

t

= 20t3>2 2

0

vA = A 20t

3>2

B rad>s laws

or

When t = 5 s

Wide

copyright

in

teaching

Web)

vA = 20 A 53>2 B = 223.61 rad>s

of

the

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instructors

States

World

and

use

United

on

learning.

is

vB rB = vA rA

the

(including

for

work

student

the

by

rA 25 b(223.61) = 55.90 rad>s ≤v = a rB A 100 work

solely

of

protected

vC = v B = ¢

work

this

assessing

is

Also, gear D is in mesh with gear C. Then

part

the

is

This

provided

integrity

of

and

vD rD = vC rC

any

courses

rC 40 b (55.90) = 22.4 rad>s ≤v = a rD C 100 sale

will

their

destroy

of

and

vD = ¢






Ans.

permitted.

Dissemination

Motion of Gears B, C, and D: Gears B and C which are mounted on the same axle will have the same angular velocity. Since gear B is in mesh with gear A, then

B D

16–23. The motor turns gear A so that its angular velocity increases uniformly from zero to 3000 rev>min after the shaft turns 200 rev. Determine the angular velocity of gear D when t = 3 s. The radii of gears A, B, C, and D are rA = 25 mm, rB = 100 mm, rC = 40 mm, and rD = 100 mm, respectively.

A C

SOLUTION Motion of Wheel A: Here, vA = a 3000

1 min 2p rad rev ba ba b = 100p rad>s min 60 s 1rev

2p rad b = 400p rad. Since the angular acceleration of gear 1 rev A is constant, it can be determined from when uA = (200 rev) a

vA 2 = (vA)0 2 + 2aA C uA - (uA)0 D (100p)2 = 02 + 2aA (400p - 0)

laws

or

aA = 39.27 rad>s2

teaching

Web)

Thus, the angular velocity of gear A when t = 3 s is

Dissemination

Wide

copyright

in

vA = A v A B 0 + a A t

not

instructors

States

World

permitted.

= 0 + 39.27(3)

by

and

use

United

on

learning.

is

of

the

= 117.81 rad>s

the

(including

for

work

student

the

Motion of Gears B, C, and D: Gears B and C which are mounted on the same axle will have the same angular velocity. Since gear B is in mesh with gear A, then assessing

is

work

solely

of

protected

vB rB = vB rA provided

integrity

of

and

work

this

rA 25 b (117.81) = 29.45 rad>s ≤v = a rB A 100 and

any

courses

part

the

is

This

v C = vB = ¢

vD = ¢






sale

vD rD = vC rC

will

their

destroy

of

Also, gear D is in mesh with gear C. Then

rC 40 b (29.45) = 11.8 rad>s ≤v = a rD C 100

Ans.

B D

*16–24. The gear A on the drive shaft of the outboard motor has a radius rA = 0.5 in. and the meshed pinion gear B on the propeller shaft has a radius rB = 1.2 in. Determine the angular velocity of the propeller in t = 1.5 s, if the drive shaft rotates with an angular acceleration a = (400t3) rad>s2, where t is in seconds. The propeller is originally at rest and the motor frame does not move.

A

2.20 in. P

SOLUTION Angular Motion: The angular velocity of gear A at t = 1.5 s must be determined first. Applying Eq. 16–2, we have dv = adt 1.5 s

L0

dv =

400t3 dt

L0

laws

or

s vA = 100t4 |1.5 = 506.25 rad>s 0

teaching in

Ans. World

Dissemination

Wide

copyright

rA 0.5 b(506.25) = 211 rad>s v = a rB A 1.2

sale

will

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destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

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and

use

United

on

learning.

is

of

the

not

instructors

States

vB =

Web)

However, vA rA = vB rB where vB is the angular velocity of propeller. Then,






permitted.

vA

B

16–25. For the outboard motor in Prob. 16–24, determine the magnitude of the velocity and acceleration of point P located on the tip of the propeller at the instant t = 0.75 s.

A

2.20 in. P

SOLUTION Angular Motion: The angular velocity of gear A at t = 0.75 s must be determined first. Applying Eq. 16–2, we have dv = adt 0.75 s

vA

L0

dv =

400t3 dt

L0

Web)

laws

or

s vA = 100t4 |0.75 = 31.64 rad>s 0

Wide

copyright

in

teaching

The angular acceleration of gear A at t = 0.75 s is given by

by

and

use

United

on

learning.

is

of

the

However, vA rA = vB rB and aA rA = aB rB where vB and aB are the angular velocity and acceleration of propeller. Then,

the

(including

for

work

student

the

rA 0.5 b (31.64) = 13.18 rad>s v = a rB A 1.2

this

part

the

is

This

provided

integrity

of

and

work

aB =

assessing

is

rA 0.5 a = a b(168.75) = 70.31 rad>s2 rB A 1.2

work

solely

of

protected

vB =

their

destroy

of

and

any

courses

Motion of P: The magnitude of the velocity of point P can be determined using Eq. 16–8. will

2.20 b = 2.42 ft>s 12 sale

vP = vB rP = 13.18 a

Ans.

The tangential and normal components of the acceleration of point P can be determined using Eqs. 16–11 and 16–12, respectively. ar = aB rP = 70.31 a

2.20 b = 12.89 ft>s2 12

an = v2B rP = A 13.182 B a

2.20 b = 31.86 ft>s2 12

The magnitude of the acceleration of point P is aP = 2a2r + a2n = 212.892 + 31.862 = 34.4 ft>s2






Ans.

not

instructors

States

World

permitted.

Dissemination

aA = 400 A 0.753 B = 168.75 rad>s2

B

16–26. The pinion gear A on the motor shaft is given a constant angular acceleration a = 3 rad>s2. If the gears A and B have the dimensions shown, determine the angular velocity and angular displacement of the output shaft C, when t = 2 s starting from rest. The shaft is fixed to B and turns with it.

B C

125 mm A

SOLUTION v = v0 + a c t vA = 0 + 3(2) = 6 rad>s 1 ac t2 2

u = u0 + v 0 t + uA = 0 + 0 +

1 (3)(2)2 2

uA = 6 rad

or

vA rA = vB rB teaching

Web)

laws

6(35) = vB(125) in

vC = vB = 1.68 rad>s

Wide

copyright

Ans.

not

instructors

States

World

permitted.

Dissemination

uA rA = uB rB

use

United

on

learning.

is

of

the

6(35) = uB (125)

and

uC = uB = 1.68 rad

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

Ans.






35 mm

16–27. For a short time, gear A of the automobile starter rotates with an angular acceleration of aA = (450t2 + 60) rad>s2, where t is in seconds. Determine the angular velocity and angular displacement of gear B when t = 2 s, starting from rest. The radii of gears A and B are 10 mm and 25 mm, respectively.

A

B

SOLUTION Motion of Gear A: Applying the kinematic equation of variable angular acceleration, dvA =

L

t

vA

dvA =

L0

L0

aAdt

L

A 450t2 + 60 B dt

vA

vA 0 = 150t3 + 60t 2

t 0

vA = A 150t + 60t B rad>s or

3

teaching

Web)

laws

When t = 2 s,

Dissemination

Wide

copyright

in

vA = 150(2)3 + 60(2) = 1320 rad>s

World

permitted.

vA dt

on United by

and

use

A 150t3 + 60t B dt

student

L0

the

(including

for

work

t

of

0 is

work

solely

uA

uA 0 = 37.5t4 + 30t2 2

protected

L0

duA =

the

t

uA

learning.

is

of

the

not

instructors

L

States

duA =

L

B rad

this

provided

integrity

of

and

work

2

assessing

uA = A 37.5t + 30t 4

courses

part

the

is

This

When t = 2 s

will

their

destroy

of

and

any

uA = 37.5(2)4 + 30(2)2 = 720 rad sale

Motion of Gear B: Since gear B is meshed with gear A, Fig. a, then vp = vA rA = vB rB vB = vA ¢

rA ≤ rB

= (1320) ¢

0.01 ≤ 0.025

= 528 rad>s uB = uA ¢ = 720

rA ≤ rB 0.01 0.025

= 288 rad






Ans.

Ans.

*16–28. For a short time, gear A of the automobile starter rotates with an angular acceleration of aA = (50v1>2) rad>s2, where v is in rad>s. Determine the angular velocity of gear B when t = 1 s. Orginally (vA)0 = 1 rad>s when t = 0. The radii of gears A and B are 10 mm and 25 mm, respectively.

A

B

SOLUTION Motion of Gear A: We have

L t

L0

dvA L aA

dt =

vA

dt = t

t0 =

L1

50vA 1>2

vA 1 vA 1>2 2 25 1

1 1 v 1>2 25 25 A or

t=

dvA

in

teaching

Web)

laws

vA = (25t + 1)2

Dissemination

Wide

copyright

When t = 1 s, vA = 676 rad>s

of

the

not

instructors

States

World

permitted.

Motion of Gear B: Since gear B is meshed with gear A, Fig. a, then

and

use

United

on

learning.

is

vp = vA rA = vB rB

work

student

the

by

rA ≤ rB solely

of

protected

the

(including

for

vB = vA ¢

this

assessing

is

work

0.01 b 0.025 provided

integrity

of

and

work

= 676 a

sale

will

their

destroy

of

and

any

courses

part

the

is

This

= 270 rad>s






Ans.

16–29. A mill in a textile plant uses the belt-and-pulley arrangement shown to transmit power. When t = 0 an electric motor is turning pulley A with an angular velocity of vA = 5 rad>s. If this pulley is subjected to a constant angular acceleration 2 rad>s2, determine the angular velocity of pulley B after B turns 6 revolutions. The hub at D is rigidly connected to pulley C and turns with it.

B 4 in.

3 in. D

SOLUTION

5 in.

When uB = 6 rev; 4(6) = 3 uC

4.5 in.

uC = 8 rev

A

8(5) = 4.5(uA)

ω A = 5 rad/s

uA = 8.889 rev (vA)22 = (vA)21 + 2aC[(uA)2 - (uA)1]

or

(vA)22 = (5)2 + 2(2)[(8.889)(2p) - 0]

teaching

Web)

laws

(vA)2 = 15.76 rad>s

Wide

copyright

in

15.76(4.5) = 5vC

instructors

States

World

permitted.

Dissemination

vC = 14.18 rad>s

United

on

learning.

is

of

the

not

14.18(3) = 4(vB)2

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

(vB)2 = 10.6 rad s






Ans.

C

16–30. A tape having a thickness s wraps around the wheel which is turning at a constant rate V. Assuming the unwrapped portion of tape remains horizontal, determine the acceleration of point P of the unwrapped tape when the radius of the wrapped tape is r. Hint: Since vP = vr, take the time derivative and note that dr>dt = v(s>2p).

s P aP r

SOLUTION vP = vr dvP dv dr = r + v a = dt dt dt Since

v

dv = 0, dt

a = va

dr b dt

In one revolution r is increased by s, so that

laws

or

2p s = u ¢r teaching

Web)

Hence, in

s u 2p

Dissemination

Wide

copyright

¢r =

on

learning.

is

of

the

not

instructors

States

World

permitted.

dr s = v dt 2p

and

use

United

s 2 v 2p

work

student

the

by

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

a =






16–31. Due to the screw at E, the actuator provides linear motion to the arm at F when the motor turns the gear at A. If the gears have the radii listed in the figure, and the screw at E has a pitch p = 2 mm, determine the speed at F when the motor turns A at vA = 20 rad>s. Hint: The screw pitch indicates the amount of advance of the screw for each full revolution.

F

D

SOLUTION

E

B

C A

vA r A = v B r B vC rC = vD rD Thus, rA r C 10 15 20 = 1 rad>s v = rB r D A 50 60

vF =

1 rad > s 1 rev (2 mm) = 0.318 mm >s 2p rad

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

vD =






rA ⫽ 10 mm rB ⫽ 50 mm rC ⫽ 15 mm rD ⫽ 60 mm

*16–32. The driving belt is twisted so that pulley B rotates in the opposite direction to that of drive wheel A. If A has a constant angular acceleration of aA = 30 rad>s2, determine the tangential and normal components of acceleration of a point located at the rim of B when t = 3 s, starting from rest.

200 mm vA

Motion of Wheel A: Since the angular acceleration of wheel A is constant, its angular velocity can be determined from vA = (vA)0 + aCt = 0 + 30(3) = 90 rad>s Motion of Wheel B: Since wheels A and B are connected by a nonslip belt, then vBrB = vArA

laws

or

rA 200 bv = a b (90) = 144 rad>s rB A 125

in

teaching

Web)

and Dissemination

Wide

copyright

aBrB = aArA

the

not

instructors

States

World

permitted.

rA 200 ba = a b (30) = 48 rad>s2 rB A 125 by

and

use

United

on

learning.

is

of

aB = a

Ans.

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

Thus, the tangential and normal components of the acceleration of point P located at the rim of wheel B are (ap)t = aBrB = 48(0.125) = 6 m>s2

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

(ap)n = vB 2rB = (144 2)(0.125) = 2592 m>s2






125 mm

vB A

SOLUTION

vB = a

B

Ans.

16–33. The driving belt is twisted so that pulley B rotates in the opposite direction to that of drive wheel A. If the angular displacement of A is uA = (5t3 + 10t2) rad, where t is in seconds, determine the angular velocity and angular acceleration of B when t = 3 s.

200 mm vA

Motion of Wheel A: The angular velocity and angular acceleration of wheel A can be determined from duA = A 15t2 + 20t B rad>s dt

aA =

dvA = A 30t + 20 B rad>s dt laws

or

and

teaching

Web)

When t = 3 s,

Dissemination

Wide

copyright

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vA = 15 A 32 B + 20(3) = 195 rad>s

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rA 200 ba = a b (110) = 176 rad>s2 rB A 125 sale

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aB = a






125 mm

vB A

SOLUTION

vA =

B

Ans.

16–34. The rope of diameter d is wrapped around the tapered drum which has the dimensions shown. If the drum is rotating at a constant rate of v, determine the upward acceleration of the the block. Neglect the small horizontal displacement of the block.

L r1 r2

SOLUTION

ω d

v = vr

dr dv r + v dt dt

= v(

dr ) dt

r = r1 + (

r2 - r1 ) dx L in

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dr = (

r2 - r1 )x L or

=

d(vr) dt

laws

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1 r2 - r1 du dr = ( ) d( ) dt 2p L dt

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16–35. z

If the shaft and plate rotates with a constant angular velocity of v = 14 rad>s, determine the velocity and acceleration of point C located on the corner of the plate at the instant shown. Express the result in Cartesian vector form.

A v

0.6 m

a 0.2 m C

D

SOLUTION We will first express the angular velocity v of the plate in Cartesian vector form. The unit vector that defines the direction of v is uOA =

- 0.3i + 0.2j + 0.6k 2( - 0.3)2 + 0.22 + 0.62

x

0.4 m

3 2 6 = - i + j + k 7 7 7

B

2 6 3 v = vuOA = 14 a - i + j + k b = [- 6i + 4j + 12k] rad>s 7 7 7

or

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For convenience, rC = [ -0.3i + 0.4j] m is chosen. The velocity and acceleration of point C can be determined from

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vC = v * rC

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= ( -6i + 4j + 12k) * ( -0.3i + 0.4j)

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= [-4.8i - 3.6j - 1.2k] m>s

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aC = a * rC + V * (V * rc)

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= 0 + ( -6i + 4j + 12k) * [( -6i + 4j + 12k) * ( -0.3i + 0.4j)] = [38.4i - 64.8j + 40.8k]m s2





0.3 m 0.3 m

Thus,


O

0.4 m

Ans.

y

*16–36. z

At the instant shown, the shaft and plate rotates with an angular velocity of v = 14 rad>s and angular acceleration of a = 7 rad>s2. Determine the velocity and acceleration of point D located on the corner of the plate at this instant. Express the result in Cartesian vector form.

A v

0.6 m

a 0.2 m C

D

SOLUTION We will first express the angular velocity v of the plate in Cartesian vector form. The unit vector that defines the direction of v and a is uOA =

O

0.4 m

- 0.3i + 0.2j + 0.6k

x

3 2 6 = - i + j + k 7 7 7

2( - 0.3)2 + 0.22 + 0.62

0.3 m 0.3 m 0.4 m B

Thus, 2 6 3 v = vuOA = 14 a - i + j + k b = [- 6i + 4j + 12k] rad>s 7 7 7

Dissemination

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For convenience, rD = [- 0.3i + 0.4j] m is chosen. The velocity and acceleration of point D can be determined from

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3 2 6 a = auOA = 7 a - i + j + k b = [-3i + 2j + 6k] rad>s 7 7 7

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vD = v * rD

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= (- 6i + 4j + 12k) * (0.3i - 0.4j) by

= [4.8i + 3.6j + 1.2k]m>s

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= ( - 3i + 2j + 6k) * ( -0.3i + 0.4j) + (-6i + 4j + 12k) * [(-6i + 4j + 12k) * (-0.3i + 0.4j)] any

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= [ - 36.0i + 66.6j - 40.2k]m s2






y

16–37. The rod assembly is supported by ball-and-socket joints at A and B. At the instant shown it is rotating about the y axis with an angular velocity v = 5 rad>s and has an angular acceleration a = 8 rad>s2. Determine the magnitudes of the velocity and acceleration of point C at this instant. Solve the problem using Cartesian vectors and Eqs. 16–9 and 16–13.

z

C 0.3 m

0.4 m

B A A 0.4 m

SOLUTION

x

vC = v * r vC = 5j * ( -0.4i + 0.3k) = {1.5i + 2k} m>s vC = 21.52 + 2 2 = 2.50 m>s

Ans.

a C = a * r - v2r = 8j * ( -0.4i + 0.3k) -52 ( -0.4i + 0.3k) = {12.4i - 4.3k} m>s2 a C = 212.4 2 + (-4.3)2 = 13.1 m>s2

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laws

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V

y

16–38. Rotation of the robotic arm occurs due to linear movement of the hydraulic cylinders A and B. If this motion causes the gear at D to rotate clockwise at 5 rad>s, determine the magnitude of velocity and acceleration of the part C held by the grips of the arm.

4 ft 2 ft

45

C

SOLUTION D

Motion of Part C: Since the shaft that turns the robot’s arm is attached to gear D, then the angular velocity of the robot’s arm vR = vD = 5.00 rad>s. The distance of part C from the rotating shaft is rC = 4 cos 45° + 2 sin 45° = 4.243 ft. The magnitude of the velocity of part C can be determined using Eq. 16–8. vC = vR rC = 5.00(4.243) = 21.2 ft>s

B

A

Ans.

The tangential and normal components of the acceleration of part C can be determined using Eqs. 16–11 and 16–12 respectively. at = arC = 0

laws

or

an = v2R rC = A 5.002 B (4.243) = 106.07 ft>s2

Ans.

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aC = 2a2t + a2n = 202 + 106.072 = 106 ft>s2

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The magnitude of the acceleration of point C is






3 ft

16–39. The bar DC rotates uniformly about the shaft at D with a constant angular velocity V. Determine the velocity and acceleration of the bar AB, which is confined by the guides to move vertically.

C A

V D

B U l

SOLUTION y = l sin u # # y = vy = l cos uu $ # $ y = a y = l(cos uu - sin uu) $ $ Here vy = vAB , ay = aAB, and u = v, u = a = 0. Ans.

aAB = l C cos u(0) - sin u(v)2 D = - v2 l sin u

Ans.

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laws

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vAB = l cos u (v) = v l cos u






*16–40. The mechanism is used to convert the constant circular motion v of rod AB into translating motion of rod CD and the attached vertical slot. Determine the velocity and acceleration of CD for any angle u of AB .

B D

l V

u x

SOLUTION x = l cos u # # x = vx = - l sin uu $ # # x = a x = - l(sin uu + cos u u2) # $ Here vx = vCD, ax = aCD, and u = v, u = a = 0. vCD = - l sin u (v) = - vl sin u

Ans.

a CD = -l C sin u(0) + cos u(v)2 D = - v2 l cos u

Ans.

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Negative signs indicate that both vCD and aCD are directed opposite to positive x.






A C

16–41. At the instant u = 50°, the slotted guide is moving upward with an acceleration of 3 m>s2 and a velocity of 2 m>s. Determine the angular acceleration and angular velocity of link AB at this instant. Note: The upward motion of the guide is in the negative y direction.

B 300 mm V, A

y U

A

SOLUTION y = 0.3 cos u # # y = vy = - 0.3 sin uu $ # ## y = a y = -0.3 A sin uu + cos uu2 B

v a

# # $ Here vy = - 2 m>s, ay = - 3 m>s2, and u = v, u = v, u = a, u = 50°. v = 8.70 rad>s

Ans.

- 3 = -0.3[sin 50°(a) + cos 50°(8.70)2]

a = - 50.5 rad>s 2

Ans.

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laws

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-2 = -0.3 sin 50°(v)






2 m/s 3 m/s2

16–42. The mechanism shown is known as a Nuremberg scissors. If the hook at C moves with a constant velocity of v, determine the velocity and acceleration of collar A as a function of u. The collar slides freely along the vertical guide.

A

L/2

u L/2

u

B

SOLUTION

C

x = 3L sin u # # v = x = 3L cos u u y = L cos u # # y = - L sin u u # # y L sin u u # = v 3L cos u u # y = (v tan u)>3 T

Ans.

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# v 1 v v $ b ba y = (sec2u u) = a 2 3 3 cos u 3L cos u v2 T 9L cos3 u

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$ y =






v

16–43. The crankshaft AB is rotating at a constant angular velocity of v = 150 rad>s. Determine the velocity of the piston P at the instant u = 30°.

B 0.75 ft

0.2 ft U

A

V x

SOLUTION x = 0.2 cos u + 2(0.75)2 - (0- 2 sin u)2 $ $ 1 1 # x = -0.2 sin uu + C (0.75)2 -(0.2 sin u)2 D - 2( -2)(0.2 sin u)(0.2 cos u)u 2 (0.2)2v sin 2 u 1 vP = -0.2v sin u - a b 2 2(0.75)2 - (0.2 sin u)2 At u = 30°, v = 150 rad>s (0.2)2(150) sin 60° 1 vP = -0.2(150) sin 30° - a b 2 2(0.75)2 - (0.2 sin 30°)2 or

vP = -18.5 ft>s = 18.5 ft>s ;

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laws

Ans.






P

*16–44. v

Determine the velocity and acceleration of the follower rod CD as a function of u when the contact between the cam and follower is along the straight region AB on the face of the cam. The cam rotates with a constant counterclockwise angular velocity V.

A r O

u

D

C B

SOLUTION Position Coordinate: From the geometry shown in Fig. a, xC =

r = r sec u cos u

Time Derivative: Taking the time derivative, # # vCD = xC = r sec u tan uu

(1)

laws

or

# Here, u = + v since v acts in the positive rotational sense of u. Thus, Eq. (1) gives vCD = rv sec u tan u :

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The time derivative of Eq. (1) gives $ # # # $ aCD = xC = r{sec u tan uu + u[sec u(sec2uu) + tan u(sec u tan uu)]} $ # aCD = r[sec u tan u u + (sec3 u + sec u tan2 u)u2]

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# $ Since u = v is constant, u = a = 0. Then, solely

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a CD = r[sec u tan u(0) + (sec3 u + sec u tan2 u)v2] assessing

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= rv2 A sec3 u + sec u tan2 u B :

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16–45. Determine the velocity of rod R for any angle u of the cam C if the cam rotates with a constant angular velocity V. The pin connection at O does not cause an interference with the motion of A on C.

V

r1

C

r2

SOLUTION

O R

Position Coordinate Equation: Using law of cosine. (1)

Time Derivatives: Taking the time derivative of Eq. (1).we have 0 = 2x

dx du dx - 2r1 ¢ -x sin u + cos u ≤ dt dt dt

(2)

du dx and v = . From Eq.(2), dt dt laws

or

0 = xv - r1(v cos u - xv sin u) in

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r1xv sin u r1 cos u - x

(3)

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v =

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However, the positive root of Eq.(1) is

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x = r1 cos u + 2r21 cos 2u + r22 + 2r1r2

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Substitute into Eq.(3),we have

the

r21v sin 2u

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+ r1v sin u

2 2r21 cos2u + r22 + 2r# 1r2

Ans.

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v = -

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Note: Negative sign indicates that v is directed in the opposite direction to that of positive x.






A x

(r1 + r2)2 = x2 + r12 - 2r1x cos u

However v =

u

16–46. The bridge girder G of a bascule bridge is raised and lowered using the drive mechanism shown. If the hydraulic cylinder AB shortens at a constant rate of 0.15 m>s, determine the angular velocity of the bridge girder at the instant u = 60°.

G

B C

A

u 3m

SOLUTION

5m

Position Coordinates: Applying the law of cosines to the geometry shown in Fig. a, s2 = 32 + 52 - 2(3)(5) cos A 180°- u B s2 = 34 - 30 cos A 180° - u B However, cos A 180°-u B = - cos u. Thus, s2 = 34 + 30 cos u Time Derivatives: Taking the time derivative,

laws

or

# # 2ss = 0 + 30 A - sin uu B # # ss = - 15 sin uu

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(1)

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# # When u = 60°, s = 234 + 30 cos 60° = 7 m. Also, s = - 0.15 m>s since s is directed towards the negative sense of s. Thus, Eq. (1) gives # 7 A - 0.15 B = - 15 sin 60°u # v = u = 0.0808 rad>s Ans.

16–47. v

The circular cam of radius r is rotating clockwise with a constant angular velocity V about the pin at O, which is at an eccentric distance e from the center of the cam. Determine the velocity and acceleration of the follower rod A as a function of u.

e

O

u

SOLUTION Position Coordinates: From the geometry shown in Fig. a, xA = e cos u + r

(1)

Time Derivatives: Taking the time derivative, # # vA = xA = - e sin uu

(2)

# Since v acts in the negative rotational sense of u, then u = - v. Thus, Eq. (2) gives vA = - e sin u( -v) = ev sin u :

or

Ans.

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Web)

laws

Taking the time derivative of Eq. (2) gives $ # $ aA = xA = - e A sin uu + cos uu2 B

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(3)

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$ Since v is constant, u = a = 0. Then Eq. (3) gives

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aA = - e[sin u(0) + cos u(v2)]

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= - ev2 cos u = ev2 cos u ;

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The negative sign indicates that aA acts towards the negative sense of xA.






Ans.

r

A

*16–48. Peg B mounted on hydraulic cylinder BD slides freely along the slot in link AC. If the hydraulic cylinder extends at a constant rate of 0.5 m>s, determine the angular velocity and angular acceleration of the link at the instant u = 45°.

C

B

0.5 m/s u D

SOLUTION 0.6 m

Position Coordinate: From the geometry shown in Fig. a, yC = 0.6 tan u m Time Derivatives: Taking the time derivative, # # vC = yC = (0.6 sec2 uu) m>s

(1)

laws

or

Here, vC = 0.5 m>s since vC acts in the positive sense of yC. When u = 45°, Eq. (1) gives # 0.5 = A 0.6 sec2 45° B u # Ans. vAB = u = 0.4167 rad>s = 0.417 rad>s

permitted.

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The time derivative of Eq. (1) gives $ # $ aC = yC = 0.6 A sec2 uu + 2 sec u sec u tan uu2 B $ # aC = 0.6 sec2 u A u + 2 tan uu2 B

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Since vC is constant, aC = 0. Thus, Eq. (2) gives $ 0 = u + 2 tan 45° A 0.41672 B $ aAB = u = - 0.3472 rad>s2 = 0.347 rad>s2d

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provided © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





integrity

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and

work

The negative sign indicates that aAB acts counterclockwise.

Ans.

A

16–49. Bar AB rotates uniformly about the fixed pin A with a constant angular velocity V. Determine the velocity and acceleration of block C, at the instant u = 60°.

B

L

V

L A

u

SOLUTION L cos u + L cos f = L

C

cos u + cos f = 1 # # sin u u] + sin f f = 0 $ $ # # cos u(u)2 + sin uu + sinf f + cos f ( f)2 = 0

L

(1) (2)

When u = 60°, f = 60°, # # thus, u = -f = v (from Eq. (1)) $ u = 0

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$ f = -1.155v2 (from Eq.(2))

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Also, sC = L sin f - L sin u # # vC = L cos f f - L cos u u # $ $ # aC = -L sin f (f)2 + L cos f (f) - L cos u(u) + L sin u(u)2

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At u = 60°, f = 60°

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sC = 0

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vC = L(cos 60°)( - v) - L cos 60°(v) = - Lv = Lv c

integrity

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aC = - L sin 60°( -v)2 + L cos 60°(-1.155v2) + 0 + L sin 60°(v)2 any

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aC = - 0.577 Lv2 = 0.577 Lv2 c






16–50. The block moves to the left with a constant velocity v0. Determine the angular velocity and angular acceleration of the bar as a function of u.

a u x

SOLUTION Position Coordinate Equation: From the geometry, x =

a = a cot u tan u

(1)

Time Derivatives: Taking the time derivative of Eq. (1),we have du dx = - a csc2 u dt dt dx du = -v0. Also, = v. dt dt laws

or

Since y0 is directed toward negative x, then

(2)

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From Eq.(2),

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Here, a =

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v0 du (2 sin u cos u) a dt

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courses

v0 du sin2 u. Substitute these values into = a dt destroy

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and

However, 2 sin u cos u = sin 2u and v =

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Eq.(3) yields a =






v0 v0 2 v0 sin 2u a sin2u b = a b sin 2u sin2 u a a a

Ans.

v0

16–51. The bar is confined to move along the vertical and inclined planes. If the velocity of the roller at A is vA = 6 ft>s when u = 45°, determine the bar’s angular velocity and the velocity of roller B at this instant.

vA

A u 5 ft

SOLUTION 30⬚

sB cos 30° = 5 sin u B

sB = 5.774 sin u # s

B

# = 5.774 cos u u

vB

(1)

5 cos u = sA + sB sin 30° # # # -5 sin u u = sA + sB sin 30°

(2)

teaching

Web)

laws

or

Combine Eqs.(1) and (2): # # - 5 sin u u = - 6 + 5.774 cos u (u)(sin 30°) # # - 3.536u = - 6 + 2.041u # v = u = 1.08 rad>s

Dissemination

Wide

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Ans.

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permitted.

From Eq.(1): not

instructors

# vB = sB = 5.774 cos 45°(1.076) = 4.39 ft>s

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*16–52. Arm AB has an angular velocity of V and an angular acceleration of A. If no slipping occurs between the disk and the fixed curved surface, determine the angular velocity and angular acceleration of the disk.

C

A r

ω ', α '

ω ,α

SOLUTION

B

ds = (R - r) du = - r df (R - r) ¢

R

df du ≤ = -r ¢ ≤ dt dt

(R - r) v r

Ans.

a¿ = -

(R - r) a r

Ans.

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laws

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v¿ = -






16–53. If the wedge moves to the left with a constant velocity v, determine the angular velocity of the rod as a function of u.

L v

u

SOLUTION Position Coordinates:Applying the law of sines to the geometry shown in Fig. a, xA L = sin(f - u) sin A 180° - f B xA =

L sin(f - u) sin A 180° - f B

However, sin A 180° - f B = sinf. Therefore, L sin (f - u) sin f laws

or

xA =

Time Derivative: Taking the time derivative,

Dissemination

Wide

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teaching

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# L cos (f - u)(- u) # xA = sin f

not

instructors

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(1)

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v sin f L cos (f - u)

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Since point A is on the wedge, its velocity is vA = - v. The negative sign indicates that vA is directed towards the negative sense of xA. Thus, Eq. (1) gives






f

16–54. The slotted yoke is pinned at A while end B is used to move the ram R horizontally. If the disk rotates with a constant angular velocity V, determine the velocity and acceleration of the ram. The crank pin C is fixed to the disk and turns with it.

R

B

V

u

SOLUTION (1)

x = l tan f However

r s s = = sin f sin (180° - u) sin u d = s cos f - r cos u

A

r sin u s

sin f =

d + r cos u s

cos f =

not and

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(d + r cos u)4

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(d + r cos u)3

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lr sin u(2r2 - d2 + rd cos u)

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x= a = lrv c ¶

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# # (d + r cos u)2( -d sin uu) - (r + d cos u)(2)(d + r cos u)( -r sin uu)






Ans.

permitted.

Ans. World

v

instructors

(d + r cos u)2

States

lr(r + d cos u)

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r sin u sin f s lr sin u b = l§ ¥ = From Eq. (1) x = l a cos f d + r cos u d + r cos u s # # # (d + r cos u)(lr cos uu) - (lr sin u)(-r sin uu) xª = v = Where u = v (d + r cos u)2 =

f

d

r

C l

16–55. The Geneva wheel A provides intermittent rotary motion vA for continuous motion vD = 2 rad>s of disk D. By choosing d = 100 22 mm, the wheel has zero angular velocity at the instant pin B enters or leaves one of the four slots. Determine the magnitude of the angular velocity V A of the Geneva wheel at any angle u for which pin B is in contact with the slot.

d ⫽ 100 2 mm

A 100 mm · vD ⫽ u

f

u

100 mm

D

SOLUTION tan f =

B

0.1 sin u 0.1 A 22 - cos u B

# sec2 f f =

sin u

=

22 - cos u # ª) A 22 - cos u B (cos uu) - sin u(sin uu

A 22 - cos u B

22 cos u - 1 # u A 22 - cos u B 2

=

2

(1)

From the geometry: r 2 = (0.1 sin u)2 + [0.1 A 22 - cos u B ]2 = 0.01 A 3 - 2 22 cos u B 0.01 A 3 - 2 22 cos u B

(3 - 222 cos u)

=

[0.1 A 22 - cosu B ]

2

A 22 - cos u B 2

teaching

Web)

[0.1 A 22 - cos u B ]

2

=

or

r2

laws

sec2 f =

Wide

copyright

in

From Eq. (1)

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Dissemination

22 cos u - 1 # u A 22 - cos u B 2

the

not

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is

fª =

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A 22 - cos u B

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protected

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is

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# 22 cos u - 1 # u f = 3 - 2 22 cos u






Ans.

· vA ⫽ f

*16–56. At the instant shown, the disk is rotating with an angular velocity of V and has an angular acceleration of A. Determine the velocity and acceleration of cylinder B at this instant. Neglect the size of the pulley at C.

A 3 ft V, A

u

C

5 ft

SOLUTION

B

s = 232 + 52 - 2(3)(5) cos u # 1 1 # vB = s = (34 - 30 cos u)- 2(30 sin u)u 2 15 v sin u

# # 15 v cos uu + 15v sin u

-

225 v2 sin2 u

Ans.

3

(34 - 30 cos u) 2

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(34 - 30 cos u)

1 2

3

(34 - 30 cos u) 2 or

15 (v2 cos u + a sin u)

+

Web)

234 - 30 cos u

# 1 a - b(15v sin u) a 30 sin uu b 2

laws

# aB = s =

=

Ans.

1

(34 - 30 cos u) 2

teaching

vB =






16–57. If h and u are known, and the speed of A and B is vA = vB = v, determine the angular velocity V of the body and the direction f of vB.

vA u A

V

SOLUTION

vB f

vB = vA + v * rB>A -v cos fi + v sin fj = v cos ui + v sin uj + ( -vk) * ( -hj) + ) (:

- v cos f = v cos u - vh

(1)

(+ c )

v sin f = v sin u

(2)

From Eq. (2),

f = u

From Eq. (1),

v =

Ans.

2v cos u h

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Ans.






h

B

16–58. If the block at C is moving downward at 4 ft/s, determine the angular velocity of bar AB at the instant shown.

C 3 ft

A

30°

ω AB

SOLUTION

2 ft

Kinematic Diagram: Since link AB is rotating about fixed point A, then vB is always directed perpendicular to link AB and its magnitude is vB = vAB rAB = 2vAB. At the instant shown, vB is directed towards the negative y axis. Also, block C is moving downward vertically due to the constraint of the guide. Then vc is directed toward negative y axis. Velocity Equation: Here, rC>A = {3 cos 30°i + 3 sin 30°j} ft = {2.598i + 1.50j} ft. Applying Eq. 16–16, we have vC = vB + vBC * rC>B

or

- 4j = - 2vAB j + (vBCk) * (2.598i + 1.50j)

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laws

-4j = - 1.50vBCi + (2.598vBC - 2vAB)j

Wide

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Equating i and j components gives

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Dissemination

vBC = 0

0 = - 1.50vBC -4 = 2.598(0) - 2vAB

not

vAB = 2.00 rad s

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Ans.






B

vC = 4 ft/s

16–59. The velocity of the slider block C is 4 ft>s up the inclined groove. Determine the angular velocity of links AB and BC and the velocity of point B at the instant shown.

B

C

1 ft

SOLUTION For link BC

45⬚

A

vC = {- 4 cos 45°i + 4 sin 45°j} ft>s

vB = -vB i

v = vBC k

rB>C = {1i} ft vC = vB + v * rC>B - 4 cos 45°i + 4 sin 45°j = -vBi + (vBC k) * (1i) - 4 cos 45°i + 4 sin 45°j = -vB i + vBCj

or

Equating the i and j components yields: Ans.

vBC = 2.83 rad>s

Ans. Wide

copyright

-4 cos 45° = vBC

in

teaching

Web)

laws

vB = 2.83 ft>s

-4 cos 45° = - vB

instructors

States

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permitted.

Dissemination

For link AB: Link AB rotates about the fixed point A. Hence

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vB = vAB rAB

and

vAB = 2.83 rad>s

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2.83 = vAB(1)






vC

1 ft

4 ft/s

*16–60. The epicyclic gear train consists of the sun gear A which is in mesh with the planet gear B. This gear has an inner hub C which is fixed to B and in mesh with the fixed ring gear R. If the connecting link DE pinned to B and C is rotating at vDE = 18 rad>s about the pin at E, determine the angular velocities of the planet and sun gears.

100 mm 600 mm

A E

B C D

SOLUTION

200 mm vD E

vD = rDE vDE = (0.5)(18) = 9 m>s c The velocity of the contact point P with the ring is zero.

R

vD = vP + v * rD>P 9j = 0 + ( -vB k) * ( - 0.1i) vB = 90 rad>s

b

Ans.

Let P¿ be the contact point between A and B.

or

vP¿ = vP + v * rP¿>P teaching

Web)

laws

vP¿ j = 0 + (- 90k) * ( -0.4i)

Dissemination

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vP¿ = 36 m>s c

permitted.

d

World

Ans. the

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vP¿ 36 = 180 rad>s = rA 0.2

of

vA =






18 rad/s 300 mm

16–61. The shaper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at C. Determine the velocity of the slider block C at the instant u = 60°, if link AB is rotating at 4 rad>s.

125 mm C 45

B

SOLUTION vAB

vC = vB + v * rC>B

4 rad/s 300 mm

-vCi = -4(0.3) sin 30°i + 4(0.3) cos 30°j + vk * ( -0.125 cos 45°i + 0.125 sin 45°j)

u

-vC = - 1.0392 - 0.008839v A

0 = 0.6 - 0.08839v Solving, v = 6.79 rad > s vC = 1.64 m>s

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Ans.






16–62. If the flywheel is rotating with an angular velocity of vA = 6 rad>s, determine the angular velocity of rod BC at the instant shown.

1.5 m A 0.3 m

vA ⫽ 6 rad/s

O C B

SOLUTION

60⬚ 0.6 m D

Rotation About a Fixed Axis: Flywheel A and rod CD rotate about fixed axes, Figs. a and b. Thus, the velocity of points B and C can be determined from vB = vA * rB = (-6k) * ( -0.3j) = [-1.8i] m>s vC = vCD * rC = (vCDk) * (0.6 cos 60° i + 0.6 sin 60° j) = - 0.5196vCD i + 0.3vCD j

or

General Plane Motion: By referring to the kinematic diagram of link BC shown in Fig. c and applying the relative velocity equation, we have

teaching

Web)

laws

vB = vC + vBC * rB>C

Wide

copyright

in

- 1.8i = - 0.5196vCD i + 0.3vCD j + (vBC k) * ( -1.5i)

instructors

States

World

permitted.

Dissemination

- 1.8i = - 0.5196vCD i + (0.3vCD - 1.5vBC)j

use

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not

Equating the i and j components

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- 1.8 = - 0.5196vCD

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0 = 0.3vCD - 1.5vBC

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provided

integrity

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and

vCD = 3.46 rad>s

any

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Ans.

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and

vBC = 0.693 rad>s






16–63. If the angular velocity of link AB is vAB = 3 rad>s, determine the velocity of the block at C and the angular velocity of the connecting link CB at the instant u = 45° and f = 30°.

C θ = 45°

3 ft

A

ω A B = 3 rad/s

SOLUTION

2 ft φ = 30° B

vC = vB + vC>B

B vC R = C 6 ;

30°c

S + D vCB (3) T 45°b

+ ) (:

-vC = 6 sin 30° - vCB (3) cos 45°

(+ c )

0 = - 6 cos 30° + vCB (3) sin 45° d

Ans. or

vCB = 2.45 rad>s vC = 2.20 ft>s ;

teaching

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laws

Ans.

Wide

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Also,

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Dissemination

vC = vB + v * rC>B

on

learning.

is

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the

not

-vC i = (6 sin 30°i - 6 cos 30°j) + (vCB k) * (3 cos 45°i + 3 sin 45°j) -vC = 3 - 2.12vCB

(+ c )

0 = - 5.196 + 2.12vCB

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vCB = 2.45 rad s

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vC = 2.20 ft s ;






Ans.

*16–64. Pinion gear A rolls on the fixed gear rack B with an angular velocity v = 4 rad>s. Determine the velocity of the gear rack C.

C

A 0.3 ft

v

B

SOLUTION vC = vB + vC>B + ) (;

vC = 0 + 4(0.6) vC = 2.40 ft>s

Ans.

Also: vC = vB + v * rC>B -vC i = 0 + (4k) * (0.6j) vC = 2.40 ft>s

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16–65. The pinion gear rolls on the gear racks. If B is moving to the right at 8 ft>s and C is moving to the left at 4 ft>s, determine the angular velocity of the pinion gear and the velocity of its center A.

C

A V

0.3 ft

B

SOLUTION vC = vB + vC>B + ) (:

- 4 = 8 - 0.6(v) v = 20 rad>s

Ans.

vA = vB + vA>B + ) (:

vA = 8 - 20(0.3) vA = 2 ft>s :

Ans.

or

Also:

teaching

Web)

laws

vC = vB + v * rC>B

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- 4i = 8i + (vk) * (0.6j)

States

World

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Dissemination

-4 = 8 - 0.6v

United

on

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the

Ans.

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vA = vB + v * rA>B

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vAi = 8i + 20k * (0.3j)

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vA = 2 ft>s :






Ans.

not

instructors

v = 20 rad>s

16–66. Determine the angular velocity of the gear and the velocity of its center O at the instant shown.

A 45 4 ft/s

SOLUTION General Plane Motion: Applying the relative velocity equation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a, vB = vC + v * rB>C 3i = -4i +

A - vk B * A 2.25j B

3i = A 2.25v - 4 B i Equating the i components yields 3 = 2.25v - 4

(1)

v = 3.111 rad>s

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Web)

laws

or

Ans. (2)

Wide

copyright

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For points O and C,

World

permitted.

Dissemination

vO = vC + v * rO>C the

not

instructors

States

A -3.111k B * A 1.5j B

on

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= - 4i +

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= [0.6667i] ft>s

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vO = 0.667 ft>s :






Ans.

0.75 ft O 1.50 ft

3 ft/s

16–67. Determine the velocity of point A on the rim of the gear at the instant shown.

A 45 4 ft/s

SOLUTION General Plane Motion: Applying the relative velocity equation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a, vB = vC + v * rB>C 3i = - 4i +

A -vk B * A 2.25j B

3i = A 2.25v - 4 B i

(1)

v = 3.111 rad>s

(2) teaching

Web)

laws

3 = 2.25v - 4

or

Equating the i components yields

Wide

copyright

in

For points A and C,

World

permitted.

Dissemination

vA = vC + v * rA>C

learning.

is

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States

A vA B x i + A vA B y j = - 4i + A - 3.111k B * A - 1.061i + 2.561j B

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A vA B x i + A vA B y j = 3.9665i + 3.2998j

the

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Equating the i and j components yields

A vA B y = 3.2998 ft>s work

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A vA B x = 3.9665 ft>s

part

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Thus, the magnitude of vA is

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destroy

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courses

vA = 2 A vA B x 2 + A vA B y 2 = 23.96652 + 3.29982 = 5.16 ft>s

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and its direction is u = tan - 1 C






A vA B y A vA B x

S = tan - 1 ¢

3.2998 ≤ = 39.8° 3.9665

Ans.

0.75 ft O 1.50 ft

3 ft/s

*16–68. Part of an automatic transmission consists of a fixed ring gear R, three equal planet gears P, the sun gear S, and the planet carrier C, which is shaded. If the sun gear is rotating at vS = 6 rad>s, determine the angular velocity vC of the planet carrier. Note that C is pin connected to the center of each of the planet gears.

C

P

S

4 in.

SOLUTION

vC

vD = vA + vD>A

2 in.

24 = 0 + 4(vP) b b vP = 6 rad>s

R P

vE = vA + vE>A vE = 0 + 6(2) b b vE = 12 in.>s 12 = 2 rad>s 6

Ans.

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Dissemination

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laws

or

vC =






P

vS vP

16–69. If the gear rotates with an angular velocity of v = 10 rad>s and the gear rack moves at vC = 5 m>s, determine the velocity of the slider block A at the instant shown.

A

0.5 m

SOLUTION General Plane Motion: Referring to the diagram shown in Fig. a and applying the relative velocity equation, vC

= - 5i + ( -10k) * (0.075j) = [ -4.25i] m>s Then, applying the relative velocity equation to link AB shown in Fig. b, vA = vB + vAB * rA>B laws

or

vA j = - 4.25i + ( - vAB k) * ( -0.5 cos 60° i + 0.5 sin 60° j)

Wide

copyright

in

teaching

Web)

vA j = (0.4330vAB - 4.25)i + 0.25vAB j

States

World

permitted.

Dissemination

Equating the i and j components, yields

(1)

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the

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instructors

0 = 0.4330vAB - 4.25

(2)

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vA = 0.25vAB

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vA = 2.45 m>s c





10 rad/s

75 mm

5 m/s C

vB = vC + v * rB>C


v 60

Ans.

16–70. If the slider block C is moving at vC = 3 m>s, determine the angular velocity of BC and the crank AB at the instant shown.

B 0.5 m 60⬚

1m

A 45⬚

vC ⫽ 3 m/s

SOLUTION Rotation About a Fixed Axis: Referring to Fig. a, vB = vAB * rB = ( -vAB k) * (0.5 cos 60° i + 0.5 sin 60° j) = 0.4330vAB i - 0.25vAB j General Plane Motion: Applying the relative velocity equation and referring to the kinematic diagram of link BC shown in Fig. b, vB = vC + vBC * rB>C laws

or

0.4330vAB i - 0.25vAB j = - 3j + ( - vBC k) * ( -1 cos 45° i + 1 sin 45° j)

Wide

copyright

in

teaching

Web)

0.4330vAB i - 0.25vAB j = 0.7071vBC i + (0.7071vBC - 3)j

States

World

permitted.

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0.4330vAB = 0.7071vBC

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-0.25vAB = 0.7071vBC - 3

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vBC = 2.69 rad>s

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vAB = 4.39 rad>s






C

Ans.

16–71. The two-cylinder engine is designed so that the pistons are connected to the crankshaft BE using a master rod ABC and articulated rod AD. If the crankshaft is rotating at v = 30 rad>s, determine the velocities of the pistons C and D at the instant shown.

D

C 250 mm 250 mm

45°

50 mm

SOLUTION

A 45°

vC = vB + vC>B

50 mm

+ (0.25) v vC = 1.5 ; c30° 45°d + (;) vC cos 45° = 1.5 - vC(0.25)(cos 30°) (+ T)

vC sin 45° = 0 + vC(0.25) (sin 30°)

vC = 0.776 m>s

Ans.

v C = 4.39 rad>s vA = vB + vA>B or Web) Wide

copyright

vD = vA + vD>A

[0.05(4.39) = 0.2195] 45°a

laws

+

teaching

:

in

= 1.5 ;

vA

instructors

States

World

permitted.

Dissemination

vD = 1.5 + 0.21 95 + v¿(0 .25) Q Q 45°d 45°d the

not

vD = - 1.5 sin 45°

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(R+)

Ans.

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vD = 1.06 m>s a

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vB = v * rB>E

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vC = vB + vBC * rC>B

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- vC cos 45° = - 1.5 - vBC (0.2165)

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- vC cos 45°i - vC sin 45°j = (30k) * (0.05j) + (vBC k) * (0.25 cos 60°i + 0.25 sin 60°j)

- vC sin 45° = 0.125vBC vC = 0.776 m>s

Ans.

v BC = 4.39 rad>s vA = vB + vBC * rA>B vD = vA + vAD * rD>A vD cos 45°i - vD sin 45°j = (30k) * (0.05j) + ( -4.39k) * (0.05 cos 45°i + 0.05 sin 45°j) + (vAD k) * (- 0.25 cos 45°i + 0.25 sin 45°j) vD cos 45° = - 1.5 + 0.1552 - vAD (0.1768) - vD sin 45° = 0.1552 - 0.1768vAD vAD = 3.36 rad s vD = 1.06 m s






Ans.

60° B ω = 30 rad/s E

*16–72. Determine the velocity of the center O of the spool when the cable is pulled to the right with a velocity of v. The spool rolls without slipping.

A

O R

SOLUTION Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point P is zero. The kinematic diagram of the spool is shown in Fig. a. General Plane Motion:Applying the relative velocity equation and referring to Fig. a, vB = vP + v * rB>D vi = 0 + ( -vk) * C (R - r)j D vi = v(R - r)i

or

Equating the i components, yields

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laws

v R - r

v =

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v = v(R - r)

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vO = vP + v * rO>P

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vO = ¢






Ans.

r

v

16–73. Determine the velocity of point A on the outer rim of the spool at the instant shown when the cable is pulled to the right with a velocity of v. The spool rolls without slipping.

A

O R

SOLUTION Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point P is zero. The kinematic diagram of the spool is shown in Fig. a. General Plane Motion:Applying the relative velocity equation and referring to Fig. a, vB = vP + v * rB>D vi = 0 + ( -vk) * C (R - r)j D vi = v(R - r)i

or

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v R - r

v =

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vA = vP + v * rA>P

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Ans.

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vA = ¢

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16–74. If crank AB rotates with a constant angular velocity of vAB = 6 rad>s, determine the angular velocity of rod BC and the velocity of the slider block at the instant shown. The rod is in a horizontal position.

0.5 m 0.3 m

B

C 60⬚

30⬚ vAB ⫽ 6 rad/s

A

SOLUTION Rotation About a Fixed Axis: Referring to Fig. a, vB = vAB * rB = (6k) * (0.3 cos 30° i + 0.3 sin 30° j) = [ -0.9i + 1.559j] General Plane Motion: Applying the relative velocity equation to the kinematic diagram of link BC shown in Fig. b, vB = vC + vBC * rB>C laws

or

(- 0.9i + 1.559j) = (- vC cos 60° i - vC sin 60° j) + ( -vBC k) * ( -0.5i)

Wide

copyright

in

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- 0.9i + 1.559j = - 0.5vC i + (0.5vBC - 0.8660vC)j

States

World

permitted.

Dissemination

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(1)

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-0.9 = - 0.5vC

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1.559 = 0.5vBC - 0.8660vC

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Ans.

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vC = 1.80 m>s

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vBC = 6.24 rad>s






Ans.

16–75. If the slider block A is moving downward at vA = 4 m>s, determine the velocity of block B at the instant shown.

E

B

300 mm 4

250 mm

D

400 mm

SOLUTION

30° A

C

vB = 4 T + v (0.55) AB : 3 + ) (: vB = 0 + vAB (0.55)( ) 5 4 0 = - 4 + vAB (0.55)( ) 5

Solving, vAB = 9.091 rad>s vB = 3.00 m>s

Ans.

Also: laws

or

vB = vA + vAB * rB>A in

teaching

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-4 3 (0.55)i + (0.55) j } 5 5

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vB i = - 4j + ( -vAB k) * {

permitted.

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vB = vAB (0.33)

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0 = - 4 + 0.44vAB by

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vB = 3.00 m>s






vA = 4 m/s

300 mm

vB = vA + vB>A

(+ c )

3 5

Ans.

*16–76. If the slider block A is moving downward at vA = 4 m>s, determine the velocity of block C at the instant shown.

E

B

300 mm 4

250 mm

3 5

D

400 mm

vA ⫽ 4 m/s

300 mm 30⬚ A

C

SOLUTION

General Plane Motion: Applying the relative velocity equation by referring to the kinematic diagram of link AB shown in Fig. a, vB = vA + V AB * rB>A 4 3 vB i = - 4j + ( -vAB k) * c - 0.55 a b i + 0.55a bj d 5 5 vB i = 0.33vAB i + (0.44vAB - 4) j Equating j component, 0 = 0.44vAB - 4

vAB = 9.091 rad>sb

teaching

Web)

laws

or

Using the result of vAB, in

vD = vA + V AB * rD>A

States

World

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Dissemination

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copyright

3 4 = - 4j + ( - 9.091k) * c -0.3 a b i + 0.3 a bj d 5 5

use

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= {1.636i - 1.818j} m>s

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vC = vD + V CD * rC>D assessing

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vC i = (1.636i - 1.818j) + ( -vCD k) * (-0.4 cos 30° i - 0.4 sin 30° j)

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this

vC i = (1.636 - 0.2vCD)i + (0.3464vCD - 1.818)j

and

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0 = 0.3464vCD - 1.818

vC = 1.636 - 0.2(5.249) = 0.587 m>s :






Ans.

16–77. The planetary gear system is used in an automatic transmission for an automobile. By locking or releasing certain gears, it has the advantage of operating the car at different speeds. Consider the case where the ring gear R is held fixed, vR = 0, and the sun gear S is rotating at vS = 5 rad>s. Determine the angular velocity of each of the planet gears P and shaft A.

40 mm vR P

vS

R

S

SOLUTION

A

80 mm

vA = 5(80) = 400 mm>s ; vB = 0 vB = vA + v * rB>A 0 = -400i + (vp k) * (80j)

40 mm

0 = -400i - 80vp i vP = -5 rad>s = 5 rad>s

Ans.

vC = vB + v * rC>B

teaching

Web)

laws

or

vC = 0 + ( -5k) * (-40j) = - 200i in

200 = 1.67 rad>s 120

Ans.

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Dissemination

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vA =






16–78. If the ring gear D rotates counterclockwise with an angular velocity of vD = 5 rad>s while link AB rotates clockwise with an angular velocity of vAB = 10 rad>s, determine the angular velocity of gear C.

D 0.5 m A 0.125 m vAB

SOLUTION

10 rad/s C

Rotation About a Fixed Axis: Since link AB and gear D rotate about a fixed axis, Fig. a, the velocity of the center B and the contact point of gears D and C is 0.375 m

vB = vAB rB = 10(0.375) = 3.75 m>s vP = vD rP = 5(0.5) = 2.5 m>s General Plane Motion: Applying the relative velocity equation and referring to the kinematic diagram of gear C shown in Fig. b, vB = vP + vC * rB>P

laws

or

-3.75i = 2.5i + (vC k) * (0.125j)

in

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-3.75i = (2.5 - 0.125vC)i

Dissemination

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-3.75 = 2.5 - 0.125vC

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vC = 50 rad>s

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Ans.






B

16–79. The differential drum operates in such a manner that the rope is unwound from the small drum B and wound up on the large drum A. If the radii of the large and small drums are R and r, respectively, and for the pulley it is (R + r)>2, determine the speed at which the bucket C rises if the man rotates the handle with a constant angular velocity of v. Neglect the thickness of the rope.

A B

SOLUTION Rotation About a Fixed Axis: Since the datum rotates about a fixed axis, Fig. a, we obtain vB = vr and vA = vR. General Plane Motion: Applying the relative velocity equation and referring to the kinematic diagram of pulley shown in Fig. b,

C

vA = vB + V P * rA>B vRj = - vrj + ( - vPk) * ( - 2rPi) vRj = (-vr + 2vPrP)j Equating the j components, yields or

vR = - vr + 2vPrP teaching

Web)

laws

v(R + r) 2rP

Wide

copyright

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vP =

permitted.

Dissemination

Since rP = (R + r)>2, then

on

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World

v(R + r) = v (R + r)

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vP =

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Using this result, we have

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vC = vB + V P * rC>B

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v = c (R - r) d j 2

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v(R + r) dj 2

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= c -vr +

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solely

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= - vrj + ( -v k) * ( - rP i)

Thus, vC =






v (R - r) c 2

Ans.

*16–80. Mechanical toy animals often use a walking mechanism as shown idealized in the figure. If the driving crank AB is propelled by a spring motor such that vAB = 5 rad>s, determine the velocity of the rear foot E at the instant shown. Although not part of this problem, the upper end of the foreleg has a slotted guide which is constrained by the fixed pin at G.

G

3 in. C

B

1 in.

2.5 in. 0.5 in. 50° A ω AB = 5 rad/s

D

SOLUTION

2 in.

vC = vB + vC>B

60°

vC = 2.5 + 3v T c30° c 50° + ) (:

E

F

yC cos 30° = 2.5 sin 50° + 0

vC = 2.21 in.>s 2.21 = 2.21 rad>s 1

vE = (2.21) (2) = 4.42 in.>s

Ans. laws

or

vEC =

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vB = vAB * rB>A

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vC = vEC * rC>D

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vC = vB + v * rC>B

the

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(vEC k) * (cos 60°i + sin 60°j) = (- 5k) * (0.5 cos 50°i + 0.5 sin 50°j) + (vk) * ( -3i)

this

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-0.866vEC = 1.915

the

is

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and

work

0.5vEC = - 1.607 - 3v

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vEC = - 2.21 rad s v = - 0.167 rad s vE = 2(2.21) = 4.42 in. s






45°

Ans.

16–81. In each case show graphically how to locate the instantaneous center of zero velocity of link AB. Assume the geometry is known.

B

V

A

A

(a) V

V A

SOLUTION

B

a) (c)

Dissemination

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c)






(b)

C B

16–82. Determine the angular velocity of link AB at the instant shown if block C is moving upward at 12 in.>s.

A

VAB C 5 in.

45⬚

4 in. 30⬚

B

SOLUTION rIC-B rIC - C 4 = = sin 45° sin 30° sin 105° rIC-C = 5.464 in. rIC-B = 2.828 in. vC = vB C(rIC - C) 12 = vB C(5.464) vB C = 2.1962 rad>s

laws

or

vB = vB C(rIC-B)

in

teaching

Web)

= 2.1962(2.828) = 6.211 in.>s

Dissemination

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copyright

vB = vAB rAB

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World

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6.211 = vAB(5)

Ans.

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vAB = 1.24 rad>s






16–83. At the instant shown, the disk is rotating at v = 4 rad>s. Determine the velocities of points A, B, and C.

D

C

A

SOLUTION

0.15 m

The instantaneous center is located at point A. Hence, vA = 0 rC>IC = 20.152 + 0.152 = 0.2121 m

Ans.

E

rB>IC = 0.3 m

vB = v rB>IC = 4(0.3) = 1.2 m>s

Ans.

vC = v rC>IC = 4(0.2121) = 0.849 m>s

Ans.

destroy

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O

v ⫽ 4 rad/s B

*16–84. If link CD has an angular velocity of vCD = 6 rad>s, determine the velocity of point B on link BC and the angular velocity of link AB at the instant shown.

0.3 m 0.3 m C

B E

0.6 m 30

A

SOLUTION Rotation About Fixed Axis: Referring to Fig. a and b, vC = vCD rC = 6(0.6) = 3.60 m>s ; vB = vAB rB = vAB(1.2) 60° b

(1)

General Plane Motion: The location of IC for link BC is indicated in Fig. c. From the geometry of this figure, rC>IC = 0.6 tan 30° = 0.3464 m rB>IC =

0.6 = 0.6928 m cos 30°

Thus, the angular velocity of link BC can be determined from laws

or

3.60 = 10.39 rad>s 0.3464

Web)

rC>IC

=

teaching

vC

in

vBC =

vB = vBC rB>IC = 10.39 (0.6928) = 7.20 m>s

Ans. is

of

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States

World

60° b

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Substitute this result into Eq. (1),

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7.20 = vAB (1.2)

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the

vAB = 6 rad>s d






Ans.

permitted.

Dissemination

Wide

copyright

Then

vCD

6 rad/s D

16–85. If link CD has an angular velocity of vCD = 6 rad>s, determine the velocity of point E on link BC and the angular velocity of link AB at the instant shown.

0.3 m 0.3 m C

B E

0.6 m 30

A

SOLUTION vC = vCD (rCD) = (6)(0.6) = 3.60 m>s vBC =

vC 3.60 = 10.39 rad>s = rC>IC 0.6 tan 30°

vB = vBC rB>IC = (10.39) a vAB =

vB = rAB

0.6 b = 7.20 m>s cos 30°

7.20 = 6 rad>s 0.6 b a sin 30°

d

Ans.

vE = vBC rE>IC = 10.39 2(0.6 tan 30°)2 + (0.3)2 = 4.76 m>s

laws

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Ans.

0.3 b = 40.9° b 0.6 tan 30°

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permitted.

Dissemination

Wide

copyright

u = tan - 1 a






vCD

6 rad/s D

16–86. At the instant shown, the truck travels to the right at 3 m>s, while the pipe rolls counterclockwise at v = 6 rad>s without slipping at B. Determine the velocity of the pipe’s center G.

G 1.5 m B

SOLUTION Kinematic Diagram: Since the pipe rolls without slipping, then the velocity of point B must be the same as that of the truck, i.e; yB = 3 m>s. Instantaneous Center: rB>IC must be determined first in order to locate the the instantaneous center of zero velocity of the pipe. yB = vrB>IC 3 = 6(rB>IC) rB>IC = 0.5 m

laws

or

Thus, rG>IC = 1.5 - rB>IC = 1.5 - 0.5 = 1.00 m. Then yG = vrG>IC = 6(1.00) = 6.00 m>s ;

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Ans.






3 m/s

16–87. If crank AB is rotating with an angular velocity of vAB = 6 rad>s, determine the velocity of the center O of the gear at the instant shown.

0.6 m

B

C O

0.4 m vAB 60

A

SOLUTION Rotation About a Fixed Axis: Referring to Fig. a, vB = vAB rB = 6(0.4) = 2.4 m>s General Plane Motion: Since the gear rack is stationary, the IC of the gear is located at the contact point between the gear and the rack, Fig. b. Thus, vO and vC can be related using the similar triangles shown in Fig. b, vg =

vC vO = rC>IC rO>IC

laws

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vO vC = 0.2 0.1

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The location of the IC for rod BC is indicated in Fig. c. From the geometry shown,

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vC = vBC rC>IC 2vO = 2(1.039) vO = 1.04 m>s :






Ans.

6 rad/s

0.1 m 0.1 m

*16–88. If link AB is rotating at vAB = 6 rad>s, determine the angular velocities of links BC and CD at the instant u = 60°.

A 250 mm

ω AB = 6 rad/s

30° C

B 300 mm

400 mm

SOLUTION rIC - B = 0.3 cos 30° = 0.2598 m

θ

rIC - C = 0.3 cos 60° = 0.1500 m vBC =

1.5 = 5.774 = 5.77 rad>s 0.2598

Ans.

vC = 5.774(0.15) = 0.8661 m>s 0.8661 = 2.17 rad s 0.4

Ans.

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vCD =






D

16–89. The oil pumping unit consists of a walking beam AB, connecting rod BC, and crank CD. If the crank rotates at a constant rate of 6 rad>s, determine the speed of the rod hanger H at the instant shown. Hint: Point B follows a circular path about point E and therefore the velocity of B is not vertical.

9 ft 1.5 ft

9 ft E A

B

9 ft

10 ft 3 ft 6 rad/s H

D C

SOLUTION 1.5 b = 9.462° and 9 rBE = 292 + 1.52 = 9.124 ft. Since crank CD and beam BE are rotating about fixed points D and E, then vC and vB are always directed perpendicular to crank CD and beam BE, respectively. The magnitude of vC and vB are yC = vCD rCD = 6(3) = 18.0 ft>s and yB = vBE rBE = 9.124vBE. At the instant shown, vC is directed vertically while vB is directed with an angle 9.462° with the vertical. u = tan - 1 a

geometry,

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Instantaneous Center: The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC. From the geometry 10 = 60.83 ft sin 9.462°

rC>IC =

10 = 60.0 ft tan 9.462° destroy

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The angular velocity of link BC is given by vBC =

yC rC>IC

=

18.0 = 0.300 rad>s 60.0

Thus, the angular velocity of beam BE is given by yB = vBC rB>IC 9.124vBE = 0.300(60.83) vBE = 2.00 rad>s The speed of rod hanger H is given by yH = vBErEA = 2.00(9) = 18.0 ft>s






Ans.

16–90. vB ⫽ 10 ft/s B

Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point D at this instant.

D E

45 C A

SOLUTION 1.6 - x x = 5 10 5x = 16 - 10x x = 1.06667 ft v =

10 = 9.375 rad>s 1.06667

rIC-D = 2(0.2667)2 + (0.8)2 - 2(0.2667)(0.8) cos 135° = 1.006 ft

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sin f sin 135° = 0.2667 1.006 in

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f = 10.80°

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vC = 0.2667(9.375) = 2.50 ft>s

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vD = 1.006(9.375) = 9.43 ft>s

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u = 45° + 10.80° = 55.8°






0.8 ft 30 F vA

5 ft/s

16–91. vB ⫽ 10 ft/s B

Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point E at this instant.

D E

45 C A

SOLUTION x 1.6 - x = 5 10 5x = 16 - 10x x = 1.06667 ft v =

10 = 9.375 rad>s 1.06667

vC = v(rIC - C)

or

= 9.375(1.06667 - 0.8) laws

= 2.50 ft>s

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vE = v(rIC - E)

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= 9.3752(0.8)2 + (0.26667)2

not

= 7.91 ft>s

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0.8 ft 30 F vA ⫽ 5 ft/s

*16–92. Knowing that the angular velocity of link AB is vAB = 4 rad>s, determine the velocity of the collar at C and the angular velocity of link CB at the instant shown. Link CB is horizontal at this instant.

350 mm C

SOLUTION

B

45⬚

rIC-B rIC-C 0.350 = = sin 75° sin 45° sin 60°

60⬚

rIC-B = 0.2562 m rIC-C = 0.3138 m vCB =

A

2 = 7.8059 = 7.81 rad>s 0.2562

Ans.

vC = 7.8059(0.3138) = 2.45 m>s

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500 mm

VAB

16–93. If the collar at C is moving downward to the left at vC = 8 m>s, determine the angular velocity of link AB at the instant shown. 350 mm C

SOLUTION

B

45⬚

rIC-B rIC-C 0.350 = = sin 75° sin 45° sin 60°

60⬚

rIC-B = 0.2562 m rIC-C = 0.3138 m vCB =

A

8 = 25.494 rad>s 0.3138

vB = 25.494(0.2562) = 6.5315 m>s 6.5315 = 13.1 rad>s 0.5

or

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laws

vAB =






500 mm

VAB

16–94. If the roller is given a velocity of vA = 6 ft>s to the right, determine the angular velocity of the rod and the velocity of C at the instant shown.

C

B 6 ft

SOLUTION v =

4 ft

6 ft>s = 1.125 rad>s = 1.12 rad>s 5.334 ft

Ans.

vC = (1.125 rad>s)(3.003 ft) = 3.38 ft>s

Ans. 60⬚

vA ⫽ 6 ft/s

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A






16–95. As the car travels forward at 80 ft/s on a wet road, due to slipping, the rear wheels have an angular velocity v = 100 rad>s. Determine the speeds of points A, B, and C caused by the motion.

80 ft/s C B 1.4 ft

SOLUTION 80 = 0.8 ft 100

r =

vA = 0.6(100) = 60.0 ft s :

Ans.

vC = 2.2(100) = 220 ft s ;

Ans. 60.3°

Ans.

b

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laws

or

vB = 1.612(100) = 161 ft s






A 100 rad/s

*16–96. Determine the angular velocity of the double-tooth gear and the velocity of point C on the gear.

0.3 m C B

SOLUTION General Plane Motion: The location of the IC can be found using the similar triangles shown in Fig. a. rA>IC 4

0.45 - rA>IC =

rA>IC = 0.18 m

6

Then, y = 0.3 - rA>IC = 0.3 - 0.18 = 0.12 m

laws

or

and

in

teaching

Web)

rC>IC = 20.32 + 0.122 = 0.3231 m 0.12 b = 21.80° 0.3 learning.

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Dissemination

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copyright

f = tan - 1 a

the

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Thus, the angular velocity of the gear can be determined from

work

student

vA 4 = = 22.22 rad>s = 22.2 rad>s rA>IC 0.18

the

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v =

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vC = vrC>IC = 22.2(0.3231) = 7.18 m>s any

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f = 90° - f = 90° - 21.80° = 68.2°






A

vA ⫽ 4 m/s

Ans.

0.15 m vB ⫽ 6 m/s

16–97. The wheel is rigidly attached to gear A, which is in mesh with gear racks D and E. If D has a velocity of vD = 6 f t>s to the right and the wheel rolls on track C without slipping, determine the velocity of gear rack E.

D vD

6 ft/s

1.5 ft A 0.75 ft

O C C

vE E

SOLUTION General Plane Motion: Since the wheel rolls without slipping on track C, the IC is located there, Fig. a. Here, rD>IC = 2.25 ft

rE>IC = 0.75 ft

Thus, the angular velocity of the gear can be determined from v =

vD 6 = 2.667 rad>s = rD>IC 2.25

Then, vE = vrE>IC = 2.667(0.75) = 2 ft>s ;

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Ans.






16–98. The wheel is rigidly attached to gear A, which is in mesh with gear racks D and E. If the racks have a velocity of vD = 6 ft>s and vE = 10 ft>s, show that it is necessary for the wheel to slip on the fixed track C. Also find the angular velocity of the gear and the velocity of its center O.

D vD

6 ft/s

1.5 ft A 0.75 ft

O C C

vE E

SOLUTION General Plane Motion: The location of the IC can be found using the similar triangles shown in Fig. a, rD>IC 6

=

3 - rD>IC

rD>IC = 1.125 ft

10

Thus, rO>IC = 1.5 - rD>IC = 1.5 - 1.125 = 0.375ft rF>IC = 2.25 - rD>IC = 2.25 - 1.125 = 1.125ft

laws

or

Thus, the angular velocity of the gear is Web)

vD 6 = = 5.333 rad>s = 5.33 rad>s rD>IC 1.125

in

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v =

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The velocity of the contact point F between the wheel and the track is

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the

vF = vrF>IC = 5.333(1.125) = 6 ft>s ; by

Since vF Z 0, the wheel slips on the track

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The velocity of center O of the gear is

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vO = vrO>IC = 5.333(0.375) = 2ft>s ;

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Ans.






16–99. The epicyclic gear train is driven by the rotating link DE, which has an angular velocity vDE = 5 rad>s. If the ring gear F is fixed, determine the angular velocities of gears A, B, and C.

C E

30 mm 40 mm

B vDE ⫽ 5 rad/s A

SOLUTION vE = 0.

= 0.8 m>s

0 vC = 0.

26.7 rad>s

vP = (0.

6.7) = 1.6 m>s

F

Ans.

1.6 = x x

or

x = 0.05 ; = 28.75 rad>s

Ans. in

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Web)

0

laws

vB =

Dissemination

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copyright

vP¿ = 28.75(0.08 - 0.05565) = 0.700 m>s ;

permitted.

0.700 = 14.0 rad>s 0.05

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vA =






D

50 mm

*16–100. The similar links AB and CD rotate about the fixed pins at A and C. If AB has an angular velocity vAB = 8 rad>s, determine the angular velocity of BDP and the velocity of point P.

B

300 mm

300 mm

300 mm

300 mm 60°

60° A

C

ωAB = 8 rad/s 700 mm

SOLUTION Kinematic Diagram: Since link AB and CD is rotating about fixed points A and C. then vB and vD are always directed perpendicular to link AB and CD respectively. The magnitude of vB and vD are vB = vAB rAB = 8(0.3) = 2.40 m>s and vD = vCD rCD = 0.3 vCD. At the instant shown. vB and vD are directed at 30° with the horizontal.

P

Instantaneous Center: The instantaneous center of zero velocity of link BDP at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vD. From the geometry. 0.3 = 0.600 m cos 60° or

rB>IC =

in

teaching

Web)

laws

rP>IC = 0.3 tan 60° + 0.7 = 1.220 m

Dissemination

Wide

copyright

The angular velocity of link BDP is given by

permitted.

vB 2.40 = 4.00 rad s = rB>IC 0.600

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vBDP =

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Thus, the velocity of point P is given by

work

vP = vBDP rP IC = 4.00(1.220) = 4.88 m s ;

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D

16–101. If rod AB is rotating with an angular velocity vAB = 3 rad>s, determine the angular velocity of rod BC at the instant shown.

C 3 ft 4 ft

B 2 ft vAB

3 rad/s A

45

SOLUTION 4 sin 60° - 2 sin 45° b = 43.10°. 3 Since links AB and CD is rotating about fixed points A and D, then vB and vC are always Kinematic Diagram: From the geometry, u = sin - 1 a

directed perpendicular to links AB and CD, respectively. The magnitude of vB and vC are yB = vAB rAB = 3(2) = 6.00 ft>s and yC = vCDrCD = 4vCD. At the instant

laws

or

shown, vB is directed at an angle of 45° while vC is directed at 30°

Dissemination

Wide

copyright

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Instantaneous Center: The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC. Using law of sines, we have 3 sin 75°

rB>IC = 3.025 ft

=

3 sin 75°

rC>IC = 0.1029 ft

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The angular velocity of link BC is given by

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integrity

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yB 6.00 = = 1.983 rad>s = 1.98 rad>s rB>IC 3.025 This

vBC =

Ans.

60

D

16–102. If rod AB is rotating with an angular velocity vAB = 3 rad>s, determine the angular velocity of rod CD at the instant shown.

C 3 ft 4 ft

B 2 ft vAB

3 rad/s A

45

SOLUTION 4sin 60° - 2 sin 45° b = 43.10°. 3 Since links AB and CD is rotating about fixed points A and D, then vB and vC are Kinematic Diagram: From the geometry. u = sin - 1 a

always directed perpendicular to links AB and CD, respectively. The magnitude of vB and vC are yB = vAB rAB = 3(2) = 6.00 ft>s and yC = vCD rCD = 4vCD. At the

Dissemination

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Instantaneous Center: The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC. Using law of sines, we have

Web)

laws

or

instant shown, vB is directed at an angle of 45° while vC is directed at 30°.

3 sin 75°

rB>IC = 3.025 ft

=

3 sin 75°

rC>IC = 0.1029 ft

the

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=

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sin 103.1°

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rB>IC

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The angular velocity of link BC is given by

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yB 6.00 = 1.983 rad>s = rB>IC 3.025

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Thus, the angular velocity of link CD is given by

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vBC =

4vCD = 1.983(0.1029) vCD = 0.0510 rad>s






Ans.

60

D

16–103. At a given instant the top end A of the bar has the velocity and acceleration shown. Determine the acceleration of the bottom B and the bar’s angular acceleration at this instant.

vA ⫽ 5 ft/s aA ⫽ 7 ft/s2

A

10 ft

SOLUTION

60 B

v =

5 = 1.00 rad>s 5

aB = aA + aB>A aB = 7 + 10 + a(10) : a 30° T h 30° + ) (:

aB = 0 - 10 sin 30° + a(10) cos 30°

(+ c)

0 = - 7 + 10 cos 30° + a(10) sin 30° Ans.

aB = - 7.875 ft>s2 = 7.88 ft>s2 ;

Ans. teaching

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laws

or

a = - 0.3321 rad>s2 = 0.332 rad>s2 b

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aBi = - 7j - (1)2(10 cos 60°i - sin 60°j) + (ak) * (10 cos 60°i - 10 sin 60°j)

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States

World

permitted.

Dissemination

aB = aA - v2rB>A + a * rB>A

aB = - 10 cos 60° + a(10 sin 60°)

(+ c)

0 = - 7 + 10 sin 60° + a(10 cos 60°)

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a = -0.3321 rad>s2 = 0.332 rad>s2 b

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aB = - 7.875 ft>s2 = 7.88 ft>s2 ;






Ans.

*16–104. At a given instant the bottom A of the ladder has an acceleration aA = 4 ft>s2 and velocity vA = 6 ft>s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant. 16 ft

SOLUTION v =

6 = 0.75 rad>s 8

30⬚

A

aB = aA + (aB>A)n + (aB>A)t aB = 4 + (0.75)2 (16) + a(16) ; 30° d 30° f T + ) (;

0 = 4 + (0.75)2(16) cos 30° - a(16) sin 30°

(+ T )

aB = 0 + (0.75)2(16) sin 30° + a(16) cos 30°

or

Solving, Ans.

aB = 24.9 ft>s2 T

Ans. Dissemination

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laws

a = 1.47 rad>s2

instructors

States

World

permitted.

Also:

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United

on

learning.

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aB = aA + a * rB>A - v2rB>A

for

work

student

the

by

and

- aBj = - 4i + (ak) * (16 cos 30°i + 16 sin 30°j) - (0.75)2(16 cos 30°i + 16 sin 30°j)

solely

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protected

the

(including

0 = - 4 - 8a - 7.794

this

assessing

is

work

- aB = 13.856a - 4.5

Ans.

part

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provided

integrity

of

and

work

a = 1.47 rad>s2

any

courses

Ans.

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and

aB = 24.9 ft>s2 T






B

16–105. At a given instant the top B of the ladder has an acceleration aB = 2 ft>s2 and a velocity of vB = 4 ft>s, both acting downward. Determine the acceleration of the bottom A of the ladder, and the ladder’s angular acceleration at this instant. 16 ft

SOLUTION 4 = 0.288675 rad>s 16 cos 30°

v =

30⬚

A

aA = aB + a * rA>B - v2rA>B - aAi = - 2j + (ak) * ( - 16 cos 30°i - 16 sin 30°j) - (0.288675)2( -16 cos 30°i - 16 sin 30°j) - aA = 8a + 1.1547

Ans.

aA = -0.385 ft>s2 = 0.385 ft>s2 :

Ans.

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teaching

Web)

laws

a = -0.0962 rad>s2 = 0.0962 rad>s2b

or

0 = -2 - 13.856 a + 0.6667






B

16–106. Crank AB is rotating with an angular velocity of vAB = 5 rad>s and an angular acceleration of aAB = 6 rad>s2. Determine the angular acceleration of BC and the acceleration of the slider block C at the instant shown.

0.5 m

C 0.3 m

SOLUTION Angular Velocity: Since crank AB rotates about a fixed axis, Fig. a, vB = vAB rB = 5(0.3) = 1.5 m>s : The location of the IC for link BC is indicated in Fig. b. From the geometry of this figure, rB>IC = 0.5 tan 45° = 0.5 m Then, or

vB 1.5 = = 3 rad>s rB>IC 0.5 in

teaching

Web)

laws

vBC =

States

World

permitted.

Dissemination

Wide

copyright

Acceleration and Angular Acceleration: Since crank AB rotates about a fixed axis, Fig. c,

learning.

is

of

the

not

instructors

a B = aAB * rB - vAB2rB by

and

use

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on

= ( -6k) * (0.3j) - 52(0.3j)

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= [1.8i - 7.5j] m>s2

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Using this result and applying the relative acceleration equation by referring to Fig. d, integrity

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aC = aB + aBC * rC>B - vBC2 rC>B part

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aC cos 45°i + aC sin 45°j = (1.8i - 7.5j) + (aBCk) * (0.5i) - 32(0.5i)

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and

any

courses

0.7071aCi + 0.7071aCj = - 2.7i + (0.5aBC - 7.5)j sale

will

Equating the i and j components, 0.7071aC = - 2.7

(1)

0.7071aC = 0.5aBC -7.5

(2)

Solving Eqs. (1) and (2), aC = - 3.818 m>s2 = 3.82 m>s2 b

Ans.

aBC = 9.60 rad>s2d

Ans.

The negative sign indicates that aC acts in the opposite sense to that shown in Fig. c.






45°

B

vAB ⫽ 5 rad/s aAB ⫽ 6 rad/s2 A

16–107. At a given instant, the slider block A has the velocity and deceleration shown. Determine the acceleration of block B and the angular acceleration of the link at this instant.

B 300 mm

45°

SOLUTION vAB =

vB rA>IC

A v A = 1.5 m/s

1.5 = 7.07 rad>s = 0.3 cos 45°

aB = aA + a * rB>A - v2 rB>A -aB j = 16i + (ak) * (0.3 cos 45°i + 0.3 sin 45° j) - (7.07)2 (0.3 cos 45°i + 0.3 sin 45°j) + b a: (+ T)

0 = 16 - a (0.3) sin 45° - (7.07)2 (0.3) cos 45° aB = 0 - a(0.3) cos 45° + (7.07)2 (0.3) sin 45°

Ans.

aB = 5.21 m s2

Ans.

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Web)

laws

aA>B = 25 4 rad>s2 d

or

Solving:






aA = 16 m/s2

*16–108. As the cord unravels from the cylinder, the cylinder has an angular acceleration of a = 4 rad>s2 and an angular velocity of v = 2 rad>s at the instant shown. Determine the accelerations of points A and B at this instant. A

α = 4 rad/s2 ω = 2 rad/s

aC = 4(0.75) = 3 ft>s2

0.75 ft

T

aA = aC + a * rA>C - v2rA>C aA = - 3j + 4k * (0.75j) - (2)2 (0.75j) aA = { - 3i - 6j} ft>s2 aA = 2(-3)2 + ( - 6)2 = 6.71 ft>s2

Ans.

6 u = tan - 1 a b = 63.4°d 3

Ans.

laws

or

aB = aC + a * rB>C - v2 rB>C in

teaching

Web)

aB = - 3j + 4k * (- 0.75i) - (2)2 ( - 0.75i) Dissemination

Wide

copyright

aB = {3i - 6j} ft>s2

permitted.

aB = 2(3)2 + ( -6)2 = 6.71 ft>s2

the

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learning.

6 f = tan - 1 a b = 63.4°c 3






C

B

SOLUTION

Ans.

16–109. The hydraulic cylinder is extending with a velocity of vC = 3 ft>s and an acceleration of aC = 1.5 ft>s2. Determine the angular acceleration of links BC and AB at the instant shown.

B C 1.5 ft 45⬚

SOLUTION Angular Velocity: Since link AB rotates about a fixed axis, Fig. a, then vB = vAB rB = vAB (1.5) The location of the IC for link BC is indicated in Fig. b. From the geometry of this figure, 3 = 4.243 ft cos 45°

rB>IC =

rC>IC = 3 tan 45° = 3 ft Then

or

vC 3 = = 1 rad>s rC>IC 3 teaching

Web)

laws

vBC =

Wide

copyright

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and

World

permitted.

Dissemination

vB = vBC rB>IC

is

of

the

not

instructors

States

vAB(1.5) = (1)(4.243)

the

by

and

use

United

on

learning.

vAB = 2.828 rad>s

of

protected

the

(including

for

work

student

Acceleration and Angular Acceleration: Since crank AB rotates about a fixed axis, Fig. c, then assessing

is

work

solely

a B = aAB * rB - vAB2rB provided

integrity

of

and

work

this

= (aAB k) * (- 1.5 cos 45°i + 1.5 sin 45°j) - 2.8282( -1.5 cos 45°i + 1.5 sin 45°j)

and

any

courses

part

the

is

This

= (8.485 - 1.061aAB)i - (8.485 + 1.061aAB)j

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Using this result and applying the relative acceleration equation by referring to Fig. d, aB = aC + aBC * rB>C - vBC2 rB>C

(8.485 - 1.061aAB)i - (8.485 + 1.061aAB)j = - 1.5i + (aBCk) * (-3i) - 12( -3i) (8.485 - 1.061aAB)i - (8.485 + 1.061aAB)j = 1.5i - 3aBC j Equating the i and j components, yields 8.485 - 1.061aAB = 1.5

(1)

-(8.485 + 1.061aAB) = - 3aBC

(2)

Solving Eqs. (1) and (2),






vC ⫽ 3 ft/s aC ⫽ 1.5 ft/s2 D

3 ft

aAB = 6.59 rad>s2

Ans.

aBC = 5.16 rad>s2

Ans.

A

16–110. At a given instant the wheel is rotating with the angular motions shown. Determine the acceleration of the collar at A at this instant.

A 60° 500 mm

ω = 8 rad/s α = 16 rad/s2

150 mm

B

SOLUTION

30°

Using instantaneous center method: 8(0.15) vB vAB = = = 4.157 rad>s rB>IC 0.5 tan 30° aA = aB + aA>B = 2.4 + 9.6 + (4.157 )2(0.5) 60°d a60° c30°

aA ;

+ ) (:

+ a(0.5) a60°

-aA = 2.4 cos 60° + 9.6 cos 30° - 8.65 cos 60° - a(0.5) sin 60°

( + c ) 0 = 2.4 sin 60° - 9.6 sin 30° - 8.65 sin 60° + a(0.5) cos 60° a = 40.8 rad>s2 d or

aA = 12.5 m>s2 ;

teaching

Web)

laws

Ans.

Wide

copyright

in

Also:

instructors

States

World

permitted.

Dissemination

aA = aB + a * rA>B - v2 rA>B

use

United

on

learning.

is

of

the

not

aA i = (8)2(0.15)(cos 30°)i - (8)2(0.15) sin 30°j + (16)(0.15) sin 30°i + (16)(0.15) cos 30° j

student

the

by

and

+ (ak) * (0.5 cos 60°i + 0.5 sin 60°j) - (4.157 2) (0.5 cos 60°i + 0.5 sin 60°j)

of

protected

the

(including

for

work

aA = 8.314 + 1.200 - 0.433 a - 4.326

this

assessing

is

work

solely

0 = - 4.800 + 2.0785 + 0.25a - 7.4935

the

is

This

provided

integrity

of

and

work

a = 40.8 rad s2 d

part

aA = 12.5 m s2 ;

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of

and

any

courses

Ans.






16–111. Crank AB rotates with the angular velocity and angular acceleration shown. Determine the acceleration of the slider block C at the instant shown.

0.4 m B

vAB ⫽ 4 rad/s aAB ⫽ 2 rad/s2 30⬚ 0.4 m

A

SOLUTION Angular Velocity: Since crank AB rotates about a fixed axis, Fig. a, vB = vAB rB = 4(0.4) = 1.6 m>s The location of the IC for link BC is indicated in Fig. b. From the geometry of this figure, rB>IC = 0.4 m Then

or

vB 1.6 = = 4 rad>s rB>IC 0.4 laws

vBC =

Wide

copyright

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teaching

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Acceleration and Angular Acceleration: Since crank AB rotates about a fixed axis, Fig. a States

World

permitted.

Dissemination

aB = aAB * rB - vAB2rB learning.

is

of

the

not

instructors

= (- 2k) * (0.4 cos 30°i + 0.4 sin 30°j) - 4 2(0.4 cos 30°i + 0.4 sin 30°j)

student

the

by

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= [- 5.143i - 3.893j] m>s2

the

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for

work

Using this result and applying the relative acceleration equation by referring to Fig. c,

assessing

is

work

solely

of

protected

aC = aB + aBC * rC>B - vBC2 rC>B

provided

integrity

of

and

work

this

aCi = (- 5.143i - 3.893j) + (aBCk) * (0.4 cos 30°i - 0.4 sin 30°j) - 42(0.4 cos 30°i - 0.4 sin 30°j)

and

any

courses

part

the

is

This

aCi = (0.2aBC - 10.69)i + (0.3464aBC - 0.6928)j

0 = 0.3464aBC - 0.6928

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aC = 0.2aBC - 10.69

will

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Equating the i and j components, yields

(1) (2)

Solving Eqs. (1) and (2), aBC = 2 rad>s2 aC = - 10.29 m>s2 = 10.3 m>s2 ;






Ans.

30⬚ C

*16–112. D

The wheel is moving to the right such that it has an angular velocity v = 2 rad>s and angular acceleration a = 4 rad>s2 at the instant shown. If it does not slip at A, determine the acceleration of point B.

B 4 rad/s2 2 rad/s

a v

1.45 ft

60

30

C

SOLUTION Since no slipping aC = ar = 4(1.45) = 5.80 ft>s2 aB = aC + aB>C

A

aB = 5.80 + (2)2(1.45) + 4(1.45) : c 30° g 30° + ) (:

(aB)x = 5.80 + 5.02 + 2.9 = 13.72

(+ c )

(aB)y = 0 - 2.9 + 5.02 = 2.12

laws

or

aB = 2(13.72)2 + (2.12)2 = 13.9 ft>s2 2.123 b = 8.80° a u 13.72

in

teaching

Web)

Ans. Wide

copyright

u = tan - 1 a

instructors

States

World

permitted.

Dissemination

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United

on

learning.

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of

the

not

aB = aC + a * rB>C - v2rB>C

student

the

by

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use

aB = 5.80i + (-4k) * ( - 1.45 cos 30°i + 1.45 sin 30°j) - (2)2( - 1.45 cos 30°i + 1.45 sin 30°j)

protected

the

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aB = {13.72i + 2.123j} ft>s2

integrity

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2.123 b = 8.80° a u 13.72

This

u = tan - 1 a

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is

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of

aB = 2(13.72)2 + (2.123)2 = 13.9 ft>s2






Ans.

16–113. v ⫽ 3 rad/s a ⫽ 8 rad/s2

The disk is moving to the left such that it has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point B.

D 30⬚

45⬚

0.5 m B

A

SOLUTION aC = 0.5(8) = 4 m>s2 aB = aC + aB>C

+ B A:

(aB)x = -4 + 4.5 cos 30° + 4 sin 30° = 1.897 m>s2

A+cB

(aB)y = 0 + 4.5 sin 30° - 4 cos 30° = -1.214 m>s2 Ans.

1.214 b = 32.6° c 1.897

Ans.

teaching

in

Dissemination

Wide

copyright

u = tan - 1 a

Web)

laws

aB = 2(1.897)2 + ( - 1.214)2 = 2.25 m>s2

or

aB = c 4 d + D (3)2 (0.5) T + D (0.5) (8 ) T ; f 30° a 30°

instructors

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World

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United

on

learning.

is

of

the

not

aB = aC + a * rB>C - v2 rB>C

student

the

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(aB)x i + (aB)y j = - 4i + (8k) * ( - 0.5 cos 30°i - 0.5 sin 30°j) - (3)2 (-0.5 cos 30°i - 0.5 sin 30°j) (aB)x = -4 + 8(0.5 sin 30°) + (3)2(0.5 cos 30°) = 1.897 m>s2

A+cB

(aB)y = 0 - 8(0.5 cos 30°) + (3)2 (0.5 sin 30°) = - 1.214 m>s2 integrity

of provided

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is

Ans.

and

any

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1.214 b = 32.6° c 1.897

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u = tan - 1 a

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aB = 2(1.897)2 + ( - 1.214)2 = 2.25 m>s2






Ans.

C

16–114. v ⫽ 3 rad/s a ⫽ 8 rad/s2

The disk is moving to the left such that it has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point D.

D 30⬚

45⬚

0.5 m B

SOLUTION aC = 0.5(8) = 4 m>s2 aD = aC + aD>C aD = c 4 d + D (3)2 (0.5 )T + D 8 (0.5) T ; 45° b e 45° + B A:

(aD)x = - 4 - 4.5 sin 45° - 4 cos 45° = -10.01 m>s2

A+cB

(aD)y = 0 - 4.5 cos 45° + 4 sin 45° = - 0.3536 m>s2 0.3536 b = 2.02° d 10.01

or

Ans. laws

u = tan - 1 a

teaching

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aD = 2( -10.01)2 + (- 0.3536)2 = 10.0 m>s2

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aD = aC + a * rD>C - v2 rD>C

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(aD)x i + (aD)y j = - 4i + (8k) * (0.5 cos 45°i + 0.5 sin 45°j) - (3)2 (0.5 cos 45°i + 0.5 sin 45°j) (aD)x = - 4 - 8(0.5 sin 45°) - (3)2(0.5 cos 45°) = - 10.01 m>s2

A+cB

(aD)y = +8(0.5 cos 45°) - (3)2 (0.5 sin 45°) = -0.3536 m>s2 this

assessing

is

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the

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provided

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0.3536 b = 2.02° d 10.01

Ans.

part

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u = tan - 1 a

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aD = 2( -10.01)2 + ( -0.3536)2 = 10.0 m>s2






Ans.

A

C

16–115. A cord is wrapped around the inner spool of the gear. If it is pulled with a constant velocity v, determine the velocities and accelerations of points A and B. The gear rolls on the fixed gear rack.

B 2r A

r

G v

SOLUTION Velocity analysis: v r

v =

vB = vrB>IC =

v (4r) = 4v : r

vA = v rA>IC =

v A 2(2r)2 + (2r)2 B = 2 22v r

Ans.

Ans.

a45°

or

Acceleration equation: From Example 16–3, Since aG = 0, a = 0 rA>G = - 2r i

teaching

Web)

laws

rB>G = 2r j

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copyright

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aB = aG + a * rB>G - v2rB>G

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World

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Dissemination

2v2 v 2 j = 0 + 0 - a b (2rj) = r r

Ans.

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2v2 T r

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work

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aB =

solely

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a A = aG + a * rA>G - v2rA>G

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2v2 : r

and

aA =

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2v2 v 2 i = 0 + 0 - a b ( - 2ri) = r r






*16–116. At a given instant, the gear racks have the velocities and accelerations shown. Determine the acceleration of point A.

a v

2 ft/s2 6 ft/s

SOLUTION Velocity Analysis: The angular velocity of the gear can be obtained by using the method of instantaneous center of zero velocity. From similar triangles, yD yC v = = rD>IC rC>IC 6 rD>IC

=

2

A 0.25 ft

(1)

rC>IC

B

Where rD>IC + rC>IC = 0.5

(2)

a v

Solving Eqs.(1) and (2) yields rD>IC = 0.375 ft Thus,

yD 6 = 16.0 rad>s = rD>IC 0.375

Dissemination

Wide

copyright

in

teaching

Acceleration Equation: The angular acceleration of the gear can be obtained by analyzing the angular motion of points C and D. Applying Eq. 16–18 with rD>C = { -0.5i} ft, we have

Web)

laws

or

v =

rC>IC = 0.125 ft

not

instructors

States

World

permitted.

aD = aC + a * rD>C - v2 rD>C

use

United

on

learning.

is

of

the

(aD)ni + 2j = -(aC)ni - 3j + (- ak) * ( -0.5i) - 16.02 (-0.5i)

for

work

student

the

by

and

(aD)ni + 2j = -(aC)ni + (0.5a - 3)j + 128i

of

protected

the

(including

Equating the j components, we have a = 10.0 rad>s2 this

assessing

is

work

solely

2 = 0.5 a - 3

any

destroy

of

and

aA = aC + a * rA>C - v2 rA>C

courses

part

the

is

This

provided

integrity

of

and

work

The acceleration of point A can be obtained by analyzing the angular motion of points A and C. Applying Eq. 16–18 with rA>C = {- 0.25i} ft, we have

sale

will

their

aAj = - (aC)ni - 3j + (-10.0k) * ( - 0.25i) - 16.02 (-0.25i) Equating the i and j components, we have aA = 0.500 ft>s2 T (aC)n = 64 m>s2 d






Ans.

3 ft/s2 2 ft/s

16–117. At a given instant, the gear racks have the velocities and accelerations shown. Determine the acceleration of point B.

a ⫽ 2 ft/s2 v ⫽ 6 ft/s

A

SOLUTION 0.25 ft

Angular Velocity: The method of IC will be used. The location of IC for the gear is indicated in Fig. a. using the similar triangle, 2 6 = rD>IC 0.5 - rD>IC

B

rD>IC = 0.125 ft a ⫽ 3 ft/s2 v ⫽ 2 ft/s

Thus, v =

vD rD>IC

=

2 = 16 rad>s b 0.125 ft

laws

or

Acceleration and Angular Acceleration: Applying the relative acceleration equation for points C and D by referring to Fig. b, in

teaching

Web)

aC = aD + A * rC>D - v2rC>D Dissemination

Wide

copyright

(aC)ni + 2j = - (aD)ni - 3j + ( - ak) * ( -0.5i) - 162(-0.5i)

of

the

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World

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(aC)ni + 2j = [128 - (aD)n]i + (0.5a - 3)j

and

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learning.

is

Equating j component,

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for

work

student

the

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2 = 0.5a - 3

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Using this result, the relative acceleration equation applied to points A and C, Fig. b, gives

provided

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of

and

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this

aC = aA + A * rC>A - v2rC>A

any their

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and

(aC)ni + 2j = 64i + (aA + 2.5)j

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This

(aC)ni + 2j = aAj + (- 10k) * ( -0.25i) - 162( -0.25i)

2 = aA + 2.5

sale

will

Equating j component,

aA = - 0.5 ft>s2 = 0.5 ft>s2 T

Using this result to apply the relative acceleration equation to points A and B, aB = aA + A * rB>A - v2rB>A - (aB)ti + (aB)nj = - 0.5j + ( -10k) * ( - 0.25j) - 162(- 0.25j) -(aB)ti + (aB)n j = - 2.5i + 63.5j Equating i and j components, (aB)t = 2.50 ft>s2

(aB)n = 63.5 ft>s2

Thus, the magnitude of aB is aB = 2(aB)2t + (aB)2n = 22.502 + 63.52 = 63.55 ft>s2

Ans.

and its direction is u = tan - 1 c © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





(aB)n 63.5 d = tan - 1 a b = 87.7° ub (aB)t 2.50

Ans.

16–118. At a given instant gears A and B have the angular motions shown. Determine the angular acceleration of gear C and the acceleration of its center point D at this instant. Note that the inner hub of gear C is in mesh with gear A and its outer rim is in mesh with gear B.

5 in. A

vA ⫽ 4 rad/s aA ⫽ 8 rad/s2

20 in. D 5 in.

SOLUTION

B

aP = aP¿ + aP>P¿ + ) (:

vB ⫽ 1 rad/s aB ⫽ 6 rad/s2

120 = -40 + a(15)

a = 10.67 rad>s2 d

Ans.

aP = aD + aP>D + ) (:

120 = (aD)t + (10.67)(10)

(aD)t = 13.3 in. >s2 : vP = vP¿ + vP>P¿ or

20 = -20 + v(15)

Web)

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v = 2.667 rad>s

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vD = -20 + 10(2.667)

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vD = vP + vD>P

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vD = 6.67 in. >s

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(6.67)2 = 4.44 in. >s2 c 10 is

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(aD)n =

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4.44 ) = 18.4° 13.3

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u = tan - 1 (

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(4.44)2 + (13.3)2 = 14.1 in. s2

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aD =






C

10 in.

16–119. The wheel rolls without slipping such that at the instant shown it has an angular velocity V and angular acceleration A. Determine the velocity and acceleration of point B on the rod at this instant.

A 2a B

SOLUTION vB = vA + vB/A (Pin) + v = ; B

1 Qv 22aR + 2av¿ a b 2 22

+c O = v¿ =

1

1 22

Qv22aR + 2av¿ a

23 b 2

v 23

vB = 1.58 va

Ans.

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a A = aO + aA/O (Pin)

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(a A)x + (aA)y = aa + a(a) + v2a

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(a A)x = aa - v2a

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a B = aA + aB/A (Pin)

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v 2 23 1 a B = aa - v2a + 2a(a¿)a b - 2a a b 2 2 23 2 1 b a b 2 23

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v

the part

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23

b + 2a a

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O = -aa + 2aa¿ a

a B = 1.58aa - 1.77v2a






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a¿ = 0.577a - 0.1925v2

Ans.

O a

V, A

*16–120. The center O of the gear and the gear rack P move with the velocities and accelerations shown. Determine the angular acceleration of the gear and the acceleration of point B located at the rim of the gear at the instant shown.

B

vO ⫽ 3 m/s aO ⫽ 6 m/s2 vP ⫽ 2 m/s aP ⫽ 3 m/s2

O

A

SOLUTION Angular Velocity: The location of the IC is indicated in Fig. a. Using similar triangles, 3 2 = rO>IC 0.15 - rO>IC

rO>IC = 0.09 m

Thus, v =

vO 3 = = 33.33 rad>s rO>IC 0.09

Acceleration and Angular Acceleration: Applying the relative acceleration equation to points O and A and referring to Fig. b,

laws

or

aA = aO + a * rA>O - v2rA>O

in

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-3i + (aA)n j = 6i + ( -ak) * (- 0.15j) - 33.332( -0.15j)

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- 3i + (aA)n j = (6 - 0.15a)i + 166.67j

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Equating the i components,

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- 3 = 6 - 0.15a

Ans.

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a = 60 rad>s2

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Using this result, the relative acceleration equation is applied to points O and B, Fig. b, which gives

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aB = aO + a * rB>O - v2rB>O

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(aB)ti - (aB)n j = 6i + ( - 60k) * (0.15j) - 33.332(0.15j) sale

will

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(aB)ti - (aB)n j = 15i - 166.67j Equating the i and j components, (aB)t = 15 m>s2

(aB)n = 166.67 m>s2

Thus, the magnitude of aB is aB = 3(aB)t 2 + (aB)n 2 = 3152 + 166.672 = 167 m>s2 and its direction is u = tan-1 c






(aB)n 166.67 d = tan-1 a b = 84.9° c (aB)t 15

150 mm

Ans.

P

16–121. The tied crank and gear mechanism gives rocking motion to crank AC, necessary for the operation of a printing press. If link DE has the angular motion shown, determine the respective angular velocities of gear F and crank AC at this instant, and the angular acceleration of crank AC.

C F

100 mm

50 mm

v DE ⫽ 4 rad/s

75 mm

SOLUTION

B

100 mm

Velocity analysis:

G 30

yD = vDErD>E = 4(0.1) = 0.4 m>s c 150 mm

vB = vD + vB>D yB = 0.4 + (vG)(0.075) c

a 30°

+ ) (:

T

yB cos 30° = 0,

yB = 0

(+ c ) vG = 5.33 rad>s vAC = 0

Ans.

or

yC = 0,

laws

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vFrF = vGrG

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100 b = 10.7 rad>s 50

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vF = 5.33 a

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Acceleration analysis:

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(aD)n = (4)2(0.1) = 1.6 m>s2 :

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(aD)t = (20)(0.1) = 2 m>s2 c

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(aB)n + (aB)t = (aD)n + (aD)t + (aB>D)n + (aB>D)t provided

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= 1.6 + 2 + (5.33)2(0.075) + aG (0.075) :

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courses

a 30°

This

0 + (a B)t

(aB)t sin 30° = 0 + 2 + 0 + aG (0.075)

+ ) (:

(aB)t cos 30° = 1.6 + 0 + (5.33)2(0.075) + 0 sale

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(+ c )

Solving, (aB)t = 4.31 m>s2,

aG = 2.052 rad>s2b

Hence, aAC =






E D

(a B)t 4.31 = 28.7 rad>s2b = rB>A 0.15

Ans.

A

aDE ⫽ 20 rad/s2

16–122. Pulley A rotates with the angular velocity and angular acceleration shown. Determine the angular acceleration of pulley B at the instant shown.

50 mm

vA aA

40 rad/s 5 rad/s2

SOLUTION

50 mm

Angular Velocity: Since pulley A rotates about a fixed axis,

B

vC = vA rA = 40(0.05) = 2 m>s c

125 mm

The location of the IC is indicated in Fig. a. Thus, vB =

E

vC 2 = 11.43 rad>s = rC>IC 0.175

Acceleration and Angular Acceleration: For pulley A,

laws

or

(aC)t = aArA = 5(0.05) = 0.25 m>s2 c

in

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Using this result and applying the relative acceleration equation to points C and D by referring to Fig. b,

permitted.

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aD = aC + aB * rD>C - vB 2rD>C

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World

(aD)n i = (aC)n i + 0.25j + ( -aB k) * (0.175i)-11.432(0.175i)

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(aD)n i = [(aC)n - 22.86]i + (0.25 - 0.175aB)j

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0 = 0.25 - 0.175aB

Ans.

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aB = 1.43 rad>s2






16–123. Pulley A rotates with the angular velocity and angular acceleration shown. Determine the acceleration of block E at the instant shown.

50 mm

vA aA

40 rad/s 5 rad/s2

SOLUTION

50 mm

Angular Velocity: Since pulley A rotates about a fixed axis,

B

vC = vArA = 40(0.05) = 2 m>s c

125 mm

The location of the IC is indicated in Fig. a. Thus, vB =

E

vC 2 = 11.43 rad>s = rC>IC 0.175

Acceleration and Angular Acceleration: For pulley A,

laws

or

(aC)t = aA rA = 5(0.05) = 0.25 m>s2 c

in

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Using this result and applying the relative acceleration equation to points C and D by referring to Fig. b,

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aD = aC + aB * rD>C - vB 2rD>C

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(aD)n i = (aC)n i + 0.25j + ( - aBk) * (0.175i) - 11.432(0.175i)

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learning.

(aD)n i = [(aC)n - 22.86]i + (0.25 - 0.175aB)j

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Equating the j components,

the

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This

provided

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aB = 1.429 rad>s = 1.43 rad>s2

this

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0 = 0.25 - 0.175a B

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aE = aC + aB * rE>C - vB 2rE>C

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Using this result, the relative acceleration equation applied to points C and E, Fig. b, gives

aE j = [(aC)n i + 0.25j] + (- 1.429k) * (0.125i) - 11.432(0.125i) aE j = [(aC)n - 16.33]i + 0.0714j Equating the j components, aE = 0.0714 m>s2 c






Ans.

*16–124. At a given instant, the gear has the angular motion shown. Determine the accelerations of points A and B on the link and the link’s angular acceleration at this instant.

B 60⬚ 3 in.

v ⫽ 6 rad/s a ⫽ 12 rad/s2

2 in.

SOLUTION

yA = vrA>IC = 6(1) = 6 in.>s aO = -12(3)i = { - 36i} in.>s2

rA>O = {- 2j} in.

a = {12k} rad>s2

aA = a0 + a * rA>O - v2rA>O = -36i + (12k) * ( -2j) - (6)2( - 2j) = { -12i + 72j} in.>s2 aA = 2(- 12)2 + 722 = 73.0 in.>s2

or

Ans. laws

72 b = 80.5° b 12

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Ans.

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For link AB

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6 = 0 q

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student

rB>A = {8 cos 60°i + 8 sin 60°j} in.

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aAB = - aAB k

protected

aB = aB i

the

(including

rA>IC

for

yA

United

on

learning.

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the

not

instructors

The IC is at q , so vAB = 0, i.e., vAB =

integrity

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aB = aA + aAB * rB>A - v2 rB>A

part

the

is

This

provided

aB i = (- 12i + 72j) + ( - aAB k) * (8 cos 60°i + 8 sin 60°j) - 0 aB = - 12 + 8 sin 60°(18) = 113 in.>s2 :

(+ c)

0 = 72 - 8 cos 60°aAB

Ans.





their

destroy

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+ B A:


8 in. A

For the gear

u = tan-1 a

O

sale

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aAB = 18 rad>s2 b

Ans.

16–125. The ends of the bar AB are confined to move along the paths shown. At a given instant, A has a velocity of vA = 4 ft>s and an acceleration of a A = 7 ft>s2. Determine the angular velocity and angular acceleration of AB at this instant.

B

2 ft 60

2 ft

SOLUTION vA aA

vB = vA + vB>A 30°

4 ft/s 7 ft/s2

= 4+ v(4.788) T h 51.21°

vB b

+ ) (:

-vB cos 30° = 0 - v(4.788) sin 51.21°

(+ c )

vB sin 30° = - 4 + v(4.788) cos 51.21°

vB = 20.39 ft>s

30° b

v = 4.73 rad>sd

Ans.

laws

or

a B = a A + a B>A = 7 + 107.2 + 4.788(a) d 51.21° T h 51.21°

Web)

2 07.9 60° d

teaching

+

in

at 30° b

at cos 30° + 207.9 cos 60° = 0 + 107.2 cos 51.21° + 4.788a(sin 51.21°)

(+ c)

at sin 30° - 207.9 sin 60° = - 7 - 107.2 sin 51.21° + 4.788a(cos 51.21°) on

learning.

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at(0.866) - 3.732a = -36.78

the

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at (0.5) - 3a = 89.49 assessing

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at = - 607 ft>s2

Ans.

part

the

is

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provided

integrity

of

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this

a = -131 rad>s2 = 131 rad>s2b

destroy sale

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vB = vA = v * rB>A

of

and

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courses

Also:

-vB cos 30°i + vB sin 30°j = - 4j + (vk) * (3i + 3.732j) -vB cos 30° = - v(3.732) vB sin 30° = -4 + v(3) v = 4.73 rad>sd

Ans.

vB = 20.39 ft>s a B = a A - v2rB>A + a * rB>A (-at cos 30°i + at sin 30°j) + ( -207.9 cos 60°i - 207.9 sin 60°j) = - 7j - (4.732)2(3i + 3.732j) + (ak) * (3i + 3.732j) -at cos 30° - 207.9 cos 60° = - (4.732)2(3) - a(3.732) at sin 30° - 207.9 sin 60° = - 7 -(4.732)2(3.732) + a(3) at = - 607 ft>s2 a = -131 rad>s2 = 131 rad>s2b © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

A

16–126. At a given instant, the cables supporting the pipe have the motions shown. Determine the angular velocity and angular acceleration of the pipe and the velocity and acceleration of point B located on the pipe.

v

a

5 ft/s

2 ft

v = 0.25 rad>s b

Ans.

vB = 5.00 ft>s T

Ans.

aB = aA + aB>A 1.5 + (aB)x = 2 + (aA)x + (a) (4) + (0.25) 2 (4) :

T

:

T

1.5 = - 2 + a(4) laws

or

(+ T )

;

c

a = 0.875 rad>s2 d

in

teaching

Web)

Ans.

Dissemination

Wide

copyright

aB = aO + aB>O

World

:

of

the

not

T

T

permitted.

= aO + 0.875(2) + (0.25)2 (2)

instructors

:

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States

1.5 + (aB)x

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+ ) (a ) = (0.25)2(2) = 0.125 ft>s2 (: B x

Ans.

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Ans.

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of

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this

assessing

is

1.5 b = 85.2° 0.125

protected

the

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student

aB = 2(1.5)2 + (0.125)2 = 1.51 ft>s2 u = tan - 1 a

part

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5 = 0.25 rad>s 20

Ans.

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v =

vB = 5.00 ft>s T

Ans.

aB = aA + a * rB>A - v2rB>A - 1.5j + (aB)x i = 2j - (aA)x i + (ak) * ( - 4i) - (0.25)2(- 4i) - 1.5 = 2 - 4a a = 0.875 rad>s2 d

Ans.

aB = aO + a * rB>O - v2rB>O - 1.5j + (aB)x i = -aO j + (ak) * ( - 2i) - (0.25)2 ( - 2i) (aB)x = (0.25)2(2) = 0.125 aB = 2(1.5)2 + (0.125)2 = 1.51 ft>s2 u = tan - 1






1.5 0.125

O

B

( + T ) 5 = 6 - v (4)

= 85.2°

cu

Ans. Ans.

6 ft/s

a

2 ft/s2

1.5 ft/s2

SOLUTION vB = vA + vB>A

v

A

16–127. The slider block moves with a velocity of vB = 5 ft>s and an acceleration of aB = 3 ft>s2. Determine the angular acceleration of rod AB at the instant shown. 1.5 ft A vB aB

30 2 ft B

SOLUTION Angular Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, rB>IC = 2 sin 30° = 1 ft

rA>IC = 2 cos 30° = 1.732 ft

Thus, vAB =

vB rB>IC

=

5 = 5 rad>s 1 laws

or

Then

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copyright

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vA = vAB rA>IC = 5(1.732) = 8.660 ft>s

not

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World

permitted.

Acceleration and Angular Acceleration: Since point A travels along the circular

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

slot, the normal component of its acceleration has a magnitude of vA 2 8.6602 = 50 ft>s2 and is directed towards the center of the circular (aA)n = = r 1.5 slot. The tangential component is directed along the tangent of the slot. Applying

assessing

is

work

solely

of

protected

the relative acceleration equation and referring to Fig. b,

provided

integrity

of

and

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this

aA = aB + aAB * rA>B - vAB 2 rA>B

courses

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is

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50i - (aA)t j = 3i + (aAB k) * ( - 2 cos 30°i + 2 sin 30°j) - 52(- 2 cos 30° i + 2 sin 30°j)


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50i - (aA)t j = (46.30 - aAB)i + (1.732aAB + 25)j

50 = 46.30 - aAB aAB = -3.70 rad>s2 = 3.70 rad>s2 b






Ans.

5 ft/s 3 ft/s2

*16–128. The slider block moves with a velocity of vB = 5 ft>s and an acceleration of aB = 3 ft>s2. Determine the acceleration of A at the instant shown. 1.5 ft A vB aB

30 2 ft B

SOLUTION Angualr Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, rB>IC = 2 sin 30° = 1 ft

rA>IC = 2 cos 30° = 1.732 ft

Thus, vB 5 = 5 rad>s = rB>IC 1 or

vAB =

teaching

Web)

laws

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permitted.

Dissemination

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copyright

in

vA = vAB rA>IC = 5 A 1.732 B = 8.660 ft>s

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Acceleration and Angular Acceleration: Since point A travels along the circular

of

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the

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for

work

student

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on

learning.

is

slot, the normal component of its acceleration has a magnitude of vA 2 8.6602 = = 50 ft>s2 and is directed towards the center of the circular A aA B n = r 1.5 slot. The tangential component is directed along the tangent of the slot. Applying assessing

is

work

solely

the relative acceleration equation and referring to Fig. b, provided

integrity

of

and

work

this

aA = aB + aAB * rA>B - vAB 2 rA>B

A -2cos 30°i + 2 sin 30°j B -52 A -2 cos 30°i + 2 sin 30°j B and

any

courses

part

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50i - A aA B t j = 3i + A aAB k B *

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50i - A aA B t j = A 46.30 - aAB B i - A 1.732aAB + 25 B j Equating the i and j components, 50 = 46.30-aAB - A aA B t = - A 1.732aAB + 25 B Solving, aAB = - 3.70 rad>s2

A aA B t = 18.59 ft>s2 T Thus, the magnitude of aA is aA = 4A aA B t 2 + A aA B n 2 = 218.592 + 502 = 53.3ft>s2

Ans.

and its direction is u = tan-1 C






A aA B t A aA B n

S = tan-1 a

18.59 b = 20.4° c 50

Ans.

5 ft/s 3 ft/s2

16–129. Ball C moves along the slot from A to B with a speed of 3 ft>s, which is increasing at 1.5 ft>s2, both measured relative to the circular plate. At this same instant the plate rotates with the angular velocity and angular deceleration shown. Determine the velocity and acceleration of the ball at this instant.

z v a

6 rad/s 1.5 rad/s2

B

SOLUTION C

Reference Frames: The xyz rotating reference frame is attached to the plate and coincides with the fixed reference frame XYZ at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vO = aO = 0

A

# v = a = [-1.5k] rad>s2

v = [6k] rad>s

2 ft

x

y

For the motion of ball C with respect to the xyz frame, (vrel)xyz = (-3 sin 45°i - 3 cos 45°j) ft>s = [-2.121i - 2.121j] ft>s (arel)xyz = ( -1.5 sin 45°i - 1.5 cos 45°j) ft>s2 = [- 1.061i - 1.061j] ft>s2

laws

or

From the geometry shown in Fig. b, rC>O = 2 cos 45° = 1.414 ft. Thus,

copyright

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rC>O = (- 1.414 sin 45°i + 1.414 cos 45°j)ft = [-1i + 1j] ft

permitted.

Dissemination

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Velocity: Applying the relative velocity equation,

is

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vC = vO + v * rC>O + (vrel)xyz

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= 0 + (6k) * (- 1i + 1j) + ( -2.121i - 2.121j)

Ans.

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= [-8.12i - 8.12j] ft>s

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the

Acceleration: Applying the relative acceleration equation, we have

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# aC = aO + v * rC>O + v * (v * rC>O) + 2v * (vrel)xyz + (a rel)xyz part

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= 0 + (1.5k) * ( -1i + 1j) + (6k) * [(6k) * ( - 1i + 1j)] + 2(6k) * ( - 2.121i - 2.121j) + ( -1.061i - 1.061j) any

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= [61.9i - 61.0j]ft>s2






2 ft

45

16–130. The crane’s telescopic boom rotates with the angular velocity and angular acceleration shown. At the same instant, the boom is extending with a constant speed of 0.5 ft>s, measured relative to the boom. Determine the magnitudes of the velocity and acceleration of point B at this instant.

60 ft B vAB aAB

SOLUTION

30

Reference Frames: The xyz rotating reference frame is attached to boom AB and coincides with the XY fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xy frame with respect to the XY frame is vA = aA = 0

# vAB = a = [-0.01k] rad>s2

vAB = [-0.02k] rad>s

For the motion of point B with respect to the xyz frame, we have rB>A = [60j] ft

(vrel)xyz = [0.5j] ft>s

(arel)xyz = 0

Velocity: Applying the relative velocity equation, laws

or

vB = vA + vAB * rB>A + (v rel)xyz in

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Web)

= 0 + ( -0.02k) * (60j) + 0.5j

States

World

permitted.

Dissemination

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copyright

= [1.2i + 0.5j] ft > s

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Thus, the magnitude of vB, Fig. b, is

Ans.

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Acceleration: Applying the relative acceleration equation,

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vB = 21.22 + 0.52 = 1.30 ft>s

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# aB = aA + vAB * rB>A + vAB * (vAB * rB>A) + 2vAB * (vrel)xyz + (a rel)xyz

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Thus, the magnitude of aB, Fig. c, is

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= [0.62i - 0.024 j] ft>s2

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this

= 0 + ( -0.01k) * (60j) + (- 0.02k) * [(-0.02k) * (60j)] + 2( -0.02k) * (0.5j) + 0

aB = 20.622 + ( - 0.024)2 = 0.6204 ft>s2






Ans.

0.02 rad/s 0.01 rad/s2 A

16–131. While the swing bridge is closing with a constant rotation of 0.5 rad>s, a man runs along the roadway at a constant speed of 5 ft>s relative to the roadway. Determine his velocity and acceleration at the instant d = 15 ft.

d

z O

y

x v

SOLUTION Æ = {0.5k} rad>s Æ = 0 rm>o = {-15 j} ft (vm>o)xyz = {- 5j} ft>s (am>o)xyz = 0 vm = vo + Æ * rm>o + (vm>o)xyz vm = 0 + (0.5k) * ( -15j) - 5j vm = {7.5i - 5j} ft>s

laws

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Ans.

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am = aO + Æ * rm>O + Æ * (Æ * rm>O) + 2Æ * (vm>O)xyz + (am>O)xyz Dissemination

Wide

copyright

am = 0 + 0 + (0.5k) * [(0.5k) * (-15j)] + 2(0.5k) * ( -5j) + 0

permitted.

am = {5i + 3.75j} ft>s2

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0.5 rad/s

*16–132. While the swing bridge is closing with a constant rotation of 0.5 rad>s, a man runs along the roadway such that when d = 10 ft he is running outward from the center at 5 ft>s with an acceleration of 2 ft>s2, both measured relative to the roadway. Determine his velocity and acceleration at this instant.

d

z O

y

x v

SOLUTION Æ = {0.5k} rad>s Æ = 0 rm>o = {- 10 j} ft (vm>O)xyz = {- 5j} ft>s (am>O)xyz = {- 2j} ft>s2 vm = vo + Æ * rm>o + (vm>o)xyz vm = 0 + (0.5k) * (- 10j) - 5j or

vm = {5i - 5j} ft>s # am = aO + Æ * rm>O + Æ * (Æ * rm>O) + 2Æ * (vm>O)xyz + (am>O)xyz

Wide

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Dissemination

am = 0 + 0 + (0.5k) * [(0.5k) * ( -10j)] + 2(0.5k) * ( -5j) - 2j instructors

am = {5i + 0.5j} ft>s2

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0.5 rad/s

16–133. Collar C moves along rod BA with a velocity of 3 m>s and an acceleration of 0.5 m>s2, both directed from B towards A and measured relative to the rod. At the same instant, rod AB rotates with the angular velocity and angular acceleration shown. Determine the collar’s velocity and acceleration at this instant.

z

0.5 m B

x v ⫽ 6 rad/s a ⫽ 1.5 rad/s2

SOLUTION

y

# vAB = a = [1.5k] rad>s2

vB = v = [6k] rad>s

For the motion of collar C with respect to the xyz frame, we have (arel)xyz = [0.5j] m>s2

(vrel)xyz = [3j] m>s

or

rC>B = [0.5j] m

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vC = vB + vAB * rC>B + (vrel)xyz

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= [ -3i + 3j] m>s

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# aC = aB + vAB * rC>B + vAB * (vAB * rC>B) + 2vAB * (vrel)xyz + (arel)xyz

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= 0 + (1.5k) * (0.5j) + (6k) * (6k * 0.5j) + 2(6k) * (3j) + (0.5j)

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= [- 36.75i - 17.5j] m>s2






Ans.

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States

World

permitted.

Dissemination

= 0 + (6k) * (0.5j) + 3j

C A

Reference Frames: The xyz rotating reference frame is attached to rod AB and coincides with the XYZ reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vB = aB = 0

3 m/s 0.5 m/s2

16–134. y

Block A, which is attached to a cord, moves along the slot of a horizontal forked rod. At the instant shown, the cord is pulled down through the hole at O with an acceleration of 4 m>s2 and its velocity is 2 m>s. Determine the acceleration of the block at this instant. The rod rotates about O with a constant angular velocity v = 4 rad>s.

A V

Motion of moving reference. vO = 0 aO = 0 Æ = 4k # Æ = 0 Motion of A with respect to moving reference.

laws

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rA>O = 0.1i teaching

Web)

vA>O = -2i

Dissemination

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copyright

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aA>O = -4i States

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# aA = aO + Æ * rA>O + Æ * (Æ * rA>O) + 2Æ * (vA>O)xyz + (aA>O)xyz by

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a A = {- 5.60i - 16j} m>s2






O 100 mm

SOLUTION

= 0 + 0 + (4k) * (4k * 0.1i) + 2(4k * ( - 2i)) - 4i

x

Ans.

16–135. A girl stands at A on a platform which is rotating with a constant angular velocity v = 0.5 rad>s. If she walks at a constant speed of v = 0.75 m>s measured relative to the platform, determine her acceleration (a) when she reaches point D in going along the path ADC, d = 1 m; and (b) when she reaches point B if she follows the path ABC, r = 3 m.

O A r

SOLUTION

D B

(a) # aD = aO + Æ * rD>O + Æ * (Æ * rD>O) + 2Æ * (vD>O)xyz + (aD>O)xyz Motion of moving reference

x

(1)

Motion of D with respect to moving reference

aO = 0

rD>O = {1i} m

Æ = {0.5k} rad>s # Æ = 0

(vD>O)xyz = {0.75j} m>s (aD>O)xyz = 0 laws

or

Substitute the data into Eq.(1): in

teaching

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aB = 0 + (0) * (1i) + (0.5k) * [(0.5k) * (1i)] + 2(0.5k) * (0.75j) + 0

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Ans.

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= { -1i} m>s2

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# aB = aO + Æ * rB>O + Æ * (Æ * rB>O) + 2Æ * (vB>O)xyz + (aB>O)xyz

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(2)

Motion of B with respect to moving reference solely

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Motion of moving reference

rB>O = {3i} m this

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ao = 0

(vB>O)xyz = {0.75j} m>s part

the

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Æ = {0.5k} rad>s

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(aB>O)xyz = -(aB>O)n i + (aB>O)t j 2

= -(0.75 3 )i

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Æ = 0

= {-0.1875i} m>s Substitute the data into Eq.(2): aB = 0 + (0) * (3i) + (0.5k) * [(0.5k) * (3i)] + 2(0.5k) * (0.75j) + (- 0.1875i) = { -1.69i} m>s2






Ans.

y d

C

*16–136. A girl stands at A on a platform which is rotating with an angular acceleration a = 0.2 rad>s2 and at the instant shown has an angular velocity v = 0.5 rad>s. If she walks at a constant speed v = 0.75 m>s measured relative to the platform, determine her acceleration (a) when she reaches point D in going along the path ADC, d = 1 m; and (b) when she reaches point B if she follows the path ABC, r = 3 m.

O A r

SOLUTION # aD = aO + Æ * rD>O + Æ * (Æ * rD>O) + 2Æ * (vD>O)xyz + (aD>O)xyz

x

(1)



aO = 0

rD>O = {1i} m

Æ = {0.5k} rad>s # Æ = {0.2k} rad>s2

(vD>O)xyz = {0.75j} m>s (aD>O)xyz = 0 laws

or

Substitute the data into Eq.(1):

Ans.

permitted.

Dissemination

copyright

= {-1i + 0.2j} m>s2

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aB = 0 + (0.2k) * (1i) + (0.5k) * [(0.5k) * (1i)] + 2(0.5k) * (0.75j) + 0

not

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(b)

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of

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# aB = aO + Æ * rB>O + Æ * (Æ * rB>O) + 2Æ * (vB>O)xyz + (aB>O)xyz by

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(2)

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Motion of B with respect to moving reference

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rB>O = {3i} m

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aO = 0

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This

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integrity

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(vB>O)xyz = {0.75j} m>s

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(aB>O)xyz = -(aB>O)n i + (a B>O)t j = - A 0.75 3 B i 2

sale

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and

Æ = {0.5k} rad>s # Æ = {0.2k} rad>s2

= {-0.1875i} m>s Substitute the data into Eq.(2): aB = 0 + (0.2k) * (3i) + (0.5k) * [(0.5k) * (3i)] + 2(0.5k) * (0.75j) + ( -0.1875i) = {-1.69i + 0.6j} m s2






D B

(a)

Ans.

y d

C

16–137. At the instant shown, rod AB has an angular velocity vAB = 3 rad>s and an angular acceleration aAB = 5 rad>s2. Determine the angular velocity and angular acceleration of rod CD at this instant. The collar at C is pin-connected to CD and slides over AB.

vAB aAB

A

3 rad/s 5 rad/s2

60 C 0.75 m B 0.5 m

SOLUTION rC>A = (0.75 sin 60°)i - (0.75 cos 60°)j

D

rC>A = {0.6495i - 0.375j} m vC = vCD * rC>D = (vCDk) * (0.5j) = { -0.5vCDi} m>s a C = aCD * rCD - v2CD rCD = (aCDk) * (0.5j) - v2CD(0.5j) laws

or

aC = { -0.5 aCDi - v2CD(0.5)j} m>s2

Wide

copyright

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vC = vA + Æ * rC>A + (vC>A)xyz

World

permitted.

Dissemination

-0.5vCDi = 0 + (3k) * (0.6495i - 0.375j) + vC>A sin 60°i - vC>A cos 60°j

learning.

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-0.5vCD = 1.125 + 0.866vC>A

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0 = 1.9485 - 0.5vC>A

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vC>A = 3.897 m>s

Ans.

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vCD = - 9.00 rad>s = 9.00 rad>sb # a C = a A + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (a C>A)xyz

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This

a C = 0 + (5k) * (0.6495i - 0.375j) + (3k) * [(3k) * (0.6495i - 0.375j)]

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of

+ 2(3k) * [3.897(0.866)i - 0.5(3.897)j] + 0.866a C>A i - 0.5a C>A j sale

0.5 aCD i - ( -9.00)2(0.5)j = 0 + 1.875i + 3.2475j - 5.8455i + 3.375j + 11.6910i + 20.2488j + 0.866a C>A i - 0.5a C>A j 0.5 aCD = 7.7205 + 0.866a C>A -40.5 = 26.8713 - 0.5aC>A aC>A = 134.7 m>s2 aCD = 249 rad>s2b






Ans.

16–138. Collar B moves to the left with a speed of 5 m>s, which is increasing at a constant rate of 1.5 m>s2, relative to the hoop, while the hoop rotates with the angular velocity and angular acceleration shown. Determine the magnitudes of the velocity and acceleration of the collar at this instant.

A


450 mm

SOLUTION 200 mm

Reference Frames: The xyz rotating reference frame is attached to the hoop and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vA = aA = 0

B

# v = a = [- 3k] rad>s2

v = [ -6k] rad>s

For the motion of collar B with respect to the xyz frame, rB>A = [- 0.45j] m (vrel)xyz = [- 5i] m>s (vrel)xyz2

52 = 125 m>s2. Thus, 0.2

=

laws

r

or

The normal components of (arel)xyz is [(arel)xyz]n =

in

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(arel)xyz = [- 1.5i + 125j] m>s

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vB = vA + v * rB>A + (vrel)xyz

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= 0 + ( -6k) * ( -0.45j) + ( - 5i)

the

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= [- 7.7i] m>s

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Thus,

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vB = 7.7 m>s ;

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aB = aA + v * rB>A + v * (v * rB>A) + 2v * (vrel)xyz + (arel)xyz sale

= 0 + (- 3k) * (- 0.45j) + ( - 6k) * [( -6k) * ( -0.45j)] + 2( - 6k) * (- 5i) + ( - 1.5i + 125j) = [ - 2.85i + 201.2j] m>s2 Thus, the magnitude of aB is therefore aB = 32.852 + 201.22 = 201 m>s2






Ans.

16–139. D

Block B of the mechanism is confined to move within the slot member CD. If AB is rotating at a constant rate of vAB = 3 rad>s, determine the angular velocity and angular acceleration of member CD at the instant shown.

100 mm

A

SOLUTION vAB

Coordinate Axes: The origin of both the fixed and moving frames of reference are located at point C. The x, y, z moving frame is attached to and rotates with rod CD since peg B slides along the slot in member CD.

B 3 rad/s

vCD, aCD

Kinematic Equation: Applying Eqs. 16–24 and 16–27, we have 30

vB = vC + Æ * rB>C + (vB>C)xyz # aB = aC + Æ * rB>C + Æ * (Æ * rB>C) + 2Æ * (vB>C)xyz + (aB>C)xyz

(1) (2)

vC = 0

rB>C = {0.2i} m

aC = 0

(vB>C)xyz = (vB>C)xyz i

Æ = -vCD k # Æ = -aCD k

(a B>C)xyz = (a B>C)xyz i

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= {- 0.2598i - 0.150j} m>s

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vB = vAB * rB>A = -3k * (0.05i - 0.08660j)

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aB = aAB * rB>A - v2ABrB>A = 0 - 32 (0.05i - 0.08660j)

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= {- 0.450i + 0.7794j} m>s

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Substitute the above data into Eq.(1) yields

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vB = vC + Æ * rB>C + (vB>C)xyz sale

-0.2598i - 0.150j = 0 + (-vCDk) * 0.2i + (vB>C)xyz i -0.2598i - 0.150j = (vB>C)xyzi - 0.2vCD j Equating i and j components, we have (vB>C)xyz = - 0.2598 m>s vCD = 0.750 rad>s

Ans.

Substitute the above data into Eq.(2) yields # aB = aC + Æ * rB>C + Æ * (Æ * rB>C) + 2Æ * (vB>C)xyz + (aB>C)xyz -0.450i + 0.7794j = 0 + ( -aCDk) * 0.2i + ( -0.750k) * [( -0.750k) * 0.2i] + 2(- 0.750k) * (-0.2598i) + (aB>C)xyz i -0.450i + 0.7794j = C (aB>C)xyz - 0.1125 D i + (0.3897 - 0.2aCD)j Equating i and j components, we have (a B>C)xyz = - 0.3375 m>s2 © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





aCD = -1.95 rad>s2 = 1.95 rad>s2

Ans.

not

instructors

States

World

The velocity and acceleration of peg B can be determined using Eqs. 16–9 and 16–14 with rB>A = {0.1 cos 60°i - 0.1 sin 60°j} m = {0.05i - 0.08660j} m.

permitted.

Dissemination

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Motion of C with respect to moving reference

laws


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C

200 mm

*16–140. At the instant shown rod AB has an angular velocity vAB = 4 rad>s and an angular acceleration aAB = 2 rad>s2. Determine the angular velocity and angular acceleration of rod CD at this instant.The collar at C is pin connected to CD and slides freely along AB.

vAB aAB

A 60

0.75 m

Coordinate Axes: The origin of both the fixed and moving frames of reference are located at point A. The x, y, z moving frame is attached to and rotate with rod AB since collar C slides along rod AB. Kinematic Equation: Applying Eqs. 16–24 and 16–27, we have vC = vA + Æ * rC>A + (vC>A)xyz # + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (v C>A)xyz + (a C>A)xyz

(1) (2)

Motion of C with respect to moving reference rC>A = 50.75i6m or

Motion of moving reference vA = 0 aA = 0 Æ = 4k rad>s # Æ = 2k rad>s2

laws

(vC>A)xyz = (yC>A)xyz i

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(a C>A)xyz = (aC>A)xyz i

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learning.

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States

World

The velocity and acceleration of collar C can be determined using Eqs. 16–9 and 16–14 with rC>D = { -0.5 cos 30°i - 0.5 sin 30°j }m = { -0.4330i - 0.250j} m.

protected

the

(including

for

= -0.250vCDi + 0.4330vCDj

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by

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use

vC = vCD * rC>D = -vCDk * ( -0.4330i - 0.250j)

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aC = a CD * rC>D - v2CD rC>D

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is

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= -aCD k * ( -0.4330i - 0.250j) - v2CD( -0.4330i - 0.250j)

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part

= A 0.4330v2CD - 0.250 aCD B i + A 0.4330aCD + 0.250v2CD B j

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Substitute the above data into Eq.(1) yields v C = vA + Æ * rC>A + (vC>A)xyz -0.250 vCD i + 0.4330vCDj = 0 + 4k * 0.75i + (yC>A)xyz i -0.250vCD i + 0.4330vCD j = (yC>A)xyz i + 3.00j Equating i and j components and solve, we have (yC>A)xyz = - 1.732 m>s vCD = 6.928 rad>s = 6.93 rad>s






0.5 m C

D B

SOLUTION

aC = aA

4 rad/s 2 rad/s2

Ans.

16–140. continued Substitute the above data into Eq.(2) yields # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz

C 0.4330 A 6.9282 B - 0.250 aCD D i + C 0.4330aCD + 0.250 A 6.9282 B D j = 0 + 2k * 0.75i + 4k * (4k * 0.75i) + 2 (4k) * ( -1.732i) + (aC>A)xyz i (20.78 - 0.250aCD)i + (0.4330 aCD + 12)j = C (aC>A)xyz - 12.0 D i - 12.36j Equating i and j components, we have (aC>A)xyz = 46.85 m>s2 d

Ans.

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aCD = - 56.2 rad>s2 = 56.2 rad>s2






16–141. The “quick-return” mechanism consists of a crank AB, slider block B, and slotted link CD. If the crank has the angular motion shown, determine the angular motion of the slotted link at this instant.

D

100 mm vAB aAB

3 rad/s 9 rad/s2

30 A

SOLUTION vB = 3(0.1) = 0.3 m>s (aB)t = 9(0.1) = 0.9 m>s

30 2

(aB )n = (3)2 (0.1) = 0.9 m>s2

vCD, aCD

vB = vC + Æ * rB>C + (vB>C)xyz 0.3 cos 60°i + 0.3 sin 60°j = 0 + (vCDk) * (0.3i) + vB>C i vB>C = 0.15 m>s vCD = 0.866 rad>s d # aB = aC + Æ * rB>C + Æ * (Æ * rB>C) + 2Æ * (vB>C)xyz + (aB>C)xyz

Web)

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Ans.

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in

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0.9 cos 60°i - 0.9 cos 30°i + 0.9 sin 60°j + 0.9 sin 30°j = 0 + (aCD k) * (0.3i)

States

World

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Dissemination

+(0.866k) * (0.866k * 0.3i) + 2(0.866k * 0.15i) + aB>C i

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-0.3294i + 1.2294j = 0.3aCD j - 0.225i + 0.2598j + aB>C i

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aB>C = - 0.104 m>s2

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aCD = 3.23 rad>s2 d






Ans.

C

300 mm

B

16–142. At the instant shown, the robotic arm AB is rotating counter clockwise at v = 5 rad>s and has an angular acceleration a = 2 rad>s2. Simultaneously, the grip BC is rotating counterclockwise at v¿ = 6 rad>s and a¿ = 2 rad>s2, both measured relative to a fixed reference. Determine the velocity and acceleration of the object heldat the grip C.

125 mm

y

ω ,α

SOLUTION

A

vC = vB + Æ * rC>B + (vC>B)xyz # aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz

(1) (2)

Motion of C with respect to moving reference rC>B = {0.125 cos 15°i + 0.125 sin 15°j} m

Æ = {6k} rad>s # Æ = {2k} rad>s2

(vC>B)xyz = 0 (aC>B)xyz = 0

laws

or

Motion of B:

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vB = v * rB>A Dissemination

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= (5k) * (0.3 cos 30°i + 0.3 sin 30°j)

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= {- 0.75i + 1.2990j} m>s

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aB = a * rB>A - v2rB>A

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= (2k) * (0.3 cos 30°i + 0.3 sin 30°j) - (5)2(0.3 cos 30°i + 0.3 sin 30°j)

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= { -6.7952i - 3.2304j} m>s2

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vC = (- 0.75i + 1.2990j) + (6k) * (0.125 cos 15°i + 0.125 sin 15°j) + 0 = { -0.944i + 2.02j} m>s

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aC = (- 6.79527i - 3.2304j) + (2k) * (0.125 cos 15°i + 0.125 sin 15°j) + (6k) * [(6k) * (0.125 cos 15°i + 0.125 sin 15°j)] + 0 + 0 = { -11.2i - 4.15j} m s2






15°

300 mm

30°


C

B

Ans.

ω, α

x

16–143. Peg B on the gear slides freely along the slot in link AB. If the gear’s center O moves with the velocity and acceleration shown, determine the angular velocity and angular acceleration of the link at this instant.

150 mm vO aO

B

3 m/s 1.5 m/s2

600 mm O

150 mm

SOLUTION Gear Motion: The IC of the gear is located at the point where the gear and the gear rack mesh, Fig. a. Thus, vO 3 = 20 rad>s = v = rO>IC 0.15 Then, vB = vrB>IC = 20(0.3) = 6 m>s : aO 1.5 = = 10 rad>s. By referring to Fig. b, r 0.15

Since the gear rolls on the gear rack, a =

aB = aO + a * rB>O - v2 rB>O (aB)t i - (aB)n j = 1.5i + ( - 10k) * 0.15j - 202(0.15j) (aB)t i - (aB)n j = 3i - 60j Thus, or

(aB)n = 60 m>s2

Dissemination

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Reference Frame: The x¿y¿z¿ rotating reference frame is attached to link AB and coincides with the XYZ fixed reference frame, Figs. c and d. Thus, vB and aB with respect to the XYZ frame is

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(aB)t = 3 m>s2

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vB = [6 sin 30°i - 6 cos 30° j] = [3i - 5.196j] m>s

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aB = (3 sin 30° - 60 cos 30°)i + (-3 cos 30° - 60 sin 30°)j

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= [ -50.46i - 32.60j] m>s2

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For motion of the x¿y¿z¿ frame with reference to the XYZ reference frame, # vA = aA = 0 vAB = -vABk vAB = - aAB k provided

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For the motion of point B with respect to the x¿y¿z¿ frame is (vrel)x¿y¿z¿ = (vrel)x¿y¿z¿ j

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rB>A = [0.6j]m

(arel)x¿y¿z¿ = (arel)x¿y¿z¿ j

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Velocity: Applying the relative velocity equation, sale

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vB = vA + vAB * rB>A + (vrel)x¿y¿z¿ 3i - 5.196j = 0 + ( -vABk) * (0.6j) + (vrel)x¿y¿z¿ j 3i - 5.196j = 0.6vAB i + (vrel)x¿y¿z¿j Equating the i and j components yields vAB = 5 rad>s

3 = 0.6vAB

Ans.

(vrel)x¿y¿z¿ = - 5.196 m>s Acceleration: Applying the relative acceleration equation. # aB = aA + vAB * rB>A + vAB * (vAB * rB>A) + 2vAB * (vrel)x¿y¿z¿ + (a rel)x¿y¿z¿ -50.46i - 32.60j = 0 + ( - aABk) * (0.6j) + ( -5k) * [(- 5k) * (0.6j)] + 2( -5k) * ( -5.196j) + (arel)x¿y¿z¿j -50.46i - 32.60j = (0.6aAB - 51.96)i + C (arel)x¿y¿z¿ - 15 D j Equating the i components, - 50.46 = 0.6a AB - 51.96 aAB = 2.5 rad>s2 © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

A

*16–144. The cars on the amusement-park ride rotate around the axle at A with a constant angular velocity vA>f = 2 rad>s, measured relative to the frame AB. At the same time the frame rotates around the main axle support at B with a constant angular velocity vf = 1 rad>s. Determine the velocity and acceleration of the passenger at C at the instant shown.

y

D 8 ft 8 ft A

x

SOLUTION vC = vA + Æ * rC>A + (vC>A)xyz # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz

C vA/f ⫽ 2 rad/s

(1)

15 ft 30

(2)

vf ⫽ 1 rad/s

Motion of moving refernce

Motion of C with respect to moving reference

Æ = {3k} rad>s # Æ = 0

rC>A = {- 8i} ft (vC>A)xyz = 0

or

(a C>A)xyz = 0

teaching

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Motion of A:

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vA = v * rA>B

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= (1k) * ( - 15 cos 30°i + 15 sin 30°j)

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= {- 7.5i - 12.99j} ft>s the

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aA = a * rA>B - v2 rA>B

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= 0 - (1)2(- 15 cos 30°i + 15 sin 30°j)

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= {12.99i - 7.5j} ft>s2

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= {-7.5i - 37.0j }ft>s

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vC = ( - 7.5i - 12.99j) + (3k) * ( -8i) + 0

aC = (12.99i - 7.5j) + 0 + (3k) * [(3k) * (-8i) + 0 + 0] = {85.0i - 7.5j}ft>s2






Ans.

B

16–145. The cars on the amusement-park ride rotate around the axle at A with a constant angular velocity vA>f = 2 rad>s, measured relative to the frame AB. At the same time the frame rotates around the main axle support at B with aconstant angular velocity vf = 1 rad>s. Determine the velocity and acceleration of the passenger at D at the instant shown.

y

D 8 ft 8 ft A

SOLUTION

C ωA = 2 rad/s /f

vD = vA + Æ * rD>A + (vD>A)xyz aD = aA

(1)

# + v * rD>A + Æ * (Æ * rD>A) + 2Æ * (vD>A)xyz + (aD>A)xyz


30°

(2)

rD>A = {8j} ft (vD>A)xyz = 0 (aD>A)xyz = 0

laws

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Motion of A:

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vA = v * rA>B Dissemination

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= (1k) * (-15 cos 30°i + 15 sin 30°j)

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= {- 7.5i - 12.99j} ft>s

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aA = a * rA>B - v2rA>B

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= 0 - (1)2(-15 cos 30°i + 15 sin 30°j)

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= {12.99i - 7.5j} ft>s2

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= { -31.5i - 13.0j} ft>s

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vD = ( -7.5i - 12.99j) + (3k) * (8j) + 0

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aD = (12.99i - 7.5j) + 0 + (3k) * [(3k) * (8j)] + 0 + 0 = {13.0i - 79.5j} ft s2






Ans.

15 ft

ω f = 1 rad/s


Æ = {3k} rad>s # Æ = 0

x

B

16–146. If the slotted arm AB rotates about the pin A with a constant angular velocity of vAB = 10 rad>s, determine the angular velocity of link CD at the instant shown.

B D

600 mm

vAB ⫽ 10 rad/s

SOLUTION

30⬚ A

Reference Frame: The xyz rotating reference frame is attached to link AB and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vA = 0

#

vAB = 0

vAB = [10k] rad>s

For the motion of point D relative to the xyz frame, we have rD>A = [0.6i] m

(vrel)xyz = (vrel)xyz i

Since link CD rotates about a fixed axis, vD can be determined from

laws

or

vD = vCD * rD

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= (vCD k) * (0.45 cos 15° i + 0.45 sin 15° j)

permitted.

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= - 0.1165vCD i + 0.4347vCD j

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vD = vA + vAB * rD>A + (vrel)xyz

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- 0.1165vCD i + 0.4347vCD j = 0 + (10k) * (0.6i) + (vrel)xyz i

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- 0.1165vCD i + 0.4347vCD j = (vrel)xyzi + 6j

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- 0.1165vCD = (vrel)xyz

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0.4347vCD = 6 sale

Solving, vCD = 13.80 rad>s = 13.8 rad>s (vrel)xyz = - 1.608 m>s






Ans.

C

450 mm 45⬚

16–147. At the instant shown, boat A travels with a speed of 15 m>s, which is decreasing at 3 m>s2, while boat B travels with a speed of 10 m>s, which is increasing at 2 m>s2. Determine the velocity and acceleration of boat A with respect to boat B at this instant.

30 m 15 m/s 50 m

SOLUTION

A

B 3 m/s2

50 m 10 m/s 2 m/s2

Reference Frame: The xyz rotating reference frame is attached to boat B and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Since

vB = [-10j] m>s

aA = [ -4.5i - 3j] m>s2

aB = [2i - 2j] m>s2 laws

vA = [15j] m>s

or

boats A and B move along the circular paths, their normal components of vA 2 vB 2 152 102 = 4.5 m>s2 and (aB)n = = 2 m>s2. = = acceleration are (aA)n = r r 50 50 Thus, the motion of boats A and B with respect to the XYZ frame are

v = [0.2k] rad>s

World

vB 10 = 0.2 rad>s = r 50

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the

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# v = [0.04k] rad>s2

(aB)t 2 # = 0.04 rad>s2 = v = r 50

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rA>B = [-20i] m

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vA = vB + v * rA>B + (vrel)xyz sale

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15j = - 10j + (0.2k) * ( - 20i) + (vrel)xyz 15j = - 14j + (vrel)xyz (vrel)xyz = [29j] m>s

Ans.

Acceleration: Applying the relative acceleration equation, # aA = aB + v * rA>B + v * (v * rA>B) + 2v * (vrel)xyz + (arel)xyz (- 4.5i - 3j) = (2i - 2j) + (0.04k) * (-20i) + (0.2k) * C (0.2k) * ( -20i) D + 2(0.2k) * (29j) + (arel)xyz -4.5i - 3j = - 8.8i - 2.8j + (arel)xyz (arel)xyz = [4.3i - 0.2j] m>s2






Ans.

*16–148. At the instant shown, boat A travels with a speed of 15 m>s, which is decreasing at 3 m>s2, while boat B travels with a speed of 10 m>s, which is increasing at 2 m>s2. Determine the velocity and acceleration of boat B with respect to boat A at this instant.

30 m 15 m/s 50 m

SOLUTION

A

B 3 m/s2

50 m 10 m/s 2 m/s2

Reference Frame: The xyz rotating reference frame is attached to boat A and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Since

vB = [-10j] m>s

aA = [-4.5i - 3j] m>s2

aB = [2i - 2j] m>s2 laws

vA = [15j] m>s

or

boats A and B move along the circular paths, their normal components of vA 2 vB 2 152 102 = 4.5 m>s2 and (aB)n = = 2 m>s2. = = acceleration are (aA)n = r r 50 50 Thus, the motion of boats A and B with respect to the XYZ frame are

Dissemination

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copyright

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Also, the angular velocity and angular acceleration of the xyz reference frame with respect to the XYZ reference frame are vA 15 = 0.3 rad>s = r 50

World

permitted.

v = [0.3k] rad>s

# v = [-0.06k] rad>s2 the

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(aA)t 3 # = v = = 0.06 rad>s2 r 50

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rB>A = [20i] m

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vB = vA + v * rB>A + (vrel)xyz sale

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-10j = 15j + (0.3k) * (20i) + (vrel)xyz - 10j = 21j + (vrel)xyz (vrel)xyz = [-31j] m>s

Ans.

Acceleration: Applying the relative acceleration equation, # aB = aA + v * rB>A + v(v * rB>A) + 2v * (vrel)xyz + (arel)xyz (2i - 2j) = ( -4.5i - 3j) + ( - 0.06k) * (20i) + (0.3k) * C (0.3k) * (20i) D + 2(0.3k) * ( -31j) + (arel)xyz 2i - 2j = 12.3i - 4.2j + (arel)xyz (arel)xyz = [-10.3i + 2.2j] m>s2






Ans.

16–149. vA ⫽ 3 m/s aA ⫽ 1.5 m/s2

If the piston is moving with a velocity of vA = 3 m>s and acceleration of aA = 1.5 m>s2, determine the angular velocity and angular acceleration of the slotted link at the instant shown. Link AB slides freely along its slot on the fixed peg C.

A

30⬚ C 0.5 m

SOLUTION

B

Reference Frame: The xyz reference frame centered at C rotates with link AB and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is aAB = - aABk

vAB = - vABk

vC = aC = 0

The motion of point A with respect to the xyz frame is rA>C = [ -0.5i] m

(arel)xyz = (arel)xyzi

(vrel)xyz = (vrel)xyz i

The motion of point A with respect to the XYZ frame is

or

vA = 3 cos 30°i + 3 sin 30°j = [2.598i + 1.5j] m>s

teaching

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laws

aA = 1.5 cos 30°i + 1.5 sin 30°j = [1.299i + 0.75j] m>s

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Dissemination

vA = vC + vAB * rA>C + (vrel)xyz

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2.598i + 1.5j = 0 + (- vAB k) * ( -0.5i) + (vrel)xyz i

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2.598i + 1.5j = (vrel)xyzi + 0.5vAB j

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Equating the i and j components,

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(vrel)xyz = 2.598 m>s provided

integrity

vAB = 3 rad>s

Ans.

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0.5vAB = 1.5

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Acceleration: Applying the relative acceleration equation, sale

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# aA = aC + vAB * rA>C + vAB * (vAB * rA>C) + 2vAB * (vrel)xyz + (arel)xyz 1.299i + 0.75j = 0 + (- aABk) * (- 0.5i) + (- 3k) * [( -3k) * ( - 0.5i)] + 2(- 3k) * (2.598i) + (arel)xyzi 1.299i + 0.75j = C 4.5 + (arel)xyz D i + (0.5aAB - 15.59)j Equating the j components, 0.75 = 0.5aAB - 15.59 aAB = 32.68 rad>s2 = 32.7 rad>s2






Ans.

16–150. B

The two-link mechanism serves to amplify angular motion. Link AB has a pin at B which is confined to move within the slot of link CD. If at the instant shown, AB (input) has an angular velocity of vAB = 2.5 rad>s, determine the angular velocity of CD (output) at this instant.

D

150 mm

C 30 45

SOLUTION

vAB

rBA 0.15 m = sin 120° sin 45° rBA = 0.1837 m vC = 0 aC = 0 Æ = -vDCk # Æ = -aDCk laws

or

rB>C = {- 0.15 i} m in

teaching

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(vB>C)xyz = (yB>C)xyzi Dissemination

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copyright

(aB>C)xyz = (aB>C)xyzi

is

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States

World

permitted.

vB = vAB * rB>A = ( -2.5k) * ( -0.1837 cos 15°i + 0.1837 sin 15°j)

by

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learning.

= {0.1189i + 0.4436j} m>s

(including

for

work

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the

vB = vC + Æ * rB>C + (vB>C)xyz

is

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0.1189i + 0.4436j = 0 + (- vDCk) * ( - 0.15i) + (vB>C)xyz i

sale

vDC = 2.96 rad>s b






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(vB>C)xyz = 0.1189 m>s

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assessing

0.1189i + 0.4436j = (vB>C)xyz i + 0.15vDC j Solving:

A

Ans.

2.5 rad/s

16–151. The gear has the angular motion shown. Determine the angular velocity and angular acceleration of the slotted link BC at this instant. The peg at A is fixed to the gear.

A

2 ft

0.5 ft 0.7 ft

SOLUTION

B

yA = (1.2)(2) = 2.4 ft>s ; aO = 4(0.7) = 2.8 ft>s2 a A = a O + a A>O aA = 2.8 + 4(0.5) + (2)2(0.5) ; ; T a A = 4.8 + 2 ; T vA = vB + Æ * rA>B + (vA>B)xyz

in

teaching

Web)

laws

or

4 3 -2.4i = 0 + (Æk) * (1.6i + 1.2j) + yA>B a b i + yA>B a b j 5 5

Dissemination

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copyright

-2.4i = 1.6Æj - 1.2Æi + 0.8yA>B i + 0.6yA>B j

of

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-2.4 = -1.2Æ + 0.8yA>B

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0 = 1.6Æ + 0.6yA>B

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Solving,

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vBC = Æ = 0.720 rad>sd

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a A = a B + Æ * rA>B + Æ * (Æ * rA>B) + 2Æ * (vA>B)xyz + (a A>B)xyz # -4.8i - 2j = 0 + (Æ k) * (1.6i + 1.2j) + (0.72k) * (0.72k) * (1.6i + 1.2j)) sale

will

+2(0.72k) * [- (0.8)(1.92)i - 0.6(1.92)j] + 0.8aB>A i + 0.6aB>A j # # -4.8i - 2j = 1.6Æ j - 1.2Æ i - 0.8294i - 0.6221j - 2.2118j + 1.6589i + 0.8aB>A i + 0.6aB>A. # -4.8 = -1.2Æ - 0.8294 + 1.6589 + 0.8aB/A # -2 = 1.6Æ - 0.6221 - 2.2118 + 0.6aB/A # -4.6913 = - Æ + 0.667aB/A # 0.5212 = Æ + 0.357a B/A # Ans. aBC = Æ = 2.02 rad/s2 d aB/A = -4.00 ft>s2






C

O


*16–152. vB

The Geneva mechanism is used in a packaging system to convert constant angular motion into intermittent angular motion. The star wheel A makes one sixth of a revolution for each full revolution of the driving wheel B and the attached guide C. To do this, pin P, which is attached to B, slides into one of the radial slots of A, thereby turning wheel A, and then exits the slot. If B has a constant angular velocity of vB = 4 rad>s, determine V A and AA of wheel A at the instant shown.

B

The circular path of motion of P has a radius of rP = 4 tan 30° = 2.309 in. Thus, vP = - 4(2.309)j = -9.238j aP = -(4)2(2.309)i = - 36.95i

laws

or

Thus,

in

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vP = vA + Æ * rP>A + (vP>A)xyz

permitted.

Dissemination

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- 9.238j = 0 + (vA k) * (4j) - vP>A j

is

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# aP = aA + Æ * rP>A + Æ * (Æ * rP>A) + 2Æ * (vP>A)xyz + (aP>A)xyz

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aA = 9.24 rad>s2 d


4 in.

A u

-36.95 = -4aA

C P

SOLUTION

Solving,

4 rad/s

Ans.

30

17–1. z

Determine the moment of inertia Iy for the slender rod. The rod’s density r and cross-sectional area A are constant. Express the result in terms of the rod’s total mass m. l

SOLUTION

A x

Iy =

LM

x 2 dm

l

= =

L0

x 2 (r A dx)

1 r A l3 3

m = rAl

laws

or

Thus, 1 m l2 3

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Iy =






y

17–2. z

The solid cylinder has an outer radius R, height h, and is made from a material having a density that varies from its center as r = k + ar2, where k and a are constants. Determine the mass of the cylinder and its moment of inertia about the z axis.

R

h

SOLUTION Consider a shell element of radius r and mass dm = r dV = r(2p r dr)h R

m =

(k + ar2)(2p r dr)h

L0

m = 2p h(

aR4 kR2 + ) 2 4

m = p h R2(k +

aR2 ) 2

Ans.

laws

or

dI = r2 dm = r2(r)(2p r dr)h

teaching in

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copyright

L0

Web)

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Iz =

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Iz =






Ans.

17–3. y

Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

R x

SOLUTION 2p

Iz =

L0

r A(R du)R2 = 2p r A R3 2p

m =

L0

r A R du = 2p r A R

Thus, I z = m R2

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Ans.






*17–4. Determine the moment of inertia of the semiellipsoid with respect to the x axis and express the result in terms of the mass m of the semiellipsoid. The material has a constant density r.

y

y2 x2 ⫹ 2⫽1 2 a b b x

SOLUTION dIx = m =

y2dm 2

Lv

a

rdV

a

2

L0

2

x

2

a

b dx

2 rpab2 3 a

2

1 x 2 4 rp b a1 - 2 b dx 2 L0 a

teaching

Web)

Ix =

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=

rpb a 1 -

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=

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4 rpab4 15

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=

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2 mb2 5

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Ix =






Ans.

17–5. y

The sphere is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the sphere. The material has a constant density r.

x2 + y2 = r2

x

SOLUTION dIx =

y2 dm 2

dm = r dV = r(py2 dx) = r p(r2 - x2) dx dIx =

1 r p(r2 - x2)2 dx 2 r

8 pr r5 15 laws

=

1 2 2 2 r p(r - x ) dx L- r 2

or

Ix =

r

2

Ix =

2 m r2 5

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r p(r - x ) dx copyright

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Ans.

17–6. Determine the mass moment of inertia Iz of the cone formed by revolving the shaded area around the z axis. The density of the material is r. Express the result in terms of the mass m of the cone.

z z ⫽ hr (r ⫺ y)

SOLUTION

h

Differential Element: The mass of the disk element shown shaded in Fig. a is ro ro 2 dm = r dV = rpr2dz. Here, r = y = ro z. Thus, dm = rp aro - zb dz. The h h mass moment of inertia of this element about the z axis is ro 4 1 1 1 1 dIz = dmr2 = (rpr2dz)r2 = rpr4dz = rparo - zb dz 2 2 2 2 h

y

x

Mass: The mass of the cone can be determined by integrating dm. Thus, h ro 2 rparo - z b dz m = dm = h L L0

laws

or

h ro 3 1 h 1 = rpc aro - z b a - b d 2 = rpro2h ro 3 h 3 0

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Mass Moment of Inertia: Integrating dIz, we obtain h ro 4 1 dIz = rp aro - z b dz Iz = h L L0 2

the

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instructors

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permitted.

h ro 3 1 1 h 1 rpc aro - z b a - b d 2 = rpro4h ro 2 5 h 10 0 United

on

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=

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From the result of the mass, we obtain rpro2h = 3m. Thus, Iz can be written as the

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the

1 1 3 A rpro2h B ro2 = (3m)ro2 = mro2 10 10 10

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Iz =






Ans.

r

17–7. y

The solid is formed by revolving the shaded area around the y axis. Determine the radius of gyration ky . The specific weight of the material is g = 380 lb>ft3.

3 in. y3

x

SOLUTION

3 in.

The moment of inertia of the solid : The mass of the disk element 1 dm = rpx2 dy = 81 rpy6 dy. dIy =

1 dmx2 2

=

1 A rpx2 dy B x2 2

=

1 1 rpx4 dy = rpy12 dy 2 2(94) 3

or

1 y12 dy rp 2(94) L0

teaching

Web)

L

dIy =

laws

Iy =

Dissemination

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in

= 29.632r

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of

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learning. United

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use student

the

Lm

1 rp y6 dy = 12.117r 81 L0

by

dm =

the

3

m =

the

(including

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work

29.632r Iy = = 1.56 in. Am A 12.117r

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ky =






9x

Ans.

*17–8. y

The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of inertia Iy . The specific weight of concrete is g = 150 lb>ft3.

6 in.

y

SOLUTION d Iy = =

1 1 (dm)(10)2 - (dm)x2 2 2 1 1 [pr(10)2 dy](10)2 - prx2 dyx2 2 2 8

Iy = =

8

1 9 2 (10)4 dy a b y2dy R pr B 2 L0 L0 2 1 2 p(150) 3

32.2(12)

2

9 2

1 3

B (10)4(8) - a b a b (8)3 R

or

= 324.1 slug # in2 laws

Iy = 2.25 slug # ft 2

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Ans.






4 in.

2 2 8 in. 9x

x

17–9. z

Determine the moment of inertia Iz of the torus. The mass of the torus is m and the density r is constant. Suggestion: Use a shell element.

R

SOLUTION

a

dm = 2p(R - x)(2z¿r dx) dlz = (R - x)2 dm = 4pr[(R3 - 3R2 x + 3Rx2 - x3) 2a2 - x2 dx] a

Iz = 4pr[R3

L-a

a

2a2 - x2 dx - 3R 2

L-a

a

x3 2a 2 - x 2 dx + 3R

a

2a 2 - x 2 dx -

L-a

L-a

3

x 2a 2 - x 2

3 2 a) 4

= 2p2 rRa2 (R2 +

laws

or

Since m = rV = 2pRrpa2 3 2 a ) 4

in

teaching

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Ans.

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Iz = m(R2 +






17–10. O

Determine the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through point O. The slender rod has a mass of 10 kg and the sphere has a mass of 15 kg.

450 mm

SOLUTION

A

Composite Parts: The pendulum can be subdivided into two segments as shown in Fig. a. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated.

100 mm B

Moment of Inertia: The moment of inertia of the slender rod segment (1) and the sphere segment (2) about the axis passing through their center of mass can be 1 2 computed from (IG)1 = ml2 and (IG)2 = mr2. The mass moment of inertia of 12 5 each segment about an axis passing through point O can be determined using the parallel-axis theorem. IO = ©IG + md2 1 2 (10)(0.452) + 10(0.2252) d + c (15)(0.12) + 15(0.552) d 12 5

or

= c

laws

= 5.27 kg # m2

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Ans.






17–11. The slender rods have a weight of 3 lb>ft. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point A.

A 2 ft

1 ft

SOLUTION Ans.

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IA

1.5 ft

3(3) 1 3(3) 1 3(3) d (3)2 + c d(3)2 + c d(2)2 = 2.17 slug # ft2 = c 3 32.2 12 32.2 32.2






1.5 ft

*17–12. Determine the moment of inertia of the solid steel assembly about the x axis. Steel has a specific weight of gst = 490 lb>ft3.

0.25 ft 0.5 ft

2 ft

SOLUTION Ix =

1 3 3 m (0.5)2 + m (0.5)2 m (0.25)2 2 1 10 2 10 3

1 3 1 3 1 490 = c p(0.5)2(3)(0.5)2 + a b p(0.5)2 (4)(0.5)2 a bp(0.25)2(2)(0.25)2 d a b 2 10 3 10 2 32.2 = 5.64 slug # ft2

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3 ft

x

17–13. The wheel consists of a thin ring having a mass of 10 kg and four spokes made from slender rods, each having a mass of 2 kg. Determine the wheel’s moment of inertia about an axis perpendicular to the page and passing through point A.

500 mm

SOLUTION A

IA = Io + md3 = c2 c

1 (4)(1)2 d + 10(0.5)2 d + 18(0.5)2 12

= 7.67 kg # m2

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Ans.






17–14. If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A.

4 ft

1 ft O

SOLUTION A

Composite Parts: The wheel can be subdivided into the segments shown in Fig. a. The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a 3 distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2). 2 Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O. IO = a

100 1 20 20 15 b (4 2) + 8c a b(32) + a b(2.52) d + a b(12) 32.2 12 32.2 32.2 32.2

laws

or

= 84.94 slug # ft2

instructors

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permitted.

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The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found using the parallel-axis theorem 100 20 15 IA = IO + md 2, where m = + 8a b + = 8.5404 slug and d = 4 ft. 32.2 32.2 32.2 Thus, is

of

the

not

IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2

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Ans.






17–15. Determine the moment of inertia about an axis perpendicular to the page and passing through the pin at O. The thin plate has a hole in its center. Its thickness is 50 mm, and the material has a density r = 50 kg>m3.

O

150 mm

SOLUTION IG =

1.40 m

1 1 C 50(1.4)(1.4)(0.05) D C (1.4)2 + (1.4)2 D - C 50(p)(0.15)2(0.05) D (0.15)2 12 2

= 1.5987 kg # m2 IO = IG + md2 m = 50(1.4)(1.4)(0.05) - 50(p)(0.15)2(0.05) = 4.7233 kg IO = 1.5987 + 4.7233(1.4 sin 45°)2 = 6.23 kg # m2

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Ans.






1.40 m

*17–16. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg>m2.

200 mm

O

200 mm

SOLUTION Composite Parts: The plate can be subdivided into two segments as shown in Fig. a. Since segment (2) is a hole, it should be considered as a negative part. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated.

200 mm

Mass Moment of Inertia: The moment of inertia of segments (1) and (2) are computed as m1 = p(0.2 2)(20) = 0.8p kg and m2 = (0.2)(0.2)(20) = 0.8 kg. The moment of inertia of the plate about an axis perpendicular to the page and passing through point O for each segment can be determined using the parallel-axis theorem. IO = ©IG + md2 laws

or

1 1 = c (0.8p)(0.22) + 0.8p(0.22) d - c (0.8)(0.22 + 0.22) + 0.8(0.22) d 2 12

Web)

= 0.113 kg # m2

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Ans.






17–17. The assembly consists of a disk having a mass of 6 kg and slender rods AB and DC which have a mass of 2 kg/m. Determine the length L of DC so that the center of mass is at the bearing O. What is the moment of inertia of the assembly about an axis perpendicular to the page and passing through O?

0.8 m

0.5 m

D

0.2 m

L A

O

B C

SOLUTION Measured from the right side, y =

6(1.5) + 2(1.3)(0.65) = 0.5 6 + 1.3(2) + L(2)

L = 6.39 m IO =

Ans.

1 1 1 (6)(0.2)2 + 6(1)2 + (2)(1.3)(1.3)2 + 2(1.3)(0.15)2 + (2)(6.39)(6.39)2 + 2(6.39)(0.5)2 2 12 12

IO = 53.2 kg # m2

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Ans.






17–18. The assembly consists of a disk having a mass of 6 kg and slender rods AB and DC which have a mass of 2 kg/m. If L = 0.75 m, determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through O.

0.8 m

0.5 m

D

0.2 m

L A

O

B C

SOLUTION IO =

1 1 1 (6)(0.2)2 + 6(1)2 + (2)(1.3)(1.3)2 + 2(1.3)(0.15)2 + (2)(0.75)(0.75)2 + 2(0.75)(0.5)2 2 12 12

IO = 6.99 kg # m2

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17–19. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg>m. The thin plate has a mass of 12 kg>m2. Determine the location y of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

0.4 m

0.4 m

A

B O

_ y

G

SOLUTION y =

1.5(3)(0.75) + p(0.3)2(12)(1.8) - p(0.1)2(12)(1.8)

C

1.5(3) + p(0.3)2(12) - p(0.1)2(12) + 0.8(3)

0.1 m

= 0.8878 m = 0.888 m IG =

Ans. 0.3 m

1 (0.8)(3)(0.8)2 + 0.8(3)(0.8878)2 12 +

1 (1.5)(3)(1.5)2 + 1.5(3)(0.75 - 0.8878)2 12

laws

or

1 + [p(0.3)2(12)(0.3)2 + [p(0.3)2(12)](1.8 - 0.8878)2 2 teaching

Web)

1 [p(0.1)2(12)(0.1)2 - [p(0.1)2(12)](1.8 - 0.8878)2 2

in

-

Ans.

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IG = 5.61 kg # m2






1.5 m

*17–20. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg>m2. Determine the moment of inertia of the pendulum about an axis perpendicular to the page and passing through the pin at O.

0.4 m

0.4 m

A

B O

–y

G

SOLUTION Io =

1 1 1 [3(0.8)](0.8)2 + [3(1.5)](1.5)2 + [12(p)(0.3)2](0.3)2 12 3 2

+ [12(p)(0.3)2](1.8)2 -

C 0.1 m

1 [12(p)(0.1)2](0.1)2 - [12(p)(0.1)2](1.8)2 2

0.3 m

= 13.43 = 13.4 kg # m2

Ans.

Also, from the solution to Prob. 17–16, m = 3(0.8 + 1.5) + 12[p(0.3)2 - p(0.1)2] = 9.916 kg

laws

or

Io = IG + m d2 teaching

Web)

= 5.61 + 9.916(0.8878)2 in

= 13.4 kg # m2

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Ans.






1.5 m

17–21. The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location y of the center of mass G of the pendulum; then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

O

y 2m

G

SOLUTION y =

0.5 m

© ym 1(3) + 2.25(5) = = 1.781 m = 1.78 m ©m 3 + 5

Ans. 1m

IG = ©IG + md 2 =

1 1 (3)(2)2 + 3(1.781 - 1)2 + (5)(0.52 + 12) + 5(2.25 - 1.781)2 12 12

= 4.45 kg # m2

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17–22. 20 mm

Determine the moment of inertia of the overhung crank about the x axis. The material is steel having a destiny of r = 7.85 Mg>m3.

30 mm 90 mm 50 mm x

180 mm

20 mm

SOLUTION mc = 7.85(10 ) A (0.05)p(0.01) 3

2

x¿

B = 0.1233 kg

30 mm 20 mm

mr = 7.85(103)((0.03)(0.180)(0.02)) = 0.8478 kg

50 mm

1 Ix = 2 c (0.1233)(0.01)2 + (0.1233)(0.06)2 d 2 + c

1 (0.8478) A (0.03)2 + (0.180)2 B d 12

= 0.00325 kg # m2 = 3.25 g # m2

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30 mm

17–23. 20 mm

Determine the moment of inertia of the overhung crank about the x¿ axis. The material is steel having a destiny of r = 7.85 Mg>m3.

30 mm 90 mm 50 mm x

180 mm

20 mm

SOLUTION mc = 7.85 A 10

3

x¿

B A (0.05)p(0.01) B = 0.1233 kg

30 mm

2

mp = 7.85 A 103 B A (0.03)(0.180)(0.02) B = 0.8478 kg

20 mm

50 mm

1 1 Ix = c (0.1233)(0.01)2 d + c (0.1233)(0.02)2 + (0.1233)(0.120)2 d 2 2 + c

1 (0.8478) A (0.03)2 + (0.180)2 B + (0.8478)(0.06)2 d 12

= 0.00719 kg # m2 = 7.19 g # m2

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30 mm

*17–24. The door has a weight of 200 lb and a center of gravity at G. Determine how far the door moves in 2 s, starting from rest, if a man pushes on it at C with a horizontal force F = 30 lb. Also, find the vertical reactions at the rollers A and B.

6 ft

3 ft

200 )a 30 = ( 32.2 G

+ ©F = m(a ) ; : x G x

aG = 4.83 ft>s2 a+©MA = ©(Mk)A;

NB(12) - 200(6) + 30(9) = (

200 )(4.83)(7) 32.2

NB = 95.0 lb + c ©Fy = m(aG)y ;

Ans.

NA + 95.0 - 200 = 0 NA = 105 lb

Web)

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1 2 a t 2 G

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s = s0 + v0t +

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1 (4.83)(2)2 = 9.66 ft 2


12 ft

F

SOLUTION

s = 0 + 0 +

B

G

C

+ ) (:

6 ft

A

5 ft

17–25. The door has a weight of 200 lb and a center of gravity at G. Determine the constant force F that must be applied to the door to push it open 12 ft to the right in 5 s, starting from rest. Also, find the vertical reactions at the rollers A and B.

6 ft

3 ft

+ )s = s + v t + 1 a t2 (: 0 0 G 2 1 aG(5)2 2

12 = 0 + 0 + ac = 0.960 ft>s2 F =

200 (0.960) 32.2

F = 5.9627 lb = 5.96 lb

Ans.

teaching

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200 (0.960)(7) 32.2 in

NB(12) - 200(6) + 5.9627(9) =

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NB = 99.0 lb

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NA + 99.0 - 200 = 0

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+ c ©Fy = m(aG)y ;

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NA = 101 lb






12 ft

F

SOLUTION

a + ©MA = ©(Mk)A ;

B

G

C

+ ©F = m(a ) ; : x G x

6 ft

A

Ans.

5 ft

17–26. The uniform pipe has a weight of 500 lb/ft and diameter of 2 ft. If it is hoisted as shown with an acceleration of 0.5 ft>s2, determine the internal moment at the center A of the pipe due to the lift.

0.5 ft/s2

5 ft A 1 ft 5 ft

SOLUTION Pipe: + c ©Fy = m ay ;

T-10 000 =

10 000 (0.5) 32.2

T = 10 155.27 lb Cables: + c ©Fy = 0;

10 155.27 - 2P cos 45° = 0 P = 7 180.867 lb

Web)

laws

or

) = 5 077.64 lb

teaching

1 22

-5000 (0.5)(5) 32.2 in

Px = Py = 7 180.867(

MA + 5000(5) - 5077.64(5) - 5077.64(1) =

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a + ©MA = ©(Mk)A ;

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MA = 5077.6 lb # ft = 5.08(103) lb # ft

5 ft

5 ft

5 ft

17–27. The drum truck supports the 600-lb drum that has a center of gravity at G. If the operator pushes it forward with a horizontal force of 20 lb, determine the acceleration of the truck and the normal reactions at each of the four wheels. Neglect the mass of the wheels.

20 lb G 2 ft

SOLUTION + ©F = m(a ) ; ; x G x

20 = (

A

600 )a 32.2 G Ans. 600 (1.0733)(2) 32.2

20(4) - 600(0.5) + 2NB (1.5) = NB = 86.7 lb

+ c ©Fy = m(aG)y ;

Ans.

2NA + 2(86.7) - 600 = 0 NA = 213 lb

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B

0.5 ft 1 ft

aG = 1.0733 ft>s2 = 1.07 ft>s2 a + ©MA = ©(Mk)A ;

4 ft

*17–28. 1 ft

If the cart is given a constant acceleration of a = 6 ft>s2 up the inclined plane, determine the force developed in rod AC and the horizontal and vertical components of force at pin B. The crate has a weight of 150 lb with center of gravity at G, and it is secured on the platform, so that it does not slide. Neglect the platform’s weight.

G

1 ft C 30⬚

SOLUTION A

Equations of Motion: FAC can be obtained directly by writing the moment equation of motion about B,

30⬚

+ ©MB = ©(Mk)B; 150(2) - FAC sin 60°(3) = - a

150 150 b (6) cos 30°(1) - a b(6) sin 30°(2) 32.2 32.2 FAC = 135.54 lb = 136 lb

Ans.

Using this result and writing the force equations of motion along the x and y axes, 150 b (6) cos 30° 32.2 Bx = 43.57 lb = 43.6 lb

Ans.

150 (6) sin 30° 32.2 By = 46.59 lb = 46.6 lb

Ans.

or

135.54 cos 60° - Bx = a

laws

+ ©F = m(a ) ; : x G x

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By + 135.54 sin 60° - 150 =

copyright

+ c ©Fy = m(aG)y;






2 ft

a B

P

17–29. 1 ft

If the strut AC can withstand a maximum compression force of 150 lb before it fails, determine the cart’s maximum permissible acceleration. The crate has a weight of 150 lb with center of gravity at G, and it is secured on the platform, so that it does not slide. Neglect the platform’s weight.

G

1 ft C 30⬚

SOLUTION A

Equations of Motion: FAC in terms of a can be obtained directly by writing the moment equation of motion about B.

30⬚

+ ©MB = ©(Mk)B; 150(2) - FAC sin 60°(3) = - a

150 150 b a cos 30°(1) - a ba sin 30°(2) 32.2 32.2

FAC = (3.346a + 115.47) lb Assuming AC is about to fail, FAC = 150 = 3.346a + 115.47 a = 10.3 ft>s2

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2 ft

a B

P

17–30. The drop gate at the end of the trailer has a mass of 1.25 Mg and mass center at G. If it is supported by the cable AB and hinge at C, determine the tension in the cable when the truck begins to accelerate at 5 m>s2. Also, what are the horizontal and vertical components of reaction at the hinge C?

B

A

30⬚

G C

1.5 m 45⬚

SOLUTION a + ©MC = ©(Mk)C ;

T sin 30°(2.5) - 12 262.5(1.5 cos 45°) = 1250(5)(1.5 sin 45°) T = 15 708.4 N = 15.7 kN

+ ©F = m(a ) ; ; x G x

Ans.

- Cx + 15 708.4 cos 15° = 1250(5) Cx = 8.92 kN

+ c ©Fy = m(aG)y ;

Ans.

Cy - 12 262.5 - 15 708.4 sin 15° = 0 Cy = 16.3 kN

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1m

17–31. The pipe has a length of 3 m and a mass of 500 kg. It is attached to the back of the truck using a 0.6-m-long chain AB. If the coefficient of kinetic friction at C is mk = 0.4, determine the acceleration of the truck if the angle u = 10° with the road as shown.

B 3m

θ = 10°

C

SOLUTION f = sin - 1 ¢

0.4791 ≤ = 52.98° 0.6

+ ©F = m(a ) ; : x G x

T cos 52.98° - 0.4NC = 500aG

+ c ©Fy = m(aG)y ;

NC - 500(9.81) + T sin 52.98° = 0

a + ©MC = ©(Mk)C;

- 500(9.81)(1.5 cos 10°) + T sin (52.98° - 10°)(3) = -500aG(0.2605) T = 3.39 kN NC = 2.19 kN aG = 2.33 m>s2

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A 1m

*17–32. The mountain bike has a mass of 40 kg with center of mass at point G1, while the rider has a mass of 60 kg with center of mass at point G2. Determine the maximum deceleration when the brake is applied to the front wheel, without causing the rear wheel B to leave the road. Assume that the front wheel does not slip. Neglect the mass of all the wheels.

G2

1.25 m

G1 0.4 m A

SOLUTION

B 0.4 m

Equations of Motion: Since the rear wheel B is required to just leave the road, NB = 0. Thus, the acceleration a of the bike can be obtained directly by writing the moment equation of motion about point A. + ©MA = (Mk)A; - 40(9.81)(0.4) - 60(9.81)(0.6) = - 40a(0.4) - 60a(1.25) a = 5.606 m>s2 = 5.61 m>s2

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0.4 m 0.2 m

17–33. The mountain bike has a mass of 40 kg with center of mass at point G1, while the rider has a mass of 60 kg with center of mass at point G2. When the brake is applied to the front wheel, it causes the bike to decelerate at a constant rate of 3 m>s2. Determine the normal reaction the road exerts on the front and rear wheels. Assume that the rear wheel is free to roll. Neglect the mass of all the wheels.

G2

1.25 m

G1 0.4 m A

SOLUTION

B 0.4 m

Equations of Motion: NB can be obtained directly by writing the moment equation of motion about point A. + ©MA = (Mk)A; NB(1) - 40(9.81)(0.4) - 60(9.81)(0.6) = - 60(3)(1.25) - 40(3)(0.4) NB = 237.12 N = 237 N

Ans.

Using this result and writing the force equations of motion along the y axis, NA + 237.12 - 40(9.81) - 60(9.81) = 0 NA = 743.88 N = 744 N

Ans.

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+ c ©Fy = m(aG)y;






0.4 m 0.2 m

17–34. The trailer with its load has a mass of 150 kg and a center of mass at G. If it is subjected to a horizontal force of P = 600 N, determine the trailer’s acceleration and the normal force on the pair of wheels at A and at B. The wheels are free to roll and have negligible mass.

G P

1.25 m 0.25 m

SOLUTION

0.75 m

Equations of Motion: Writing the force equation of motion along the x axis, + ©F = m(a ) ; : x G x

600 = 150a

a = 4 m>s2 :

Ans.

Using this result to write the moment equation about point A, a + ©MA = (Mk)A ;

150(9.81)(1.25) - 600(0.5) - NB(2) = -150(4)(1.25) NB = 1144.69 N = 1.14 kN

Ans.

Using this result to write the force equation of motion along the y axis, or

NA + 1144.69 - 150(9.81) = 150(0) laws

+ c ©Fy = m(a G)y ;

NA = 326.81 N = 327 N

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Ans.

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B

0.25 m

1.25 m

A

600 N

0.5 m

17–35. At the start of a race, the rear drive wheels B of the 1550-lb car slip on the track. Determine the car’s acceleration and the normal reaction the track exerts on the front pair of wheels A and rear pair of wheels B. The coefficient of kinetic friction is mk=0.7, and the mass center of the car is at G. The front wheels are free to roll. Neglect the mass of all the wheels.

0.75 ft G A

SOLUTION Equations of Motion: Since the rear wheels B are required to slip, the frictional force developed is FB = msNB = 0.7NB. 1550 a 32.2

+ ©F = m(a ) ; ; x G x

0.7NB =

+ c ©Fy = m(aG)y;

NA + NB - 1550 = 0

(2)

a + ©MG = 0;

NB(4.75) - 0.7NB(0.75) - NA(6) = 0

(3)

(1)

Solving Eqs. (1), (2), and (3) yields a = 13.2 ft>s2

or

Ans.

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NB = 909.54 lb = 910 lb

laws

NA = 640.46 lb = 640 lb






B 6 ft

4.75 ft

*17–36. Determine the maximum acceleration that can be achieved by the car without having the front wheels A leave the track or the rear drive wheels B slip on the track. The coefficient of static friction is ms = 0.9. The car’s mass center is at G, and the front wheels are free to roll. Neglect the mass of all the wheels.

0.75 ft G A

B 6 ft

4.75 ft

SOLUTION Equations of Motion: + ©F = m(a ) ; ; x G x + c ©Fy = m(a G)y;

FB =

1550 a 32.2

(1)

NA + NB - 1550 = 0

a + ©MG = 0;

(2)

NB(4.75) - FB(0.75) - NA(6) = 0

(3)

If we assume that the front wheels are about to leave the track, NA = 0. Substituting this value into Eqs. (2) and (3) and solving Eqs. (1), (2), (3), a = 203.93 ft>s2

or

FB = 9816.67 lb

laws

NB = 1550 lb

Wide

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Since FB 7 (FB)max = msNB = 0.9(1550) lb = 1395 lb, the rear wheels will slip. Thus, the solution must be reworked so that the rear wheels are about to slip. permitted.

(4) not

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Dissemination

FB = msNB = 0.9NB

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is

of

the

Solving Eqs. (1), (2), (3), and (4) yields

and

NB = 923.08 lb for

work

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the

by

NA = 626.92 lb

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a = 17.26 ft>s2 = 17.3 ft>s2






Ans.

,

17–37. If the 4500-lb van has front-wheel drive, and the coefficient of static friction between the front wheels A and the road is ms = 0.8, determine the normal reactions on the pairs of front and rear wheels when the van has maximum acceleration. Also, find this maximum acceleration. The rear wheels are free to roll. Neglect the mass of the wheels.

G 2.5 ft B

SOLUTION Equations of Motion: The maximum acceleration occurs when the front wheels are about to slip. Thus, FA = msNA = 0.8 NA. Referring to the free-body diagram of the van shown in Fig. a, we have + ©F = m(a ) ; : x G x

0.8NA = a

4500 ba 32.2 max

(1)

+ c ©Fy = m(aG)y;

NA + NB - 4500 = 0

(2)

+ ©MG = 0;

NA(3.5) + 0.8NA(2.5) - NB(6) = 0

(3)

Solving Eqs. (1), (2), and (3) yields NB = 2152.171 lb = 2.15 kip

Ans. laws

or

NA = 2347.82 lb = 2.35 kip amax = 13.44 ft>s2 = 13.4 ft>s2

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Ans.






6 ft

A 3.5 ft

17–38. If the 4500-lb van has rear-wheel drive, and the coefficient of static friction between the front wheels B and the road is ms = 0.8, determine the normal reactions on the pairs of front and rear wheels when the van has maximum acceleration. The front wheels are free to roll. Neglect the mass of the wheels.

G 2.5 ft B

SOLUTION Equations of Motion: The maximum acceleration occurs when the rear wheels are about to slip. Thus, FB = msNB = 0.8 NB. Referring to Fig. a, + ©F = m(a ) ; : x G x

0.8NB = a

4500 ba 32.2 max

+ c ©Fy = m(aG)y;

NA + NB - 4500 = a

+ ©MG = 0;

NA(3.5) + 0.8NB(2.5) - NB(6) = 0

(1) 4500 b(0) 32.2

(2) (3)

Solving Eqs. (1), (2), and (3) yields amax = 12.02 ft>s2 = 12.0 ft>s2

Ans. or

NB = 2.10 kip

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NA = 2.40 kip






6 ft

A 3.5 ft

17–39. A

The uniform bar of mass m is pin connected to the collar, which slides along the smooth horizontal rod. If the collar is given a constant acceleration of a, determine the bar’s inclination angle u. Neglect the collar’s mass. L

SOLUTION Equations of Motion: Writing the moment equation of motion about point A, + ©MA = (Mk)A;

mg sin ua

L L b = ma cos u a b 2 2

a u = tan-1 a b g

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Ans.






u

a

*17–40. The lift truck has a mass of 70 kg and mass center at G. If it lifts the 120-kg spool with an acceleration of 3 m>s2 , determine the reactions on each of the four wheels. The loading is symmetric. Neglect the mass of the movable arm CD.

0.7 m

SOLUTION a + ©MB = ©(Mk)B ;

D

C

70(9.81)(0.5) + 120(9.81)(0.7) - 2NA(1.25)

G

= -120(3)(0.7)

0.4 m

NA = 567.76 N = 568 N

A

Ans.

B 0.75 m

+ c ©Fy = m(a G)y ;

2(567.76) + 2NB - 120(9.81) - 70(9.81) = 120(3) NB = 544 N

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Ans.






0.5 m

17–41. The lift truck has a mass of 70 kg and mass center at G. Determine the largest upward acceleration of the 120-kg spool so that no reaction on the wheels exceeds 600 N. 0.7 m

SOLUTION Assume NA = 600 N. a + ©MB = ©(Mk)B ;

G

70(9.81)(0.5) + 120(9.81)(0.7) - 2(600)(1.25) = -120a(0.7)

0.4 m A

a = 3.960 m>s2 + c ©Fy = m(a G)y ;

2(600) + 2NB - 120(9.81) - 70(9.81) = 120(3.960) OK

Thus a = 3.96 m>s2

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Ans.





B 0.75 m

NB = 570 N 6 600 N


D

C

0.5 m

17–42. The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration? The coefficient of static friction between the crate and the cart is ms = 0.5.

0.6 m

F

1m

SOLUTION 15

Equations of Motion: Assume that the crate slips, then Ff = ms N = 0.5N. a + ©MA = ©(Mk)A ;

50(9.81) cos 15°(x) - 50(9.81) sin 15°(0.5) = 50a cos 15°(0.5) + 50a sin 15°(x)

(1)

+Q©Fy¿ = m(a G)y¿ ;

N - 50(9.81) cos 15° = -50a sin 15°

(2)

R+ ©Fx¿ = m(aG)x¿ ;

50(9.81) sin 15° - 0.5N = -50a cos 15°

(3)

Solving Eqs. (1), (2), and (3) yields x = 0.250 m or

N = 447.81 N

a = 2.01 m>s2

teaching

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laws

Ans. in

Since x 6 0.3 m , then crate will not tip. Thus, the crate slips.

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Ans.






17–43. 1 ft

Determine the acceleration of the 150-lb cabinet and the normal reaction under the legs A and B if P = 35 lb. The coefficients of static and kinetic friction between the cabinet and the plane are ms = 0.2 and mk = 0.15, respectively. The cabinet’s center of gravity is located at G. P

Equations of Equilibrium: The free-body diagram of the cabinet under the static condition is shown in Fig. a, where P is the unknown minimum force needed to move the cabinet. We will assume that the cabinet slides before it tips. Then, FA = msNA = 0.2NA and FB = msNB = 0.2NB. + ©F = 0; : x

P - 0.2NA - 0.2NB = 0

(1)

+ c ©Fy = 0;

NA + NB - 150 = 0

(2)

+ ©MA = 0;

NB(2) - 150(1) - P(4) = 0

(3)

NB = 135 lb laws

or

NA = 15 lb

in

teaching

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Since P 6 35 lb and NA is positive, the cabinet will slide.

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NA + NB - 150 = 0

(4) (5)

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+ ©F = m(a ) ; : x G x

150 ba 32.2 by

35 - 0.15NA - 0.15NB = a

the

+ ©F = m(a ) ; : x G x

work

solely

NB(1) - 0.15NB(3.5) - 0.15NA(3.5) - NA(1) - 35(0.5) = 0

(6)

this

assessing

is

+ ©MG = 0;

the

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Solving Eqs. (4), (5), and (6) yields

part

a = 2.68 ft>s2 and

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NB = 123 lb sale

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NA = 26.9 lb






Ans.

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Equations of Motion: Since the cabinet is in motion, FA = mkNA = 0.15NA and FB = mkNB = 0.15NB. Referring to the free-body diagram of the cabinet shown in Fig. b,

3.5 ft

A

Solving Eqs. (1), (2), and (3) yields P = 30 lb

G 4 ft

SOLUTION

1 ft

B

*17–44. The assembly has a mass of 8 Mg and is hoisted using the boom and pulley system. If the winch at B draws in the cable with an acceleration of 2 m>s2, determine the compressive force in the hydraulic cylinder needed to support the boom. The boom has a mass of 2 Mg and mass center at G.

6m

2m

SOLUTION

G

sB + 2sL = l aB = - 2aL

4m

2 = - 2aL

B

C

60

aL = - 1 m>s

2

A

1m

D

Assembly: + c ©Fy = m ay ;

2m

2T - 8(103)(9.81) = 8(103)(1)

laws

or

T = 43.24 kN

teaching

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Boom: in

FCD(2) - 2(103)(9.81)(6 cos 60°) - 2(43.24)(103)(12 cos 60°) = 0

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Ans.






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FCD = 289 kN

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a + ©MA = 0;

17–45. The 2-Mg truck achieves a speed of 15 m>s with a constant acceleration after it has traveled a distance of 100 m, starting from rest. Determine the normal force exerted on each pair of front wheels B and rear driving wheels A. Also, find the traction force on the pair of wheels at A. The front wheels are free to roll. Neglect the mass of the wheels.

G 0.75 m A 2m

SOLUTION Kinematics: The acceleration of the truck can be determined from v2 = v02 + 2ac(s - s0) 152 = 0 + 2a(100 - 0) a = 1.125 m>s2 Equations of Motion: NB can be obtained directly by writing the moment equation of motion about point A. + ©MA = (Mk)A;

NB(3.5) - 2000(9.81)(2) = - 2000(1.125)(0.75) NB = 10 729.29 N = 10.7 kN

Ans. laws

or

Using this result and writing the force equations of motion along the x and y axes, Web)

NA + 10 729.29 - 2000(9.81) = 0

teaching

+ c ©Fy = m(aG)y;

Ans. in

FA = 2000(1.125) = 2250 N = 2.25 kN NA = 8890.71 N = 8.89 kN

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Ans.





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+ ©F = m(a ) ; : x G x


B 1.5 m

17–46. Determine the shortest time possible for the rear-wheel drive, 2-Mg truck to achieve a speed of 16 m>s with a constant acceleration starting from rest. The coefficient of static friction between the wheels and the road surface is ms = 0.8. The front wheels are free to roll. Neglect the mass of the wheels.

G 0.75 m A 2m

SOLUTION Equations of Motion: The maximum acceleration of the truck occurs when its rear wheels are on the verge of slipping. Thus, FA = msNA = 0.8NA. Referring to the free-body diagram of the truck shown in Fig. a, we can write + ©F = m(a ) ; : x G x

0.8NA = 2000a

(1)

+ c ©Fy = m(aG)y;

NA + NB - 2000(9.81) = 0

(2)

+ ©MG = 0;

NB(1.5) + 0.8NA(0.75) - NA(2) = 0

(3)

Solving Eqs. (1), (2), and (3) yields a = 4.059 m>s2

NB = 9471.72 N

or

NA = 10 148.28 N

teaching

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laws

Kinematics: Since the acceleration of the truck is constant, we can apply in

v = v0 + at

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+ B A:

Ans.

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t = 3.94 s





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Dissemination

16 = 0 + 4.059t


B 1.5 m

17–47. 0.5 ft

The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. If the acceleration is a = 20 ft>s2, determine the maximum height h of G2 of the rider so that the snowmobile’s front skid does not lift off the ground. Also, what are the traction (horizontal) force and normal reaction under the rear tracks at A?

a G2 G1 1 ft

SOLUTION

1.5 ft

Equations of Motion: Since the front skid is required to be on the verge of lift off, NB = 0. Writing the moment equation about point A and referring to Fig. a, a + ©MA = (Mk)A ;

150 250 (20)(hmax) + (20)(1) 32.2 32.2

250(1.5) + 150(0.5) =

hmax = 3.163 ft = 3.16 ft

Ans.

Writing the force equations of motion along the x and y axes, + ©F = m(a ) ; ; x G x

FA =

250 150 (20) + (20) 32.2 32.2 FA = 248.45 lb = 248 lb

laws

or

Ans. Web)

NA - 250 - 150 = 0 teaching

+ c ©Fy = m(a G)y ;

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NA = 400 lb






h A

*17–48. The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. If h = 3 ft, determine the snowmobile’s maximum permissible acceleration a so that its front skid does not lift off the ground. Also, find the traction (horizontal) force and the normal reaction under the rear tracks at A.

0.5 ft

a G2 G1 1 ft

SOLUTION

1.5 ft

Equations of Motion: Since the front skid is required to be on the verge of lift off, NB = 0. Writing the moment equation about point A and referring to Fig. a, a + ©MA = (Mk)A ;

150 250 a b (3) + a a b(1) 32.2 max 32.2 max

250(1.5) + 150(0.5) = a a max = 20.7 ft>s2

Ans.

Writing the force equations of motion along the x and y axes and using this result, we have FA =

150 250 (20.7) + (20.7) 32.2 32.2 or

+ ©F = m(a ) ; ; x G x

laws

FA = 257.14 lb = 257 lb

teaching

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Ans. Wide

copyright

in

+ c ©Fy = m(a G)y ; NA - 150 - 250 = 0 NA = 400 lb

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permitted.

Dissemination

Ans.

their


h A

17–49. A

If the cart’s mass is 30 kg and it is subjected to a horizontal force of P = 90 N, determine the tension in cord AB and the horizontal and vertical components of reaction on end C of the uniform 15-kg rod BC.

30⬚ 1m

C

SOLUTION Equations of Motion: The acceleration a of the cart and the rod can be determined by considering the free-body diagram of the cart and rod system shown in Fig. a. + ©F = m(a ) ; : x G x

a = 2 m>s2

90 = (15 + 30)a

The force in the cord can be obtained directly by writing the moment equation of motion about point C by referring to Fig. b. + ©MC = (Mk)C;

FAB sin 30°(1) - 15(9.81) cos 30°(0.5) = - 15(2) sin 30°(0.5) FAB = 112.44 N = 112 N

or

Ans.

teaching

Web)

laws

Using this result and applying the force equations of motion along the x and y axes, in

-Cx + 112.44 sin 30° = 15(2)

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+ ©F = m(a ) ; : x G x

Ans.

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Dissemination

Cx = 26.22 N = 26.2 N the

not

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Cy + 112.44 cos 30° - 15(9.81) = 0

on

learning.

is

of

+ c ©Fy = m(aG)y;

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Cy = 49.78 N = 49.8 N






Ans.

30⬚

B

P

17–50. A

If the cart’s mass is 30 kg, determine the horizontal force P that should be applied to the cart so that the cord AB just becomes slack. The uniform rod BC has a mass of 15 kg.

30⬚ 1m

30⬚

C

SOLUTION Equations of Motion: Since cord AB is required to be on the verge of becoming slack, FAB = 0. The corresponding acceleration a of the rod can be obtained directly by writing the moment equation of motion about point C. By referring to Fig. a. + ©MC = ©(MC)A;

-15(9.81) cos 30°(0.5) = - 15a sin 30°(0.5) a = 16.99 m>s2

Using this result and writing the force equation of motion along the x axis and referring to the free-body diagram of the cart and rod system shown in Fig. b, P = (30 + 15)(16.99) or

+ B ©F = m(a ) ; A: x G x

Ans.

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= 764.61 N = 765 N






B

P

17–51. at

The pipe has a mass of 800 kg and is being towed behind the truck. If the acceleration of the truck is a t = 0.5 m>s2, determine the angle u and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

B

A G 0.4 m


- 0.1NC + T cos 45° = 800(0.5)

+ c ©Fy = may ;

NC - 800(9.81) + T sin 45° = 0

a + ©MG = 0;

- 0.1NC(0.4) + T sin f(0.4) = 0

C

NC = 6770.9 N T = 1523.24 N = 1.52 kN 0.1(6770.9) 1523.24

f = 26.39°

u = 45° - f = 18.6°

Ans.

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or

sin f =

Ans.






u

45

*17–52. at

The pipe has a mass of 800 kg and is being towed behind a truck. If the angle u = 30°, determine the acceleration of the truck and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

B

A


T cos 45° - 0.1NC = 800a

+ c ©Fy = may ;

NC - 800(9.81) + T sin 45° = 0

a + ©MG = 0;

u

G 0.4 m C

T sin 15°(0.4) - 0.1NC(0.4) = 0

NC = 6161 N Ans.

a = 1.33 m>s2

Ans.

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T = 2382 N = 2.38 kN






45

17–53. The arched pipe has a mass of 80 kg and rests on the surface of the platform. As it is hoisted from one level to the next, a = 0.25 rad>s2 and v = 0.5 rad>s at the instant u = 30°. If it does not slip, determine the normal reactions of the arch on the platform at this instant.

500 mm G A 1m

θ

B

200 mm

ω, α 1m

SOLUTION + c ©Fy = m(aG)y ;

NA + NB - 80(9.81) = 20 sin 60° - 20 cos 60° NA + NB = 792.12

a + ©MA = ©(Mk)A ;

NB(1) - 80(9.81)(0.5) = 20 cos 60°(0.2) + 20 sin 60°(0.5) - 20 cos 60°(0.5) + 20 sin 60°(0.2) Ans.

NA = 391 N

Ans.

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NB = 402 N






17–54. The arched pipe has a mass of 80 kg and rests on the surface of the platform for which the coefficient of static friction is ms = 0.3. Determine the greatest angular acceleration a of the platform, starting from rest when u = 45°, without causing the pipe to slip on the platform.

500 mm G A 1m

θ

B

200 mm

ω, α 1m

SOLUTION aG = (aG)t = (1)(a) a + ©MA = ©(Mk)A ;

NB(1) - 80(9.81)(0.5) = 80(1a)(sin 45°)(0.2) + 80(1a)(cos 45°)(0.5)

+ ©F = m(a ) ; ; x G x

0.3NA + 0.3NB = 80(1a) sin 45°

+ c ©Fy = m(aG)y ;

NA + NB - 80(9.81) = 80(1a) cos 45°

Solving, a = 5.95 rad>s2

Ans.

or

NA = 494 N

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NB = 628 N






17–55. 1 ft

At the instant shown, link CD rotates with an angular velocity of vCD = 8 rad>s. If link CD is subjected to a couple moment of M = 650 lb # ft, determine the force developed in link AB and the angular acceleration of the links at this instant. Neglect the weight of the links and the platform. The crate weighs 150 lb and is fully secured on the platform.

D

M ⫽ 650 lb⭈ft

Equilibrium: Since the mass of link CD can be neglected, Dt can be obtained directly by writing the moment equation of equilibrium about point C using the free-body diagram of link CD, Fig. a, Dt(4) - 650 = 0

Dt = 162.5 lb

150 C a(4) D 32.2

a = 8.72 rad>s2

Ans.

150 (256) 32.2

©Fn = m(aG)n;

Dn + FAB + 150 =

a ©MG = 0;

Dn(1) - FAB(2) + 162.5(1) = 0

(1) teaching

Web)

laws

or

162.5 =

in

(2) Wide

copyright States

World

permitted.

Dissemination

Solving Eqs. (1) and (2), we obtain

Ans.

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FAB = 402 lb

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Dn = 641 lb






4 ft A

C 3 ft

Equations of Motion: Since the crate undergoes curvilinear translation, (aG)n = v2rG = 82(4) = 256 ft>s2 and (aG)t = arG = a(4). Referring to the freebody diagram of the crate, Fig. b, we have ©Ft = m(aG)t;

B

vCD ⫽ 8 rad/s

SOLUTION

a + ©MC = 0;

G

1 ft

*17–56. A

Determine the force developed in the links and the acceleration of the bar’s mass center immediately after the cord fails. Neglect the mass of links AB and CD. The uniform bar has a mass of 20 kg.

45⬚ 0.4 m D

B 0.6 m

SOLUTION Equations of Motion: Since the bar is still at rest at the instant the cord fails, vG = 0. v2G = 0. Referring to the free-body diagram of the bar, Fig. a, Thus, (aG)n = r ©Fn = m(aG)n;

TAB + TCD - 20(9.81) cos 45° + 50 cos 45° = 0

©Ft = m(aG)t;

20(9.81) sin 45° + 50 sin 45° = 20(aG)t

+ ©MG = 0;

TCD cos 45°(0.3) - TAB cos 45°(0.3) = 0

Solving, TAB = TCD = 51.68 N = 51.7 N

or

Ans.

in

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Web)

laws

(aG)t = 8.704 m>s2

Wide

copyright

Since (aG)n = 0, then

Dissemination

aG = (aG)t = 8.70 m>s2 R

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Ans.






C

P ⫽ 50 N

17–57. The 10-kg wheel has a radius of gyration kA = 200 mm. If the wheel is subjected to a moment M = 15t2 N # m, where t is in seconds, determine its angular velocity when t = 3 s starting from rest. Also, compute the reactions which the fixed pin A exerts on the wheel during the motion.

M A

SOLUTION + ©F = m(a ) ; : x G x

Ax = 0

+ c ©Fy = m(aG)y;

Ay - 10(9.81) = 0

c + ©MA = Ia a;

5t = 10(0.2)2a a =

dv = 12.5t dt 3

v =

L0

12.5 2 (3) 2

12.5t dt =

v = 56.2 rad>s

or

Ans. Ans.

Ay = 98.1 N

Ans.

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Ax = 0






17–58. The 80-kg disk is supported by a pin at A. If it is released from rest from the position shown, determine the initial horizontal and vertical components of reaction at the pin. A 1.5 m

SOLUTION + ©F = m(a ) ; ; x G x

Ax = 0

+ c ©Fy = m(aG)y ;

Ay - 80(9.81) = - 80(1.5)(a)

c + ©MA = IAa;

3 80(9.81)(1.5) = [ (80)(1.5)2]a 2

Ans.

a = 4.36 rad>s2 Ay = 262 N

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Ans.






17–59. The uniform slender rod has a mass m. If it is released from rest when u = 0°, determine the magnitude of the reactive force exerted on it by pin B when u = 90°.

L 3 A B

u

2 L 3

SOLUTION C

Equations of Motion: Since the rod rotates about a fixed axis passing through point L L B, (aG)t = a rG = a a b and (aG)n = v2rG = v2 a b . The mass moment of inertia 6 6 1 2 mL . Writing the moment equation of motion about of the rod about its G is IG = 12 point B, - mg cos u a

+ ©MB = ©(Mk)B;

a =

L L L 1 b = - m c a a b d a b - a mL2 ba 6 6 6 12

3g cos u 2L

laws

or

1 This equation can also be obtained by applying ©MB = IBa, whereIB = mL2 + 12 2 1 L ma b = mL2. Thus, 6 9 in

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L 1 b = - a mL2 b a 6 9

Wide

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-mg cos u a

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Dissemination

3g cos u 2L

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Using this result and writing the force equation of motion along the n and t axes,

is

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not

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a =

States

+ ©MB = IBa;

work

student

the

by

3g L cos u b a b d 2L 6

(including

for

mg cos u - Bt = mc a

©Ft = m(aG)t;

assessing

1 mv2L + mg sin u 6

this

integrity

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(2)

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Bn =

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L bd 6 provided

Bn - mg sin u = m c v2 a

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©Fn = m(aG)n;

(1)

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the

3 mg cos u 4

of

Bt =

sale

Kinematics: The angular velocity of the rod can be determined by integrating L

vdv = v

L0

adu u

vdv =

v =

L

3g cos u du 2L L0

3g sin u BL

When u = 90°, v =

3g . Substituting this result and u = 90° into Eqs. (1) and (2), AL

3 mg cos 90° = 0 4 3g 1 3 Bn = ma b(L) + mg sin 90° = mg 6 L 2

Bt =

FA = 3At2 + An2 =






2 3 3 02 + a mg b = mg C 2 2

Ans.

*17–60. ω

The drum has a weight of 80 lb and a radius of gyration kO = 0.4 ft. If the cable, which is wrapped around the drum, is subjected to a vertical force P = 15 lb, determine the time needed to increase the drum’s angular velocity from v1 = 5 rad>s to v2 = 25 rad>s. Neglect the mass of the cable.

0.5 ft O

SOLUTION c + ©MO = IO a;

15(0.5) = [

80 (0.4)2] a 32.2

P

(1)

a = 18.87 rad>s2 (c + )v = v0 + a t 25 = 5 + 18.87 t t = 1.06 s

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Ans.






17–61. Cable is unwound from a spool supported on small rollers at A and B by exerting a force of T = 300 N on the cable in the direction shown. Compute the time needed to unravel 5 m of cable from the spool if the spool and cable have a total mass of 600 kg and a centroidal radius of gyration of kO = 1.2 m. For the calculation, neglect the mass of the cable being unwound and the mass of the rollers at A and B. The rollers turn with no friction.

T

300 N 1.5 m 30

0.8 m O

A

B

SOLUTION 1m

Equations of Motion: The mass moment of inertia of the spool about point O is given by IO = mk2O = 600 A 1.2 2 B = 864 kg # m2. Applying Eq. 17–16, we have a + ©MO = IO a;

- 300(0.8) = - 864a

a = 0.2778 rad>s2

5 s Kinematics: Here, the angular displacement u = = = 6.25 rad. Applying r 0.8 1 2 equation u = u0 + v0 t + at , we have 2 (c +)

6.25 = 0 + 0 +

1 (0.2778)t2 2 or

t = 6.71 s

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Ans.






17–62. The 10-lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb # ft>rad, so that the torque developed is M = 15u2 lb # ft, where u is in radians. If the bar is released from rest when it is vertical at u = 90°, determine its angular velocity at the instant u = 0°.

1 ft u 1 ft O

SOLUTION c + ©MO = IOa; - 5u = [

1 10 ( )(2)2]a 12 32.2

- 48.3 u = a a du = v dv o

-

Lp2

48.3 u du =

v

L0

v dv

1 48.3 p 2 ( ) = v2 2 2 2 v = 10.9 rad/s

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Ans.






17–63. The 10-lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb # ft>rad, so that the torque developed is M = 15u2 lb # ft, where u is in radians. If the bar is released from rest when it is vertical at u = 90°, determine its angular velocity at the instant u = 45°.

1 ft u 1 ft O

SOLUTION c + ©MO = IOa;

5u = [

1 10 ( )(2)2]a 12 32.2

a = - 48.3u a du = v dv -

p 4

Lp2

v

48.3u du =

L0

v dv

p p 1 - 24.15 a( )2 - ( )2 b = v2 4 2 2 v = 9.45 rad s

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*17–64. If shaft BC is subjected to a torque of M = (0.45t1>2) N # m, where t is in seconds, determine the angular velocity of the 3-kg rod AB when t = 4 s, starting from rest. Neglect the mass of shaft BC.

B 300 mm

C

A

SOLUTION Equations of Motion: The mass moment of inertia of the rod about the z axis is 1 Iz = IG + md2 = (3) A 0.32 B + 3 A 0.152 B = 0.09 kg # m2. Writing the moment 12 equation of motion about the z axis, 0.45t1>2 = 0.09a

©Mz = Iza;

a = 5t1>2 rad>s2

Kinematics: The angular velocity of the rod can be obtained by integration. L

a dt

v

L0

dv =

t 1>2

L0

5t dt or

dv =

laws

L

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v = A 3.333t3>2 B rad>s

permitted.

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When t = 4 s,

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Ans.






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v = 3.333 A 43>2 B = 26.7 rad>s

M ⫽ (0.45 t1/2) N⭈m

17–65. 0.6 m

Determine the vertical and horizontal components of reaction at the pin support A and the angular acceleration of the 12-kg rod at the instant shown, when the rod has an angular velocity of v = 5 rad>s.

A v ⫽ 5 rad/s

SOLUTION Equations of Motion: Since the rod rotates about a fixed axis passing through

point A, (aG)t = arG = a(0.3) and (aG)n = v2rG = A 52 B (0.3) = 7.5 m>s2. The mass 1 mL2 = moment of inertia of the rod about its mass center is IG = 12 1 (12) A 0.62 B = 0.36 kg # m2. Writing the moment equation of motion about point A 12 and referring to Fig. a, - 12(9.81)(0.3) = - 0.36a - 12[a(0.3)](0.3) a = 24.525 rad>s2 = 24.5 rad>s2

Ans. 1 2 ml = 3 laws

This result can also be obtained by applying ©MA = IAa, where IA =

or

+ ©MA = ©(Mk)A;

teaching in

-12(9.81)(0.3) = - 1.44a

Ans. is

of

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not

instructors

States

World

a = 24.525 rad>s2 = 24.5 rad>s2

for

work

student

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by

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United

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learning.

Using this result to write the force equations of motion along the x and y axes, we have Ax = 12(7.5) = 90 N

+ c ©Fy = m(aG)y;

Ay = 12(9.81) = - 12[24.525(0.3)]

Ans.

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of provided

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Ay = 29.43 N = 29.4 N

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the

(including

+ ©F = m(a ) ; ; x G x






Ans.

permitted.

Dissemination

Wide

copyright

+ ©MA = IAa;

Web)

1 (12)(0.62) = 1.44 kg # m2. Thus, 3

B

17–66. a

The kinetic diagram representing the general rotational motion of a rigid body about a fixed axis passing through O is shown in the figure. Show that IGA may be eliminated by moving the vectors m(aG)t and m(aG)n to point P, located a distance rGP = k2G>rOG from the center of mass G of the body. Here kG represents the radius of gyration of the body about an axis passing through G. The point P is called the center of percussion of the body.

P m(aG)t G m(aG)n

rOG

m(aG)t rOG + IG a = m(aG)t rOG + A mk2G B a However, (aG)t rOG

k2G = rOG rGP and a =

m(aG)t rOG + IG a = m(aG)t rOG + (mrOG rGP) c = m(aG)t(rOG + rGP)

(aG)t d rOG

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rGP

O

SOLUTION


IG a

17–67. Determine the position rP of the center of percussion P of the 10-lb slender bar. (See Prob. 17–66.) What is the horizontal component of force that the pin at A exerts on the bar when it is struck at P with a force of F=20 lb?

A

rP 4 ft

P

SOLUTION Using the result of Prob 17–66,

rGP =

k2G rAG

B =

1 ml2 2 a bR B 12 m

1 l 6

=

l 2

Thus, 1 2 2 1 l + l = l = (4) = 2.67 ft 6 2 3 3

Ans. laws

or

rP =

Web)

1 10 a b (4)2 da 3 32.2

teaching

20(2.667) = c

in

c + ©MA = IA a;

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a = 32.2 rad>s2

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(a G)t = 2(32.2) = 64.4 ft>s2

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United

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10 b (64.4) 32.2

by

student

-Ax + 20 = a

the

+ ©F = m(a ) ; ; x G x

work

Ax = 0

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for

Ans.






F

*17–68. The disk has a mass M and a radius R. If a block of mass m is attached to the cord, determine the angular acceleration of the disk when the block is released from rest. Also,what is the velocity of the block after it falls a distance 2R starting from rest?

R

SOLUTION c + ©MO = ©(Mk) O;

mgR = 12 MR2 (a) + m (aR) R 2mg R(M + 2m)

a =

Ans.

a = aR v2 = v20 + 2 a(s - s0)

8mgR (M + 2m)

Ans.

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laws

v =

2mgR ) (2R - 0) R(M + 2m

or

v2 = 0 + 2(






17–69. The door will close automatically using torsional springs mounted on the hinges. Each spring has a stiffness k = 50 N # m>rad so that the torque on each hinge is M = 150u2 N # m, where u is measured in radians. If the door is released from rest when it is open at u = 90°, determine its angular velocity at the instant u = 0°. For the calculation, treat the door as a thin plate having a mass of 70 kg.

M 1.5 m

A

SOLUTION IAB

0.4 m

1 1 ml2 + md2 = (70)(1.2)2 + 70(0.6)2 = 33.6 kg # m2 = 12 12

©MAB = IAB a ;

2(50u) = - 33.6(a)

v

L0

0

Lp2

2.9762udu

v = 2.71 rad>s

will

their

destroy

of

and

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is

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provided

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laws

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Ans.






M

1.2 m

a = - 2.9762u

vdv = adu vdv = -

θ

B

0.4 m

17–70. The door will close automatically using torsional springs mounted on the hinges. If the torque on each hinge is M = ku, where u is measured in radians, determine the required torsional stiffness k so that the door will close 1u = 0°2 with an angular velocity v = 2 rad>s when it is released from rest at u = 90°. For the calculation, treat the door as a thin plate having a mass of 70 kg.

M 1.5 m

A

SOLUTION ©MA = IAa;

2M = - B

0.4 m

1 (70)(1.2)2 + 70(0.6)2 R (a) 12

ku = - 16.8a a du = v dv 0

Lp2

udu = 16.8

2

L0

v dv

laws

or

16.8 k p 2 ( ) = (2)2 2 2 2

teaching

Web)

k = 27.2 N # m>rad

will

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Ans.






M

1.2 m

M = - 16.8a

-k

θ

B

0.4 m

17–71. The pendulum consists of a 10-kg uniform slender rod and a 15-kg sphere. If the pendulum is subjected to a torque of M = 50 N # m, and has an angular velocity of 3 rad>s when u = 45°, determine the magnitude of the reactive force pin O exerts on the pendulum at this instant.

O u

M ⫽ 50 N⭈m

600 mm A

100 mm

SOLUTION Equations of Motion: Since the pendulum rotates about a fixed axis passing through point O, [(aG)OA]t = a(rG)OA = a(0.3), [(aG)B]t = a(rG)B = a(0.7), [(aG)OA]n = v2(rG)OA = (32)(0.3) = 2.7 m>s2, and [(aG)B]n = v2(rG)B = (32)(0.7) = 6.3 m>s2. The mass moment of inertia of the rod and sphere about their respective 1 1 (IG)OA = ml2 = (10)(0.62) = 0.3 kg # m2 mass centers are and 12 12 2 2 (IG)B = mr2 = (15)(0.12) = 0.06 kg # m2. Writing the moment equation of 5 5 motion about point O, we have

B

+ ©MO = ©(Mk)O; -10(9.81) cos 45°(0.3) - 15(9.81) cos 45°(0.7) - 50 =

laws

or

-10[a(0.3)](0.3) - 0.3a - 15[a(0.7)](0.7) - 0.06a

in

teaching

Web)

a = 16.68 rad>s2

of

the

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is

-10(9.81) cos 45°(0.3) - 15(9.81) cos 45°(0.7) - 50 = - 8.61a and

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+ ©MO = IOa;

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a = 16.68 rad>s2

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Using this result to write the force equations of motion along the n and t axes, this

assessing

10(9.81) cos 45° + 15(9.81) cos 45° + Ot integrity

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©Ft = m(aG)t;

any their

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and

Ot = 51.81 N

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= 10[16.68(0.3)] + 15[16.68(0.7)]

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On - 10(9.81) sin 45° - 15(9.81) sin 45° = 10(2.7) + 15(6.3) sale

©Fn = m(aG)n;

On = 294.92 N Thus, FO = 4Ot 2 + On2 = 451.812 + 294.922 = 299.43 N = 299 N






Ans.

permitted.

Dissemination

Wide

copyright

This result can also be obtained by applying ©MO = IOa, where IO = ©IG + md2 = 2 1 (10)(0.62) + 10(0.32) + (15)(0.12) + 15(0.72) = 8.61 kg # m2. Thus, 12 5

*17–72. ␻

The disk has a mass of 20 kg and is originally spinning at the end of the strut with an angular velocity of v = 60 rad>s. If it is then placed against the wall, where the coefficient of kinetic friction is mk = 0.3, determine the time required for the motion to stop. What is the force in strut BC during this time?

B

150 mm


FCB sin 30° - NA = 0

+ c ©Fy = m(aG)y ;

FCB cos 30° - 20(9.81) + 0.3NA = 0

a + ©MB = IB a;

60⬚ C

1 0.3NA (0.15) = c (20)(0.15)2 da 2 NA = 96.6 N FCB = 193 N

Ans.

a = 19.3 rad>s2 v = v0 + ac t laws

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t = 3.11 s






A

17–73. The slender rod of length L and mass m is released from rest when u = 0°. Determine as a function of u the normal and the frictional forces which are exerted by the ledge on the rod at A as it falls downward. At what angle u does the rod begin to slip if the coefficient of static friction at A is m?

A u L

SOLUTION Equations of Motion: The mass moment inertia of the rod about its mass center is 1 given by IG = mL2. At the instant shown, the normal component of acceleration 12 L of the mass center for the rod is (a G)n = v2rG = v2 a b . The tangential component 2 L of acceleration of the mass center for the rod is (aG)t = ars = aa b . 2 L L 1 L b = - a mL2 b a - m caa b d a b 2 12 2 2 3g cos u 2L

teaching

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a =

or

-mg cos u a

laws

a + ©MA = ©(Mk)O ;

in

3g L cos ua b d 2L 2

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mg cos u - NA = m c

copyright

+b©Ft = m(aG)t ;

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mg cos u 4

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NA =

and

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L bd 2

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Ff - mg sin u = mc v2 a

the

a + ©Fn = m(aG)n ;

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Kinematics:Applying equation v dv = a du, we have this

integrity

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3g cos u du L0° 2L any

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3g sin u L

3g sin u into Eq. (1) gives L Ff =

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5mg sin u 2

Ans.

If the rod is on the verge of slipping at A, Ff = mNA. Substitute the data obtained above, we have 5mg mg sin u = ma cos u b 2 4 u = tan-1 a






m b 10

Ans.

17–74. The 5-kg cylinder is initially at rest when it is placed in contact with the wall B and the rotor at A. If the rotor always maintains a constant clockwise angular velocity v = 6 rad>s, determine the initial angular acceleration of the cylinder. The coefficient of kinetic friction at the contacting surfaces B and C is mk = 0.2. B 125 mm

SOLUTION

v 45

Equations of Motion: The mass moment of inertia of the cylinder about point O is 1 1 given by IO = mr 2 = (5)(0.1252) = 0.0390625 kg # m2. Applying Eq. 17–16, 2 2 we have + ©F = m(a ) ; : x G x

NB + 0.2NA cos 45° - NA sin 45° = 0

(1)

+ c ©Fy = m(a G)y ;

0.2NB + 0.2NA sin 45° + NA cos 45° - 5(9.81) = 0

(2)

a + ©MO = IO a;

0.2NA (0.125) - 0.2NB (0.125) = 0.0390625a

C

(3)

Solving Eqs. (1), (2), and (3) yields; or

NB = 28.85 N

laws

NA = 51.01 N

Web)

a = 14.2 rad>s2

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A

17–75. The wheel has a mass of 25 kg and a radius of gyration kB = 0.15 m. It is originally spinning at v1 = 40 rad>s. If it is placed on the ground, for which the coefficient of kinetic friction is mC = 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of reaction which the pin at A exerts on AB during this time? Neglect the mass of AB.

0.4 m A 0.3 m B

0.2 m

V

SOLUTION

C

IB = mk2B = 25(0.15)2 = 0.5625 kg # m2 + c ©Fy = m(aG)y ; + ©F = m(a ) ; : x G x a + ©MB = IBa;

A 35 B FAB + NC - 25(9.81) = 0

(1)

A 45 B FAB = 0

(2)

0.5NC -

0.5NC(0.2) = 0.5625(- a)

(3)

Solvings Eqs. (1),(2) and (3) yields: FAB = 111.48 N

NC = 178.4 N

or

a = - 31.71 rad>s2 Ans.

Ay = 35 FAB = 0.6(111.48) = 66.9 N

Ans. Wide

copyright

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Ax = 45FAB = 0.8(111.48) = 89.2 N

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v = v0 + ac t

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0 = 40 + ( -31.71) t

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t = 1.26 s






Ans.

*■17–76. A 40-kg boy sits on top of the large wheel which has a mass of 400 kg and a radius of gyration kG = 5.5 m. If the boy essentially starts from rest at u = 0°, and the wheel begins to rotate freely, determine the angle at which the boy begins to slip. The coefficient of static friction between the wheel and the boy is ms = 0.5. Neglect the size of the boy in the calculation.

u

SOLUTION c + ©MO = ©(Mk)O ;

392.4(8 sin u) = 400(5.5)2 a + 40(8)(a)(8)

0.2141 sin u = a a du = v dv u

L0

v

0.2141 sin u du =

vdv

L0

u

1 2 v 2

-0.2141 cos u 2 = 0

v = 0.4283(1 - cos u) laws

or

2

Web)

392.4 cos u - N = 40(v2)(8)

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b ©Fy¿ = m(aG)y¿;

in

+

392.4 sin u - 0.5 N = 40(8)(a)

Wide Dissemination

= m(aG)x¿ ;

copyright

+R©Fx¿

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permitted.

N = 392.4 cos u - 137.05(1 - cos u) = 529.45 cos u - 137.05

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the

392.4 sin u - 0.5(529.45 cos u - 137.05) = 320(0.2141 sin u)

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323.89 sin u - 264.73 cos u + 68.52 = 0

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(including

- sin u + 0.8173 cos u = 0.2116

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Solve by trial and error part

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u = 29.8°

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Note: The boy will loose contact with the wheel when N = 0, i.e. sale

will

N = 529.45 cos u - 137.05 = 0 u = 75.0° 7 29.8° Hence slipping occurs first.






Ans.

8m

17–77. M ⫽ 200(1 ⫺ e⫺0.2 t ) N⭈m

Gears A and B have a mass of 50 kg and 15 kg, respectively. Their radii of gyration about their respective centers of mass are kC = 250 mm and kD = 150 mm. If a torque of M = 200(1 - e-0.2t) N # m, where t is in seconds, is applied to gear A, determine the angular velocity of both gears when t = 3 s, starting from rest.

300 mm C

SOLUTION

B

rA Equations of Motion: Since gear B is in mesh with gear A, aB = a baA = rB 0.3 a b a = 1.5aA.The mass moment of inertia of gears A and B about their respective 0.2 A

A

centers are IC = mA kC 2 = 50(0.252) = 3.125 kg # m2 and ID = mB kD2 = 15(0.152) = 0.3375 kg # m2. Writing the moment equation of motion about the gears’ center using the free-body diagrams of gears A and B, Figs. a and b, a + ©MC = ICaA;

F(0.3) - 200(1 - e-0.2t) = - 3.125aA

(1)

F(0.2) = 0.3375(1.5aA)

(2)

and

laws

or

a + ©MD = IDaB;

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Web)

Eliminating F from Eqs. (1) and (2) yields

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Kinematics: The angular velocity of gear A can be determined by integration.

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aA = 51.49(1 - e-0.2t) rad>s2

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aA dt

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dvA =

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t

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vA

51.49(1 - e-0.2t)dt L0 L0 vA = 51.49(t + 5e-0.2t - 5) rad>s

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dvA =

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When t = 3 s,

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vA = 51.49(3 + 5e-0.2(3) - 5) = 38.31 rad>s = 38.3 rad>s sale

will

Then vB = a

rA 0.3 bv = a b (38.31) rB A 0.2

= 57.47 rad>s = 57.5 rad>s






200 mm D

Ans.

17–78. Block A has a mass m and rests on a surface having a coefficient of kinetic friction mk. The cord attached to A passes over a pulley at C and is attached to a block B having a mass 2m. If B is released, determine the acceleration of A. Assume that the cord does not slip over the pulley. The pulley can be approximated as a thin disk of radius r and mass 14m. Neglect the mass of the cord.

r A

SOLUTION

B

Block A: + ©F = ma ; : x x

T1 - mk mg = ma

(1)

2mg - T2 = 2ma

(2)

Block B: + T ©Fy = may; Pulley C: 1 1 a T2r - T1r = c a m br2 d a b r 2 4 1 ma 8

(3) laws

T2 - T1 =

or

c + ©MC = IG a;

25 a 8

permitted.

(2 - mk)g =

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1 ma 8

Dissemination

2mg - 2ma - (ma + mk mg) =

copyright

in

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Substituting Eqs. (1) and (2) into (3),

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8 (2 - mk)g 25

Ans.

the

a =

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not

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States

1 ma + 3ma 8

of

2mg - mk mg =

C

17–79. The two blocks A and B have a mass of 5 kg and 10 kg, respectively. If the pulley can be treated as a disk of mass 3 kg and radius 0.15 m, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley. r O

SOLUTION

A

Kinematics: Since the pulley rotates about a fixed axis passes through point O, its angular acceleration is a =

B

a a = = 6.6667a r 0.15

The mass moment of inertia of the pulley about point O is 1 1 Io = Mr2 = (3)(0.152) = 0.03375 kg # m2 2 2 Equation of Motion: Write the moment equation of motion about point O by referring to the free-body and kinetic diagram of the system shown in Fig. a, 5(9.81)(0.15) - 10(9.81)(0.15) laws

or

a + ©Mo = ©(Mk)o;

teaching

Web)

= - 0.03375(6.6667a) - 5a(0.15) - 10a(0.15) in

a = 2.973 m>s2 = 2.97 m>s2

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*17–80. The two blocks A and B have a mass mA and mB , respectively, where mB 7 mA . If the pulley can be treated as a disk of mass M, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley. r O

SOLUTION A

a = ar B

c + ©MC = © (Mk)C ;

1 mB g(r) - mA g(r) = a Mr2 b a + mB r2 a + mA r2 a 2 g(mB - mA)

a =

1 r a M + mB + mA b 2 g(mB - mA)

a =

Ans.

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laws

or

1 a M + mB + mA b 2






17–81. Determine the angular acceleration of the 25-kg diving board and the horizontal and vertical components of reaction at the pin A the instant the man jumps off. Assume that the board is uniform and rigid, and that at the instant he jumps off the spring is compressed a maximum amount of 200 mm, v = 0, and the board is horizontal. Take k = 7 kN>m.

1.5 m

1.5 m

A k

SOLUTION a + a MA = IAa; + c a Ft = m(aG)t ; + ; a Fn = m(aG)n ;

1 1.5(1400 - 245.25) = c (25)(3)2 da 3 1400 - 245.25 - Ay = 25(1.5a) Ax = 0

Ans.

A y = 289 N

Ans.

a = 23.1 rad>s2

Ans.

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Ax = 0

or

Solving,






17–82. The lightweight turbine consists of a rotor which is powered from a torque applied at its center. At the instant the rotor is horizontal it has an angular velocity of 15 rad/s and a clockwise angular acceleration of 8 rad>s2. Determine the internal normal force, shear force, and moment at a section through A. Assume the rotor is a 50-m-long slender rod, having a mass of 3 kg/m.

10 m A 25 m

SOLUTION + ©F = m(a ) ; ; n G n + T ©Ft = m(aG)t ;

NA = 45(15)2 (17.5) = 177kN

Ans.

VA + 45(9.81) = 45(8)(17.5) VA = 5.86 kN

c + ©MA = ©(Mk)A ;

Ans.

MA + 45(9.81)(7.5) = c

1 (45)(15)2 d(8) + [45(8)(17.5)](7.5) 12

MA = 50.7 kN # m

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Ans.






17–83. The two-bar assembly is released from rest in the position shown. Determine the initial bending moment at the fixed joint B. Each bar has a mass m and length l.

A

l

B

l

SOLUTION

C

Assembly: IA =

1 2 1 l (m)(l)2 + m(l2 + ( )2) ml + 3 12 2

= 1.667 ml2 c + ©MA = IA a;

l mg( ) + mg(l) = (1.667ml2)a 2 0.9 g l or

a =

teaching

Web)

laws

Segment BC: in

l>2 1 l l ml2 d a + m(l 2 + ( )2)1>2 a( )( ) l 2 2 2 12 2 l + (2)

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permitted.

Dissemination

M = c

copyright

c + ©MB = ©(Mk)B;

on

learning.

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of

the

not

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World

0.9g 1 2 1 ml a = ml2 ( ) 3 3 l

and

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M =

by

M = 0.3gml

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Ans.






*17–84. The armature (slender rod) AB has a mass of 0.2 kg and can pivot about the pin at A. Movement is controlled by the electromagnet E, which exerts a horizontal attractive force on the armature at B of FB = 10.2110-32l-22 N, where l in meters is the gap between the armature and the magnet at any instant. If the armature lies in the horizontal plane, and is originally at rest, determine the speed of the contact at B the instant l = 0.01 m. Originally l = 0.02 m.

l B

E

150 mm

SOLUTION Equation of Motion: The mass moment of inertia of the armature about point A is 1 given by IA = IG + m r2G = (0.2) A 0.152 B + 0.2 A 0.0752 B = 1.50 A 10-3 B kg # m2 12 Applying Eq. 17–16, we have 0.2(10-3)

a + ©MA = IAa;

l2

A

(0.15) = 1.50 A 10-3 B a 0.02 l2 or

a =

not

instructors

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Dissemination

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laws

Kinematic: From the geometry, l = 0.02 - 0.15 u. Then dl = - 0.15 du or dv dl v . Also, v = hence dv = . Substitute into equation vdv = adu, du = 0.15 0.15 0.15 we have

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dv dl v ¢ ≤ = a¢ ≤ 0.15 0.15 0.15

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vdv = - 0.15 adl is

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0.02 dl l2

- 0.15

this

L0.02 m

integrity

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L0

0.01 m

vdv =

assessing

v

Ans.

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v = 0.548 m s






17–85. The bar has a weight per length of w. If it is rotating in the vertical plane at a constant rate V about point O, determine the internal normal force, shear force, and moment as a function of x and u.

O V U L

SOLUTION

x

a = v2 a L -

x u bh z

Forces: wx 2 x v aL - b u = N u + S au + wx T h g z h

(1)

Moments:

or

x Ia = M - S a b 2 laws

1 Sx 2

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Web)

(2) Wide

copyright

in

O = M -

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permitted.

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Solving (1) and (2),

not

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v2 x A L - B + cos u R g 2

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N = wx B

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the

S = wx sin u

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the

1 wx2 sin u 2

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M =






Ans.

17–86. A force F = 2 lb is applied perpendicular to the axis of the 5-lb rod and moves from O to A at a constant rate of 4 ft>s. If the rod is at rest when u = 0° and F is at O when t = 0, determine the rod’s angular velocity at the instant the force is at A. Through what angle has the rod rotated when this occurs? The rod rotates in the horizontal plane.

O u 4 ft/s

4 ft

F ⫽ 2 lb

SOLUTION IO = a + a MO = IOa;

A

1 1 5 mR2 = a b (4)2 = 0.8282 slug # ft2 3 3 32.2 2(4t) = 0.8282(a) a = 9.66t dv = adt v

L0

t

#

dv =

L0

9.66t dt

teaching

Web)

laws

or

v = 4.83t2

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in

When t = 1 s, v = 4.83(1)2 = 4.83 rad>s States

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du = v dt u

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4.83t2 dt

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L0

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du =

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the

u = 1.61 rad = 92.2°






Ans.

17–87. A

The 15-kg block A and 20-kg cylinder B are connected by a light cord that passes over a 5-kg pulley (disk). If the system is released from rest, determine the cylinder’s velocity after its has traveled downwards 2 m. Neglect friction between the plane and the block, and assume the cord does not slip over the pulley.

100 mm O

SOLUTION Equations of Motion: Since the pulley rotates about a fixed axis passing through a a = = 10a. The mass moment of inertia point O, its angular acceleration is a = r 0.1 1 2 1 of the pulley about point O is IO = mr = (5)(0.12) = 0.025 kg # m2. Writing the 2 2 moment equation of equilibrium about point O and realizing that the moment of 15(9.81) N and NA cancel out, we have + ©MO = ©(Mk)O;

B

- 20(9.81)(0.1) = - 15a(0.1) - 20a(0.1) - 0.025(10a) a = 5.232 m>s2

or

Kinematics: Since the angular acceleration is constant, v2 = v0 2 + 2ac(s - s0)

teaching

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laws

(+ T)

Wide

copyright

in

v2 = 02 + 2(5.232)(2 - 0)

Ans.

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Dissemination

v = 4.57 m>s T






*17–88. A

The 15-kg block A and 20-kg cylinder B are connected by a light cord that passes over a 5-kg pulley (disk). If the system is released from rest, determine the cylinder’s velocity after its has traveled downwards 2 m. The coefficient of kinetic friction between the block and the horizontal plane is mk = 0.3. Assume the cord does not slip over the pulley.

100 mm O

SOLUTION Equations of Motion: Since the block is in motion, FA = mk NA = 0.3NA. Referring to the free-body diagram of block A shown in Fig. a, + c ©Fy = m(aG)y;

NA - 15(9.81) = 0

+ ©F = m(a ) ; : x G x

T1 - 0.3(147.15) = 15a

B

NA = 147.15 N (1)

Referring to the free-body diagram of the cylinder, Fig. b, + c ©Fy = m(aG)y;

T2 - 20(9.81) = - 20a

(2)

a a = = 10a. r 0.1 1 2 The mass moment of inertia of the pulley about O is IO = mr = 2 1 (5)(0.12) = 0.025 kg # m2. Writing the moment equation of motion about point O 2 using the free-body diagram of the pulley shown in Fig. c, Dissemination

Wide

copyright

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laws

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Since the pulley rotates about a fixed axis passing through point O, a =

permitted.

(3) World

T1(0.1) - T2(0.1) = - 0.025(10a)

is

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the

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instructors

States

+ ©MO = IOa;

T1 = 104.967 N work

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T2 = 115.104 N

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v2 = v02 + 2ac(s - s0)

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v = 4.027 m>s = 4.03 m>s T






Ans.

17–89. The “Catherine wheel” is a firework that consists of a coiled tube of powder which is pinned at its center. If the powder burns at a constant rate of 20 g>s such as that the exhaust gases always exert a force having a constant magnitude of 0.3 N, directed tangent to the wheel, determine the angular velocity of the wheel when 75% of the mass is burned off. Initially, the wheel is at rest and has a mass of 100 g and a radius of r = 75 mm. For the calculation, consider the wheel to always be a thin disk.

r C 0.3 N

SOLUTION Mass of wheel when 75% of the powder is burned = 0.025 kg Time to burn off 75 % =

0.075 kg = 3.75 s 0.02 kg>s

m(t) = 0.1 - 0.02 t Mass of disk per unit area is

or

0.1 kg m = = 5.6588 kg>m2 A p(0.075 m)2 laws

r0 =

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0.1 - 0.02t A p(5.6588)

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0.3r =

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+ ©MC = ICa;

3

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a = 0.6 A 2p(5.6588) B [0.1 - 0.02t]- 2 3

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a = 2.530[0.1 - 0.02t]- 2 dv = a dt v

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t

dv = 2.530

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3

[0.1 - 0.02t]- 2 dt 1

v = 253 C (0.1 - 0.02t)- 2 - 3.162 D For t = 3.75 s, v = 800 rad s






Ans.

17–90. If the disk in Fig. 17–21a rolls without slipping, show that when moments are summed about the instantaneous center of zero velocity, IC, it is possible to use the moment equation ©MIC = IICa, where IIC represents the moment of inertia of the disk calculated about the instantaneous axis of zero velocity.

SOLUTION c + ©MlC = ©(MK)lC;

©MlC = IGa + (maG)r

Since there is no slipping, aG = ar Thus, ©MIC = A IG + mr2 B a By the parallel–axis thoerem, the term in parenthesis represents IIC. Thus, ©MIC = IICa

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Q.E.D.






17–91. The 20-kg punching bag has a radius of gyration about its center of mass G of kG = 0.4 m. If it is initially at rest and is subjected to a horizontal force F = 30 N, determine the initial angular acceleration of the bag and the tension in the supporting cable AB.

A 1m B 0.3 m G

SOLUTION

0.6 m

+ ©F = m(a ) ; : x G x

30 = 20(aG)x

+ c ©Fy = m(aG)y;

T - 196.2 = 20(aG)y

a + ©MG = IGa;

F

30(0.6) = 20(0.4)2a a = 5.62 rad>s2

Ans.

(aG)x = 1.5 m>s2 aB = aG + aB>G aB i = (aG)y j + (aG)xi - a(0.3)i or

(aG)y = 0

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T = 196 N






*17–92. FA ⫽ 100 lb

The uniform 150-lb beam is initially at rest when the forces are applied to the cables. Determine the magnitude of the acceleration of the mass center and the angular acceleration of the beam at this instant.

FB ⫽ 200 lb

A

B

12 ft

SOLUTION Equations of Motion: The mass moment of inertia of the beam about its mass center 1 1 150 ml2 = a b A 122 B = 55.90 slug # ft2. is IG = 12 12 32.2 + ©F = m(a ) ; : x G x

200 cos 60° =

150 (a ) 32.2 G x

(aG)x = 21.47 ft>s2 + c ©Fy = m(aG)y;

150 (a ) 32.2 G y

100 + 200 sin 60° - 150 = (aG)y = 26.45 ft>s2

200 sin 60°(6) - 100(6) = 55.90a or

+ ©MG = IGa;

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a = 7.857 rad>s2 = 7.86 rad>s2

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Thus, the magnitude of aG is aG = 3(aG)x2 + (aG)y2 = 321.472 + 26.452 = 34.1 ft>s2

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60⬚

17–93. B

The rocket has a weight of 20 000 lb, mass center at G, and radius of gyration about the mass center of kG = 21 ft when it is fired. Each of its two engines provides a thrust T = 50 000 lb. At a given instant, engine A suddenly fails to operate. Determine the angular acceleration of the rocket and the acceleration of its nose B.

30 ft

G

SOLUTION a + ©MG = IGa;

50 000(1.5) =

20 000 (21)2a 32.2

a = 0.2738 rad>s2 = 0.274 rad>s2 + c ©Fy = m(aG)y;

Ans.

20 000 a 32.2 G

50 000 - 20 000 =

A

aG = 48.3 ft>s2

1.5 ft

aB = aG + aB>G

1.5 ft

or

Since v = 0 laws

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aB = 48.3j - 0.2738(30)i

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aB = 2(48.3)2 + (8.214)2 = 49.0 ft>s2

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48.3 = 80.3° ub 8.214

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u = tan - 1






permitted.

Dissemination

Wide

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= 48.34j - 8.214i

T

17–94. The wheel has a weight of 30 lb and a radius of gyration of kG = 0.6 ft. If the coefficients of static and kinetic friction between the wheel and the plane are ms = 0.2 and mk = 0.15, determine the wheel’s angular acceleration as it rolls down the incline. Set u = 12°.

G 1.25 ft

SOLUTION

u

+ b©Fx = m(aG)x ; +a©Fy

= m(aG)y ;

a + ©MG = IG a;

30 ba 30 sin 12° - F = a 32.2 G N - 30 cos 12° = 0 F(1.25) = c a

30 b (0.6)2 da 32.2

Assume the wheel does not slip. aG = (1.25)a

laws

or

Solving:

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F = 1.17 lb

Dissemination

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copyright

N = 29.34 lb

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aG = 5.44 ft>s2

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a = 4.35 rad>s2 by

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Fmax = 0.2(29.34) = 5.87 lb 7 1.17 lb






17–95. The wheel has a weight of 30 lb and a radius of gyration of kG = 0.6 ft. If the coefficients of static and kinetic friction between the wheel and the plane are ms = 0.2 and mk = 0.15, determine the maximum angle u of the inclined plane so that the wheel rolls without slipping.

G 1.25 ft

SOLUTION

u

Since wheel is on the verge of slipping: + b©Fx = m(aG)x ;

30 sin u - 0.2N = a

+ a©Fy = m(aG)y ;

N - 30 cos u = 0

a + ©MC = IG a;

0.2N(1.25) = c a

30 b (1.25a) 32.2

(1) (2)

30 b (0.6)2 da 32.2

(3)

Substituting Eqs.(2) and (3) into Eq. (1),

laws

or

30 sin u - 6 cos u = 26.042 cos u

in

teaching

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30 sin u = 32.042 cos u

Ans.

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u = 46.9°






permitted.

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tan u = 1.068

*17–96. The spool has a mass of 100 kg and a radius of gyration of kG = 0.3 m. If the coefficients of static and kinetic friction at A are ms = 0.2 and mk = 0.15, respectively, determine the angular acceleration of the spool if P = 50 N.

P 250 mm

G

A


50 + FA = 100aG

+ c ©Fy = m(aG)y ;

NA - 100(9.81) = 0

c + ©MG = IG a;

50(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4a a = 1.30 rad>s2 aG = 0.520 m>s2

Ans. NA = 981 N

FA = 2.00 N

Since (FA)max = 0.2(981) = 196.2 N 7 2.00 N

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400 mm

17–97. Solve Prob. 17–96 if the cord and force P = 50 N are directed vertically upwards.

P 250 mm

G

A

SOLUTION + ©F = m(a )x; F = 100a : x G A G + c ©Fy = m(aG)y;

NA + 50 - 100(9.81) = 0

c + ©MG = IG a;

50(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4 a a = 0.500 rad>s2 aG = 0.2 m>s2

Ans. FA = 20 N

NA = 931 N

OK

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laws

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Since (FA)max = 0.2(931) = 186.2 N 7 20 N






400 mm

17–98. The spool has a mass of 100 kg and a radius of gyration kG = 0.3 m. If the coefficients of static and kinetic friction at A are ms = 0.2 and mk = 0.15, respectively, determine the angular acceleration of the spool if P = 600 N.

P 250 mm

G

A


600 + FA = 100aG

+ c ©Fy = m(aG)y;

NA - 100(9.81) = 0

c + ©MG = IG a;

600(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4a a = 15.6 rad>s2 aG = 6.24 m>s2

Ans. NA = 981 N

FA = 24.0 N

Since (FA)max = 0.2(981) = 196.2 N 7 24.0 N

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400 mm

17–99. The upper body of the crash dummy has a mass of 75 lb, a center of gravity at G, and a radius of gyration about G of kG = 0.7 ft. By means of the seat belt this body segment is assumed to be pin-connected to the seat of the car at A. If a crash causes the car to decelerate at 50 ft>s2, determine the angular velocity of the body when it has rotated to u = 30°.

θ

G

SOLUTION

1.9 ft A

75 75 (a ) R (1.9) ≤ (0.7)2 R a + B 32.2 32.2 G t

75(1.9 sin u) = B ¢

a + ©MA = ©(Mk)A;

50 ft/s2

+b aG = aA + (aG>A)n + (aG>A)t (aG)t = - 50 cos u + 0 + (a )(1.9) 142.5 sin u = 1.1413a - 221.273 cos u + 8.4084a 142.5 sin u + 221.273 cos u = 9.5497a v dv = a du v

30°

L0

laws

or

(14.922 sin u + 23.17 cos u)du

teaching

Web)

L0

v dv =

permitted.

Dissemination

Wide

copyright

in

1 2 v = - 14.922(cos 30° - cos 0°) + 23.17(sin 30° - sin 0°) 2 States

World

v = 5.21 rad s

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*17–100. A uniform rod having a weight of 10 lb is pin supported at A from a roller which rides on a horizontal track. If the rod is originally at rest, and a horizontal force of F = 15 lb is applied to the roller, determine the acceleration of the roller. Neglect the mass of the roller and its size d in the computations.

d

A

F

2 ft

SOLUTION Equations of Motion: The mass moment of inertia of the rod about its mass center is 10 1 1 ml2 = a b(2 2) = 0.1035 slug # ft2. At the instant force F is given by IG = 12 12 32.2 applied, the angular velocity of the rod v = 0. Thus, the normal component of acceleration of the mass center for the rod (aG)n = 0. Applying Eq. 17–16, we have ©Ft = m(a G)t;

15 = a

a + ©MA = ©(Mk)A ;

10 ba a G = 48.3 ft>s2 32.2 G 0 = a

10 b(48.3)(1) - 0.1035 a 32.2 laws

or

a = 144.9 rad >s2

Wide

copyright

in

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Kinematics: Since v = 0, (aG>A)n = 0. The acceleration of roller A can be obtain by analyzing the motion of points A and G. Applying Eq. 16–17, we have

instructors

States

World

permitted.

Dissemination

aG = aA + (aG>A)t + (aG>A)n the

is

of United

on

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;

and by

48.3 = aA - 144.9

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work

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+ ) (:

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use

:

not

c 48.3 d = c aA d + c 144.9(1) d + c0 d

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aA = 193 ft>s2






Ans.

17–101. Solve Prob. 17–100 assuming that the roller at A is replaced by a slider block having a negligible mass. The coefficient of kinetic friction between the block and the track is mk=0.2. Neglect the dimension d and the size of the block in the computations.

d

A

F

2 ft

SOLUTION Equations of Motion: The mass moment of inertia of the rod about its mass center is 1 1 10 ml2 = a b (2 2) = 0.1035 slug # ft2. At the instant force F is given by IG = 12 12 32.2 applied, the angular velocity of the rod v = 0. Thus, the normal component of acceleration of the mass center for the rod (aG)n = 0. Applying Eq. 17–16, we have ©Fn = m(aG)n ;

10 - N = 0

N = 10.0 lb

©Ft = m(aG)t ;

15 - 0.2(10.0) = a

a + ©MA = ©(Mk)A ;

0 = a

10 ba 32.2 G

aG = 41.86 ft>s2

laws

or

10 b (41.86)(1) - 0.1035a 32.2

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aG = aA + (aG>A)t + (aG>A)n ;

work

student

(including

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for

:

the

c 41. 8 6 d = c a A d + c125.58(1) d + C 0 D

the

41.86 = aA - 125.58 is

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aA = 167 ft>s2






permitted.

Dissemination

Kinematics: Since v = 0, (aG>A)n = 0. The acceleration of block A can be obtain by analyzing the motion of points A and G. Applying Eq. 16–17, we have

Wide

copyright

in

teaching

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a = 125.58 rad >s2

17–102. The 2-kg slender bar is supported by cord BC and then released from rest at A. Determine the initial angular acceleration of the bar and the tension in the cord.

C

30° B

A 300 mm


T cos 30° = 2(aG)x

+ c ©Fy = m(aG)y ;

T sin 30° - 19.62 = 2(aG)y

a + ©MG = IGa ;

T sin 30°(0.15) = [

1 (2)(0.3)2]a 12

aB = aG + aB>G

(aB) sin 30° = (aG)x

(+ c)

(aB) cos 30° = - (aG)y - a (0.15)

laws

+ ) (:

or

aB sin 30°i - aB cos 30°j = (aG)xi + (aG)y j + a (0.15)j

copyright

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permitted.

Dissemination

Wide

1.7321(aG)x = - (aG)y - 0.15a

learning.

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(aG)x = 2.43 m s2

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(aG)y = - 8.41 m s2

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a = 28.0 rad s2






Ans.

not

States

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T = 5.61 N

17–103. If the truck accelerates at a constant rate of 6 m>s2, starting from rest, determine the initial angular acceleration of the 20-kg ladder. The ladder can be considered as a uniform slender rod. The support at B is smooth.

C 1.5 m

B 2.5 m

60

SOLUTION Equations of Motion: We must first show that the ladder will rotate when the acceleration of the truck is 6 m>s2. This can be done by determining the minimum acceleration of the truck that will cause the ladder to lose contact at B, NB = 0. Writing the moment equation of motion about point A using Fig. a, a + ©MA = ©(Mk)A;

20(9.81) cos 60°(2) = 20amin (2 sin 60°) amin = 5.664 m>s2

teaching

Web)

laws

or

Since a min 6 6 m>s2, the ladder will in the fact rotate.The mass moment of inertia about 1 1 its mass center is IG = ml2 = (20) A 4 2 B = 26.67 kg # m2. Referring to Fig. b, 12 12 Wide of

the

not

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World

(1)

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learning.

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Kinematics: The acceleration of A is equal to that of the truck. Thus, a A = 6 m>s2 ; . Applying the relative acceleration equation and referring to Fig. c, the

(including

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aG = aA + a * rG>A - v2 rG>A

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the

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of

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(aG)x i + (aG)y j = (2 sin 60° a - 6)i + aj

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(aG)x i + (aG)y j = - 6i + ( -ak) * ( -2 cos 60° i + 2 sin 60° j) - 0

and

any

courses

Equating the i and j components, (2)

sale

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destroy

of

(a G)x = 2 sin 60° a - 6 (a G)y = a

(3)

Substituting Eqs. (2) and (3) into Eq. (1), a = 0.1092 rad>s2 = 0.109 rad>s2






Ans.

permitted.

Dissemination

- 20(a G)y (2 cos 60°) - 26.67a

in

20(9.81) cos 60°(2) = -20(a G)x (2 sin 60°)

copyright

a + ©MA = ©(Mk)A;

A

*17–104. P

If P = 30 lb, determine the angular acceleration of the 50-lb roller. Assume the roller to be a uniform cylinder and that no slipping occurs.

1.5 ft 30⬚

SOLUTION Equations of Motion: The mass moment of inertia of the roller about its mass center 1 1 50 is IG = mr2 = a b A 1.52 B = 1.7469 slug # ft2. We have 2 2 32.2 50 a 32.2 G

+ ©F = m(a ) ; : x G x

30 cos 30° - Ff =

+ c ©Fy = m(aG)y;

N - 50 - 30 sin 30° = 0

+ ©MG = IGa;

Ff(1.5) = 1.7469a

(1) N = 65 lb (2)

Since the roller rolls without slipping, aG = ar = a(1.5)

or

(3)

teaching

Web)

laws

Solving Eqs. (1) through (3) yields a = 7.436 rad>s2 = 7.44 rad>s2

Wide

copyright

in

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

aG = 11.15 ft>s2

Ff = 8.660 lb






17–105. P

If the coefficient of static friction between the 50-lb roller and the ground is ms = 0.25, determine the maximum force P that can be applied to the handle, so that roller rolls on the ground without slipping. Also, find the angular acceleration of the roller. Assume the roller to be a uniform cylinder.

1.5 ft 30⬚

SOLUTION Equations of Motion: The mass moment of inertia of the roller about its mass center 1 1 50 b A 1.52 B = 1.7469 slug # ft2. We have is IG = mr2 = a 2 2 32.2 50 a 32.2 G

+ ©F = m(a ) ; : x G x

P cos 30° - Ff =

+ c ©Fy = m(aG)y;

N - P sin 30° - 50 = 0

(2)

+ ©MG = IGa;

Ff(1.5) = 1.7469a

(3)

(1)

Since the roller is required to be on the verge of slipping, (4)

Ff = msN = 0.25N

(5) teaching

Web)

laws

or

aG = ar = a(1.5)

a = 18.93 rad>s2 = 18.9 rad>s2

Ans.

instructors

States

World

Dissemination

P = 76.37 lb = 76.4 lb

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

is the

(including

for

work

student

the

by

and

use

United

on

learning.

Ff = 22.05 lb

the

not

aG = 28.39 ft>s2

of

N = 88.18 lb






permitted.

Wide

copyright

in

Solving Eqs. (1) through (5) yields

17–106. The spool has a mass of 500 kg and a radius of gyration kG = 1.30 m. It rests on the surface of a conveyor belt for which the coefficient of static friction is ms = 0.5 and the coefficient of kinetic friction is mk = 0.4. If the conveyor accelerates at a C = 1 m>s2, determine the initial tension in the wire and the angular acceleration of the spool. The spool is originally at rest.

0.8 m G aC

SOLUTION + : a Fx = m(aG)x; + c a Fy = m(aG)y; c + a MG = IGa;

-Fs + T = 500aG Ns - 500(9.81) = 0 Fs(1.6) - T(0.8) = 500(1.30)2a ap = aG + ap>G (ap)yj = aG i - 0.8ai aG = 0.8a

or

Ns = 4905 N

teaching

Web)

laws

Assume no slipping in

ac 1 = = 1.25 rad>s 0.8 0.8

permitted.

Dissemination

Wide

Ans.

copyright

a =

is

of

the

not

instructors

States

World

aG = 0.8(1.25) = 1 m>s2

learning.

T = 2.32 kN

on United

and work

student

the

(including

for

work

solely

of

protected

this

assessing

is

the

is

This

provided

integrity

of

and

work

(Fs)max = 0.5(4.905) = 2.45 7 1.82

sale

will

their

destroy

of

and

any

courses

part

(No slipping occurs)






Ans.

use by

the

Fs = 1.82 kN Since

1.6 m

17–107. The spool has a mass of 500 kg and a radius of gyration kG = 1.30 m. It rests on the surface of a conveyor belt for which the coefficient of static friction is ms = 0.5. Determine the greatest acceleration a C of the conveyor so that the spool will not slip. Also, what are the initial tension in the wire and the angular acceleration of the spool? The spool is originally at rest.

0.8 m G aC

SOLUTION + : a Fx = m(aG)x; + c a Fy = m(aG)y; c + a MG = IGa,

T - 0.5NS = 500aG Ns - 500(9.81) = 0 0.5Ns(1.6) - T(0.8) = 500(1.30)2a

ap = aC + ap>G (ap)yj = aGi - 0.8ai aG = 0.8a

or

Solving;

teaching

Web)

laws

Ns = 4905 N T = 3.13 kN

Wide

copyright

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Ans. Ans.

instructors

States

World

permitted.

Dissemination

a = 1.684 rad>s

and

use

United

on

learning.

is

of

the

not

aG = 1.347 m>s2

for

work

student

the

by

Since no slipping

solely

of

protected

the

(including

aC = aG + aC>G

this

assessing

is

work

ac = 1.347i - (1.684) (1.6)i

any

destroy

of

and

Also,

courses

part

the

is

This

provided

integrity

of

and

work

aC = 1.35 m>s2

will

their

0.5Ns(0.8) = [500(1.30)2 + 500(0.8)2]a sale

c + a MIC = IICa; Since NS = 4905 N

a = 1.684 rad>s






1.6 m

Ans.

*17–108. The semicircular disk having a mass of 10 kg is rotating at v = 4 rad>s at the instant u = 60°. If the coefficient of static friction at A is ms = 0.5, determine if the disk slips at this instant.

v 0.4 m

O

G

u A

SOLUTION Equations of Motion:The mass moment of inertia of the semicircular disk about its center 1 of mass is given by IG = (10) A 0.4 2 B - 10 (0.16982) = 0.5118 kg # m2. From the 2 geometry, rG>A = 20.16982 + 0.4 2 - 2(0.1698) (0.4) cos 60° = 0.3477 m Also, using sin 60° sin u = law of sines, , u = 25.01°. Applying Eq. 17–16, we have 0.1698 0.3477 a + ©MA = ©(Mk)A ;

10(9.81)(0.1698 sin 60°) = 0.5118a + 10(a G)x cos 25.01°(0.3477) + 10(a G)y sin 25.01°(0.3477)

or

(1) laws

+ ©F = m(a ) ; ; x G x

Ff = 10(a G)x

teaching

Web)

(2) in

N - 10(9.81) = - 10(aG)y

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

Kinematics:Assume that the semicircular disk does not slip at A, then (a A)x = 0. Here, rG>A = {-0.3477 sin 25.01°i + 0.3477 cos 25.01°j} m = { -0.1470i + 0.3151j} m. Applying Eq. 16–18, we have

(including

for

work

student

a G = a A + a * rG>A - v2rG>A

assessing

is

work

solely

of

protected

the

-(aG)x i - (aG)y j = 6.40j + ak * ( -0.1470i + 0.3151j) - 42( -0.1470i + 0.3151j)

part

the

is and

any

courses

Equating i and j components, we have

This

provided

integrity

of

and

work

this

-(a G)x i - (aG)y j = (2.3523 - 0.3151 a) i + (1.3581 - 0.1470a)j

(4)

their

destroy

of

(a G)x = 0.3151a - 2.3523 sale

will

(a G)y = 0.1470a - 1.3581

(5)

Solving Eqs. (1), (2), (3), (4), and (5) yields: a = 13.85 rad>s2

(aG)x = 2.012 m>s2 Ff = 20.12 N

(aG)y = 0.6779 m>s2

N = 91.32 N

Since Ff 6 (Ff)max = msN = 0.5(91.32) = 45.66 N, then the semicircular disk does not slip.






Ans.

permitted.

Dissemination

Wide

(3)

copyright

+ c Fy = m(a G)y ;

4 (0.4) ——— m 3p

17–109. The 500-kg concrete culvert has a mean radius of 0.5 m. If the truck has an acceleration of 3 m>s2, determine the culvert’s angular acceleration. Assume that the culvert does not slip on the truck bed, and neglect its thickness.

0.5m

SOLUTION Equations of Motion: The mass moment of inertia of the culvert about its mass center is IG = mr 2 = 500 A 0.52 B = 125 kg # m2. Writing the moment equation of motion about point A using Fig. a, a + ©MA = ©(Mk)A ;

0 = 125a - 500a G(0.5)

(1)

Kinematics: Since the culvert does not slip at A, (aA)t = 3 m>s2. Applying the relative acceleration equation and referring to Fig. b, aG = a A + a * rG>A - v2rG>A a Gi - 3i + (a A)n j + (ak * 0.5j) - v2(0.5j)

teaching

Web)

laws

or

aGi = (3 - 0.5a)i + C (aA)n - 0.5v2 D j

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(2) not

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World

permitted.

Dissemination

a G = 3 - 0.5a

use

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on

learning.

is

of

the


work

student

the

by

and

aG = 1.5 m>s2 :

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of

and

any

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the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

a = 3 rad>s2






3 m/s2

4m

Ans.

17–110. v0

The 10-lb hoop or thin ring is given an initial angular velocity of 6 rad>s when it is placed on the surface. If the coefficient of kinetic friction between the hoop and the surface is mk = 0.3, determine the distance the hoop moves before it stops slipping.

6 in. O

SOLUTION + c ©Fy = m(aG)y ; + ©F = m(a ) ; ; x G x c + ©MG = IGa;

N -10 = 0 0.3(10) =

N = 10 lb

A

6 0.3(10) A 12 B =

10 32.2

B aG

aG = 9.66 ft>s2

10 A 32.2 B A 126 B 2 a

a = 19.32 rad>s2

When slipping ceases, vG = v r = 0.5v

(1)

v = v0 + at

(a + )

v = 6 + (- 19.32)t + B A;

(2)

or

vG = (vG)0 + aGt laws

vG = 0 + 9.66t in

teaching

Web)

(3)

Dissemination

Wide

copyright

Solving Eqs. (1) to (3) yields:

World

permitted.

v = 3 rad>s

on

learning.

is

of

the

s = s0 + (vG)0 t + 12 aG t2

and

use

United

+ B A;

not

instructors

vG = 1.5 ft>s

States

t = 0.1553 s

for

work

student

the

by

= 0 + 0 + 12 (9.66)(0.1553)2

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

= 0.116 ft = 1.40 in.






6 rad/s

Ans.

17–111. A long strip of paper is wrapped into two rolls, each having a mass of 8 kg. Roll A is pin supported about its center whereas roll B is not centrally supported. If B is brought into contact with A and released from rest, determine the initial tension in the paper between the rolls and the angular acceleration of each roll. For the calculation, assume the rolls to be approximated by cylinders.

A 90 mm 90 mm

SOLUTION For roll A. T(0.09) = 12 (8)(0.09)2 aA

c + ©MA = IA a;

(1)

For roll B a + ©MO = ©(Mk)O;

8(9.81)(0.09) = 12 (8)(0.09)2 aB + 8aB (0.09)

(2)

+ c ©Fy = m(aG)y ;

T - 8(9.81) = - 8aB

(3)

Kinematics:

laws

or

aB = aO + (aB>O)t + (aB>O)n

permitted.

Dissemination

copyright

(4)

Wide

in

teaching

Web)

c aBd = c aOd + caB (0.09) d + [0] T T T aB = aO + 0.09aB A+TB

is

of

the

not

instructors

States

World

also,

learning.

aO = aA (0.09)

and

use

United

on

(5)

the

by

A+TB

the

(including

for

work

student

Solving Eqs. (1)–(5) yields:

Ans.

assessing

is

work

solely

of

protected

aA = 43.6 rad>s2

Ans.

part

the

is and

any

courses

T = 15.7 N

This

provided

integrity

of

and

work

this

aB = 43.6 rad>s2

aO = 3.92 m>s2 sale

will

their

destroy

of

aB = 7.85 m>s2






Ans.

B

*17–112. v

The circular concrete culvert rolls with an angular velocity of v = 0.5 rad>s when the man is at the position shown. At this instant the center of gravity of the culvert and the man is located at point G, and the radius of gyration about G is kG = 3.5 ft. Determine the angular acceleration of the culvert. The combined weight of the culvert and the man is 500 lb.Assume that the culvert rolls without slipping, and the man does not move within the culvert.

4 ft O 0.5 ft

SOLUTIONS Equations of Motion: The mass moment of inertia of the system about its mass 500 (3.52) = 190.22 slug # ft2. Writing the moment equation of center is IG = mkG2= 32.2 motion about point A, Fig. a, + ©MA = ©(Mk)A ; - 500(0.5) = -

500 500 (a ) (4) (a ) (0.5) - 190.22a (1) 32.2 G x 32.2 G y

Kinematics: Since the culvert rolls without slipping, a0 = ar = a(4) : laws

or

Applying the relative acceleration equation and referrring to Fig. b, in

teaching

Web)

aG = aO + a * rG>O - v2rG>A Dissemination

Wide

copyright

(aG)xi - (aG)y j = 4ai + ( -ak) * (0.5i) - (0.52)(0.5i)

not

instructors

States

World

permitted.

(aG)xi - (aG)y j = (4a - 0.125)i - 0.5aj

and

use

United

on

learning.

is

of

the

Equation the i and j components,

(2) work

student

the

by

(aG)x = 4a - 0.125

(3)

solely

of

protected

the

(including

for

(aG)y = 0.5a

this

assessing

is

work

Subtituting Eqs. (2) and (3) into Eq. (1), provided

integrity

of

and

work

500 500 (4a - 0.125)(4) (0.5a)(0.5) - 190.22a 32.2 32.2 any

Ans.

sale

will

their

destroy

of

and

a = 0.582 rad>s2

courses

part

the

is

This

-500(0.5) = -






G

17–113. v0

The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor. Determine the initial angular acceleration of the disk and the acceleration of its mass center. The coefficient of kinetic friction between the disk and the floor is mk.

r

SOLUTION Equations of Motion. Since the disk slips, the frictional force is Ff = mkN. The mass 1 moment of inertia of the disk about its mass center is IG = mr2. We have 2 + c ©Fy = m(aG)y;

N - mg = 0

N = mg

+ ©F = m(a ) ; ; x G x

mk(mg) = maG

aG = mkg ;

+ ©MG = IGa;

1 -mk(mg)r = a mr2 ba 2

a =

Ans.

2mkg r

sale

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of

and

any

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part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.






17–114. v0

The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor. Determine the time before it starts to roll without slipping. What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the disk and the floor is mk.

r

SOLUTION Equations of Motion: Since the disk slips, the frictional force is Ff = mkN. The mass 1 moment of inertia of the disk about its mass center is IG = mr2. 2 + c ©Fy = m(aG)y; N - mg = 0 N = mg + ©F = m(a ) ; ; x G x

mk(mg) = maG

+ ©MG = IGa;

1 -mk(mg)r = - a m r2 b a 2

aG = mkg 2mkg r

a =

Kinematics: At the instant when the disk rolls without slipping, vG = vr. Thus, or

vG = (vG)0 + aGt teaching

Web)

vr mkg

in

(1) Wide

permitted.

Dissemination

t =

laws

vr = 0 + mkgt copyright

+ B A;

on

learning.

is

of

the

not

instructors

States

World

and

the

by

and

use

United

v = v0 + at

work

student

2mkg bt r

the

(including

for

(2)

assessing

is

work

solely

of

v = v0 + a -

protected

(a + )

provided

integrity

of

and

work

this

Solving Eqs. (1) and (2) yields v0r 3mkg

the part any

courses sale

will

their

destroy

of

t =

is

This

1 v 3 0

and

v =






Ans.

17–115. ω

The 16-lb bowling ball is cast horizontally onto a lane such that initially v = 0 and its mass center has a velocity v = 8 ft>s. If the coefficient of kinetic friction between the lane and the ball is mk = 0.12, determine the distance the ball travels before it rolls without slipping. For the calculation, neglect the finger holes in the ball and assume the ball has a uniform density.

8 ft/s G 0.375 ft

SOLUTION 16 a 32.2 G

+ ©F = m(a ) ; : x G x

0.12NA =

+ c ©Fy = m(aG)y ;

NA - 16 = 0

a + ©MG = IG a;

2 16 0.12NA(0.375) = c a b (0.375)2 da 5 32.2

Solving, NA = 16 lb;

aG = 3.864 ft>s2;

a = 25.76 rad>s2

laws

or

When the ball rolls without slipping v = v(0.375), Web)

v = v0 + ac t in

teaching

(a+)

States

World

permitted.

Dissemination

Wide

copyright

v = 0 + 25.76t 0.375

learning.

is

of

the

not

instructors

v = 9.660t by

and

use

United

on

v = v0 + ac t

the

+ B A;

the

(including

for

work

student

9.660t = 8 - 3.864t

integrity

of

and

work

provided

the part

1 2 a t 2 c

is

s = s0 + v0 t +

This

+ B A;

this

assessing

is

work

solely

of

protected

t = 0.592 s

destroy

of

and

any

courses

1 (3.864)(0.592)2 2 sale

will

their

s = 0 + 8(0.592) s = 4.06 ft






Ans.

*17–116. The uniform beam has a weight W. If it is originally at rest while being supported at A and B by cables, determine the tension in cable A if cable B suddenly fails. Assume the beam is a slender rod.

A

B

SOLUTION + c ©Fy = m(aG)y; c + ©MA = IAa;

Wa

L 1 W W L L b = c a b L2 d a + a b aa b g 4 4 12 g 4 1 =

L 1 L a + ba g 4 3

L b. 4 or

12 g a b 7 L

teaching

Web)

a =

laws

Since aG = a a

L –– 4

W TA - W = - a G g

in

g W 12 L W L (a)a b = W a ba ba b g g 7 4 L 4 States

World

permitted.

Dissemination

Wide

copyright

TA = W -

not

instructors

4 W 7

United

on

learning.

is

of

the

Ans.

the

by

and

use

TA =

the

(including

for

work

student

Also, W a g G

work

solely

of

protected

integrity provided

L 1 W b = c a b L2 d a 4 12 g

part

the

is

and

any

courses

TA a

This

c + ©MG = IG a;

of

and

work

this

assessing

TA - W = -

is

+ c ©Fy = m(aG)y ;

will

their

destroy

of

L a 4

sale

Since a G =

TA =

1 W a b La 3 g

W L 1 W a b La - W = - a b a g 4 3 g a =






12 g a b 7 L

TA =

g 1 W 12 a bLa b a b 3 g 7 L

TA =

4 W 7

Ans.

L –– 2

L –– 4

17–117. A cord C is wrapped around each of the two 10-kg disks. If they are released from rest, determine the tension in the fixed cord D. Neglect the mass of the cord.

D

A 90 mm C

SOLUTION For A: B

c + a MA = IA aA;

90 mm

1 T(0.09) = c (10)(0.09)2 d aA 2

(1)

1 T(0.09) = c (10)(0.09)2 d aB 2

(2)

For B: a + a MB = IBaB;

10(9.81) - T = 10aB

(3) or

+ T a Fy = m(aB)y;

teaching

Web)

laws

aB = aP + (aB>P)t + (aB>P)n

(4)

Dissemination

Wide

copyright

in

(+ T)aB = 0.09aA + 0.09aB + 0

not

instructors

States

World

permitted.

Solving.

use

United

on

learning.

is

of

the

aB = 7.85 m>s2

for

work

student

the

by

and

aA = 43.6 rad>s2

solely

of

protected

the

(including

aB = 43.6 rad>s2 work

T = 19.6 N work

this

assessing

is

Ans.

any sale

will

their

destroy

of

and

= 118 N

courses

part

the

is

This

provided

integrity

of

and

Ay = 10(9.81) + 19.62






17–118. The 500-lb beam is supported at A and B when it is subjected to a force of 1000 lb as shown. If the pin support at A suddenly fails, determine the beam’s initial angular acceleration and the force of the roller support on the beam. For the calculation, assume that the beam is a slender rod so that its thickness can be neglected.

1000 lb 5

3

4

B 8 ft

SOLUTION + ; a Fx = m(aG)x ;

500 4 1000 a b = (a ) 5 32.2 G x

+ T a Fy = m(aG)y ;

3 500 1000 a b + 500 - By = (a ) 5 32.2 G y

a + a MB = a (Mk)B;

3 500 1 500 500(3) + 1000 a b (8) = (aG)y(3) + c a b (10)2 d a 5 32.2 12 32.2

aB = aG + aB>G - aBi = - (aG)x i - (aG)yj + a(3)j or

(aG)y = a(3)

laws

(+ T)

Web)

a = 23.4 rad>s2

Wide

copyright

in

teaching

Ans.

By = 9.62 lb States

World

permitted.

Dissemination

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

By 7 0 means that the beam stays in contact with the roller support.






A 2 ft

17–119. The 30-kg uniform slender rod AB rests in the position shown when the couple moment of M = 150 N # m is applied. Determine the initial angular acceleration of the rod. Neglect the mass of the rollers.

A

0.75 m

SOLUTION M ⫽ 150 N⭈m

Equations of Motion: Here, the mass moment of inertia of the rod about its mass 1 2 1 center is IG = ml = (30)(1.52) = 5.625 kg # m2. Writing the moment 12 12 equations of motion about the intersection point A of the lines of action of NA and

0.75 m

NB and using, Fig. a, B

+ ©MA = ©(Mk)A;

-150 = 30(aG)x(0.75) - 5.625a (1)

5.625a - 22.5(aG)x = 150

Kinematics: Applying the relative acceleration equation to points A and G, Fig. b,

laws

or

aG = aA + a * rG>A - v2rG>A teaching

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(aG)xi + (aG)y j = - aAj + ( - ak) * (-0.75j) - 0

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(aG)xi + (aG)y j = - 0.75ai - aA j

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a = 6.667 rad>s2 = 6.67 rad>s2






Ans.

*17–120. The 30-kg slender rod AB rests in the position shown when the horizontal force P = 50 N is applied. Determine the initial angular acceleration of the rod. Neglect the mass of the rollers.

A

SOLUTION 1.5 m

Equations of Motion: Here, the mass moment of inertia of the rod about its mass 1 2 1 ml = (30)(1.52) = 5.625 kg # m2. Writing the moment equations center is IG = 12 12 of motion about the intersection point A of the lines of action of NA and NB and using, Fig. a, + ©MA = ©(Mk)A;

B

-50(0.15) = 30(aG)x(0.75) - 5.625a

P ⫽ 50 N

(1)

5.625a - 22.5(aG)x = 75

Kinematics: Applying the relative acceleration equation to points A and G, Fig. b,

or

aG = aA + a * rG>A - v2rG>A

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laws

(aG)xi + (aG)y j = - aA j + ( - ak) * (-0.75j) - 0

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(aG)xi + (aG)y j = - 0.75ai - aA j

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a = 3.333 rad>s2 = 3.33 rad>s2






Ans.

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Dissemination


18–1. At a given instant the body of mass m has an angular velocity V and its mass center has a velocity vG. Show that its kinetic energy can be represented as T = 12IICv2, where IIC is the moment of inertia of the body determined about the instantaneous axis of zero velocity, located a distance rG>IC from the mass center as shown.

IC rG/IC

G vG

SOLUTION T =

1 1 my2G + IG v2 2 2 1 1 m(vrG>IC)2 + IG v2 2 2

=

1 A mr2G>IC + IG B v2 2

=

1 I v2 2 IC

However mr2G>IC + IG = IIC Q.E.D.

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=

where yG = vrG>IC






V

18–2. The wheel is made from a 5-kg thin ring and two 2-kg slender rods. If the torsional spring attached to the wheel’s center has a stiffness k = 2 N # m>rad, and the wheel is rotated until the torque M = 25 N # m is developed, determine the maximum angular velocity of the wheel if it is released from rest.

0.5 m O

M

SOLUTION Kinetic Energy and Work: The mass moment of inertia of the wheel about point O is IO = mRr 2 + 2 ¢

1 m l2 ≤ 12 r

= 5(0.52) + 2c

1 (2)(12) d 12

= 1.5833 kg # m2 Thus, the kinetic energy of the wheel is

or

1 1 I v2 = (1.5833) v2 = 0.79167 v2 2 O 2

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T =

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Since the wheel is released from rest, T1 = 0. The torque developed is M = ku = 2u. Here, the angle of rotation needed to develop a torque of M = 25 N # m is

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u = 12.5 rad

2u = 25

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The wheel achieves its maximum angular velocity when the spacing is unwound that M is when the wheel has rotated u = 12.5 rad. Thus, the work done by q is the

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2u du

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= 156.25 J

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12.5 rad 2

T1 + © u 1 - 2 = T2 0 + 156.25 = 0.79167 v2 v = 14.0 rad/s






Ans.

18–3. The wheel is made from a 5-kg thin ring and two 2-kg slender rods. If the torsional spring attached to the wheel’s center has a stiffness k = 2 N # m>rad, so that the torque on the center of the wheel is M = 12u2 N # m, where u is in radians, determine the maximum angular velocity of the wheel if it is rotated two revolutions and then released from rest.

0.5 m O

M

SOLUTION Io = 2c

1 (2)(1)2 d + 5(0.5)2 = 1.583 12

T1 + ©U1 - 2 = T2 4p

0 +

L0

2u du =

1 (1.583) v2 2

(4p)2 = 0.7917v2 v = 14.1 rad/s

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Ans.






*18–4. The 50-kg flywheel has a radius of gyration of k0 = 200 mm about its center of mass. If it is subjected to a torque of M = (9u1>2) N # m, where u is in radians, determine its angular velocity when it has rotated 5 revolutions, starting from rest.

M ⫽ (9 u1/2) N⭈m

O

SOLUTION Kinetic Energy and Work: The mass moment inertia of the flywheel about its mass center is IO = mkO2= 50(0.22) = 2 kg # m2. Thus, the kinetic energy of the flywheel is T =

1 1 I v2 = (2)v2 = v2 2 O 2

Since the wheel is initially at rest, T1 = 0 . Referring to Fig. a, W, Ox, and Oy do no work while M does positive work. When the wheel rotates u = (5 rev) ¢

2p rad ≤ = 10p, the work done by M is 1 rev 10p or teaching

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10p 0

Wide Dissemination

= 6u3>2 `

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Mdu =

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T1 + ©U1 - 2 = T2

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0 + 1056.52 = v2

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v = 32.5 rad>s






Ans.

18–5. The spool has a mass of 60 kg and a radius of gyration kG = 0.3 m. If it is released from rest, determine how far its center descends down the smooth plane before it attains an angular velocity of v = 6 rad>s. Neglect friction and the mass of the cord which is wound around the central core.

0.3 m

SOLUTION

0.5 m

G

T1 + ©U1 - 2 = T2 0 + 60(9.81) sin 30°(s) =

1 1 C 60(0.3)2 D (6)2 + (60) C 0.3(6) D 2 2 2

A 30⬚

s = 0.661 m

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Ans.






18–6. Solve Prob. 18–5 if the coefficient of kinetic friction between the spool and plane at A is mk = 0.2.

0.3 m

SOLUTION

0.5 m

G

sG sA = 0.3 (0.5 - 0.3) A

sA = 0.6667sG +a©Fy

= 0;

30

NA - 60(9.81) cos 30° = 0 NA = 509.7 N T1 + ©U1 - 2 = T2 1 C 60(0.3)2 D (6)2 2 or

0 + 60(9.81) sin 30°(sG) - 0.2(509.7)(0.6667sG) =

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1 (60) C (0.3)(6) D 2 2 in

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sG = 0.859 m

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Dissemination

Ans.






18–7. v

The double pulley consists of two parts that are attached to one another. It has a weight of 50 lb and a radius of gyration about its center of kO = 0.6 ft. If it rotates with an angular velocity of 20 rad>s clockwise, determine the kinetic energy of the system. Assume that neither cable slips on the pulley.

20 rad/s

0.5ft

1 ft O

SOLUTION T =

1 1 1 I v2 + mA v2A + mB v2B 2 O O 2 2

T =

1 50 1 20 1 30 a (0.6)2 b (20)2 + a b C (20)(1) D 2 + a b C (20)(0.5) D 2 2 32.2 2 32.2 2 32.2

B 30 lb A 20 lb

= 283 ft # lb

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Ans.






*18–8. v ⫽ 20 rad/s

The double pulley consists of two parts that are attached to one another. It has a weight of 50 lb and a centroidal radius of gyration of kO = 0.6 ft and is turning with an angular velocity of 20 rad> s clockwise. Determine the angular velocity of the pulley at the instant the 20-lb weight moves 2 ft downward.

0.5 ft

1 ft O

SOLUTION Kinetic Energy and Work: Since the pulley rotates about a fixed axis, vA = vrA = v(1) and vB = vrB = v(0.5). The mass moment of inertia of the B 30 lb

50 ≤ (0.62) = 0.5590 slug # ft2. Thus, the 32.2

pulley about point O is IO = mkO 2 = ¢

A 20 lb

kinetic energy of the system is T =

1 1 20 1 30 (0.5590)v2 + ¢ ≤ [v(1)]2 + ¢ ≤ [v(0.5)]2 2 2 32.2 2 32.2 or

=

1 1 1 I v2 + mAvA2 + mBvB2 2 O 2 2

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Web)

laws

= 0.7065v2

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Dissemination

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copyright

in

Thus, T1 = 0.7065(202) = 282.61 ft # lb. Referring to the FBD of the system shown in Fig. a, we notice that Ox, Oy, and Wp do no work while WA does positive work and WB does negative work. When A moves 2 ft downward, the pulley rotates

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SA SB = rA rB

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SB 2 = 1 0.5

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SB = 2(0.5) = 1 ft c

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Thus, the work of WA and WB are

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UWA = WA SA = 20(2) = 40 ft # lb

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UWB = - WB SB = - 30(1) = - 30 ft # lb Principle of Work and Energy: T1 + U1 - 2 = T2 282.61 + [40 + ( -30)] = 0.7065 v2 v = 20.4 rad>s






Ans.

18–9. 400 mm

If the cable is subjected to force of P = 300 N, and the spool starts from rest, determine its angular velocity after its center of mass O has moved 1.5 m. The mass of the spool is 100 kg and its radius of gyration about its center of mass is kO = 275 mm. Assume that the spool rolls without slipping.

P ⫽ 300 N

O 200 mm

SOLUTION Kinetic Energy and Work: Referring to Fig. a, we have vO = vrO>IC = v(0.4) The mass moment of inertia of the spool about its mass center is IO = mkO2 = 100(0.2752) = 7.5625 kg # m2. Thus, the kinetic energy of the spool is 1 1 mvO2+ IOv2 2 2 1 1 = (100)[v(0.4)]2 + (7.5625)v2 2 2

T =

= 11.78125v2

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UP = Psp = 300(2.25) = 675 J

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T1 + ©U1 - 2 = T2

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v = 7.57 rad>s

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0 + 675 = 11.78125v2






Ans.

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Dissemination

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work P does positive work. When the center O of the spool moves to the right by rP>IC 0.6 SO = 1.5 m, P displaces sP = s = ¢ ≤ (1.5) = 2.25 m. Thus, the work done rO> IC O 0.4 by P is

Web)

laws

or

Since the spool is initially at rest, T1 = 0. Referring to Fig. b, W, N, and Ff do no

18–10. The two tugboats each exert a constant force F on the ship. These forces are always directed perpendicular to the ship’s centerline. If the ship has a mass m and a radius of gyration about its center of mass G of kG, determine the angular velocity of the ship after it turns 90°. The ship is originally at rest.

F

G

SOLUTION Principle of Work and Energy: The two tugboats create a couple moment of p M = Fd to rotate the ship through an angular displacement of u = rad. The mass 2 moment of inertia about its mass center is IG = mk2G. Applying Eq. 18–14, we have

–F

T1 + a U 1 - 2 = T2 0 + Mu =

1 I v2 2 G or

1 p b = A mk2G B v2 2 2 in

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0 + Fda

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Dissemination

Wide

pFd 1 kG A m

copyright

v =






d

18–11. At the instant shown, link AB has an angular velocity vAB = 2 rad>s. If each link is considered as a uniform slender bar with a weight of 0.5 lb>in., determine the total kinetic energy of the system.

vAB

A

2 rad/s

3 in. 4 in.

C

SOLUTION vBC =

6 = 1.5 rad>s 4

B 5 in.

vC = 1.5(4 22) = 8.4853 in.>s 45

rIC - G = 2(2)2 + (4)2 = 4.472 vG = 1.5(4.472) = 6.7082 in.>s

3 2 4 2 1 4(0.5) 6.7082 2 1 1 4(0.5) 1 1 3(0.5) c a b a b d (2)2 + c da b + c a b a b d(1.5)2 2 3 32.2 12 2 32.2 12 2 12 32.2 12 laws

T =

8.4853 = 1.697 rad>s 5

or

vDC =

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5 2 1 1 5(0.5) c a b a b d (1.697)2 = 0.0188 ft # lb 2 3 32.2 12

in

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D

*18–12. Determine the velocity of the 50-kg cylinder after it has descended a distance of 2 m. Initially, the system is at rest. The reel has a mass of 25 kg and a radius of gyration about its center of mass A of kA = 125 mm.

A

SOLUTION T1 + ©U1 - 2 = T2 0 + 50(9.81)(2) =

2 1 v [(25)(0.125)2] ¢ ≤ 2 0.075

+

1 (50) v2 2

v = 4.05 m>s

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75 mm

18–13. The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of kA = 6 in. If pulley B attached to the motor is subjected to a torque of M = 40(2 - e -0.1u) lb # ft, where u is in radians, determine the velocity of the 200-lb crate after it has moved upwards a distance of 5 ft, starting from rest. Neglect the mass of pulley B.

7.5 in.

M

SOLUTION Kinetic Energy and Work: Since the wheel rotates about a fixed axis , vC = vrC = v(0.375). The mass moment of inertia of A about its mass center is 50 b A 0.52 B = 0.3882 slug # ft2. Thus, the kinetic energy of the IA = mkA 2 = a 32.2 system is

1 1 200 (0.3882)v2 + a b C v(0.375) D 2 2 2 32.2

teaching

Web)

=

laws

1 1 IA v2 + mC vC 2 2 2 or

T = TA + TC =

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copyright

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= 0.6308v2

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of

0

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= 2280.93 ft # lb

33.33 rad

their

= c 40 A 2u + 10e - 0.1u B d 2

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and

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40 A 2 - e - 0.1u B du

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this

33.33 rad

MduB =

UWC = - WC sC = - 200(5) = - 1000 ft # lb Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [2280.93 - 1000] = 0.6308v2 v = 45.06 rad>s Thus, vC = 45.06(0.375) = 16.9 ft>s c






Ans.

permitted.

Dissemination

Since the system is initially at rest, T1 = 0. Referring to Fig. b, Ax, Ay, and WA do no work, M does positive work, and WC does negative work. When crate C moves 5 ft sC 5 = 13.333 rad. Then, = upward, wheel A rotates through an angle of uA = r 0.375 rA 0.625 b (13.333) = 33.33 rad u = a pulley B rotates through an angle of uB = rB A 0.25 . Thus, the work done by M and WC is UM =

A

3 in. B

4.5 in.

18–14. The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of kA = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb # ft, determine the velocity of the 200-lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

7.5 in.

M

SOLUTION Kinetic Energy and Work: Since the wheel at A rotates about a fixed axis, vC = vrC = v(0.375). The mass moment of inertia of wheel A about its mass center 50 b A 0.52 B = 0.3882 slug # ft2. Thus, the kinetic energy of the is IA = mkA 2 = a 32.2 system is

1 1 200 (0.3882)v2 + a b C v(0.375) D 2 2 2 32.2

Web)

=

laws

1 1 IA v2 + mC vC 2 2 2

or

T = TA + TC =

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copyright

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= 0.6308v2

of

the

is

2p rad b = 10p rad, the wheel rotates through an angle of 1 rev rB 0.25 uA = uB = a b (10p) = 4p. Thus, the crate displaces upwards through a rA 0.625 distance of sC = rC uA = 0.375(4p) = 1.5p ft. Thus, the work done by M and WC is assessing

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uB = (5 rev)a

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UWC = -WC sC = - 200(1.5p) = -300p ft # lb

T1 + ©U1 - 2 = T2 0 + [500p - 300p] = 0.6308v2 v = 31.56 rad>s Thus, vC = 31.56(0.375) = 11.8 ft>s c






Ans.

not

instructors

States

World

permitted.

Dissemination

Since the system is initially at rest, T1 = 0. Referring to Fig. b, Ax, Ay, and WA do no work, M does positive work, and WC does negative work. When pulley B rotates

UM = MuB = 50(10p) = 500p ft # lb

A

3 in. B

4.5 in.

18–15. The 50-kg gear has a radius of gyration of 125 mm about its center of mass O. If gear rack B is stationary, while the 25-kg gear rack C is subjected to a horizontal force of P = 150 N, determine the speed of C after the gear’s center O has moved to the right a distance of 0.3 m, starting from rest.

C

P ⫽ 150 N 150 mm O B

SOLUTION Kinetic Energy and Work: Referring to Fig. a, vC vC v = = = 3.333vC rC>IC 0.3 Then, vO = vrO>IC = (3.333vC)(0.15) = 0.5vC The mass moment of inertia of the gear about its mass center is IO = mkO2 = 50(0.1252) = 0.78125 kg # m2. Thus, the kinetic energy of the system is T = TA + TC

teaching

Web)

laws

or

1 1 1 = c mAvO 2 + IOv 2 d + mC vC2 2 2 2

States

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permitted.

Dissemination

Wide

copyright

in

1 1 1 = c (50)(0.5vC)2 + (0.78125)(3.333vC)2 d + (25) vC2 2 2 2

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Since the system is initially at rest, T1 = 0. Referring to Fig. b, WC, WA, F, and N do of

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right through a distance of sO = 0.3 m, P displaces horizontally through a distance rC>IC 0.3 sO = a b(0.3) = 0.6 m. Thus, the work done by P is of sC = rO>IC 0.15

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UP = PsD = 150(0.6) = 90 J

T1 + ©U1-2 = T2 0 + 90 = 23.090vC 2 vC = 1.97 m>s






*18–16. Gear B is rigidly attached to drum A and is supported by two small rollers at E and D. Gear B is in mesh with gear C and is subjected to a torque of M = 50 N # m. Determine the angular velocity of the drum after C has rotated 10 revolutions, starting from rest. Gear B and the drum have 100 kg and a radius of gyration about their rotating axis of 250 mm. Gear C has a mass of 30 kg and a radius of gyration about its rotating axis of 125 mm.

300 mm A

E 150 mm

SOLUTION Kinetic Energy and Work: Since gear B is in mesh with gear C and both gears rotate rB 0.2 b v = 1.333vA. The mass moment of the about fixed axes, vC = a b vA = a rC 0.15 A drum and gear C about their rotating axes are IA = mAk2 = 100(0.252) = 6.25 kg # m2 and IC = mCk2 = 30(0.1252) = 0.46875 kg # m2. Thus, the kinetic energy of the system is

1 1 (6.25)vA2 + (0.46875)(1.333vA)2 2 2

or

=

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1 1 I v 2 + ICvC 2 2 A A 2 laws

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= 3.5417vA2

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vA = 29.8 rad>s






Ans.

not

instructors

Since the system is initially at rest, T1 = 0. Referring to Fig. a, M does positive work. 2p rad b = 20p, the work done by M is When the gear C rotates u = (10 rev)a 1 rev

D C

M ⫽ 50 N⭈m

T = TA + TC

0 + 1000p = 3.5417vA2

B

200 mm

18–17. The center O of the thin ring of mass m is given an angular velocity of v0. If the ring rolls without slipping, determine its angular velocity after it has traveled a distance of s down the plane. Neglect its thickness.

v0 s r O

SOLUTION u

T1 + ©U1-2 = T2 1 1 (mr2 + mr2)v0 2 + mg(s sin u) = (mr2 + mr2)v2 2 2 g v = v0 2 + 2 s sin u A r

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Ans.






18–18. If the end of the cord is subjected to a force of P = 75 lb, determine the speed of the 100-lb block C after P has moved a distance of 4 ft, starting from rest. Pulleys A and B are identical, each of which has a weight of 10 lb and a radius of gyration of k = 3 in. about its center of mass.

4 in. B

A 4 in. P = 75 lb

SOLUTION C

Kinetic Energy and Work: Referring to Fig. a, we have vD = vArD>IC = vA(0.6667) (vG)A = vC = vArC>IC = vA(0.3333) Since pulley B rotates about a fixed axis, its angular velocity is vB =

vA(0.6667) vD = = 2vA rB 0.3333

The mass moment of inertia of pulleys A and B about their resperctive mass centers or

10 3 2 ≤ ¢ ≤ = 0.01941 slug # ft2. Thus, the kinetic 32.2 12 teaching

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are (IA)G = (IB)G = mk2 = ¢

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T = TA + TB + TC

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1 1 1 1 = B mA(vG)A2 + (IG)AvAR2 + (IG)BvB2+ mCvC2 2 2 2 2

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Since the system is initially at rest, T1 = 0. Referring to Fig. b, R 1, R 2, and WB do no work, P does positive work, and WA and WC do negative work. When P moves sD = 4 ft downward, the center of the pulley moves upward through a distance of rC>IC 0.3333 sC = s = (4) = 2 ft. Thus, the work done by WA, WC, and P is rD>IC D 0.6667 UWA = - WAsC = - 10(2) = - 20 ft # lb UWC = - WCsC = - 100(2) = - 200 ft # lb UP = PsD = 75(4) = 300 ft # lb Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [- 20 + ( - 200) + 300] = 0.2383vA2 vA = 18.32 rad>s Thus, vC = 18.32(0.3333) = 6.11 ft>s c © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

18–19. When u = 0°, the assembly is held at rest, and the torsional spring is untwisted. If the assembly is released and falls downward, determine its angular velocity at the instant u = 90°. Rod AB has a mass of 6 kg, and disk C has a mass of 9 kg.

C 450 mm B k ⫽ 20 N⭈m/rad u

SOLUTION

A

Kinetic Energy and Work: Since the rod rotates about a fixed axis, (vG)AB = vrGAB = v(0.225) and (vG)C = vrGC = v(0.525). The mass moment of the rod 1 1 and the disk about their respective mass centers are (IAB)G = ml2 = (6)(0.452) 12 12 1 1 = 0.10125 kg # m2 and (IC)G = mr2 = (9)(0.0752) = 0.0253125 kg # m2. Thus, 2 2 the kinetic energy of the pendulum is 1 1 T = © mvG 2 + IGv 2 2 2

or

1 1 1 1 = c (6)[v(0.225)]2 + (0.10125)v2 d + c (9)[v(0.525)]2 + (0.0253125)v2 d 2 2 2 2

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= 1.4555v2 in

1 1 I v2, where IO = c (6)(0.452) 2 O 12 Dissemination

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1 + 6(0.2252) d + c (9)(0.0752) + 9(0.5252) d = 2.9109 kg # m2. Thus, 2

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Since the pendulum is initially at rest, T1 = 0. Referring to Fig. a, Ox and Oy do no work, WC and WAB do positive work, and M does negative work. When u = 90°, WAB and WC displace vertically through distances of hAB = 0.225 m and hC = 0.525 m. Thus, the work done by WAB, WC, and M is

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UWAB = WABhAB = 6(9.81)(0.225) = 13.24 J

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UWC = WChC = 9(9.81)(0.525) = 46.35 J p>2

UM = -

L

Mdu = -

L0

20udu = - 24.67 J

Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [13.24 + 46.35 + ( - 24.67)] = 1.4555v2 v = 4.90 rad>s






Ans.

75 mm

*18–20. If P = 200 N and the 15-kg uniform slender rod starts from rest at u = 0°, determine the rod’s angular velocity at the instant just before u = 45°.

600 mm A 45°

u

P ⫽ 200 N B

SOLUTION Kinetic Energy and Work: Referring to Fig. a, rA>IC = 0.6 tan 45° = 0.6 m Then rG>IC = 30.32 + 0.62 = 0.6708 m

Thus,

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(vG)2 = v2rG>IC = v2(0.6708)

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1 The mass moment of inertia of the rod about its mass center is IG = ml2 12 1 = (15)(0.62) = 0.45 kg # m2. Thus, the final kinetic energy is 12

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1 1 m(vG)22 + IG v2 2 2 2

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1 1 (15)[w2(0.6708)]2 + (0.45) v2 2 2 2

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= 3.6v2 2

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Since the rod is initially at rest, T1 = 0. Referring to Fig. b, NA and NB do no work, while P does positive work and W does negative work. When u = 45°, P displaces through a horizontal distance sP = 0.6 m and W displaces vertically upwards through a distance of h = 0.3 sin 45°, Fig. c. Thus, the work done by P and W is UP = PsP = 200(0.6) = 120 J UW = - Wh = - 15(9.81)(0.3 sin 45°) = - 31.22 J Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [120 - 31.22] = 3.6v22 v2 = 4.97 rad>s






Ans.

18–21. A yo-yo has a weight of 0.3 lb and a radius of gyration kO = 0.06 ft. If it is released from rest, determine how far it must descend in order to attain an angular velocity v = 70 rad>s. Neglect the mass of the string and assume that the string is wound around the central peg such that the mean radius at which it unravels is r = 0.02 ft.

SOLUTION r

vG = (0.02)70 = 1.40 ft>s T1 + ©U1 - 2 = T2 0 + (0.3)(s) =

1 0.3 1 0.3 a b(1.40)2 + c(0.06)2 a b d(70)2 2 32.2 2 32.2

s = 0.304 ft

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O

18–22. If the 50-lb bucket is released from rest, determine its velocity after it has fallen a distance of 10 ft. The windlass A can be considered as a 30-lb cylinder, while the spokes are slender rods, each having a weight of 2 lb. Neglect the pulley’s weight.

B

3 ft

4 ft 0.5 ft

SOLUTION Kinetic Energy and Work: Since the windlass rotates about a fixed axis, vC = vArA vC vC = or vA = = 2vC. The mass moment of inertia of the windlass about its rA 0.5 mass center is IA =

C

2 1 30 1 2 a b A 0.52 B + 4 c a b A 0.52 B + A 0.752 B d = 0.2614 slug # ft2 2 32.2 12 32.2 32.2

Thus, the kinetic energy of the system is

or

T = TA + T C 1 1 I v 2 + m C vC 2 2 A 2

=

1 1 50 (0.2614)(2vC)2 + a bv 2 2 2 32.2 C instructors

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Since the system is initially at rest, T1 = 0. Referring to Fig. a, WA, Ax, Ay, and RB do no work, while WC does positive work. Thus, the work done by WC, when it displaces vertically downward through a distance of sC = 10 ft, is

vC = 19.6 ft>s






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UWC = WCsC = 50(10) = 500 ft # lb

Ans.

A

0.5 ft

18–23. The combined weight of the load and the platform is 200 lb, with the center of gravity located at G. If a couple moment of M = 900 lb # ft is applied to link AB, determine the angular velocity of links AB and CD at the instant u = 60°. The system is at rest when u = 0°. Neglect the weight of the links.

1 ft

G

M ⫽ 900 lb⭈ft

4 ft

A 2 ft

B

C

3 ft u D

SOLUTION Kinetic Energy and Work: Since the weight of the links are negligible and the crate and platform undergo curvilinear translation, the kinetic energy of the system is T =

1 1 200 b v 2 = 3.1056vG2 mvG2 = a 2 2 32.2 G

Since the system is initially at rest, T1 = 0. Referring to Fig. a, Bx, By, Dx, and Dy do no work while M does positive work and W does negative work. When u = 60°, W displaces upward through a distance of h = 4 sin 60° ft = 3.464 ft. Thus, the work done by M and W is

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p UM = Mu = 900 a b = 300p ft # lb 3

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UW = - Wh = - 200(3.464) = - 692.82 ft # lb

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T1 + ©U1-2 = T2

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0 + [300p - 692.82] = 3.1056vG2 by

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vG 8.966 = = 2.24 rad>s r 4

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vAB = vCD =






Ans.

*18–24. The tub of the mixer has a weight of 70 lb and a radius of gyration kG = 1.3 ft about its center of gravity. If a constant torque M = 60 lb # ft is applied to the dumping wheel, determine the angular velocity of the tub when it has rotated u = 90°. Originally the tub is at rest when u = 0°.

θ 0.8 ft

SOLUTION G

T1 + ©U1 - 2 = T2 M

1 70 1 70 p )(1.3)2 d (v)2 + [ ] (0.8v)2 0 + 60( ) - 70(0.8) = c ( 2 2 32.2 2 32.2 v = 3.89 rad>s

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18–25. The tub of the mixer has a weight of 70 lb and a radius of gyration kG = 1.3 ft about its center of gravity. If a constant torque M = 60 lb # ft is applied to the tub, determine its angular velocity when it has rotated u = 45°. Originally the tub is at rest when u = 0°.

u 0.8 ft

SOLUTION

G

Kinetic Energy and Work: The mass moment of inertia of the tub about point O is M

IO = mkG2+ mrG2 =

70 70 (1.32) + (0.82) 32.2 32.2

= 5.0652 slug # ft2 Thus, kinetic energy of the tub is T =

1 1 I v2 = (5.0652)v2 = 2.5326 v2 2 O 2

in

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Initially, the tub is at rest. Thus, T1 = 0. Referring to the FBD of the tub, Fig. a, we notice that Ox and Oy do no work while M does positive work and W does negative work. Thus, the work done by M and W are

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p UM = Mu = 60 a b = 15p ft # lb 4

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UW = - Wh = - 70 [0.8 (1 - cos 45°)] = - 16.40 ft # lb

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0 + [15p + ( - 16.40)] = 2.5326 v2

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v = 3.48 rad>s






18–26. Two wheels of negligible weight are mounted at corners A and B of the rectangular 75-lb plate. If the plate is released from rest at u = 90°, determine its angular velocity at the instant just before u = 0°.

A 3 ft 1.5 ft

u

SOLUTION

B

Kinetic Energy and Work: Referring Fig. a, (vG)2 = vrA>IC = v a 20.752 + 1.52 b = 1.677v2 The mass moment of inertia of the plate about its mass center is 1 1 75 b (1.52 + 32) = 2.1836 slug # ft2. Thus, the final IG = m(a2 + b2) = a 12 12 32.2 kinetic energy is

or

1 75 1 a b (1.677v2)2 + IG (2.1836)v22 2 32.2 2

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1 1 m(vG)22 + v22 2 2 laws

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= 4.3672v2 2

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v2 = 5.37 rad>s






Ans.

not

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States

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permitted.

Dissemination

Since the plate is initially at rest, T1 = 0. Referring to Fig. b, NA and NB do no work, while W does positive work. When u = 0°, W displaces vertically through a distance of h = 20.752 + 1.52 = 1.677 ft, Fig. c. Thus, the work done by W is

18–27. P ⫽ 25 lb

The 100-lb block is transported a short distance by using two cylindrical rollers, each having a weight of 35 lb. If a horizontal force P = 25 lb is applied to the block, determine the block’s speed after it has been displaced 2 ft to the left. Originally the block is at rest. No slipping occurs.

1.5 ft

SOLUTION T1 + ©U1-2 = T2 0 + 25(2) =

vB 2 vB 2 1 35 1 1 35 1 100 a b (vB)2 + 2 B a ba b + a a b (1.5)2 b a b R 2 32.2 2 32.2 2 2 2 32.2 3 vB = 5.05 ft>s

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Ans.






1.5 ft

*18–28. The hand winch is used to lift the 50-kg load. Determine the work required to rotate the handle five revolutions. The gear at A has a radius of 20 mm.

rB = 130 mm B

SOLUTION

100 mm

20(uA) = uB (130)

A

When uA = 5 rev. = 10 p uB = 4.8332 rad Thus load moves up s = 4.8332(0.1 m) = 0.48332 m U = 50(9.81)(0.48332) = 237 J

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375 mm

18–29. A motor supplies a constant torque or twist of M = 120 lb # ft to the drum. If the drum has a weight of 30 lb and a radius of gyration of kO = 0.8 ft, determine the speed of the 15-lb crate A after it rises s = 4 ft starting from rest. Neglect the mass of the cord.

M = 120 lb · ft

1.5 ft O

SOLUTION Free Body Diagram: The weight of the crate does negative work since it acts in the opposite direction to that of its displacement sw. Also, the couple moment M does positive work as it acts in the same direction of its angular displacement u. The reactions Ox, Oy and the weight of the drum do no work since point O does not displace.

A

s

teaching

Web)

laws

or

Kinematic: Since the drum rotates about point O, the angular velocity of the vA vA drum and the speed of the crate can be related by vD = = = 0.6667 vA . rD 1.5 When the crate rises s = 4 ft, the angular displacement of the drum is given by s 4 u = = = 2.667 rad . rD 1.5

is

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1 1 I v2 + mC v2C 2 O 2 is

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0 + Mu - WC sC =

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1 15 1 (0.5963)(0.6667vA)2 + a bvA2 2 2 32.2 part

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0 + 120(2.667) - 15(4) =

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vA = 26.7 ft s






permitted.

Dissemination

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in

Principle of Work and Energy: The mass moment of inertia of the drum about point 30 O is IO = mk2O = a b (0.82) = 0.5963 slug # ft2. Applying Eq. 18–13, we have 32.2

18–30. M P ⫽ 750 N

Motor M exerts a constant force of P = 750 N on the rope. If the 100-kg post is at rest when u = 0°, determine the angular velocity of the post at the instant u = 60°. Neglect the mass of the pulley and its size, and consider the post as a slender rod.

C

A 4m

SOLUTION

3m

Kinetic Energy and Work: Since the post rotates about a fixed axis, vG = vrG = v (1.5). The mass moment of inertia of the post about its mass center is 1 IG = (100)(32) = 75 kg # m2. Thus, the kinetic energy of the post is 12 T = =

u B

1 1 mvG2 + IGv2 2 2 1 1 (100)[v(1.5)]2 + (75)v2 2 2

= 150v2

in

teaching

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laws

or

1 This result can also be obtained by applying T = IBv2, where IB = 2 1 (100)(32) + 100 (1.52) = 300 kg # m2. Thus, 12 1 1 I v2 = (300)v2 = 150v2 2 B 2

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T =

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P displaces sP = A¿C - AC, where AC = 24 2 + 32 - 2(4)(3) cos 30° = 2.053 m

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and A¿C = 24 2 + 32 = 5 m. Thus, sP = 5 - 2.053 = 2.947 m. Also, W displaces vertically upwards through a distance of h = 1.5 sin 60° = 1.299 m. Thus, the work done by P and W is

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and

work

this

UP = PsP = 750(2.947) = 2210.14 J

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T1 + ©U1-2 = T2

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UW = - Wh = - 100 (9.81)(1.299) = - 1274.36 J

0 + [2210.14 - 1274.36] = 150v2 v = 2.50 rad>s






Ans.

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Since the post is initially at rest, T1 = 0. Referring to Fig. a, Bx, By, and R C do no work, while P does positive work and W does negative work. When u = 60° ,

18–31. The uniform bar has a mass m and length l. If it is released from rest when u = 0°, determine its angular velocity as a function of the angle u before it slips.

u

O

2l — 3

SOLUTION Kinetic Energy and Work: Before the bar slips, the bar rotates about the fixed axis passing through point O. The mass moment of inertia of the bar about this axis is 1 l 2 1 IO = ml2 + ma b = ml2. Thus, the kinetic energy of the bar is 12 6 9 1 1 1 1 I v2 = a ml2 bv2 = ml2v2 2 O 2 9 18

T =

Initially, the bar is at rest. Thus, T1 = 0. Referring to the FBD of the bar, Fig. a, we notice that N and Ff do no work while W does positive work which is given by

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mgl l UW = Wh = mg a sin ub = sin u 6 6

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l — 3

*18–32. The uniform bar has a mass m and length l. If it is released from rest when u = 0°, determine the angle u at which it first begins to slip. The coefficient of static friction at O is ms = 0.3.

θ

O

2l — 3

SOLUTION T1 + ©U1-2 = T2 1 1 l l 0 + m g ( sin u) = [ m l2 + m ( )2]v2 6 2 12 6 v =

3 g sin u l

l l 1 m g cos u( ) = [ m l2 + m( )2] a 6 12 6 3 g cos u 2l laws

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N = 0.75 m g cos u

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18–33. The two 2-kg gears A and B are attached to the ends of a 3-kg slender bar. The gears roll within the fixed ring gear C, which lies in the horizontal plane. If a 10-N # m torque is applied to the center of the bar as shown, determine the number of revolutions the bar must rotate starting from rest in order for it to have an angular velocity of vAB = 20 rad>s. For the calculation, assume the gears can be approximated by thin disks.What is the result if the gears lie in the vertical plane?

C 400 mm

A

SOLUTION

M = 10 N · m

Mu = T2 1 1 1 10u = 2a IGv2gear b + 2 a mgear b (0.200vAB)2 + IABv2AB 2 2 2 1 (2)(0.150)2 = 0.0225 kg # m2, 2 1 200 IAB = (3)(0.400)2 = 0.0400 kg # m2 and vgear = v , 12 150 AB 2 200 b v2AB + 2(0.200)2v2AB + 0.0200v2AB 10u = 0.0225 a 150 in

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Using mgear = 2 kg, IG =

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150 mm

150 mm

Energy equation (where G refers to the center of one of the two gears):


B

18–34. A ball of mass m and radius r is cast onto the horizontal surface such that it rolls without slipping. Determine its angular velocity at the instant u = 90°, if it has an initial speed of vG as shown.

SOLUTION

G

vG . r

Kinetic Energy and Work: Since the ball rolls without slipping, vG = vr or v =

2 2 mr . Thus, the 5

The mass moment of inertia of the ball about its mass cener is IG = kinetic energy of the ball is T =

1 1 mvG2+ IGv2 2 2

=

vG 2 1 1 2 mvG2+ a mr2 b a b r 2 2 5

=

7 mv 2 10 G

7 mvG2. Referring to the FBD of the 10 ball, Fig. a, we notice that N does no work while W does negative work. When laws

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u R

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18–35. A ball of mass m and radius r is cast onto the horizontal surface such that it rolls without slipping. Determine the minimum speed vG of its mass center G so that it rolls completely around the loop of radius R + r without leaving the track. u R r G

SOLUTION + T ©Fy = m(aG)y ;

mg = m a

v2 b R

v2 = gR T1 + ©U1 - 2 = T2 gR 1 2 2 v2G 1 1 2 1 a mr b a 2 b + mv2G - mg(2R) = a mr2 b a 2 b + m(gR) 2 5 2 2 5 2 r r 1 1 1 2 1 vG + v2G = 2gR + gR + gR 5 2 5 2 or

3 gR A7

vG = 3

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vG

*18–36. At the instant shown, the 50-lb bar rotates clockwise at 2 rad>s. The spring attached to its end always remains vertical due to the roller guide at C. If the spring has an unstretched length of 2 ft and a stiffness of k = 6 lb>ft, determine the angular velocity of the bar the instant it has rotated 30° clockwise.

C

4 ft

k

A B

SOLUTION

2 rad/s 6 ft

Datum through A. T1 + V1 = T2 + V2 1 1 1 50 1 1 50 c a b(6)2 d(2)2 + (6)(4 - 2)2 = c a b (6)2 dv2 2 3 32.2 2 2 3 32.2 +

1 (6)(7 - 2)2 - 50(1.5) 2

v = 2.30 rad>s

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18–37. At the instant shown, the 50-lb bar rotates clockwise at 2 rad>s. The spring attached to its end always remains vertical due to the roller guide at C. If the spring has an unstretched length of 2 ft and a stiffness of k = 12 lb>ft, determine the angle u, measured from the horizontal, to which the bar rotates before it momentarily stops.

C

4 ft

k

A B

SOLUTION

2 rad/s 6 ft

T1 + V1 = T2 + V2 1 1 1 1 50 c a b (6)2 d (2)2 + (12)(4 - 2)2 = 0 + (12)(4 + 6 sin u - 2)2 - 50(3 sin u) 2 3 32.2 2 2 61.2671 = 24(1 + 3 sin u)2 - 150 sin u 37.2671 = - 6 sin u + 216 sin2 u Set x = sin u, and solve the quadratic equation for the positive root: sin u = 0.4295 u = 25.4°

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18–38. The spool has a mass of 50 kg and a radius of gyration kO = 0.280 m. If the 20-kg block A is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity v = 5 rad>s. Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

0.3 m 0.2 m O

SOLUTION vA = 0.2v = 0.2(5) = 1 m>s System:

A

T1 + V1 = T2 + V2 [0 + 0] + 0 =

1 1 (20)(1)2 + [50(0.280)2](5)2 - 20(9.81) s 2 2

s = 0.30071 m = 0.301 m

Ans.

Block: laws in

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1 (20)(1)2 2

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0 + 20(9.81)(0.30071) - T(0.30071) =

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T1 + ©U1 - 2 = T2

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T = 163 N






18–39. The spool has a mass of 50 kg and a radius of gyration kO = 0.280 m. If the 20-kg block A is released from rest, determine the velocity of the block when it descends 0.5 m.

0.3 m 0.2 m O

SOLUTION Potential Energy: With reference to the datum established in Fig. a, the gravitational potential energy of block A at position 1 and 2 are V1 = (Vg)1 = WAy1 = 20 (9.81)(0) = 0

A

V2 = (Vg)2 = - WAy2 = - 20 (9.81)(0.5) = - 98.1 J vA vA = = 5vA. rA 0.2 Here, the mass moment of inertia about the fixed axis passes through point O is Kinetic Energy: Since the spool rotates about a fixed axis, v =

IO = mkO2 = 50 (0.280)2 = 3.92 kg # m2. Thus, the kinetic energy of the system is

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1 1 (3.92)(5vA)2 + (20)vA2 = 59vA2 2 2

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=

1 1 I v2 + mAvA2 2 O 2

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*18–40. G

An automobile tire has a mass of 7 kg and radius of gyration kG = 0.3 m. If it is released from rest at A on the incline, determine its angular velocity when it reaches the horizontal plane. The tire rolls without slipping.

0.4 m A 5m 30° 0.4 m B

SOLUTION nG = 0.4v Datum at lowest point. T1 + V1 = T2 + V2 0 + 7(9.81)(5) =

1 1 (7)(0.4v)2 + [7 (0.3)2]v2 + 0 2 2

v = 19.8 rad s

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18–41. The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, determine the velocity of the collar at the instant the rod becomes horizontal, i.e., u = 0°. The system is released from rest when u = 45°.

C

3 ft

SOLUTION

u B

T1 + V1 = T2 + V2

0.8 ft A

vC 2 1 1 4 1 1 b (3)2 d a b + a b(vC)2 + 0 0 + 4(1.5 sin 45°) + 1(3 sin 45°) = c a 2 3 32.2 3 2 32.2 vC = 13.3 ft>s

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18–42. The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, determine the velocity of the collar at the instant u = 30°. The system is released from rest when u = 45°.

C

3 ft

SOLUTION

u B

vB = 0.8vA vBC

0.8 ft

vB vC vG = = = 1.5 2.598 1.5

A

Thus, vB = vG = 1.5vBC

vC = 2.598vBC

vA = 1.875 vBC

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T1 + V1 = T2 + V2

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0 + 4(1.5 sin 45°) + 1(3 sin 45°) 1 1 20 1 20 c a b (0.8)2 d (1.875vBC)2 + a b (1.5 vBC)2 2 2 32.2 2 32.2 instructors

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vBC = 1.180 rad>s

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Ans.

18–43. The door is made from one piece, whose ends move along the horizontal and vertical tracks. If the door is in the open position, u = 0°, and then released, determine the speed at which its end A strikes the stop at C. Assume the door is a 180-lb thin plate having a width of 10 ft.

5 ft

3 ft

SOLUTION T 1 + V1 = T2 + V2 0 + 0 =

1 1 180 1 180 c a b (8)2 d v2 + a b (1v)2 - 180(4) 2 12 32.2 2 32.2

v = 6.3776 rad>s vC = v(5) = 6.3776(5) = 31.9 m>s

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A

C

B

u

*18–44. Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10 kg and a radius of gyration of 125 mm about its center of mass. Gear B and drum C have a combined mass of 30 kg and a radius of gyration about their center of mass of 150 mm.

100 mm 150 mm C

A

B 200 mm

SOLUTION Potential Energy: With reference to the datum shown in Fig. a, the gravitational potential energy of block D at position (1) and (2) is V1 = (Vg)1 = WD(yD)1 = 50 (9.81)(0) = 0

D

V2 = (Vg)2 = - WD(yD)2 = - 50(9.81)(2) = - 981 J vD vD = = 10vD. rD 0.1

Kinetic Energy: Since gear B rotates about a fixed axis, vB =

rB 0.2 bv = a b(10vD) = 13.33vD. rA B 0.15 The mass moment of inertia of gears A and B about their mass centers are IA = mAkA2 = 10(0.1252) = 0.15625 kg # m2 and IB = mBkB2 = 30(0.152) = 0.675 kg # m2.Thus, the kinetic energy of the system is laws

or

Also, since gear A is in mesh with gear B, vA = a

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1 1 1 (0.15625)(13.33vD)2 + (0.675)(10vD)2 + (50)vD2 2 2 2

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0 + 0 = 72.639vD2 - 981

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vD = 3.67 m>s






18–45. The disk A is pinned at O and weighs 15 lb. A 1-ft rod weighing 2 lb and a 1-ft-diameter sphere weighing 10 lb are welded to the disk, as shown. If the spring is orginally stretched 1 ft and the sphere is released from the position shown, determine the angular velocity of the disk when it has rotated 90˚.

k = 4 lb/ft

1 ft

1 ft

SOLUTION T1 + V1 = T2 + V2 [0 + 0 + 0] +

1 1 1 15 1 1 2 1 2 (4)(1)2 = [ ( )(2)2]v2 + [ ( )(1)2]v2 + ( )(v )2R 2 2 2 32.2 2 12 32.2 2 32.2 G +

1 2 10 1 10 1 p [ ( )(0.5)2]v2 + ( )(v )2s - 2(2.5) - 10(3.5) + (4)(1 + 2( ))2 2 5 32.2 2 32.2 G 2 2

Since (vG)S = 3.5v

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(vG)R = 2.5v

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O

2 ft

A

18–46. The disk A is pinned at O and weighs 15 lb. A 1-ft rod weighing 2 lb and a 1-ft-diameter sphere weighing 10 lb are welded to the disk, as shown. If the spring is originally stretched 1 ft and the sphere is released from the position shown, determine the angular velocity of the disk when it has rotated 45°.

k ⫽ 4 lb/ft

1 ft

1 ft

SOLUTION Potential Energy: From the geometry shown in Fig. a, we obtain (yG1)2 = 2.5 sin 45° ft = 1.7678 ft and (yG2)2 = 3.5 sin 45° = 2.4749 ft. With reference to the datum set in Fig. a, the initial and final gravitational potential energy of the system is (Vg)1 = W1(yG1)1 + W2(yG2)1 = 2(0) + 10(0) = 0 (Vg)2 = - W1(yG1)2 - W2(yG2)2 = - 2(1.7678) - 10(2.4749) = - 28.284 ft # lb p (2) = 2.5708 ft. 4

The initial and final stretch of the spring is s1 = 1 ft and s2 = 1 +

laws

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Thus the initial and final elastic potential energy of the system are 1 1 ks 2 = (4)(12) = 2 ft # lb 2 1 2

(Ve)2 =

1 1 ks 2 = (4)(2.5708)2 = 13.218 ft # lb 2 2 2

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Kinetic Energy: The mass moment of inertia of the disk assembly about the fixed

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2 10 10 1 2 2 a b (0.52) + a b (3.52) + a b(12) + a b (2.52) 5 32.2 32.2 12 32.2 32.2 integrity

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IO =

T =

1 1 I v2 = (5.1605)v2 = 2.5802v2 2 O 2

Since the system is at rest initally, T1 = 0 Conservation of Energy: T1 + V1 = T2 + V2 0 + (0 + 2) = 2.5802v2 + (- 28.284 + 13.218) v = 2.572 rad>s = 2.57 rad>s






O

2 ft

Ans.

A

18–47. At the instant the spring becomes undeformed, the center of the 40-kg disk has a speed of 4 m>s. From this point determine the distance d the disk moves down the plane before momentarily stopping. The disk rolls without slipping.

k = 200 N/m

0.3 m

SOLUTION

30°

Datum at lowest point. T1 + V1 = T2 + V2 4 2 1 1 1 1 c (40)(0.3)2 d a b + (40)(4)2 + 40(9.81)d sin 30° = 0 + (200)d2 2 2 0.3 2 2 100d2 - 196.2d - 480 = 0 Solving for the positive root d = 3.38 m

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*18–48. A chain that has a negligible mass is draped over the sprocket which has a mass of 2 kg and a radius of gyration of kO = 50 mm. If the 4-kg block A is released from rest from the position s = 1 m, determine the angular velocity of the sprocket at the instant s = 2 m.

100 mm O

s⫽1m

SOLUTION T1 + V1 = T2 + V2 0 + 0 + 0 =

1 1 (4)(0.1 v)2 + C 2(0.05)2 D v2 - 4(9.81)(1) 2 2

A

v = 41.8 rad>s

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18–49. Solve Prob. 18–48 if the chain has a mass of 0.8 kg>m. For the calculation neglect the portion of the chain that wraps over the sprocket.

100 mm O

s

1m

SOLUTION T1 + V1 = T2 + V2 1 1 (4)(0.1 v)2 + C 2(0.05)2 D v2 2 2

0 - 4(9.81)(1) - 2 C 0.8(1)(9.81)(0.5) D = +

A

1 (0.8)(2)(0.1 v)2 - 4(9.81)(2) - 0.8(2)(9.81)(1) 2 v = 39.3 rad>s

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18–50. The compound disk pulley consists of a hub and attached outer rim. If it has a mass of 3 kg and a radius of gyration kG = 45 mm, determine the speed of block A after A descends 0.2 m from rest. Blocks A and B each have a mass of 2 kg. Neglect the mass of the cords.

100 mm 30 mm

SOLUTION B

T1 + V1 = T2 + V2 [0 + 0 + 0] + [0 + 0] =

1 1 1 [3(0.045)2]v2 + (2)(0.03v)2 + (2)(0.1v)2 - 2(9.81)sA + 2(9.81)sB 2 2 2

sA sB = 0.03 0.1

u =

sB = 0.3 sA Set sA = 0.2 m, sB = 0.06 m

or

Substituting and solving yields,

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v = 14.04 rad>s vA = 0.1(14.04) = 1.40 m>s

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A

18–51. A spring having a stiffness of k = 300 N>m is attached to the end of the 15-kg rod, and it is unstretched when u = 0°. If the rod is released from rest when u = 0°, determine its angular velocity at the instant u = 30°. The motion is in the vertical plane.

B

u k

300 N/m

SOLUTION

0.6 m

Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the rod at positions (1) and (2) is

A

A Vg B 1 = W(yG)1 = 15(9.81)(0) = 0 A Vg B 2 = - W(yG)2 = - 15(9.81)(0.3 sin 30°) = -22.0725 J Since the spring is initially unstretched, (Ve)1 = 0. When u = 30°, the stretch of the spring is sP = 0.6 sin 30° = 0.3 m. Thus, the final elastic potential energy of the spring is

or

1 1 ks 2 = (300) A 0.32 B = 13.5 J 2 P 2 laws

A Ve B 2 =

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V1 = (Vg)1 + (Ve)1 = 0 + 0 = 0

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V2 = (Vg)2 + (Ve)2 = - 22.0725 + 13.5 = -8.5725 J

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Kinetic Energy: Since the rod is initially at rest, T1 = 0. From the geometry shown in Fig. b, rG>IC = 0.3 m. Thus, (VG)2 = v2rG>IC = v2 (0.3). The mass moment of inertia 1 1 of the rod about its mass center is IG = ml2 = (15) A 0.62 B = 0.45 kg # m2. Thus, 12 12 the final kinetic energy of the rod is

= 0.9v2 2 Conservation of Energy: T1 + V1 = T2 + V2 0 + 0 = 0.9v2 2 - 8.5725 v2 = 3.09 rad>s






Ans.

*18–52. The two bars are released from rest at the position u. Determine their angular velocities at the instant they become horizontal. Neglect the mass of the roller at C. Each bar has a mass m and length L.

B L A

C

SOLUTION Potential Energy: Datum is set at point A. When links AB and BC is at their L initial position, their center of gravity is located 2 sin u above the datum. Their L gravitational potential energy at this position is mg a sin u b . Thus, the initial and 2 final potential energies are V1 = 2 a

mgL sin u b = mgL sin u 2

V2 = 0

Kinetic Energy: When links AB and BC are in the horizontal position, then vB = vAB L laws

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1 1 (I ) v2AB + (IBC)C v2BC 2 AB A 2

1 1 1 1 a mL2 bv2 + a mL2 b v2 2 3 2 3

=

1 mL2 v2 3

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Conservation of Energy: Applying Eq. 18–18, we have T1 + V1 = T2 + V2 0 + mgL sin u =

1 mL2 v2 + 0 3

vAB = vBC = v =






3g sin u L

Ans.

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The mass moment inertia for link AB and BC about point A and C is 1 L 2 1 (IAB)A = (IBC)C = mL2 + m a b = mL2. Since links AB and CD are at rest 2 2 3 initially, the initial kinetic energy is T1 = 0. The final kinetic energy is given by

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located at point C. Thus, vB = vBCrB>IC or vABL = vBCL, hence vAB = vBC = v.

L

18–53. The two bars are released from rest at the position u = 90°. Determine their angular velocities at the instant they become horizontal. Neglect the mass of the roller at C. Each bar has a mass m and length L.

B L A

SOLUTION Potential Energy: With reference to the datum established in Fig. a, the gravitational potential energy of the system at position 1 and 2 are (V1)g = WAB(yGAB)1 + WBC(yGBC) 1 = mg a

L L b + mga b = mg L 2 2

(V2)g = WAB(yGAB)2 + WBC(yGBC)2 = 0 Kinetic Energy: Since the system is at rest initially, T1 = 0. Referring to the kinematic diagram of the system at position 2 shown in Fig. b, vB = vAB rAB = vBC rGBC>IC;

vAB(L) = vBC(L) vAB = vBC laws

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vGBC = vBC rGBC>IC = vBC a

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1 mL2 vBC2 3 Conservation of Energy:

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vAB = vBC =






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Dissemination

The mass moment of inertia of bar AB about the fixed axis passing through A is 1 IA = mL2 and the mass moment of inertia of bar BC about its mass center is 3 1 IGBC = mL2. Thus, the kinetic energy of the system is 12

L u

u C

18–54. If the 250-lb block is released from rest when the spring is unstretched, determine the velocity of the block after it has descended 5 ft. The drum has a weight of 50 lb and a radius of gyration of kO = 0.5 ft about its center of mass O.

0.375 ft k ⫽ 75 lb/ft 0.75 ft O

SOLUTION Potential Energy: With reference to the datum shown in Fig. a, the gravitational potential energy of the system when the block is at position 1 and 2 is (Vg)1 = W(yG)1 = 250(0) = 0 (Vg)2 = - W(yG)2 = - 250(5) = - 1250 ft # lb When the block descends sb = 5 ft, the drum rotates through an angle of sb 5 u = = = 6.667 rad. Thus, the stretch of the spring is x = s + s0 = rb 0.75 rspu + 0 = 0.375(6.667) = 2.5 ft. The elastic potential energy of the spring is 1 2 1 kx = (75)(2.52) = 234.375 ft # lb 2 2 laws

or

(Ve)2 =

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V2 = (Vg)2 + (Ve)2 = - 1250 + 234.375 = - 1015.625 ft # lb

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V1 = (Vg)1 + (Ve)1 = 0

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Kinetic Energy: Since the drum rotates about a fixed axis passing through point O, vb vb v = = = 1.333vb. The mass moment of inertia of the drum about its mass rb 0.75 50 center is IO = mkO 2 = a 0.52 b = 0.3882 slug # ft2. 32.2

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= 4.2271vb2 Since the system is initially at rest, T1 = 0. T1 + V1 = T2 + V2

0 + 0 = 4.2271vb 2 - 1015.625 vb = 15.5 ft>s






T

Ans.

18–55. The 6-kg rod ABC is connected to the 3-kg rod CD. If the system is released from rest when u = 0°, determine the angular velocity of rod ABC at the instant it becomes horizontal.

A 0.5 m

B 0.3 m

SOLUTION

C

Potential Energy: When rod ABC is in the horizontal position, Fig. a, 0.3 u = sin - 1 a b = 48.59°. With reference to the datum in Fig. a, the initial and final 0.4 gravitational potential energy of the system is

u 0.4 m D

V1 = (Vg)1 = W1(yG1)1 + W2(yG2)1 = 6(9.81)(0.8) + 3(9.81)(0.2) = 52.974 J V2 = (Vg)2 = W1(yG1)2 + W2(yG2)2

or

= 6(9.81)(0.4 cos 48.59°) + 3(9.81)(0.2 cos 48.59°) = 19.466 J

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1 1 m (v ) 2 + IG1(vABC)22 2 1 G1 2 2

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2 1 1 (6)c (vABC)2(0.4) d + (0.32)(vABC)2 2 2 2 part

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= 0.64vABC2

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Conservation of Energy: T1 + V1 = T2 + V2 0 + 52.974 = 0.64vABC 2 + 19.466 (vABC)2 = 7.24 rad>s






Ans.

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Kinetic Energy: Since the system is initially at rest, T1 = 0. Referring to Fig. b, (vG1)2 = (vABC)2 rG1>IC = (vABC)2(0.4). Since point C is at the IC(vC)2 = 0. Then, (vC)2 0 vCD = = = 0. The mass moment of inertia of rod ABC about its mass rC 0.4 1 center is IG1 = (6)(0.82) = 0.32 kg # m2. Thus, the final kinetic energy of the 12 system is

*18–56. If the chain is released from rest from the position shown, determine the angular velocity of the pulley after the end B has risen 2 ft. The pulley has a weight of 50 lb and a radius of gyration of 0.375 ft about its axis. The chain weighs 6 lb/ft.

0.5 ft

4 ft

SOLUTION

6 ft

Potential Energy: (yG1)1 = 2 ft, (yG 2)1 = 3 ft, (yG1)2 = 1 ft, and (yG2)2 = 4 ft. With reference to the datum in Fig. a, the gravitational potential energy of the chain at position 1 and 2 is

B

V1 = (Vg)1 = W1(yG1)1 - W2(yG2)1

A

= - 6(4)(2) - 6(6)(3) = - 156 ft # lb V2 = (Vg)2 = - W1(yG1)2 + W2(yG2)2 = - 6(2)(1) - 6(8)(4) = - 204 ft # lb

in

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laws

or

Kinetic Energy: Since the system is initially at rest, T1 = 0. The pulley rotates about a fixed axis, thus, (VG1)2 = (VG2)2 = v2 r = v2(0.5). The mass moment of inertia of 50 the pulley about its axis is IO = mkO 2 = (0.3752) = 0.2184 slug # ft2. Thus, the 32.2 final kinetic energy of the system is

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1 1 1 IOv2 2 + m1(VG1)2 2 + m2 (VG2)2 2 2 2 2 instructors

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the

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1 1 6(2) 1 6(8) 1 6(0.5)(p) (0.2184)v2 2 + c d [v2(0.5)]2 + c d[v2(0.5)]2 + c d[v2(0.5)]2 2 2 32.2 2 32.2 2 32.2 by

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T1 + V1 = T2 + V2

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v2 = 11.3 rad >s

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0 + (-156) = 0.3787v22 = ( -204)






18–57. If the gear is released from rest, determine its angular velocity after its center of gravity O has descended a distance of 4 ft. The gear has a weight of 100 lb and a radius of gyration about its center of gravity of k = 0.75 ft. 1 ft O

SOLUTION Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the gear at position 1 and 2 is V1 = (Vg)1 = W(y0)1 = 100(0) = 0 V2 = (Vg)2 = - W1(y0)2 = - 100(4) = - 400 ft # lb Kinetic Energy: Referring to Fig. b, we obtain vO = vrO / IC = v(1).The mass moment 100 (0.752) = 1.7469 kg # m2. 32.2

of inertia of the gear about its mass center is IO = mkO 2 = Thus,

or

1 1 mvO2 + IOv2 2 2 laws

T =

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1 100 1 a b [v (1)]2 + (1.7469)v2 2 32.2 2

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= 2.4262v2

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Since the gear is initially at rest, T1 = 0.

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T1 + V1 = T2 + V2

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0 + 0 = 2.4262v2 - 400

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v = 12.8 rad>s






18–58. When the slender 10-kg bar AB is horizontal it is at rest and the spring is unstretched. Determine the stiffness k of the spring so that the motion of the bar is momentarily stopped when it has rotated clockwise 90°.

A

B

1.5 m

SOLUTION T1 + V1 = T2 + V2 0 + 0 = 0 +

1 1.5 (k)(3.3541 – 1.5 )2 – 98.1 a b 2 2 k = 42.8 N>m

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k

1.5 m

C

18–59. When the slender 10-kg bar AB is horizontal it is at rest and the spring is unstretched. Determine the stiffness k of the spring so that the motion of the bar is momentarily stopped when it has rotated clockwise 45°.

A

B

1.5 m

SOLUTION Potential Energy: From the geometry shown in Fig. a, we obtain (yG)2 = 0.75 sin 45° = 0.5303 m and CB¿ = 232 + 1.52 - 2(3)(1.5) cos 45° = 2.2104. With reference to the datum established in Fig. a, the initial and final gravitational potential energy of the system is (Vg)1 = WAB(yG)1 = 0 (Vg)2 = - WAB(yG)2 = - 10(9.81)(0.5303) = - 52.025 J

or

Initially, the spring is unstretched. Thus, (Ve)1 = 0. At the final position, the spring stretches S = CB¿ - CB = 2.2104 - 1.5 = 0.7104 m. Then (Ve)1 = 0 and (Ve)2 = 1 2 1 ks = k (0.7104 2) = 0.2524k. 2 2

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V1 = (Ve)1 + (Vg)1 = 0

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Kinetic Energy: Since the bar is at rest initially and stops momentarily at the final position, T1 = T2 = 0.

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T1 + V1 = T2 + V2

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the

0 + 0 = 0 + 0.2524k - 52.025

Ans.

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k = 206.15 N>m = 206 N>m






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in

V2 = (Ve)2 + (Vg)2 = 0.2524k - 52.025

k

1.5 m

C

*18–60. If the 40-kg gear B is released from rest at u = 0°, determine the angular velocity of the 20-kg gear A at the instant u = 90°. The radii of gyration of gears A and B about their respective centers of mass are kA = 125 mm and kB = 175 mm. The outer gear ring P is fixed.

P

0.15 m A

u

B

SOLUTION

0.2 m

Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of gear B at positions (1) and (2) is V1 = (Vg)1 = WB(yGB)1 = 40(9.81)(0) = 0 V2 = (Vg)2 = - WB(yGB)2 = - 40(9.81)(0.35) = - 137.34 J Kinetic Energy: Referring to Fig. b, vP = vArA = vA(0.15). Then, vB =

vP =

rP>IC

teaching

Web)

laws

or

vA(0.15) = 0.375vA. Subsequently, vGB = vBrGB>IC = (0.375vA)(0.2) = 0.075vA. 0.4 The mass moments of inertia of gears A and B about their mass centers are and IA = mAkA2 = 20(0.1252) = 0.3125 kg # m2 IB = mBkB2 = 40(0.1752) = 2 # 1.225 kg m . Thus, the kinetic energy of the system is

Wide

copyright

in

T = TA + TB 1 1 1 I v 2 + c mBvGB2 + IBvB 2d 2 A A 2 2

=

1 1 1 (0.3125)vA2 + c (40)(0.075vA)2 + (1.225)(0.375vA)2d 2 2 2 for

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the

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= 0.3549vA2

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Since the system is initially at rest, T1 = 0.

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T1 + V1 = T2 + V2

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0 + 0 = 0.3549vA2 - 137.34 vA = 19.7 rad>s






Ans.

18–61. A uniform ladder having a weight of 30 lb is released from rest when it is in the vertical position. If it is allowed to fall freely, determine the angle u at which the bottom end A starts to slide to the right of A. For the calculation, assume the ladder to be a slender rod and neglect friction at A. 10 ft

SOLUTION

u

Potential Energy: Datum is set at point A. When the ladder is at its initial and final position, its center of gravity is located 5 ft and (5 cos u ) ft above the datum. Its initial and final gravitational potential energy are 30(5) = 150 ft # lb and 30(5 cos u ) = 150 cos u ft # lb, respectively. Thus, the initial and final potential energy are V1 = 150 ft # lb

A

V2 = 150 cos u ft # lb

or

Kinetic Energy: The mass moment inertia of the ladder about point A is 1 30 30 IA = a b (102) + a b (52) = 31.06 slug # ft2. Since the ladder is initially at 12 32.2 32.2 rest, the initial kinetic energy is T1 = 0. The final kinetic energy is given by

teaching

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laws

1 1 I v2 = (31.06)v2 = 15.53v2 2 A 2 in

T2 =

Dissemination

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copyright

Conservation of Energy: Applying Eq. 18–18, we have

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T1 + V1 = T2 + V2

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0 + 150 = 15.53v2 + 150 cos u

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v2 = 9.66(1 - cos u)

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Equation of Motion: The mass moment inertia of the ladder about its mass center is 1 30 IG = a b (102) = 7.764 slug # ft2. Applying Eq. 17–16, we have 12 32.2

any

courses

-30 sin u(5) = - 7.764a - a

30 b [a(5)](5) 32.2

destroy

of

and

+ ©MA = ©(Mk)A;

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a = 4.83 sin u c

+ ©Fx = m(aG)x;

Ax = -

30 [9.66(1 - cos u)(5)] sin u 32.2 +

Ax = -

30 [4.83 sin u(5)] cos u 32.2

30 (48.3 sin u - 48.3 sin u cos u - 24.15 sin u cos u) 32.2

= 45.0 sin u (1 - 1.5 cos u) = 0 If the ladder begins to slide, then Ax = 0 . Thus, for u>0, 45.0 sin u (1 - 1.5 cos u) = 0 u = 48.2°






Ans.

18–62. The 50-lb wheel has a radius of gyration about its center of gravity G of kG = 0.7 ft. If it rolls without slipping, determine its angular velocity when it has rotated clockwise 90° from the position shown. The spring AB has a stiffness k = 1.20 lb/ft and an unstretched length of 0.5 ft. The wheel is released from rest.

3 ft 0.5 ft

B

0.5 ft G 1 ft


1 1 50 1 50 (1.20)[2(3)2 + (0.5)2 - 0.5]2 = [ (0.7)2]v2 + ( )(1v)2 2 2 32.2 2 32.2 1 (1.20)(0.9292 - 0.5)2 2

+

v = 1.80 rad s

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k = 1.20 lb/ft

A

18–63. The uniform window shade AB has a total weight of 0.4 lb. When it is released, it winds up around the spring-loaded core O. Motion is caused by a spring within the core, which is coiled so that it exerts a torque M = 0.3(10 - 3)u lb # ft, where u is in radians, on the core. If the shade is released from rest, determine the angular velocity of the core at the instant the shade is completely rolled up, i.e., after 12 revolutions. When this occurs, the spring becomes uncoiled and the radius of gyration of the shade about the axle at O is kO = 0.9 in. Note: The elastic potential energy of the torsional spring is Ve = 12ku2, where M = ku and k = 0.3(10 - 3) lb # ft>rad.

M B

O

3 ft

A

SOLUTION T1 + V1 = T2 + V2 0 - (0.4)(1.5) +

1 0.4 0.9 2 2 1 (0.3)(10 - 3)(24p)2 = a ba b v 2 2 32.2 12

v = 85.1 rad>s

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O

*18–64. The motion of the uniform 80-lb garage door is guided at its ends by the track. Determine the required initial stretch in the spring when the door is open, u = 0°, so that when it falls freely it comes to rest when it just reaches the fully closed position, u = 90°. Assume the door can be treated as a thin plate, and there is a spring and pulley system on each of the two sides of the door.

k C u 8 ft 8 ft

A

SOLUTION sA + 2 s s = l ¢sA = - 2¢ss 8 ft = - 2¢ss ¢ss = -4 ft T1 + V1 = T2 + V2 1 1 0 + 2 c (9)s2 d = 0 - 80(4) + 2 c (9)(4 + s)2 d 2 2

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or

9s2 = - 320 + 9(16 + 8s + s2) s = 2.44 ft

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B

9 lb/ft

18–65. The motion of the uniform 80-lb garage door is guided at its ends by the track. If it is released from rest at u = 0°, determine the door’s angular velocity at the instant u = 30°. The spring is originally stretched 1 ft when the door is held open, u = 0°. Assume the door can be treated as a thin plate, and there is a spring and pulley system on each of the two sides of the door.

k ⫽ 9 lb/ft C u 8 ft 8 ft

A

SOLUTION vG = 4v sA + 2ss = l ¢sA = - 2¢ss 4 ft = - 2¢ss ¢ss = - 2 ft T1 + V1 = T2 + V2

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or

1 80 1 1 80 1 b (4v)2 + c a b(8)2 dv2 - 80(4 sin 30°) 0 + 2c (9)(1)2 d = a 2 2 32.2 2 12 32.2

Ans.

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v = 1.82 rad>s






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1 + 2 c (9)(2 + 1)2 d 2

B

18–66. The end A of the garage door AB travels along the horizontal track, and the end of member BC is attached to a spring at C. If the spring is originally unstretched, determine the stiffness k so that when the door falls downward from rest in the position shown, it will have zero angular velocity the moment it closes, i.e., when it and BC become vertical. Neglect the mass of member BC and assume the door is a thin plate having a weight of 200 lb and a width and height of 12 ft. There is a similar connection and spring on the other side of the door.

12 ft

C

D

CD2 - 11.591CD + 32 = 0 Selecting the smaller root: CD = 4.5352 ft T1 + V1 = T2 + V2

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1 0 + 0 = 0 + 2 c (k)(8 - 4.5352)2 d - 200(6) 2 k = 100 lb/ft

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sale or transmission in any form or by any means, electronic, mechanical,



6 ft

2 ft

(2)2 = (6)2 + (CD)2 - 2(6)(CD) cos 15°


15

7 ft

SOLUTION


A

B

1 ft

18–67. Determine the stiffness k of the torsional spring at A, so that if the bars are released from rest when u = 0°, bar AB has an angular velocity of 0.5 rad/s at the closed position, u = 90°. The spring is uncoiled when u = 0°. The bars have a mass per unit length of 10 kg>m.

B 4m

3m

u C A k

SOLUTION Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the system at its open and closed positions is

A Vg B 1 = WAB (yG1)1 + WBC (yG2)1 = 10(3)(9.81)(1.5) + 10(4)(9.81)(1.5) = 1030.5 J

A Vg B 2 = WAB (yG1)2 + WBC (yG2)2 = 10(3)(9.81)(0) + 10(4)(9.81)(0) = 0

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or

Since the spring is initially uncoiled, (Ve)1 = 0. When the panels are in the closed p position, the coiled angle of the spring is u = rad. Thus, 2 in

p 2 1 2 1 p2 ku = ka b = k 2 2 2 8

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V1 = A Vg)1 + (Ve)1 = 1030.5 + 0 = 1030.5 J

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p2 p2 k = k 8 8 is

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V2 = A Vg)2 + (Ve)2 = 0 +

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Kinetic Energy: Since the system is initially at rest, T1 = 0. Referring to Fig. b, (vB)2 1.5 (vB)2 = (vAB)2 rB = 0.5(3) = 1.5 m>s. Then, (vBC)2 = = 0.375 rad>s. = rB>IC 4 Subsequently, (vG)2 = (vBC)2 rG2>IC = 0.375(2) = 0.75 m>s. The mass moments of inertia of AB about point A and BC about its mass center are (IAB)A =

1 2 1 ml = [10(3)] A 32 B = 90 kg # m2 3 3

and (IBC)G2 =

1 1 ml2 = [10(4)] A 42 B = 53.33 kg # m2 12 12

Thus, T2 = =

1 1 1 (I ) (v ) 2 + c mBC(vG2)2 + (IBC)G2 (vBC)2 2 d 2 AB A AB 2 2 2 1 1 1 (90) A 0.52 B + c [10(4)] A 0.752 B + (53.33) A 0.375 2 B d 2 2 2

= 26.25 J






18–67. continued Conservation of Energy: T1 + V1 = T2 + V2 0 + 1030.5 = 26.25 +

p2 k 8

k = 814 N # m>rad

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*18–68. The torsional spring at A has a stiffness of k = 900 N # m>rad and is uncoiled when u = 0°. Determine the angular velocity of the bars, AB and BC, when u = 0°, if they are released from rest at the closed position, u = 90°. The bars have a mass per unit length of 10 kg>m.

B 4m

3m

u C A k

SOLUTION Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the system at its open and closed positions is

A Vg B 1 = WAB (yG1)1 + WBC (yG2)1 = 10(3)(9.81)(0) + 10(4)(9.81)(0) = 0

A Vg B 2 = WAB (yG1)2 + WBC (yG2)2 = 10(3)(9.81)(1.5) + 10(4)(9.81)(1.5) = 1030.05 J p rad. 2

When the panel is in the closed position, the coiled angle of the spring is u =

or

Thus,

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p 2 1 2 1 ku = (900)a b = 112.5p2 J 2 2 2

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(Ve)1 =

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The spring is uncoiled when the panel is in the open position (u = 0°). Thus,

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(Ve)2 = 0 the

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V1 = A Vg)1 + (Ve)1 = 0 + 112.5p2 = 112.5p2 J

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the

V2 = A Vg)2 + (Ve)2 = 1030.05 + 0 = 1030.05 J

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Kinetic Energy: Since the panel is at rest in the closed position, T1 = 0. Referring to Fig. b, the IC for BC is located at infinity. Thus, Ans.

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(vBC)2 = 0

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Then, sale

(vG)2 = (vB)2 = (vAB)2 rB = (vAB)2 (3) The mass moments of inertia of AB about point A and BC about its mass center are (IAB)A =

1 2 1 ml = [10(3)] A 32 B = 90 kg # m2 3 3

and (IBC)G2 =

1 1 ml2 = [10(4)] A 42 B = 53.33 kg # m2 12 12

Thus, T2 = =

1 1 (IAB)A(vAB)2 2 + mBC(vG2)2 2 2 1 1 (90)(vAB)2 2 + [10(4)] C (vAB)2 (3) D 2 2 2

= 225(vAB)2 2






18–68. continued Conservation of Energy: T1 + V1 = T2 + V2 0 + 112.5p2 = 225(vAB)2 2 + 1030.05 (vAB)2 = 0.597 rad>s

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19–1. The rigid body (slab) has a mass m and rotates with an angular velocity V about an axis passing through the fixed point O. Show that the momenta of all the particles composing the body can be represented by a single vector having a magnitude mvG and acting through point P, called the center of percussion, which lies at a distance rP>G = k2G>rG>O from the mass center G. Here kG is the radius of gyration of the body, computed about an axis perpendicular to the plane of motion and passing through G.

mvG V vG

rG/O O

SOLUTION HO = (rG>O + rP>G) myG = rG>O (myG) + IG v,

where IG = mk2G

rG>O (myG) + rP>G (myG) = rG>O (myG) + (mk2G) v

yG = vrG>O

yG v

k2G rG>O

Q.E.D.

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rP>G =

or rG>O =

or

However,

k2G yG>v

laws

rP>G =






rP/G G

P

19–2. At a given instant, the body has a linear momentum L = mvG and an angular momentum H G = IG V computed about its mass center. Show that the angular momentum of the body computed about the instantaneous center of zero velocity IC can be expressed as H IC = IIC V , where IIC represents the body’s moment of inertia computed about the instantaneous axis of zero velocity. As shown, the IC is located at a distance rG>IC away from the mass center G.

mvG G

rG/IC

SOLUTION HIC = rG>IC (myG) + IG v,

IC

where yG = vrG>IC

= rG>IC (mvrG>IC) + IG v = (IG + mr2G>IC) v = IIC v

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IGV

19–3. Show that if a slab is rotating about a fixed axis perpendicular to the slab and passing through its mass center G, the angular momentum is the same when computed about any other point P.

V P

G

SOLUTION Since yG = 0, the linear momentum L = myG = 0. Hence the angular momentum about any point P is HP = IG v Since v is a free vector, so is H P .

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*19–4. The cable is subjected to a force of P = (10t2) lb, where t is in seconds. Determine the angular velocity of the spool 3 s after P is applied, starting from rest. The spool has a weight of 150 lb and a radius of gyration of 1.25 ft about its center of gravity.

O

SOLUTION Principle of Angular Impulse and Momentum: The mass moment of inertia of the 150 (1.25)2 = 7.279 slug # ft2. spool about its mass center is IO = mkO2 = 32.2 t2

IOv1 + ©

Lt1

MOdt = IOv2

3s

0 -

L0

10t2(1)dt = 7.279v2

or

10t3 3 s 2 = 7.279v2 3 0 v2 = 12.4 rad>s

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P = (10t2) lb

1 ft

19–5. The impact wrench consists of a slender 1-kg rod AB which is 580 mm long, and cylindrical end weights at A and B that each have a diameter of 20 mm and a mass of 1 kg. This assembly is free to rotate about the handle and socket, which are attached to the lug nut on the wheel of a car. If the rod AB is given an angular velocity of 4 rad>s and it strikes the bracket C on the handle without rebounding, determine the angular impulse imparted to the lug nut.

C B

Iaxle =

300 mm

1 1 (1)(0.6 - 0.02)2 + 2 c (1)(0.01)2 + 1(0.3)2 d = 0.2081 kg # m2 12 2

Mdt = Iaxle v = 0.2081(4) = 0.833 kg # m2>s

Ans.

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L






300 mm

A

SOLUTION

19–6. The space capsule has a mass of 1200 kg and a moment of inertia IG = 900 kg # m2 about an axis passing through G and directed perpendicular to the page. If it is traveling forward with a speed vG = 800 m>s and executes a turn by means of two jets, which provide a constant thrust of 400 N for 0.3 s, determine the capsule’s angular velocity just after the jets are turned off.

T = 400 N 15° 1.5 m

vG = 800 m/s

G 1.5 m 15°

SOLUTION a+

T = 400 N

(HG)1 + © 1 MGdt = (HG)2 0 + 2[400 cos 15°(0.3)(1.5)] = 900v2 v2 = 0.386 rad s

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19–7. The airplane is traveling in a straight line with a speed of 300 km> h, when the engines A and B produce a thrust of TA = 40 kN and TB = 20 kN, respectively. Determine the angular velocity of the airplane in t = 5 s. The plane has a mass of 200 Mg, its center of mass is located at G, and its radius of gyration about G is kG = 15 m.

TA ⫽ 40 kN 8 m A G B TB ⫽ 20 kN

SOLUTION Principle of Angular Impulse and Momentum: The mass moment of inertia of the airplane about its mass center is IG = mkG2 = 200 A 103 B A 152 B = 45 A 106 B kg # m2. Applying the angular impulse and momentum equation about point G, t2

Izv1 + ©

Lt1

MGdt = IGv2

0 + 40 A 103 B (5)(8) - 20 A 103 B (5)(8) = 45 A 106 B v v = 0.0178 rad>s

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8m

*19–8. The assembly weighs 10 lb and has a radius of gyration kG = 0.6 ft about its center of mass G. The kinetic energy of the assembly is 31 ft # lb when it is in the position shown. If it is rolling counterclockwise on the surface without slipping, determine its linear momentum at this instant.

0.8 ft

1 ft G

1 ft

SOLUTION IG = (0.6)2 a

10 b = 0.1118 slug # ft2 32.2

1 1 10 a b v 2 + (0.1118) v2 = 31 2 32.2 G 2

T =

(1)

vG = 1.2 v Substitute into Eq. (1), v = 10.53 rad>s

laws

or

vG = 10.53(1.2) = 12.64 ft>s 10 (12.64) = 3.92 slug # ft>s 32.2

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Ans.

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L = mvG =






19–9. z

The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz = 300 mm, rests on the smooth horizontal plane. If the belt is subjected to a force of P = 200 N, determine the angular velocity of the wheel and the speed of its center of mass O, three seconds after the force is applied.

400 mm O

SOLUTION y

x

Principle of Angular Impulse and Momentum: The mass moment of inertia of the wheel about the z axis is Iz = mkz2 = 100(0.32) = 9 kg # m2. Applying the linear and angular impulse and momentum equations using the free-body diagram of the wheel shown in Fig. a,

P ⫽ 200 N

t2

+ ;

m(vx)1 + ©

Lt1

Fx dt = m(vx)2

0 + 200(3) = 100(vO)2 Ans.

(vO)2 = 6 m>s and t2

or

Mzdt = Izv2

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Lt1

laws

Izv1 + ©

Wide

copyright

in

0 - [200(0.4)(3)] = - 9v2

Ans.

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Dissemination

v2 = 26.7 rad>s






19–10. The 30-kg gear A has a radius of gyration about its center of mass O of kO = 125 mm. If the 20-kg gear rack B is subjected to a force of P = 200 N, determine the time required for the gear to obtain an angular velocity of 20 rad>s, starting from rest. The contact surface between the gear rack and the horizontal plane is smooth.

0.15 m O

A

B

SOLUTION Kinematics: Since the gear rotates about the fixed axis, the final velocity of the gear rack is required to be (vB)2 = v2rB = 20(0.15) = 3 m>s : Principle of Impulse and Momentum: Applying the linear impulse and momentum equation along the x axis using the free-body diagram of the gear rack shown in Fig. a, + B A:

t2

m(vB)1 + ©

Fxdt = m(vB)2 Lt1 0 + 200(t) - F(t) = 20(3) F(t) = 200t - 60

laws

or

(1)

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The mass moment of inertia of the gear about its mass center is IO = mkO2 = 30(0.1252) = 0.46875 kg # m2. Writing the angular impulse and momentum equation about point O using the free-body diagram of the gear shown in Fig. b, Dissemination

t2

MOdt = IOv2 Lt1 0 + F(t)(0.15) = 0.46875(20)

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IOv1 + ©

(2) (including

for

work

student

the

by

F(t) = 62.5

is

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solely

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the

Substituting Eq. (2) into Eq. (1) yields

Ans.

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t = 0.6125 s






P ⫽ 200 N

19–11. The 30-kg reel is mounted on the 20-kg cart. If the cable wrapped around the inner hub of the reel is subjected to a force of P = 50 N, determine the velocity of the cart and the angular velocity of the reel when t = 4 s. The radius of gyration of the reel about its center of mass O is kO = 250 mm. Neglect the size of the small wheels.

O

SOLUTION Principle of Impulse and Momentum: The mass moment of inertia of the reel about its mass center is IO = mkO 2 = 30(0.2502) = 1.875 kg # m2. Referring to Fig. a, + B A:

m C (vO)1 D x + ©

Fxdt = m C (vO)2 D x Lt1 0 + 50(4) - Ox(4) = 30v t2

Ox = 50 - 7.5v

(1)

and t2

MOdt = IOv2 Lt1 0 + 50(4)(0.15) = 1.875v laws

or

IOv1 + ©

v = 16 rad>s

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Referring to Fig. b,

Dissemination

t2

Fxdt = m(v2)x Lt1 0 + Ox(4) = 20v

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Ox = 20 N

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v = 4 m>s






P ⫽ 50 N

150 mm

*19–12. The spool has a weight of 75 lb and a radius of gyration kO = 1.20 ft. If the block B weighs 60 lb, and a force P = 25 lb is applied to the cord, determine the speed of the block in 5 s starting from rest. Neglect the mass of the cord.

P 2 ft

0.75 ft O

SOLUTION c+

(HO)1 + a

L

MO dt = (HO)2

0 - 60(0.75)(5) + 25(2)(5) = + c

B

75 (1.20)2v 32.2

60 (0.75v) d (0.75) 32.2

v = 5.679 rad>s yB = vr = (5.679)(0.75) = 4.26 ft>s

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Ans.






19–13. The slender rod has a mass m and is suspended at its end A by a cord. If the rod receives a horizontal blow giving it an impulse I at its bottom B, determine the location y of the point P about which the rod appears to rotate during the impact.

A

P

SOLUTION

l y

Principle of Impulse and Momentum: t2

IG v1 + ©

(a +)

MG dt = IG v2

Lt1

l 1 0 + I a b = c ml2 d v 2 12 + b a:

I =

I

1 mlv 6

t2

m(yAx)1 + ©

Fx dt = m(yAx)2

1 mlv = mvG 6

l v 6

yG =

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0 +

Lt1

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Kinematics: Point P is the IC.

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yB = v y

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l v 6

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Ans.

B

19–14. ω0

If the ball has a weight W and radius r and is thrown onto a rough surface with a velocity v0 parallel to the surface, determine the amount of backspin, V 0, it must be given so that it stops spinning at the same instant that its forward velocity is zero. It is not necessary to know the coefficient of friction at A for the calculation.

v0

r

A

SOLUTION + B A;

m(nGx)1 + ©

L

Fx dt = m(vGx)2 W n - Ft = 0 g 0

a ( +)

(HG)1 + ©

L

(1)

MG dt = (HG)2

2 W - a r2 b v0 + Ft(r) = 0 5 g

(2)

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or

Eliminate F t between Eqs. (1) and (2): Web)

W n0 g t

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2 W 2 r v0 = 5 g

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v0 = 2.5

19–15. The assembly shown consists of a 10-kg rod AB and a 20-kg circular disk C. If it is subjected to a torque of M = (20t3>2) N # m , where t is it in seconds, determine its angular velocity when t = 3 s . When t = 0 the assembly is rotating at v1 = { -6k} rad>s .

z

M ⫽ (20t 3/2) N⭈m

0.45 m A

SOLUTION

x

Principle of Angular Impulse and Momentum: The mass moment of inertia of 1 1 the assembly about the z axis is Iz = (10) A 0.452 B + c (20) A 0.152 B + 20 A 0.62 B d = 3 2 8.10 kg # m2 . Using the free-body diagram of the assembly shown in Fig. a, Lt1

Mzdt = Izv2 3s

8.10( - 6) +

L0

- 48.6 + 8t5>2 2

20t

3>2

dt = 8.10v2

3s

= 8.10v2 0

v2 = 9.40 rad>s

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0.15 m

C

t2

Izv1 + ©

B

y

*19–16. The frame of a tandem drum roller has a weight of 4000 lb excluding the two rollers. Each roller has a weight of 1500 lb and a radius of gyration about its axle of 1.25 ft. If a torque of M = 300 lb # ft is supplied to the rear roller A, determine the speed of the drum roller 10 s later, starting from rest.

1.5 ft

SOLUTION

M ⫽ 300 lb⭈ft

Principle of Impulse and Momentum: The mass moments of inertia of the rollers 1500 about their mass centers are IC = ID = A 1.252 B = 72.787 slug # ft2. Since the 32.2 v v = = 0.6667v. Using the free-body rollers roll without slipping, v = r 1.5 diagrams of the rear roller and front roller, Figs. a and b, and the momentum diagram of the rollers, Fig. c, t2

MAdt = (HA)2

Lt1

1500 v(1.5) + 72.787(0.6667v) 32.2

0 + 300(10) - Cx(10)(1.5) = Cx = 200 - 7.893v

(1) teaching

Web)

laws

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copyright

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Dissemination

MBdt = (HB)2

on

learning. United

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by

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1500 v(1.5) + 72.787(0.6667v) 32.2

0 + Dx(10)(1.5) =

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Lt1

Wide

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(HB)1 + ©

(2)

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Dx = 7.893v solely

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Referring to the free-body diagram of the frame shown in Fig. d, work

Fxdt = m C (vG)x D 2

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m C (vG)x D 1 + ©

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4000 v 32.2

any

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Substituting Eqs. (1) and (2) into Eq. (3),

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0 + Cx(10) - Dx(10) =

(200 - 7.893v)(10) - 7.893v(10) = v = 7.09 ft>s






1.5 ft A

4000 v 32.2 Ans.

B

19–17. A motor transmits a torque of M = 0.05 N # m to the center of gear A. Determine the angular velocity of each of the three (equal) smaller gears in 2 s starting from rest. The smaller gears (B) are pinned at their centers, and the masses and centroidal radii of gyration of the gears are given in the figure.

mA M

0.05 N m

SOLUTION 40 mm

Gear A:

A

(c + )

(HA)1 + ©

L

MA dt = (HA)2 B

0 - 3(F)(2)(0.04) + 0.05(2) = [0.8(0.031)2] vA

mB 20 mm

Gear B: (a + )

(HB)1 + ©

L

MB dt = (HB)2

laws

or

0 + (F)(2)(0.02) = [0.3(0.015)2] vB

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Since 0.04vA = 0.02 vB, or vB = 2 vA, then solving,

Dissemination

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copyright

F = 0.214 N

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vA = 63.3 rad>s

is

vB = 127 rad>s

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kB

0.3 kg 15 mm

0.8 kg kA

31 mm

19–18. The man pulls the rope off the reel with a constant force of 8 lb in the direction shown. If the reel has a weight of 250 lb and radius of gyration kG = 0.8 ft about the trunnion (pin) at A, determine the angular velocity of the reel in 3 s starting from rest. Neglect friction and the weight of rope that is removed. 60

1.25 ft A

SOLUTION a+

(HA)1 + ©

L

MA dt = (HA)2

0 + 8(1.25)(3) = c

250 (0.8)2 d v 32.2

v = 6.04 rad>s

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19–19. The double pulley consists of two wheels which are attached to one another and turn at the same rate. The pulley has a mass of 15 kg and a radius of gyration kO = 110 mm. If the block at A has a mass of 40 kg, determine the speed of the block in 3 s after a constant force F = 2 kN is applied to the rope wrapped around the inner hub of the pulley. The block is originally at rest. Neglect the mass of the rope.

200 mm

(HO)1 + ©

A

L

MO dt = (HO)2

0 + 2000(0.075)(3) - 40(9.81)(0.2)(3) = 15(0.110)2v + 40(0.2v) (0.2) v = 120.4 rad>s vA = 0.2(120.4) = 24.1 m>s

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75 mm

F

SOLUTION (c +)

O

*19–20. The cable is subjected to a force of P = 20 lb, and the spool rolls up the rail without slipping. Determine the angular velocity of the spool in 5 s, starting from rest. The spool has a weight of 100 lb and a radius of gyration about its center of gravity O of kO = 0.75 ft.

P ⫽ 20 lb

1 ft

0.5 ft O

SOLUTION Kinematics: Referring to Fig. a, vO = vrO>IC = v(0.5)

30⬚

Principle of Angular Impulse and Momentum: The mass moments of inertia of the 100 (0.752) = 1.747 slug # ft2. Writing spool about its mass center is IO = mkO 2 = 32.2 the angular impulse and momentum equation about point A shown in Fig. b, t2

MAdt = (HA)2 100 [v(0.5)](0.5) 32.2

Web)

0 + 100(5) sin 30°(0.5) - 20(5)(1.5) = - 1.747v -

or

Lt1

laws

(HA)1 + ©

teaching

v = 9.91 rad>s

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Ans.






19–21. The inner hub of the wheel rests on the horizontal track. If it does not slip at A, determine the speed of the 10-lb block in 2 s after the block is released from rest. The wheel has a weight of 30 lb and a radius of gyration kG = 1.30 ft. Neglect the mass of the pulley and cord.

2 ft G 1 ft A

SOLUTION Spool, (a + )

(HA)1 + ©

L

MA dt = (HA)2

0 + T(3)(2) = c

vB 30 30 (1.3)2 + (1)2d a b 32.2 32.2 3

Block, m(vy)1 + ©

L

Fy dt = m(vy)2 or

( + T)

teaching

Web)

laws

10 v 32.2 B in

0 + 10(2) - T(2) =

Ans. permitted.

Dissemination

Wide

copyright

vB = 34.0 ft>s

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T = 4.73 lb






19–22. The 1.25-lb tennis racket has a center of gravity at G and a radius of gyration about G of kG = 0.625 ft. Determine the position P where the ball must be hit so that ‘no sting’ is felt by the hand holding the racket, i.e., the horizontal force exerted by the racket on the hand is zero.

rp 1 ft G

SOLUTION Principle of Impulse and Momentum: Here, we will assume that the tennis racket is initially at rest and rotates about point A with an angular velocity of v F dt on the L racket, Fig. a. The mass moment of inertia of the racket about its mass center is 1.25 IG = a b A 0.6252 B = 0.01516 slug # ft2. Since the racket about point A, 32.2 immediately after it is hit by the ball, which exerts an impulse of

(vG) = vrG = v(1). Referring to Fig. b, t2

+ ;

Fx dt = m(vG)2

Lt1

or

m(vG)1 + ©

Web)

laws

1.25 b C v(1) D 32.2

teaching

L

F dt = a

Wide

copyright

in

0 +

(1) World

permitted.

Dissemination

F dt = 0.03882v instructors

States

L

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learning.

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by

MA dt = (HA)2

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1.25 [v(1)](1) 32.2

F dtb rP = 0.01516v +

part

the

is

This

0.05398v rP

(2)

and

any

courses

L

F dt =

provided

integrity

of

and

work

0 + a

sale

will

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destroy

of

Equating Eqs. (1) and (2) yields 0.03882v =






0.05398v rP

rP = 1.39 ft

Ans.

P

19–23. The 100-kg reel has a radius of gyration about its center of mass G of kG = 200 mm. If the cable B is subjected to a force of P = 300 N, determine the time required for the reel to obtain an angular velocity of 20 rad>s. The coefficient of kinetic friction between the reel and the plane is mk = 0.15.

0.3 m

B

Kinematics: Referring to Fig. a, the final velocity of the center O of the spool is (vG)2 = v2rG>IC = 20(0.2) = 4 m>s ; Principle of Impulse and Momentum: The mass moment of inertia of the spool about its mass center is IG = mkG2 = 100(0.2 2) = 4 kg # m2 . Applying the linear impulse and momentum equation along the y axis, Fy dt = m(vy)2 L 0 + N(t) - 100(9.81)(t) = 0

m(vy)1 + ©

N = 981 N

Using this result to write the angular impulse and momentum equation about the IC, laws

or

t2

MIC dt = (HIC)2 Lt1 0 + 0.15(981)(t)(0.5) - 300t(0.3) = - 100(4)(0.2) - 4(20)

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(HIC)1 + ©

t = 9.74 s

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Dissemination

Wide

copyright

a+

0.1 m

P ⫽ 300 N

SOLUTION

A+cB

A G

0.2 m

*19–24. The 30-kg gear is subjected to a force of P = (20t) N, where t is in seconds. Determine the angular velocity of the gear at t = 4 s, starting from rest. Gear rack B is fixed to the horizontal plane, and the gear’s radius of gyration about its mass center O is kO = 125 mm.

150 mm O A

B

SOLUTION Kinematics: Referring to Fig. a, vO = vrO>IC = v(0.15) Principle of Angular Impulse and Momentum: The mass moment of inertia of the gear about its mass center is IO = mkO2 = 30(0.1252) = 0.46875 kg # m2 . Writing the angular impulse and momentum equation about point A shown in Fig. b, t2

(HA)1 + ©

Lt1

MA dt = (HA)2

4s or

20t(0.15)dt = 0.46875v + 30 [v(0.15)] (0.15)

teaching

Web)

L0

laws

0 +

in

4s

= 1.14375v

Wide

copyright

1.5t2 2

Ans.

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v = 21.0 rad>s






not

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permitted.

Dissemination

0

P ⫽ (20t) N

19–25. The double pulley consists of two wheels which are attached to one another and turn at the same rate. The pulley has a mass of 30 kg and a radius of gyration kO = 250 mm. If two men A and B grab the suspended ropes and step off the ledges at the same time, determine their speeds in 4 s starting from rest. The men A and B have a mass of 60 kg and 70 kg, respectively. Assume they do not move relative to the rope during the motion. Neglect the mass of the rope.

350 mm

275 mm O

SOLUTION a+

(HO)1 + © 1 MOdt = (HO)2

0 + 588.6(0.350)(4) - 686.7(0.275)(4) = 30(0.25)2v + 60(0.35v)(0.35) + 70(0.275v)(0.275) v = 4.73 rad>s Ans.

nB = 0.275(4.73) = 1.30 m>s

Ans.

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Dissemination

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nA = 0.35(4.73) = 1.66 m>s






A

B

19–26. If the shaft is subjected to a torque of M = (15t2) N # m, where t is in seconds, determine the angular velocity of the assembly when t = 3 s, starting from rest. Rods AB and BC each have a mass of 9 kg.

A

C

B

SOLUTION Principle of Impulse and Momentum: The mass moment of inertia of the rods 1 1 ml2 = (9) A 12 B = 0.75 kg # m2. Since the about their mass center is IG = 12 12 assembly rotates about the fixed axis, (vG)AB = v(rG)AB = v(0.5) and (vG)BC = v(rG)BC = va 212 + (0.5)2 b = v(1.118). Referring to Fig. a, t2

(Hz)1 + © 3s

0 +

Mz dt = (Hz)2

15t2dt = 9 C v(0.5) D (0.5) + 0.75v + 9 C v(1.118) D (1.118) + 0.75v laws

L0

Lt1

or

c+

Web)

= 15v

teaching

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0

v = 9 rad>s

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permitted.

Dissemination

Ans.






M

1m

1m

(15t2) N m

19–27. The square plate has a mass m and is suspended at its corner A by a cord. If it receives a horizontal impulse I at corner B, determine the location y of the point P about which the plate appears to rotate during the impact. A a

P

a

SOLUTION (a + )

(HG)1 + ©

L

I

m 2 (a + a2) v 0 + Ia b= 12 22 a

+ ) (:

m(vGx)1 + ©

L

Fxdt = m(vGx)2

0 + I = mvG 6I

v =

or

22am

teaching

Web)

laws

I m in

vG =

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Dissemination

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I m 22a = 6I 6 12 am

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vG y¿ = = v

work

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3 22 22 22 a a = a 6 6 3

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y =






a

a

MG dt = (HG)2

Ans.

B

y

*19–28. The crate has a mass mc. Determine the constant speed v0 it acquires as it moves down the conveyor. The rollers each have a radius of r, mass m, and are spaced d apart. Note that friction causes each roller to rotate when the crate comes in contact with it.

d A

SOLUTION The number of rollers per unit length is 1/d. Thus in one second,

v0 rollers are contacted. d v0 in t0 seconds, then the moment r

If a roller is brought to full angular speed of v = of inertia that is effected is v0 v0 1 I¿ = I a b(t0) = a m r2ba bt0 d 2 d

laws

or

Since the frictional impluse is

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F = mc sin u then

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Dissemination

MG dt = (HG)2

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a + (HG)1 + ©

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v0 v0 1 0 + (mc sin u) r t0 = c a m r2 b a b t0 d a b r 2 d

the

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mc (2 g sin u d) a b m

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v0 =






Ans.

30°

19–29. A man has a moment of inertia Iz about the z axis. He is originally at rest and standing on a small platform which can turn freely. If he is handed a wheel which is rotating at V and has a moment of inertia I about its spinning axis, determine his angular velocity if (a) he holds the wheel upright as shown, (b) turns the wheel out, u = 90°, and (c) turns the wheel downward, u = 180°. Neglect the effect of holding the wheel a distance d away from the z axis.

z u V

SOLUTION a) a (Hz)1 =

a

(H)2;

0 + Iv = IzvM + Iv

a

(H)2;

0 + Iv = IzvM + 0

a

(H)2;

0 + Iv = IzvM - Iv

vM = 0

Ans.

b) a (Hz)1 =

vM =

I v Iz

Ans.

c) 2I v Iy

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vM =

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a (Hz)1 =






19–30. Two wheels A and B have masses mA and mB, and radii of gyration about their central vertical axes of kA and kB, respectively. If they are freely rotating in the same direction at V A and V B about the same vertical axis, determine their common angular velocity after they are brought into contact and slipping between them stops.

SOLUTION (©Syst. Angular Momentum)1 = (©Syst. Angular Momentum)2 (mA k2A) vA + (mB k2B) vB = (mA k2A) v¿A + (mB k2B)vB¿ ¿ = vB ¿ = v. then Set vA

mA k2A vA + mB k2B vB

Ans.

mA k2A + mB k2B

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v =






19–31. A 150-lb man leaps off the circular platform with a velocity of vm>p = 5 ft>s, relative to the platform. Determine the angular velocity of the platform afterwards. Initially the man and platform are at rest. The platform weighs 300 lb and can be treated as a uniform circular disk.

vm/p

5 ft/s

10 ft

8 ft

SOLUTION Kinematics: Since the platform rotates about a fixed axis, the speed of point P on the platform to which the man leaps is vP = vr = v(8). Applying the relative velocity equation, vm = vP + vm>P

A+cB

vm = - v(8) + 5

(1)

Fdt L generated during the leap is internal to the system. Thus, angular momentum of the system is conserved about the axis perpendicular to the page passing through point O. The mass moment of inertia of the platform about this axis is or

Conservation of Angular Momentum: As shown in Fig. b, the impulse

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1 1 300 mr2 = a b A 102 B = 465.84 slug # ft2 2 2 32.2

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IO =

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*19–32. z

The space satellite has a mass of 125 kg and a moment of inertia Iz = 0.940 kg # m2, excluding the four solar panels A, B, C, and D. Each solar panel has a mass of 20 kg and can be approximated as a thin plate. If the satellite is originally spinning about the z axis at a constant rate vz = 0.5 rad>s when u = 90°, determine the rate of spin if all the panels are raised and reach the upward position, u = 0°, at the same instant

0.2 m 0.75 m

0.2 m

H1 = H2 1 1 2c (4)(0.15)2 d (5) = 2 c (4)(0.15)2 d v 2 2 + 2[4(0.75v)(0.75)] + c

1 (2)(1.50)2 d v 12

v = 0.0906 rad>s

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y

C

θ = 90°

A

SOLUTION c+

z

B

D x

19–33. 0.20 m 0.65 m

The 80-kg man is holding two dumbbells while standing on a turntable of negligible mass, which turns freely about a vertical axis. When his arms are fully extended, the turntable is rotating with an angular velocity of 0.5 rev>s. Determine the angular velocity of the man when he retracts his arms to the position shown. When his arms are fully extended, approximate each arm as a uniform 6-kg rod having a length of 650 mm, and his body as a 68-kg solid cylinder of 400-mm diameter. With his arms in the retracted position, assume the man as an 80-kg solid cylinder of 450-mm diameter. Each dumbbell consists of two 5-kg spheres of negligible size.

0.3 m

SOLUTION Conservation of Angular Momentum: Since no external angular impulse acts on the system during the motion, angular momentum about the axis of rotation (z axis) is conserved. The mass moment of inertia of the system when the arms are in the fully extended position is (Iz)1 = 2 c 10(0.852) d + 2 c

1 1 (6)(0.652) + 6(0.5252) d + (68)(0.2 2) 12 2

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= 19.54 kg # m2

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(Iz)2 = 2c 10(0.32) d +

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(Hz)1 = (Hz)2

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0.3 m

19–34. The 75-kg gymnast lets go of the horizontal bar in a fully stretched position A, rotating with an angular velocity of vA = 3 rad>s. Estimate his angular velocity when he assumes a tucked position B. Assume the gymnast at positions A and B as a uniform slender rod and a uniform circular disk, respectively.

G

B vA ⫽ 5 rad/s

750 mm

G

A

1.75 m

SOLUTION Conservation of Angular Momentum: Other than the weight, there is no external impulse during the motion. Thus, the angular momentum of the gymnast is conserved about his mass cener G. The mass moments of inertia of the gymnast at 1 1 ml2 = (75)(1.752) = the fully stretched and tucked positions are (IA)G = 12 12 1 1 19.14 kg # m2 and (IB)G = mr2 = (75)(0.3752) = 5.273 kg # m2. Thus, 12 2 (HA)G = (HB)G 19.14(3) = 5.273vB vB = 10.9 rad>s

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19–35. The 2-kg rod ACB supports the two 4-kg disks at its ends. If both disks are given a clockwise angular velocity 1vA21 = 1vB21 = 5 rad>s while the rod is held stationary and then released, determine the angular velocity of the rod after both disks have stopped spinning relative to the rod due to frictional resistance at the pins A and B. Motion is in the horizontal plane. Neglect friction at pin C.

0.75m C

B (VB)1

0.15 m

H1 = H2

1 1 1 2c (4) (0.15)2d(5) = 2c (4)(0.15)2d v + 2[4(0.75 v)(0.75)] + c (2)(1.50)2d v 2 2 12 v = 0.0906 rad>s

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A 0.15 m

SOLUTION c+

0.75m

(VA)1

*19–36. The 5-lb rod AB supports the 3-lb disk at its end. If the disk is given an angular velocity vD = 8 rad>s while the rod is held stationary and then released, determine the angular velocity of the rod after the disk has stopped spinning relative to the rod due to frictional resistance at the bearing A. Motion is in the horizontal plane. Neglect friction at the fixed bearing B.

3 ft

ωD A 0.5 ft

B

SOLUTION ©(HB)1 = ©(HB)2 c

3 1 5 1 3 3 1 a b (3)2 d v + c a b (0.5)2 dv + a B (0.5)2 d (8) + 0 = c a B (3v) (3) 2 32.2 3 32.2 2 32.2 32.2

v = 0.0708 rad s

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19–37. The pendulum consists of a 5-lb slender rod AB and a 10-lb wooden block. A projectile weighing 0.2 lb is fired into the center of the block with a velocity of 1000 ft>s. If the pendulum is initially at rest, and the projectile embeds itself into the block, determine the angular velocity of the pendulum just after the impact.

A

2 ft

SOLUTION

B v

Mass Moment of Inertia: The mass moment inertia of the pendulum and the embeded bullet about point A is (I2)2 =

5 1 5 2 1 10 10 0.2 a b(2 2) + (1 ) + a b (12 + 12) + (2.52) + (2.52) 12 32.2 32.2 12 32.2 32.2 32.2

Conservation of Angular Momentum: Since force F due to the impact is internal to the system consisting of the pendulum and the bullet, it will cancel out. Thus, angular momentum is conserved about point A. Applying Eq. 19–17, we have

laws

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(HA)1 = (HA)2 Web) teaching in

0.2 b(1000)(2.5) = 2.239v2 32.2 States

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a

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v2 = 6.94 rad>s

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1 ft

1 ft

= 2.239 slug # ft2

(mb yb) (rb) = (IA)2 v2

1000 ft/s

19–38. z

The 20-kg cylinder A is free to slide along rod BC. When the cylinder is at x = 0, the 50-kg circular disk D is rotating with an angular velocity of 5 rad> s. If the cylinder is given a slight push, determine the angular velocity of the disk when the cylinder strikes B at x = 600 mm. Neglect the mass of the brackets and the smooth rod.

x 0.15 m

C

A B

G 0.9 m

SOLUTION

0.3 m

Conservation of Angular Momentum: Since no external angular impulse acts on the system during the motion, angular momentum is conserved about the z axis. The mass moments of inertia of the cylinder and the disk about their centers of 1 1 m(3r2 + h2) = (20)[3(0.152) + 0.32] = 0.2625 kg # m2 and mass are (IA)G = 12 12 1 1 (ID)G = mr2 = (50)(0.92) = 20.25 kg # m2. Since the disk rotates about a fixed z 2 2 axis, (vG)A = v(rG)A = v(0.6). Referring to Fig. a,

v

(Hz)1 = (Hz)2 0.2625(5) + 20.25(5) = 20.25v + 0.2625v + 20[v(0.6)](0.6) v = 3.70 rad>s

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D

19–39. The slender bar of mass m pivots at support A when it is released from rest in the vertical position. When it falls and rotates 90°, pin C will strike support B, and pin at A will leave its support. Determine the angular velocity of the bar immediately after the impact. Assume the pin at B will not rebound.

2L 3 C L 3

SOLUTION

L 3

Conservation of Energy: With reference to the datum in Fig. a, V1 = (Vg)1 = L b and V2 = (Vg)2 = W(yG)2 = 0. Since the rod rotates about point A, 2 1 1 1 = c mL2 d v22 = mL2v22. Since the rod is initially at rest, T1 = 0. 2 3 6

W(yG)1 = mga 1 I v2 2 A 2 Then, T2 =

T1 + V1 = T2 + V2 0 + mga

L 1 b = mL2v2 2 + 0 2 6

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3g BL

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(HB)2 = (HB)3 provided

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3gL L 3g 1 1 L L a b + a mL2 b = a mL2 bv3 + mcv3 a b d a b B 4 6 12 BL 12 6 6 destroy will

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3 3g 2B L

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Conservation of Angular Momentum: Since the rod rotates about point A just 3g L 3gL before the impact, (vG)2 = v2rAG = . Also, the rod rotates about a b = BL 2 B 4 L B immediately after the impact, (vG)3 = v3rBG = v3 a b . Angular momentum is 6 conserved about point B. Thus,

B

A

*19–40. The uniform rod assembly rotates with an angular velocity of v0 on the smooth horizontal plane just before the hook strikes the peg P without rebound. Determine the angular velocity of the assembly immediately after the impact. Each rod has a mass of m.

P v0 L 2 L

SOLUTION Center of Mass: Referring to Fig. a,

©xcm x = = ©m

a

L b (m) + L(m) 2 3 = L 2m 4

Thus, the mass moment of the assembly about its mass center G is IG = c

1 L 2 1 L 2 7 mL2 + m a b d + c mL2 + ma b d = mL2 12 4 12 4 24

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Conservation of Angular Momentum: Referring to Fig. b, angular momentum is conserved about P since the sum of the angular impulses about this point is zero. 3 Here, immediately after the impact, (vG)2 = v2rPG = v2 a L b . Thus, 4

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(HP)1 = (HP)2

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7 7 3 3 mL2 b v0 = a mL2 b v2 + 2m c v2 a Lb d a L b 24 24 4 4

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L 2

19–41. A thin disk of mass m has an angular velocity V 1 while rotating on a smooth surface. Determine its new angular velocity just after the hook at its edge strikes the peg P and the disk starts to rotate about P without rebounding.

P r V1

SOLUTION H1 = H2 1 1 a mr 2bv1 = c mr 2 + mr 2dv2 2 2 1 v1 3

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v2 =






19–42. z

The vertical shaft is rotating with an angular velocity of 3 rad>s when u = 0°. If a force F is applied to the collar so that u = 90°, determine the angular velocity of the shaft. Also, find the work done by force F. Neglect the mass of rods GH and EF and the collars I and J. The rods AB and CD each have a mass of 10 kg.

0.3 m

0.3 m

D

0.3 m

E

I

u

0.1 m v

Conservation of Angular Momentum: Referring to the free-body diagram of the assembly shown in Fig. a, the sum of the angular impulses about the z axis is zero. Thus, the angular momentum of the system is conserved about the axis. The mass moments of inertia of the rods about the z axis when u = 0° and 90° are

F

Thus,

or

(Hz)1 = (Hz)2

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laws

3.8(3) = 0.2v2 v2 = 57 rad>s

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17.1 + 2( - 29.43) + UF = 324.9 UF = 367 J

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Dissemination

Principle of Work and Energy: As shown on the free-body diagram of the assembly, Fig. b, W does negative work, while F does positive work. The work of W is UW = - Wh = - 10(9.81)(0.3) = - 29.43 J. The initial and final kinetic energy of the 1 1 1 assembly is T1 = (Iz)1v1 2 = (3.8)(32) = 17.1 J and T2 = (Iz)2v2 2 = 2 2 2 1 (0.2)(572) = 324.9 J. Thus, 2 T1 + ©U1 - 2 = T2

0.1 m

J

H F

1 (10)(0.62) + 10(0.3 + 0.1)2 d = 3.8 kg # m2 12

(Iz)2 = 2 c 10(0.12) d = 0.2 kg # m2


u A

C

SOLUTION

(Iz)1 = 2 c

0.3 m

G

B

19–43. The mass center of the 3-lb ball has a velocity of (vG)1 = 6 ft>s when it strikes the end of the smooth 5-lb slender bar which is at rest. Determine the angular velocity of the bar about the z axis just after impact if e = 0.8.

z 2 ft 2 ft

B

SOLUTION

0.5 ft

Conservation of Angular Momentum: Since force F due to the impact is internal to the system consisting of the slender bar and the ball, it will cancel out. Thus, angular momentum is conserved about the z axis. The mass moment of inertia of the slender (yB)2 5 1 a b A 42 B = 0.2070 slug # ft2. Here, v2 = bar about the z axis is Iz = . 12 32.2 2 Applying Eq. 19–17, we have

r

C mb (yG)1 D (rb) = Iz v2 + C mb (yG)2 D (rb) (yB)2 3 3 b(6)(2) = 0.2070 c d + a b(yG)2(2) 32.2 2 32.2 laws

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(1)

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Coefficient of Restitution:Applying Eq. 19–20, we have in Dissemination

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(yB)2 - (yG)2 6 - 0

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0.8 =

(yB)2 - (yG)2 (yG)1 - (yB)1 of

e =

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(yB)2 = 6.943 ft>s

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Thus, the angular velocity of the slender rod is given by

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(yG)2 = 2.143 ft>s

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(yB)2 6.943 = = 3.47 rad>s 2 2 This

v2 =

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(vG)1

6 ft/s G

(Hz)1 = (Hz)2

a

O

0.5 ft

A

*19–44. A 7-g bullet having a velocity of 800 m>s is fired into the edge of the 5-kg disk as shown. Determine the angular velocity of the disk just after the bullet becomes embedded in it. Also, calculate how far u the disk will swing until it stops. The disk is originally at rest.

O u

v 30

0.2 m

SOLUTION c+

(HO)1 + ©

L

MO dt = (HO)2

0.007(800) cos 30°(0.2) + 0 =

1 (5.007)(0.2)2v + 5.007(0.2v)(0.2) 2

v = 3.23 rad>s

Ans.

T1 + V1 = T2 + V2 1 1 1 (5.007)[3.23(0.2)]2 + [ (5.007)(0.2)2](3.23)2 + 0 = 0 + 0.2(1 - cosu)(5.007)(9.81) 2 2 2 u = 32.8°

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800 m/s

19–45. The 10-lb block slides on the smooth surface when the corner D hits a stop block S. Determine the minimum velocity v the block should have which would allow it to tip over on its side and land in the position shown. Neglect the size of S. Hint: During impact consider the weight of the block to be nonimpulsive.

B

A

Conservation of Energy: If the block tips over about point D, it must at least achieve the dash position shown. Datum is set at point D. When the block is at its initial and final position, its center of gravity is located 0.5 ft and 0.7071 ft above the datum. Its initial and final potential energy are and 10(0.5) = 5.00 ft # lb 10(0.7071) = 7.071 ft # lb. The mass moment of inertia of the block about point D is 1 10 10 a b A 12 + 12 B + a b A 20.52 + 0.52 B 2 = 0.2070 slug # ft2 12 32.2 32.2

1 1 The initial kinetic energy of the block (after the impact) is ID v22 = (0.2070) v22. 2 2 Applying Eq. 18–18, we have

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T2 + V2 = T3 + V3

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1 (0.2070) v22 + 5.00 = 0 + 7.071 2

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v2 = 4.472 rad>s

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Conservation of Angular Momentum:Since the weight of the block and the normal reaction N are nonimpulsive forces, the angular momentum is conserves about point D. Applying Eq. 19–17, we have

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(HD)1 = (HD)2

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(myG)(r¿) = ID v2 provided

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10 b y d(0.5) = 0.2070(4.472) 32.2 part

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destroy

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and

y = 5.96 ft>s






C

A

B

D

S D

C

1 ft

SOLUTION

ID =

v 1 ft

19–46. The two disks each weigh 10 lb. If they are released from rest when u = 30°, determine u after they collide and rebound from each other. The coefficient of restitution is e = 0.75. When u = 0°, the disks hang so that they just touch one another.

2 ft

A

B u

1 ft

SOLUTION Ic =

1 10 10 ( )(1)2 + (1)2 = 0.46584 slug # ft2 2 32.2 32.2

T1 + V1 = T2 + V2 0 + 10(1 - cos 30°) =

1 (0.46584)v21 + 0 2

v1 = 2.398 rad>s Coefficient of restitution: v2 - (- v2) 2.398 - ( - 2.398)

= 0.75 =

(1) teaching

Web)

(vD)A1 - (vD)B1

or

(vD)B2 - (vD)A2

laws

e =

permitted.

Dissemination

Wide

copyright

in

Where, vD is the speed of point D on disk A or B. Note that (vD)B = - (vD)A and (vD)A = rv = (vD)B.

learning.

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instructors

States

World

Solving Eq.(1); v2 = 1.799 rad>s

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T1 + V1 = T2 + V2

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1 (0.46584)(1.799)2 + 0 = 0 + 10(1 - cos u) 2

Ans.

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u = 22.4°






u

1 ft

19–47. 0.3 ft

The pendulum consists of a 10-lb solid ball and 4-lb rod. If it is released from rest when u1 = 0°, determine the angle u2 after the ball strikes the wall, rebounds, and the pendulum swings up to the point of momentary rest. Take e = 0.6.

A u 2 ft

0.3 ft

SOLUTION IA =

1 4 2 10 10 ) (2)2 + ( )(0.3)2 + ( (2.3)2 = 1.8197 slug # ft2 3 32.2 5 32.2 32.2

Just before impact: T1 + V1 = T2 + V2 0 + 0 =

1 [1.8197] v2 - 4(1) - 10(2.3) 2

v = 5.4475 rad>s

laws

or

vP = 2.3(5.4475) = 12.53 ft>s

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permitted.

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vP ¿ 12.529 instructors

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vP ¿ = 7.518 ft>s

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7.518 = 3.2685 rad>s 2.3

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T1 + V1 = T2 + V2

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1 (1.8197)(3.2685)2 = 4(1)(1 - sin u2 ) + 10(2.3)(1 - sin u2 ) 2 any

courses

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u2 = 39.8°






*19–48. The 4-lb rod AB is hanging in the vertical position. A 2-lb block, sliding on a smooth horizontal surface with a velocity of 12 ft/s, strikes the rod at its end B. Determine the velocity of the block immediately after the collision. The coefficient of restitution between the block and the rod at B is e = 0.8.

A

3 ft 12 ft/s B

SOLUTION Conservation of Angular Momentum: Since force F due to the impact is internal to the system consisting of the slender rod and the block, it will cancel out. Thus, angular momentum is conserved about point A. The mass moment of inertia of the 4 1 4 slender rod about point A is IA = a b (32) + (1.52) = 0.3727 slug # ft2. 12 32.2 32.2 (vB)2 Here, v2 = . Applying Eq. 19–17, we have 3 (HA)1 = (HA)2 [mb (vb)1](rb) = IAv2 + [mb (vb)2](rb) (vB)2 2 2 b (12)(3) = 0.3727 c d + a b(vb)2 (3) 32.2 3 32.2

or

[1] laws

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(vB)2 - (vb)2 12 - 0

World

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(vB)2 - (vb)2 (vb)1 - (vB)1

States

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Coefficient of Restitution: Applying Eq. 19–20, we have

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(vb)2 = 3.36 ft s :

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(vB)2 = 12.96 ft s :






Ans.

19–49. The hammer consists of a 10-kg solid cylinder C and 6-kg uniform slender rod AB. If the hammer is released from rest when u = 90° and strikes the 30-kg block D when u = 0°, determine the velocity of block D and the angular velocity of the hammer immediately after the impact. The coefficient of restitution between the hammer and the block is e = 0.6.

u 100 mm B

SOLUTION

150 mm

Conservation of Energy: With reference to the datum in Fig. a, V1 = (Vg)1 = WAB(yGAB)1 + WC(yGC)1 = 0 and V2 = (Vg)2 = - WAB(yGAB)2 - WC(yGC)2 = -6(9.81)(0.25) - 10(9.81)(0.55) = - 68.67 J. Initially, T1 = 0. Since the hammer (vGAB)2 = v2rGAB = v2(0.25) rotates about the fixed axis, and (vGC)2 = v2rGC = v2(0.55). The mass moment of inertia of rod AB and cylinder C 1 1 ml2 = (6)(0.52) = 0.125 kg # m2 and about their mass centers is IGAB = 12 12 1 1 IC = m(3r2 + h2) = (10) C 3(0.052) + 0.152 D = 0.025 kg # m2. Thus, 12 12 1 1 1 1 I v 2 + mAB(vGAB)2 2 + IGC v22 + mC(vGC)2 2 2 GAB 2 2 2 2 or

1 1 1 1 (0.125)v22 + (6) C v2(0.25) D 2 + (0.025)v22 + (10) C v2(0.55) D 2 2 2 2 2

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=

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= 1.775 v22

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Then,

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of

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T1 + V1 = T2 + V2

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0 + 0 = 1.775v22 + ( -68.67)

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v2 = 6.220 rad>s

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(HA)1 = (HA)2

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Conservation of Angular Momentum: The angular momentum of the system is conserved point A. Then,

their

destroy

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0.125(6.220) + 6[6.220(0.25)](0.25) + 0.025(6.220) + 10[6.220(0.55)](0.55) sale

will

= 30vD(0.55) - 0.125v3 - 6[v3(0.25)](0.25) - 0.025v3 - 10[v3(0.55)](0.55) 16.5vD - 3.55v3 = 22.08






(1)

C 50 mm D

laws

T2 =

A

500 mm

19–49. continued Coefficient of Restitution: Referring to Fig. c, the components of the velocity of the impact point P just before and just after impact along the line of impact are

C (vP)x D 2 = (vGC)2 = v2rGC = 6.220(0.55) = 3.421 m>s :

C (vP)x D 3 = (vGC)3 =

and

v3rGC = v3 (0.55) ; . Thus, + :

e =

(vD)3 - C (vP)x D 3

C (vP)x D 2 - (vD)2

0.6 =

(vD)3 -

C - v3(0.55) D

3.421 - 0

(2)

(vD)3 + 0.55v3 = 2.053 Solving Eqs. (1) and (2), v3 = 0.934 rad>s

Ans.

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laws

or

(vD)3 = 1.54 m>s






19–50. The 6-lb slender rod AB is originally at rest, suspended in the vertical position. A 1-lb ball is thrown at the rod with a velocity v = 50 ft>s and strikes the rod at C. Determine the angular velocity of the rod just after the impact. Take e = 0.7 and d = 2 ft.

A

d 3 ft

C

SOLUTION

v = 50 ft/s

a + (HA)1 = (HA)2 B

1 6 1 1 b (50)(2) = c a b (3)2 d v2 + (n )(2) a 32.2 3 32.2 32.2 BL e = 0.7 =

nC - nBL 50 - 0

nC = 2v2 Thus, v2 = 7.73 rad s

or

Ans.

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permitted.

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copyright

in

teaching

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laws

nBL = - 19.5 ft s






19–51. The solid ball of mass m is dropped with a velocity v1 onto the edge of the rough step. If it rebounds horizontally off the step with a velocity v2, determine the angle u at which contact occurs. Assume no slipping when the ball strikes the step. The coefficient of restitution is e. u

v1 r v2

SOLUTION Conservation of Angular Momentum: Since the weight of the solid ball is a nonimpulsive force, then angular momentum is conserved about point A. The mass 2 moment of inertia of the solid ball about its mass center is IG = mr2. Here, 5 y2 cos u v2 = . Applying Eq. 19–17, we have r (HA)1 = (HA)2

C mb (yb)1 D (r¿) = IG v2 + C mb (yb)2 D (r–) y2 cos u 2 b + (my2)(r cos u) (my1)(r sin u) = a mr2 b a r 5 or

y2 5 = tan u y1 7 in

teaching

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laws

(1)

Dissemination

Wide

copyright

Coefficient of Restitution:Applying Eq. 19–20, we have

the

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instructors

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permitted.

0 - (yb)2 (yb)1 - 0

learning.

is

of

e =

by

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on

- (y2 sin u) -y1 cos u

for

work

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the

e =

(2)

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is

work

solely

of

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the

(including

y2 e cos u = y1 sin u

any

sale

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of

and

5 e cos u tan u = 7 sin u 7 tan2 u = e 5

courses

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provided

integrity

of

and

work

Equating Eqs. (1) and (2) yields

u = tan - 1 ¢






7 e≤ A5

Ans.

*19–52.

G

A

SOLUTION Conservation of Angular Momentum: Referring to Fig. a, the sum of the angular impulses about point A is zero. Thus, angular momentum of the wheel is conserved about this point. Since the wheel rolls without slipping, (vG)1 = v1r = v1(0.15) and (vG)2 = v2r = v2(0.15). The mass moment of inertia of the wheel about its mass center is IG = mkG2 = 50(0.1252) = 0.78125 kg # m2. Thus, (HA)1 = (HA)2 50[v1(0.15)](0.125) + 0.78125v1 = 50[v2(0.15)](0.15) + 0.78125v2 v1 = 1.109v2

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

Conservation of Energy: With reference to the datum in Fig. a, V2 = (Vg)2 = W(yG)2 = 0 and V3 = (Vg)3 = W(yG)3 = 50(9.81)(0.025) = 12.2625 J. Since the wheel is required to be at rest in the final position, T3 = 0. The initial kinetic energy 1 1 1 1 of the wheel is T2 = m(vG)22 + IGv2 2 = (50)[v2(0.15)]2 + (0.78125)(v2 2) = 2 2 2 2 0.953125v22. Then

or

(1)

learning.

is

of

the

not

instructors

T2 + V2 = T3 + V3

the

by

and

use

United

on

0.953125v2 2 + 0 = 0 + 12.2625

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protected

the

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for

work

student

v2 = 3.587 rad>s

this

assessing

is

work

solely

Substituting this result into Eq. (1), we obtain

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and

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integrity

of

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work

v1 = 3.98 rad>s






v1

150 mm

The wheel has a mass of 50 kg and a radius of gyration of 125 mm about its center of mass G. Determine the minimum value of the angular velocity v1 of the wheel, so that it strikes the step at A without rebounding and then rolls over it without slipping.

Ans.

25 mm

19–53. The wheel has a mass of 50 kg and a radius of gyration of 125 mm about its center of mass G. If it rolls without slipping with an angular velocity of v1 = 5 rad>s before it strikes the step at A, determine its angular velocity after it rolls over the step. The wheel does not loose contact with the step when it strikes it.

G

A

SOLUTION Conservation of Angular Momentum: Referring to Fig. a, the sum of the angular impulses about point A is zero. Thus, angular momentum of the wheel is conserved about this point. Since the wheel rolls without slipping, (vG)1 = v1r = (5)(0.15) = 0.75 m>s and v2 = v2r = v2(0.15). The mass moment of inertia of the wheel about its mass center is IG = mkG2 = 50(0.1252) = 0.78125 kg # m2. Thus, (HA)1 = (HA)2 50(0.75)(0.125) + 0.78125(5) = 50[v2(0.15)](0.15) + 0.78125v2 v2 = 4.508 rad>s

(1)

of

the

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States

World

permitted.

Dissemination

Wide

copyright

in

teaching

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laws

or

Conservation of Energy: With reference to the datum in Fig. a, V2 = (Vg)2 = W(yG)2 = 0 and V3 = (Vg)3 = W(yG)3 = 50(9.81)(0.025) = 12.2625 J. The initial 1 1 1 kinetic energy of the wheel is T = mvG2 + IGv2 = (50)[v(0.15)]2 + 2 2 2 1 (0.78125)v2 = 0.953125v2. Thus, T2 = 0.953125v22 = 0.953125(4.5082) = 19.37 J 2 and T3 = 0.953125v32.

and

use

United

on

learning.

is

T2 + V2 = T3 + V3

for

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student

the

by

19.37 + 0 = 0.953125v32 + 12.2625

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is

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the

(including

v3 = 2.73 rad>s






v1

150 mm

Ans.

25 mm

19–54. ω1

The disk has a mass m and radius r. If it strikes the step without rebounding, determine the largest angular velocity the disk can have and not lose contact with the step.

r 1r 8

SOLUTION (HA)1 =

1 m r2 (v1) + m(v1 r)(r - h) 2

(HA)2 =

1 m r2 (v2) + m(v2 r)(r) 2

(HA)1 = (HA)2 1 3 c m r2 + m r (r - h) d v1 = m r2 v2 2 2

laws

or

3 3 a r - h b v1 = r v2 2 2 teaching

Web)

W cos u - F = m(v22 r) in

b+ ©Fn = m an ;

r - h b - m(v22 r) r instructors

States

World

permitted.

Dissemination

Wide

copyright

F = mg a

is

of

the

not

3 r - h 2 r - h 2 2 F = mg a b - m ra b P Q v21 r r 3 for

work

student

the

by

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use

United

on

learning.

2

the

(including

1 r; also note that for maximum v1 F will approach zero. Thus 8 provided

integrity

of

and

2

v21

part

the

is

any

3 r r 2 8 r

This

2

courses

r

2 - mr 3






will

g r

sale

v1 = 1.02

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destroy

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mg

1 r 8

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Set h =

Ans.

19–55. A solid ball with a mass m is thrown on the ground such that at the instant of contact it has an angular velocity V 1 and velocity components 1vG2x1 and 1vG2y1 as shown. If the ground is rough so no slipping occurs, determine the components of the velocity of its mass center just after impact. The coefficient of restitution is e.

V1

(vG)x1 G r

SOLUTION Coefficient of Restitution (y direction):

A+TB

e =

0 - (yG)y 2

c

(yG)y2 = - e(yG)y1 = e(yG)y1

(yG)y1 - 0

Ans.

Conservation of angular momentum about point on the ground: (c + )

(HA)1 = (HA)2

2 2 - mr 2v1 + m(vG)x 1r = mr 2v2 + m(vG)x 2 r 5 5

laws

or

Since no slipping, (vG)x2 = v2 r then, Web) teaching in

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v2 =

2 v rb 5 1 copyright

5a (vG)x 1 -

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Therefore

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5 2 a (yG)x 1 - v1 rb 7 5

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Ans.

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the

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the

(yG)x 2 =






(vG)y1

*19–56. The pendulum consists of a 10-lb sphere and 4-lb rod. If it is released from rest when u = 90°, determine the angle u of rebound after the sphere strikes the floor. Take e = 0.8. 0.3 ft 2 ft

SOLUTION IA =

O 0.3 ft

4 1 2 10 10 a b(2)2 + a b (0.3)2 + a b (2.3)2 = 1.8197 slug # ft2 3 32.2 5 32.2 32.2

Just before impact: Datum through O. T1 + V1 = T2 + V2 0 + 4(1) + 10(2.3) =

1 (1.8197)v2 + 0 2

v2 = 5.4475 rad>s

teaching

Web)

laws

or

v = 2.3(5.4475) = 12.529 ft>s

Wide

copyright

in

Since the floor does not move,

World

permitted.

Dissemination

(vP) - 0 0 - ( -12.529)

of

the

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instructors

e = 0.8 =

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A+cB

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(vP)3 = 10.023 ft>s

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the

10.023 = 4.358 rad>s 2.3

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v3 =

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T3 + V3 = T4 + V4

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provided

integrity

of

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work

this

1 (1.8197)(4.358)2 + 0 = 4(1 sin u1) + 10(2.3 sin u1) 2 Ans.

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any

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u1 = 39.8°






u

20–1. z

At a given instant, the satellite dish has an angular motion # v1 = 6 rad>s and v1 = 3 rad>s2 about the z axis. At this same instant u = 25°, the angular motion about the x axis is # v2 = 2 rad>s, and v2 = 1.5 rad>s2. Determine the velocity and acceleration of the signal horn A at this instant.

V1, V1

A

SOLUTION

1.4 m

Angular Velocity: The coordinate axes for the fixed frame (X, Y, Z) and rotating frame (x, y, z) at the instant shown are set to be coincident. Thus, the angular velocity of the satellite at this instant (with reference to X, Y, Z) can be expressed in terms of i, j, k components.

V2, V2 x

v = v1 + v2 = {2i + 6k} rad>s Angular Acceleration: The angular acceleration a will be determined by investigating separately the time rate of change of each angular velocity component with respect to the fixed XYZ frame. v2 is observed to have a constant direction from the rotating xyz frame if this frame is rotating at Æ = v1 = {6k} rad>s. # Applying Eq. 20–6 with (v2)xyz = {1.5i} rad>s2. we have

teaching

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laws

or

# # v2 = (v2)xyz + v1 * v2 = 1.5i + 6k * 2i = {1.5i + 12j} rad>s2

Wide

copyright

in

Since v1 is always directed along the Z axis (Æ = 0), then

not

instructors

States

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permitted.

Dissemination

# # v1 = (v1)xyz + 0 * v1 = {3k} rad>s2

United

on

learning.

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Thus, the angular acceleration of the satellite is

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for

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# # a = v1 + v2 = {1.5i + 12j + 3k} rad>s2

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is

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the

Velocity and Acceleration: Applying Eqs. 20–3 and 20–4 with the v and a obtained above and rA = {1.4 cos 25°j + 1.4 sin 25°k} m = {1.2688j + 0.5917k} m, we have

Ans.

destroy will

their

aA = a * rA + v * (v * rA)

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and

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part

= { -7.61i - 1.18j + 2.54k} m>s

the

is

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provided

integrity

of

and

work

this

vA = v * rA = (2i + 6k) * (1.2688j + 0.5917k)

sale

= (1.3i + 12j + 3k) * (1.2688j + 0.5917k) + (2i + 6k) * [(2i + 6k) * (1.2688j + 0.5917k)] = {10.4i - 51.6j - 0.463k} m>s2






Ans.

O

u

25 y

20–2. z

Gears A and B are fixed, while gears C and D are free to rotate about the shaft S. If the shaft turns about the z axis at a constant rate of v1 = 4 rad>s, determine the angular velocity and angular acceleration of gear C.

80 mm 40 mm

V1

y S

80 mm 160 mm

C

D

B A

SOLUTION

x

The resultant angular velocity v = v1 + v2 is always directed along the instantaneous axis of zero velocity IA. v = v1 + v2 2 25

1

vj -

25

vk = 4k + v2 j

Equating j and k components 1 25

v = 4

v = - 8.944 rad>s or

-

laws

(- 8.944) = - 8.0 rad>s

Web)

25

teaching

2

25

Wide

( -8.944)k = {-8.0j + 4.0k} rad>s

Ans. Dissemination

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permitted.

25

( -8.944)j -

World

2

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the

not

instructors

v =

States

Hence

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v2 =

use

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learning.

is

For v2, Æ = v1 = {4k} rad>s.

for

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by

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# # (v2)XYZ = (v2)xyz + Æ * v2

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the

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= 0 + (4k) * ( -8j)

part

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For v1, Æ = 0.

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this

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= {32i} rad>s2

destroy

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# # (v1)XYZ = (v1)xyz + Æ * v1 = 0 + 0 = 0 sale

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# # # a = v = (v1)XYZ + (v2)XYZ a = 0 + (32i) = {32i} rad>s2






Ans.

20–3. z

The ladder of the fire truck rotates around the z axis with an angular velocity v1 = 0.15 rad>s, which is increasing at 0.8 rad>s2. At the same instant it is rotating upward at a constant rate v2 = 0.6 rad>s. Determine the velocity and acceleration of point A located at the top of the ladder at this instant.

A

30⬚

V1 40 ft

SOLUTION v = v1 + v2 = 0.15k + 0.6i = {0.6i + 0.15k} rad>s Angular acceleration: For v1, v = v1 = {0.15k} rad>s. # # (v2)XYZ = (v2)xyz + v * v2

V2 x

= 0 + (0.15k) * (0.6i) = {0.09j} rad>s2 For v1, Æ = 0. # # (v1)XYZ = (v1)xyz + v * v1 = (0.8k) + 0 = {0.8k} rad>s2

or

# # # a = v = (v1)XYZ + (v2)XYZ

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laws

a = 0.8k + 0.09j = {0.09j + 0.8k} rad>s2

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rA = 40 cos 30°j + 40 sin 30°k = {34.641j + 20k} ft

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= (0.6i + 0.15k) * (34.641j + 20k)

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= { -5.20i - 12j + 20.8k} ft>s

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aA = a * r + v * vA

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= (0.09j + 0.8k) * (34.641j + 20k) + (0.6i + 0.15k) * ( -5.20i - 12j + 20.8k)

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= { -24.1i - 13.3j - 7.20k} ft>s2






Ans.

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*20–4. z

The ladder of the fire truck rotates around the z axis with an angular velocity of v1 = 0.15 rad>s, which is increasing at 0.2 rad>s2. At the same instant it is rotating upwards at v2 = 0.6 rad>s while increasing at 0.4 rad>s2. Determine the velocity and acceleration of point A located at the top of the ladder at this instant.

A

30⬚

V1 40 ft

SOLUTION rA/O = 40 cos 30°j + 40 sin 30°k rA/O = {34.641j + 20k} ft Æ = v1 k + v2 i = {0.6i + 0.15k} rad>s

V2 x

# # # v = v1 k + v2 i + v1 k * v2 i # Æ = 0.2k + 0.4i + 0.15k * 0.6i = {0.4i + 0.09j + 0.2k} rad>s2 vA = Æ * rA/O = (0.6i + 0.15k) * (34.641j + 20k) vA = {- 5.20i - 12j + 20.8k} ft>s

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# aA = Æ * (Æ * rA/O) + v * rA/O in

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aA = (0.6i + 0.15k) * [(0.6i + 0.15k) * (34.641j + 20k)]

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+ (0.4i + 0.09j + 0.2k) * (34.641j + 20k)

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aA = {- 3.33i - 21.3j + 6.66k} ft>s2

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20–5. Gear B is connected to the rotating shaft, while the plate gear A is fixed. If the shaft is turning at a constant rate of vz = 10 rad>s about the z axis, determine the magnitudes of the angular velocity and the angular acceleration of gear B. Also, determine the magnitudes of the velocity and acceleration of point P.

z 200 mm

ω z = 10 rad/s P 50 mm B

SOLUTION vz = 10 rad>s

A

x

vy = - 10 tan 75.96° = - 40 rad>s vx = 0 v = {- 40j + 10k} rad>s v = 2 (-40)2 + (10)2 = 41.2 rad>s

Ans.

rP = {0.2j + 0.05k} m vP = v * rP = ( -40j + 10k) * (0.2j + 0.05k)

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vP = { -4i} m>s

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vP = 4.00 m>s

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# a = v = 400 rad>s2

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= 0 + (10k) * ( - 40j + 10k) = {400i} rad>s2

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= {-60j - 80k} m>s2

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aP = a * rP + v * vP = (400i) * (0.2j + 0.05k) + (- 40j + 10k) * (- 4i)

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50 mm

y

20–6. Gear A is fixed while gear B is free to rotate on the shaft S. If the shaft is turning about the z axis at vz = 5 rad>s, while increasing at 2 rad>s2, determine the velocity and acceleration of point P at the instant shown. The face of gear B lies in a vertical plane.

z Vz

P 80 mm S

A

y

80 mm

160 mm

SOLUTION Æ = {5k - 10j} rad>s # Æ = {50i - 4j + 2k} rad>s2

x

vP = Æ * rP (5k - 10j * (160j + 80k)

vP

vP = {-1600i} mm>s = {-1.60i} m>s # aP = Æ * vP + Æ * rP

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aP = {50i - 4j + 2k} * (160j + 80k) + ( -10j + 5k) * (-1600i)

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aP = {-640i - 12000j - 8000k} mm>s2 aP = {-0.640i - 12.0j - 8.00k} m>s2

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20–7. At a given instant, the antenna has an angular motion # v1 = 3 rad>s and v1 = 2 rad>s2 about the z axis. At this same instant u = 30°, the angular motion about the x axis is # v2 = 1.5 rad>s, and v2 = 4 rad>s2. Determine the velocity and acceleration of the signal horn A at this instant. The distance from O to A is d = 3 ft.

z v1 v· 1 v2

d

v· 2

A u

30

O x

SOLUTION rA = 3 cos 30°j + 3 sin 30°k = {2.598j + 1.5k} ft Æ = v1 + v2 = 3k + 1.5i v A = Æ * rA vA = (3k + 1.5i) * (2.598j + 1.5k) = -7.794i + 3.897k - 2.25j = { -7.79i - 2.25j + 3.90k} ft>s # # # Æ = v 1 + v2

Ans.

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= 4i + 4.5j + 2k

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# aA = v * rA + Æ * vA

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aA = (4i + 4.5j + 2k) * (2.598j + 1.5k) + (3k + 1.5i) * ( -7.794i - 2.25j + 3.879k)

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aA = {8.30i - 35.2j + 7.02k} ft>s2






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*20–8. The cone rolls without slipping such that at the instant # shown vz = 4 rad>s and vz = 3 rad>s2. Determine the velocity and acceleration of point A at this instant.

z Vz V· z

4 rad/s 3 rad/s2 A B

SOLUTION

x

2 ft

Angular velocity: The resultant angular velocity v = v1 + v2 is always directed along the instantaneous axis of zero velocity (y axis).

20 y

v = v1 + v2 v j = 4k + (v2 cos 20°j + v2 sin 20°k) v j = v2 cos 20°j + (4 + v2 sin 20°) k Equating j and k components: 4 + v2 sin 20° = 0

v2 = - 11.70 rad/s

laws

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v = - 11.70 cos 20° = - 10.99 rad/s

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Hence, v = {- 10.99j} rad>s

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v2 = -11.70 cos 20°j + ( -11.70 sin 20°) k = {- 10.99j - 4k} rad>s not

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# # = C (v1)xyz + Æ * v1 D + C (v2)xyz + Æ * v2 D

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# v = [3k + 0] + [(-8.2424j - 3k) + 4k * ( -10.99j - 4k)] = {43.9596i - 8.2424j} rad>s rA = 2 cos 40°j + 2 sin 40°k = {1.5321j + 1.2856k} ft vA = v * rA = ( -10.99j) * (1.5321j + 1.2856k) = {-14.1i} ft>s

Ans.

aA = a * rA + v * vA = (43.9596i - 8.2424j) * (1.5321j + 1.2856k) + (- 10.99j) * ( - 14.1i) = {- 10.6i - 56.5j - 87.9k} ft>s2






Ans.

20–9. The cone rolls without slipping such that at the instant # shown vz = 4 rad>s and vz = 3 rad>s2. Determine the velocity and acceleration of point B at this instant.

z Vz V· z

4 rad/s 3 rad/s2 A B

SOLUTION

x

2 ft

Angular velocity: The resultant angular velocity v = v1 + v2 is always directed along the instantaneous axis of zero velocity (y axis).

20 y

v = v1 + v2 v j = 4k + (v2 cos 20°j + v2 sin 20°k) v j = v2 cos 20°j + (4 + v2 sin 20°) k Equating j and k components: 4 +v2 sin 20° = 0

v2 = - 11.70 rad>s

laws

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v = - 11.70 cos 20° = - 10.99 rad>s

in

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Hence, v = { - 10.99j} rad>s

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v2 = -11.70 cos 20°j + ( -11.70 sin 20°) k = {- 10.99j - 4k} rad>s not

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Angular acceleration:

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3 B cos 20°j - 3k = { -8.2424j - 3k} rad>s2 sin 20°

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# v = [3k + 0] + [( -8.2424j - 3k) + 4k * ( -10.99j - 4k)] = {43.9596i - 8.2424j} rad>s rB = 2 sin 20°i + 2 cos 20°j + 2 sin 20° cos 20°k = -0.68404i + 1.8794j + 0.64279k v B = v * rB = ( -10.99j) * ( -0.68404i + 1.8794j + 0.64279k) = -7.0642i - 7.5176k = { -7.06i - 7.52k} ft>s

Ans.

aB = a * rB + v * vB = (43.9596i - 8.2424j) * (- 0.68404i + 1.8794j + 0.64279k) + (-10.99j) * (- 7.0642i - 7.5176k) = {77.3i - 28.3j - 0.657k} ft>s2 © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

20–10. At the instant when u = 90°, the satellite’s body is rotating with an angular velocity of v1 = 15 rad>s and angular # acceleration of v1 = 3 rad>s2. Simultaneously, the solar panels rotate with an angular velocity of v2 = 6 rad>s and # angular acceleration of v2 = 1.5 rad>s2. Determine the velocity and acceleration of point B on the solar panel at this instant.

z V1, V1

6 ft

SOLUTION

O

Here, the solar panel rotates about a fixed point O. The XYZ fixed reference frame x is set to coincide with the xyz rotating frame at the instant considered. Thus, the angular velocity of the solar panel can be obtained by vector addition of v1 and v2.

1 ft 1 ft

# # # a = v = v1 + v2

laws

or

If we set the xyz frame to have an angular velocity of Æ = v1 = [15k] rad>s, then the direction of v2 will remain constant with respect to the xyz frame, which is along the y axis. Thus,

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# # v2 = (v2)xyz + v1 * v2 = 1.5j + (15k * 6j) = [-90i + 1.5j] rad>s2 Dissemination

Since v1 is always directed along the Z axis when Æ = v1, then

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a = 3k + ( -90i + 1.5j)

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= [-90i + 1.5j + 3k] rad>s2

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When u = 90°, rOB = [ -1i + 6j] ft. Thus,

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vB = v * rOB = (6j + 15k) * ( -1i + 6j)

and aB = a * rOB + v * (v * rOB) = ( -90i + 1.5j + 3k) * ( -1i + 6j) + (6j + 15k) * [(6j + 15k) * ( -1i + 6j)]






V2, V2 y

The angular acceleration of the solar panel can be determined from

= [243i - 1353j + 1.5k] ft>s2

u B

v = v1 + v2 = [6j + 15k] rad>s

= [-90i - 15j + 6k] ft>s

A

Ans.

20–11. At the instant when u = 90°, the satellite’s body travels in the x direction with a velocity of vO = 5500i6 m>s and acceleration of a O = 550i6 m>s2. Simultaneously, the body also rotates with an angular velocity of v1 = 15 rad>s and # angular acceleration of v1 = 3 rad>s2. At the same time, the solar panels rotate with an angular velocity of v2 = 6 rad>s # and angular acceleration of v2 = 1.5 rad>s2 Determine the velocity and acceleration of point B on the solar panel.

z V1, V1

6 ft

O x

A

u B 1 ft 1 ft

SOLUTION

y

The XYZ translating reference frame is set to coincide with the xyz rotating frame at the instant considered. Thus, the angular velocity of the solar panel at this instant can be obtained by vector addition of v1 and v2. v = v1 + v2 = [6j + 15k] rad>s The angular acceleration of the solar panel can be determined from

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# a = v = v1 + v2

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# # v2 = (v2)xyz + v1 * v2 = 1.5j + (15k * 6j) = [-90i + 15j] rad>s2

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If we set the xyz frame to have an angular velocity of Æ = v1 = [15k] rad>s, then the direction of v2 will remain constant with respect to the xyz frame, which is along the y axis. Thus,

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Since v1 is always directed along the Z axis when Æ = v1, then

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# # v1 = (v1)xyz + v1 * v1 = [3k] rad>s2

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Thus, a = 3k + ( - 90i + 1.5j)

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When u = 90°, rB>O = [- 1i + 6j] ft. Since the satellite undergoes general motion, then vB = vO + v * rB>O = (500i) + (6j + 15k) * ( -1i + 6j) = [410i - 15j + 6k] ft>s

Ans.

and aB = aO + a * rB>O + v * (v * rB>O) = 50i + ( -90i + 1.5j + 3k) * (- 1i + 6j) + (6j + 15k) * [(6j + 15k) * ( -1i + 6j)] = [293i - 1353j + 1.5k] ft>s2






V2, V2

Ans.

*20–12. z

The disk B is free to rotate on the shaft S. If the shaft is turning about the z axis at vz = 2 rad>s, while increasing at 8 rad>s2, determine the velocity and acceleration of point A at the instant shown.

8 rad/s2 2 rad/s S

SOLUTION

A

x 800 mm

Angular Velocity: The coordinate axes for the fixed frame (X, Y, Z) and rotating

400 mm y

frame (x, y, z) at the instant shown are set to be coincident Thus, the angular velocity

400 mm

of the disk at this instant (with reference to X, Y, Z) can be expressed in terms of i,

j,

k

components.

The

velocity

of

the

center

of

the

disk

is

v = vz(0.8) = 2(0.8) = 1.60 m>s. Since the disk rolls without slipping, its spinning v 1.60 anngular velocity is given by vs = = = 4 rad>s and is directed towards – j. r 0.4 Thus, vs = {- 4j} rad>s.

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v = vz + vz = { - 4j + 2k} rad>s

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investigating separately the time rate of change of each angular velocity component not

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with respect to the fixed XYZ frame, vs is observed to have a constant direction from

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the rotating xyz frame if this frame is rotating at Æ = vz = {2k} rad>s.The tangential # acceleration of the center of the disk is a = vz (0.8) = 8(0.8) = 6.40 m>s2.

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Since the disk rolls without slipping, its spinning anngular acceleration is given by a 6.40 # # vs = = = 16 rad>s2 and directed towards – j. Thus, (vs)xyz = {- 16j} rad>s2. r 0.4 Applying Eq. 20–6, we have

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# # vs = (vs)xyz + vz * vs = - 16j + 2k * ( -4j) = {8i - 16j} rad>s2

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Since vz is always directed along Z axis (Æ = 0), then # # vz = (vz)xyz + 0 * vz = {8k} rad>s2 Thus, the angular acceleration of the disk is # # a = vz + vz = {8i - 16j + 8k} rad>s2 Velocity and Acceleration: Applying Eqs. 20–3 and 20–4 with the v and a obtained above and rA = {- 0.4i + 0.8j} m, we have vA = v * rA = ( - 4j + 2k) * ( - 0.4i + 0.8j) = {- 1.60i - 0.800j - 1.60k} m>s

Ans.

aA = a * rA + v * (v * rA) = (8i - 16j + 8k) * ( - 0.4i + 0.8j) + ( - 4j + 2k) * [(-4j + 2k) * ( -0.4i + 0.8j)] = {1.6i - 6.40j - 6.40k} m s2






Ans.

20–13. The disk spins about the arm with an angular velocity of vs = 8 rad>s, which is increasing at a constant rate of # vs = 3 rad>s2 at the instant shown. If the shaft rotates with a constant angular velocity of vp = 6 rad>s, determine the velocity and acceleration of point A located on the rim of the disk at this instant.

z

1.5 ft

The XYZ fixed reference frame is set to coincide with the rotating xyz reference frame at the instant considered. Thus, the angular velocity of the disk at this instant can be obtained by vector addition of vs and vp.

x

vp ⫽ 6 rad/s

v = vs + vp = [ -8j - 6k] rad>s The angular acceleration of the disk is determined from # # # a = v = vs + vp

or

If we set the xyz rotating frame to have an angular velocity of Æ = vp = [- 6k] rad>s, the direction of vs will remain unchanged with respect to the xyz rotating frame which is along the y axis. Thus,

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# # vs = (vs)xyz + vp * vs = - 3j + ( -6k) * ( -8j) = [- 48i - 3j] rad>s2

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# # vp = A vp B xyz + vp * vp = 0 + 0 = 0

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Here, rA = [1.5j + 0.5k] ft, so that

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vA = v * rA = (- 8j - 6k) * (1.5j + 0.5k) = [5i] ft>s

= ( - 48i - 3j) * (1.5j + 0.5k) + ( -8j - 6k) * [( - 8j - 6k) * (1.5j + 0.5k)] = [ - 1.5i - 6j - 32k] ft>s2






Ans.

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Dissemination

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copyright

Since vp is always directed along the Z axis where Æ = vp and since it has a # constant magnitude, A vp B xyz = 0. Thus,

and

A

O

SOLUTION

0.5 ft

vs ⫽ 8 rad/s vs ⫽ 3 rad/s2

y

20–14. The wheel is spinning about shaft AB with an angular velocity of vs = 10 rad>s, which is increasing at a constant # rate of v2 = 6 rad>s2, while the frame precesses about the z axis with an angular velocity of vp = 12 rad>s, which is # increasing at a constant rate of vp = 3 rad>s2. Determine the velocity and acceleration of point C located on the rim of the wheel at this instant.

vs ⫽ 10 rad/s z vs ⫽ 6 rad/s2 0.15 m A

C x

SOLUTION The XYZ fixed reference frame is set to coincide with the rotating xyz reference frame at the instant considered. Thus, the angular velocity of the wheel at this instant can be obtained by vector addition of vs and vp.

vp ⫽ 12 rad/s vp ⫽ 3 rad/s2

v = vs + vp = [10j + 12k] rad>s The angular acceleration of the disk is determined from # # # a = v = vs + vp

laws

or

If we set the xyz rotating frame to have an angular velocity of Æ = vp = [12k] rad>s, the direction of vs will remain unchanged with respect to the xyz rotating frame which is along the y axis. Thus,

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# # vs = (vs)xyz + vp * vs = 6j + (12k) * (10j) = [-120i + 6j] rad>s2 Dissemination

Since vp is always directed along the Z axis where Æ = vp, then

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Thus, a = 1 -120i + 6j2 + 3k = [- 120i + 6j + 3k] rad>s2

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# # vp = A vp B xyz + vp * vp = 3k + 0 = [3k] rad>s2

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Here, rC = [0.15i] m, so that

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vC = v * rC = (10j + 12k) * (0.15i) = [1.8j - 1.5k] m>s

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aC = a * rC + v * (v * rC)

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= (-120i + 6j + 3k) * (0.15i) + (10j + 12k) * [(10j + 12k) * (0.15i)] = [- 36.6i + 0.45j - 0.9k] m>s2






Ans.

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20–15. At the instant shown, the tower crane rotates about the z axis with an angular velocity v1 = 0.25 rad>s, which is increasing at 0.6 rad>s2. The boom OA rotates downward with an angular velocity v2 = 0.4 rad>s, which is increasing at 0.8 rad>s2. Determine the velocity and acceleration of point A located at the end of the boom at this instant.

z

v1

0.25 rad/s A

40 ft

SOLUTION

O

v = v1 + v2 = {-0.4 i + 0.25k} rad>s

x v2

30

0.4 rad/s

Æ = {0.25 k} rad>s y

# # v = (v)xyz + Æ * v = ( - 0.8 i + 0.6k) + (0.25k) * (-0.4 i + 0.25k) = {-0.8i - 0.1j + 0.6k} rad>s2 rA = 40 cos 30°j + 40 sin 30°k = {34.64j + 20k} ft vA = v * rA = (1 - 0.4 i + 0.25 k) * (34.64j + 20k) vA = {- 8.66i + 8.00j - 13.9k} ft>s

Ans.

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aA = a # rA + v * vA = (- 0.8i-0.1j + 0.6k) * (34.64j + 20k) + ( -0.4i + 0.25k) * (-8.66i + 8.00j - 13.9k) teaching

aA = {- 24.8i + 8.29j - 30.9k} ft>s2

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*20–16. If the top gear B rotates at a constant rate of v , determine the angular velocity of gear A, which is free to rotate about the shaft and rolls on the bottom fixed gear C.

z

B rB

ω

A

SOLUTION

O

C

Also, i vP = vA * ( - rB j + h2k) = 3 vAx 0

j vAy -rB

k vAz 3 h2

x

= (vAy h2 + vAz rB)i - (vAx h2)j - vAx rB k Thus, vrB = vAy h2 + vAz rB

laws

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vAy = a

rC rB h1v ba b h1 rC h2 + rB h1

rC rB h1 v rB h1v ba bj + a bk h1 rC h2 + rB h1 rC h2 + rB h1

y h1

vP = vk * (-rB j) = vrB i

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h2

Ans.

rC

20–17. z

When u = 0°, the radar disk rotates about the y axis with an # angular velocity of u = 2 rad>s, increasing at a constant rate $ of u = 1.5 rad>s2. Simultaneously, the disk also precesses about the z axis with an angular velocity of vp = 5 rad>s, # increasing at a constant rate of vP = 3 rad>s2. Determine the velocity and acceleration of the receiver A at this instant.

20 ft

A u

x

The XYZ fixed reference frame is set to coincide with the rotating xyz reference frame at the instant considered. Thus, the# angular velocity of the radar disk at this instant can be obtained by vector addition of u and vp. # v = u + vp = [2j + 5k] rad>s The angular acceleration of the disk is determined from $ # # v = u + vp

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If we set the xyz frame to have an angular velocity of Æ = vp = [5k] rad>s, the # direction of u will remain unchanged with respect to the xyz rotating frame which is along the y axis. Thus, $ $ # u = A u B xyz + vp * u = (1.5j) + (5k * 2j) = [ -10i + 1.5j] rad>s2 Dissemination

Since vp is always directed along the Z axis where Æ = vp, then

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vA = v * rA = (2j + 5k) * (20i) = [100j - 40k] ft>s

= ( -10i + 1.5j + 3k) * (20i) + (2j + 5k) * [(2j + 5k) * (20i)] = [ -580i + 60j - 30k] ft>s2






O

u ⫽ 2 rad/s u = 1.5 rad/s2 y

SOLUTION

and

vp = 5 rad/s vp = 3 rad/s2

Ans.

20–18. Gear A is fixed to the crankshaft S, while gear C is fixed. Gear B and the propeller are free to rotate. The crankshaft is turning at 80 rad>s about its axis. Determine the magnitudes of the angular velocity of the propeller and the angular acceleration of gear B.

z 0.1 ft

B 80 rad/s S

SOLUTION Point P on gear B has a speed of

0.4 ft

vP = 80(0.4) = 32 ft>s C

The IA is located along the points of contant of B and C vs vP = 0.1 0.4 vs = 4vP v = -vP j + vs k laws

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vP = {-40j} rad>s

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vs = 4(40) k = {160k} rad>s Thus, v = vP + vs

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Ans.

A

y

20–19. Shaft BD is connected to a ball-and-socket joint at B, and a beveled gear A is attached to its other end. The gear is in mesh with a fixed gear C. If the shaft and gear A are spinning with a constant angular velocity v1 = 8 rad>s, determine the angular velocity and angular acceleration of gear A.

y

300 mm

x

B

V1

SOLUTION g = tan-1

75 = 14.04° 300

A

b = sin-1

100 mm

100

= 18.87°

23002 + 752

75 mm

C

The resultant angular velocity v = v1 + v2 is always directed along the instantaneous axis of zero velocity IA. 8 v = sin 147.09° sin 18.87°

v = 13.44 rad>s

v = 13.44 sin 18.87° i + 13.44 cos 18.87° j = {4.35i + 12.7j} rad>s

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v2 = 6.00 rad>s in

v2 8 = sin 14.04° sin 18.87°

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= 0 + (6j) * (4.3466i + 6.7162j)

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a = 0 + ( - 26.08k) = {-26.1k} rad>s2






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= {- 26.08k} rad>s2

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Ans.

*20–20. Gear B is driven by a motor mounted on turntable C. If gear A is held fixed, and the motor shaft rotates with a constant angular velocity of vy = 30 rad>s, determine the angular velocity and angular acceleration of gear B.

z

B vy ⫽ 30 rad/s y C

0.15 m

A

SOLUTION The angular velocity v of gear B is directed along the instantaneous axis of zero velocity, which is along the line where gears A and B mesh since gear A is held fixed. From Fig. a, the vector addition gives

0.3 m

v = vy + vz 2

vj -

25

1 25

vk = 30j - vzk

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Here, we will set the XYZ fixed reference frame to coincide with the xyz rotating frame at the instant considered. If the xyz frame rotates with an angular velocity of Æ = vz = [- 15k] rad>s, then vy will always be directed along the y axis with respect to the xyz frame. Thus,

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# # vy = A vy B xyz + vz * vy = 0 + ( -15k) * (30j) = [450i] rad>s2 and

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# # vz = A vz B xyz + vz * vz = 0 + 0 = 0 Thus,

# # a = vy + vz = (450i) + 0 = [450i] rad>s2






Ans.

20–21. Gear B is driven by a motor mounted on turntable C. If gear A and the motor shaft rotate with constant angular speeds of vA = {10k} rad>s and vy = {30j} rad>s, respectively, determine the angular velocity and angular acceleration of gear B.

z

B vy ⫽ 30 rad/s y C

0.15 m

A

SOLUTION If the angular velocity of the turn-table is vz, then the angular velocity of gear B is v = vy + vz = [30j + vzk] rad>s

0.3 m

Since gear A rotates about the fixed axis (z axis), the velocity of the contact point P between gears A and B is vp = vA * rA = (10k) * (0.3j) = [- 3i] m>s Since gear B rotates about a fixed point O, the origin of the xyz frame, then rOP = [0.3j - 0.15k] m. laws

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Here, we will set the XYZ fixed reference frame to conincide with the xyz rotating frame at the instant considered. If the xyz frame rotates with an angular velocity of Æ = vz = [- 5k] rad>s, then vy will always be directed along the y axis with respect to the xyz frame. Thus, # # vy = A vy B xyz + vz * vy = 0 + (- 5k) * (30j) = [150i] rad>s2 When Æ = vz, vz is always directed along the z axis. Therefore, # # vz = A vz B xyz + vz * vz = 0 + 0 = 0 Thus, # # a = vy + vz = (150i + 0) = [150i] rad>s2






Ans.

20–22. The crane boom OA rotates about the z axis with a constant angular velocity of v1 = 0.15 rad>s, while it is rotating downward with a constant angular velocity of v2 = 0.2 rad>s. Determine the velocity and acceleration of point A located at the end of the boom at the instant shown.

A

z

V1 50 ft

SOLUTION v = v1 + v2 = {0.2j + 0.15k} rad>s

O

Let the x, y, z axes rotate at Æ = v1, then

V2

# # v = (v)xyz + v1 * v2

x

# v = 0 + 0.15k * 0.2j = {-0.03i} rad>s2 rA = C 2(110)2 - (50)2 D i + 50k = {97.98i + 50k} ft i 0 97.98

j 0.2 0

k 0.15 3 50 or

vA = v * rA = 3

laws

vA = {10i + 14.7j - 19.6 k} ft>s

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j 0 0

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aA = a * rA + v * vA

i 3 = -0.03 97.98

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aA = { -6.12i + 3j - 2k} ft>s2






110 ft

y

20–23. The differential of an automobile allows the two rear wheels to rotate at different speeds when the automobile travels along a curve. For operation, the rear axles are attached to the wheels at one end and have beveled gears A and B on their other ends. The differential case D is placed over the left axle but can rotate about C independent of the axle. The case supports a pinion gear E on a shaft, which meshes with gears A and B. Finally, a ring gear G is fixed to the differential case so that the case rotates with the ring gear when the latter is driven by the drive pinion H. This gear, like the differential case, is free to rotate about the left wheel axle. If the drive pinion is turning at vH = 100 rad>s and the pinion gear E is spinning about its shaft at vE = 30 rad>s, determine the angular velocity, vA and vB , of each axle.

VH 50 mm

H

40 mm 180 mm

G

VA

5000 = 27.78 rad>s 180

laws

or

Point O is a fixed point of rotation for gears A, E, and B.

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2866.7 = 47.8 rad>s 60

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vP ¿¿ = Æ * rP ¿¿ = (27.78j + 30k) * (40j + 60k) = {466.7i} mm>s

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466.7 = 7.78 rad>s 60

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Ans.

O A

vP = vH rH = 100(50) = 5000 mm>s

vP ¿ = Æ * rP ¿ = (27.78j + 30k) * ( -40j + 60k) = {2866.7i} mm>s

vB

C To left wheel

Æ = vG + vE = {27.78j + 30k} rad>s

E

60 mm

D

vA =

z vE

SOLUTION

vG =

From motor

B

To right wheel

y

*20–24. z

The truncated double cone rotates about the z axis at a constant rate vz = 0.4 rad>s without slipping on the horizontal plane. Determine the velocity and acceleration of point A on the cone.

A vz ⫽ 0.4 rad/s 1.5 ft

0.5 ft 30⬚

SOLUTION u = sin-1 a vs =

1 ft

0.5 b = 30° 1

2 ft

x

y

0.4 = 0.8 rad>s sin 30°

v = 0.8 cos 30° = 0.6928 rad>s v = {-0.6928j} rad>s Æ = 0.4k # # v = (v)xyz + Æ * v

or

(1) laws

= 0 + (0.4k) * ( - 0.6928j)

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# v = {0.2771i} rad>s2

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rA = (3 - 3 sin 30°)j + 3 cos 30° k

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= (1.5j + 2.598k) ft

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vA = {-1.80i} ft>s

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aA = (- 0.720j - 0.831k) ft>s2

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Ans.

20–25. Disk A rotates at a constant angular velocity of 10 rad>s. If rod BC is joined to the disk and a collar by ball-and-socket joints, determine the velocity of collar B at the instant shown. Also, what is the rod’s angular velocity V BC if it is directed perpendicular to the axis of the rod?

z B

D

E

300 mm y

SOLUTION vC = {1i} m>s

200 mm C

vB = - vBj

v

vBC = vx i + vy j + vz k

rB>C = { -0.2i + 0.6j + 0.3k} m

500 mm

100 mm x

vB = vC + vBC * rB>C i 3 -vB = 1i + vx -0.2

j vy 0.6

k vz 3 0.3

Equating i, j, and k components 1 - 0.3vy - 0.6vz = 0 laws

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(1)

0.3vx + 0.2vz = vB

(3) Dissemination

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0.6vx + 0.2vy = 0

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vBC # rB>C = (vx i + vy j + vzk) # ( -0.2i + 0.6j + 0.3k) = 0

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vz = 1.36 rad>s this

vy = -0.612 rad>s

vB = 0.333 m>s

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vx = 0.204 rad>s

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vB = {-0.333j} m>s

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vBC = {0.204i - 0.612j + 1.36k} rad>s

10 rad/s

A

Ans.

20–26. If the rod is attached with ball-and-socket joints to smooth collars A and B at its end points, determine the speed of B at the instant shown if A is moving downward at a constant speed of vA = 8 ft>s. Also, determine the angular velocity of the rod if it is directed perpendicular to the axis of the rod.

z

vA

8 ft/s

A

SOLUTION 3 ft

vA = { -8k} ft>s

7 ft

vB = v B i y

rB/A = {2i + 6j - 3k} ft 2 ft

v = {vx i + vy j + vz k} rad>s

B

vB = vA + v * rB/A j vy 6

k vz 3 -3

6 ft x

laws

or

i vB i = -8k + 3 vx 2

teaching

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Expanding and equating components yields: in

vB = -3 vy - 6vz

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0 = 3 vx + 2vz

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(vx i + vy j + vz k) # (2i + 6j - 3k) = 0 is

2vx + 6vy - 3 vz = 0

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vy = - 1.061 rad>s

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vx = 0.9796 rad>s

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(4)

vz = -1.469 rad>s v = {0.980i - 1.06j - 1.47k} rad>s

Ans.

vB = 12.0 ft>s

Ans.






20–27. If the collar at A is moving downward with an acceleration aA = 5 -5k6 ft>s2, at the instant its speed is vA = 8 ft>s, determine the acceleration of the collar at B at this instant.

z

vA

8 ft/s

A

SOLUTION aB = aB i,

3 ft

aA = -5k

7 ft

From Prob. 20–26, y

vAB = 0.9796i - 1.0612j - 1.4694k

2 ft

aB = aA + aB/A

B

aB/A = vAB * vB/A + aAB * rB/A 6 ft

vB/A = vB - vA = 12i + 8k

x

laws

or

aB/A = {0.9796i - 1.0612j - 1.4694k} * (12i + 8k) + {ax i + axj + az k} * (2i + 6j - 3k)

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aB i = -5k + {0.9796i - 1.0612j - 1.4694k} * (12i + 8k)

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+ {( -3ay - 6az)i + (2az + 3ax)j + (6ax - 2ay)k }

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3ay + 6az + aB = - 8.4897

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-3ax - 2az = -25.4696

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-6ax + 2ay = 7.7344

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aB = -96.5

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aB = { -96.5i} ft>s2






Ans.

*20–28. z

If wheel C rotates with a constant angular velocity of vC = 10 rad>s, determine the velocity of the collar at B when rod AB is in the position shown.

300 mm 100 mm A 200 mm C vC vC

SOLUTION Here, rCA = [ - 0.1i] m and vC = [- 10j] rad>s. Since wheel C rotates about a fixed axis, then x

vA = vC * rCA = (- 10j) * (- 0.1i) = [-1k] m>s

600 mm

Since rod AB undergoes general motion, vA and vB can be related using the relative velocity equation.

B 4

3

5

y

vB = vA + vAB * rB>A

laws

or

4 3 Assume vB = - vBi + vBj and vAB = C (vAB)x i + (vAB)y j + (vAB)z k D . Also, 5 5 rB>A = [0.3i + 0.6j - 0.2k] m. Thus,

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3 4 - vB i + vB j = (- 1k) + C (vAB)xi + (vAB)y j + (vAB)zk D * (0.3i + 0.6j - 0.2k) 5 5

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4 3 - vBi + vB j = C -0.2(vAB)y - 0.6(vAB)z D i + C 0.2(vAB)x + 0.3(vAB)z D j + C 0.6(vAB)x - 0.3(vAB)y - 1 D k 5 5

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4 - vB = - 0.2(vAB)y - 0.6(vAB)z 5 3 v = 0.2(vAB)x + 0.3(vAB)z 5 B 0 = 0.6(vAB)x - 0.3(vAB)y - 1

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C (vAB)xi + (vAB)yj + (vAB)zk D # (0.3i + 0.6j - 0.2k) = 0 0.3(vAB)x + 0.6(vAB)y - 0.2(vAB)z = 0

(4)

Solving Eqs. (1) through (4), (vAB)x = 1.633 rad>s

(vAB)y = - 0.06803 rad>s

(vAB)z = 2.245 rad>s

vB = 1.667 m>s Then, 4 3 vB = - (1.667)i + (1.667)j = [-1.33i + 1j] m>s 5 5






Ans.

20–29. z

At the instant rod AB is in the position shown wheel C rotates with an angular velocity of vC = 10 rad>s and has an angular acceleration of aC = 1.5 rad>s2. Determine the acceleration of collar B at this instant.

300 mm 100 mm A 200 mm C vC vC

SOLUTION Here, rCA = [ -0.1i] m and aC = [- 1.5j] rad>s2. Since wheel C rotates about a fixed axis, then x a = a * r + v * (v * r ) A

C

CA

C

C

CA

600 mm

= ( -1.5j) * ( - 0.1i) + (-10j) * [( - 10j) * (- 0.1i)] = [10i - 0.15k] m>s2

B 4

3

5

y

For general motion, a A and aB can be related using the relative acceleration equation.

or

aB = aA + aAB * rB>A + vAB * (vAB * rB>A)

States

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permitted.

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0.06803j + 2.245k] rad>s. Thus,

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4 3 Assume aB = - aBi + aB j and aAB = (aAB)x i + (aAB)y j + (aAB)z k. Also, 5 5 rB>A = [0.3i + 0.6j - 0.2k] m. Using the result of Prob. 20–28, vAB = [1.633i -

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4 3 - aBi + aBj = C 7.6871 - 0.2(aAB)y - 0.6(aAB)z D i + C 0.2(aAB)x + 0.3(aAB)z - 4.6259 D j 5 5

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4 - aB = 7.6871 - 0.2(aAB)y - 0.6(aAB)z 5 3 a = 0.2(aAB)x + 0.3(aAB)z - 4.6259 5 B

(2)

0 = 0.6(aAB)x - 0.3(aAB)y + 1.3920

(3)

The fourth equation can be obtained from the dot product of aAB # rB>A = 0

C (aAB)xi + (aAB)yj + (aAB)zk D # (0.3i + 0.6j - 0.2k) = 0 0.3(aAB)x + 0.6(aAB)y - 0.2(aAB)z = 0






(4)

20–29. continued Solving Eqs. (1) through (4), (aAB)x = - 1.342 rad>s2

(aAB)y = 1.955 rad>s2

(aAB)z = 3.851 rad>s2

aB = - 6.231 m>s2 Then, 4 3 a B = - (- 6.231)i + (-6.231)j = [4.99i - 3.74j] m>s2 5 5

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Ans.






20–30. If wheel D rotates with an angular velocity of vD = 6 rad>s, determine the angular velocity of the follower link BC at the instant shown. The link rotates about the z axis at z = 2 ft.

z 0.5 ft C B

2 ft

SOLUTION 1.5 ft

Here, rD = [ -0.25i] ft and vD = [-6j] rad>s. Since wheel D rotates about a fixed axis, then x

vA = vD * rD = ( - 6j) * ( -0.25i) = [-1.5k] ft>s

vD ⫽ 6 rad/s

vB = vBC * rCB = (vBCk) * (0.5j) = - 0.5vBC i

laws

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Since rod AB undergoes general motion, vA and vB can be related using the relative velocity equation.

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vB = vA + vAB * rB>A Wide

copyright

Here, vAB = C (vAB)x i + (vAB)y j + (vAB)z k D and rB>A = [ -3i - 1j + 2k] ft. Thus,

permitted.

Dissemination

-0.5vBCi = (-1.5k) + C (vAB)xi + (vAB)y j + (vAB)zk D * ( -3i - 1j + 2k) and

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0 = - C 2(vAB)x + 3(vAB)z D

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0 = 3(vAB)y - (vAB)x - 1.5

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C 1vAB2xi + 1vAB2y j + 1vAB2zk D # 1- 3i - 1j + 2k2 = 0 - 3(vAB)x - (vAB)y + 2(vAB)z = 0

(4)

Solving Eqs. (1) through (4), (vAB)x = - 0.1071 rad>s

(vAB)y = 0.4643 rad>s

(vAB)z = 0.07143 rad>s

vBC = - 2 rad>s Then, vBC = [ -2k] rad>s






Ans.

0.25 ft D

Also, link BC rotates about a fixed axis. Assume vBC = vBC k and rCB = [0.5j] ft. Thus,

3 ft

A

y

20–31. Rod AB is attached to the rotating arm using ball andsocket joints. If AC is rotating with a constant angular velocity of 8 rad>s about the pin at C, determine the angular velocity of link BD at the instant shown.

z

vAC

8 rad/s

1.5 ft C

3 ft

SOLUTION

A

vA = 8(1.5)j = {12j} ft>s vB = -vB k vB = vA + v * rB/A

D

- vB k = 12j + (vx i + vy j + vz k) * (3i + 2j - 6k) 0 = - 6vy - 2vz

(1)

0 = 12 + 6vx + 3vz

(2) x

or laws

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4 vB D = -( ) r 2

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vB D = {-2.00i} rad>s

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v # rB>A = 0

Solving Eqs. (1), (2), and (3), v = vxi + vyj + vzk = { -1.633i + 0.2449j - 0.7347k} rad>s






(3)

B 2 ft

- vB = 2vx - 3vy

- vB = 2 a -

y

6 ft

*20–32. Rod AB is attached to the rotating arm using ball andsocket joints. If AC is rotating about point C with an angular velocity of 8 rad>s and has an angular acceleration of 6 rad>s2 at the instant shown, determine the angular velocity and angular acceleration of link BD at this instant.

z

vAC

8 rad/s

1.5 ft C

3 ft

SOLUTION

A

See Prob. 20–31. vB D = {-2.00i} rad>s

Ans.

v = {-1.633i + 0.2449j - 0.7347k} rad>s D

vB = vA + vB/A - 4k = 12j + vB/A x

aB = aA + a * rB/A + v * (vB/A) or

k -0.7347 3 -4

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laws

j 0.2449 -12

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k i az 3 + 3 -1.633 0 -6

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j ay 2

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i - (2)2 (2)j + (aB)z k = -96i + 9j + 3 ax 3

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Dissemination

0 = -96 + ay (-6) - 2az - 9.796

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- 8 = 9 + 6ax + 3az - 6.5308

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(aB)z = ax(2) - ay (3) + 19.592

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(aB)z = 69.00 ft>s2

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69.00 = 34.5 rad>s2 2

aB D = {34.5i} rad>s2






B 2 ft

vB/A = {-12j - 4k} ft>s

aB D =

y

6 ft

Ans.

20–33. Rod AB is attached to collars at its ends by ball-and-socket joints. If collar A moves upward with a velocity of vA = 58k6ft>s, determine the angular velocity of the rod and the speed of collar B at the instant shown. Assume that the rod’s angular velocity is directed perpendicular to the rod.

z

vA 4 ft B

SOLUTION vA = {8k} ft>s

3 ft

vB = -

3 4 yBi + yB k 5 5

vAB = vx i + vy j + vz k

2.5 ft

y 2 ft

rB>A = {1.5i - 2j - 1k} ft vB = vA + vAB * rB>A x

i 3 4 - yB i + yBk = 8k + 3 vx 5 5 1.5

j vy -2

k vz 3 -1

or

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3 y 5 B

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- vy + 2vz = -

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vAB # rB>A = (vx i + vy j + vzk) # (1.5i - 2j - 1k) = 0 is

1.5vx - 2vy - vz = 0

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Solving Eqs.(1) to (4) yields:

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vz = - 0.7789 rad>s destroy

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and

vy = 1.2657rad>s

Then vAB = {1.17i + 1.27j - 0.779k} rad>s






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yB = 4.71 ft>s

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v - 2vx - 1.5vx =

vx = 1.1684 rad>s

8 ft/s

A

3 ft

20–34. Rod AB is attached to collars at its ends by ball-and-socket joints. If collar A moves upward with an acceleration of a A = 54k6ft>s2, determine the angular acceleration of rod AB and the magnitude of acceleration of collar B.Assume that the rod’s angular acceleration is directed perpendicular to the rod.

z

vA 4 ft B

SOLUTION

3 ft

From Prob. 20–33

2.5 ft

vAB = {1.1684i + 1.2657j - 0.7789k} rad>s

aAB = ax i + ay j + az k

x

aB = -

4 3 a i + aB k 5 B 5

aB = aA + aAB * rB>A + vAB * (vAB * rB>A)

teaching

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laws

or

3 4 a i + aB k = 4k + (axi + ay j + az k) * (1.5i - 2j - 1k) 5 B 5

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+(1.1684i + 1.2657j - 0.7789k)

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-ay + 2az - 5.3607 = -

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ax + 1.5az + 7.1479 = 0

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Since aAB is perpendicular to the axis of the rod,

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7.5737 - 2ax - 1.5ay =

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aAB # rB>A = (ax i + ay j + azk) # (1.5i - 2j - 1k) = 0 1.5ax - 2ay - az = 0

(4)

Solving Eqs.(1) to (4) yields: ax = -2.7794 rad>s2

ay = - 0.6285rad>s2 aB = 17.6 ft>s2

Then aAB = {- 2.78i - 0.628j - 2.91k} rad>s2






y 2 ft

rB>A = {1.5i - 2j - 1k} ft

aA = {4k} ft>s2

8 ft/s

A

az = - 2.91213 rad>s2 Ans. Ans.

3 ft

20–35. n

Solve Prob. 20–25 if the connection at B consists of a pin as shown in the figure below, rather than a ball-and-socket joint. Hint: The constraint allows rotation of the rod both about bar DE (j direction) and about the axis of the pin (n direction). Since there is no rotational component in the u direction, i.e., perpendicular to n and j where u = j : n, an additional equation for solution can be obtained from V # u = 0. The vector n is in the same direction as rB>C : rD>C.

B

D

j

u

SOLUTION vC = {1i} m>s

vB = -vBj

vBC = vx i + vy j + vz k

rB>C = {-0.2i + 0.6j + 0.3k} m vB = vC + vBC * rB>C i -vBj = 1i + 3 vx -0.2

j vy 0.6

k vz 3 0.3

laws

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Equating i, j, and k components 1 + 0.3vx - 0.6vz = 0

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0.3vx + 0.2vz = vB

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= 0.8321i + 0.5547k

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20.182 + 0.122

k 0.3 3 = {0.18i + 0.12k} m2 0.3 and

0.18i + 0.12k

n =

j 0.6 0

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i rB>C * rD>C = 3 -0.2 -0.2

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rD>C = { -0.2i + 0.3k} m

u = j * n = j * (0.8321i + 0.5547k) = 0.5547i - 0.8321k vBC # u = (vx i + vy j + vzk) # (0.5547i - 0.8321k) = 0 0.5547vx - 0.8321vz = 0

(4)

Solving Eqs. (1) to (4) yields: vx = 0.769 rad>s

vy = - 2.31 rad>s

vz = 0.513 rad>s

vB = 0.333 m>s

Then






vBC = {0.769i - 2.31j + 0.513k} rad>s

Ans.

vB = {- 0.333j} m>s

Ans.

permitted.

Dissemination

0.6vx + 0.2vz = 0

rB>C = {- 0.2i + 0.6j + 0.3k} m

E

*20–36. The rod assembly is supported at B by a ball-and-socket joint and at A by a clevis. If the collar at B moves in the x–z plane with a speed vB = 5 ft>s, determine the velocity of points A and C on the rod assembly at the instant shown. Hint: See Prob. 20–35.

z vB = 5 ft/s B 30°

SOLUTION vB = {5 cos 30°i - 5 sin 30°k} ft>s rA>B = {4j - 3k} ft

vA = - vAk

vABC = vxi + vyj + vzk

C

rC>B = {2i - 3k} ft

A

x

2 ft 4 ft

vA = vB + vABC * rA>B i -vAk = 5 cos 30°i - 5 sin 30°k + 3 vx 0

j vy 4

k vz 3 -3

(1)

3vx = 0

(2) laws

5 cos 30° - 3vy - 4vz = 0

or

Equating i, j, and k components

4vx - 5 sin 30° = - vA in

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(j) = 0

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the

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= (vxi + vyj + vzk)

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Solving Eqs. (1) to (4) yields: vA = 2.5 ft>sT

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vz = 1.083 rad>s

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vx = vy = 0

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vABC = {1.083k} rad>s

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Ans.

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vA = { -2.50k} ft>s

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vC = vB + vABC * rC>B

= {4.33i + 2.17j - 2.50k} ft s





j 0 0

k 1.083 -3 sale

i = 5 cos 30°i - 5 sin 30°k + 0 2


3 ft

Ans.

y

20–37. Solve Example 20–5 such that the x, y, z axes move with curvilinear translation, Æ = 0 in which case the collar appears to have both an angular velocity Æxyz = v1 + v2 and radial motion.

SOLUTION Relative to XYZ, let xyz have Æ = 0

# Æ = 0

rB = {- 0.5k} m vB = {2j} m>s aB = {0.75j + 8k} m>s2 Relative to xyz, let x¿ y¿ z¿ be coincident with xyz and be fixed to BD. Then Æ xyz = v1 + v2 = {4i + 5k} rad>s

# # # vxyz = v1 + v2 = {1.5i - 6k} rad>s2 laws

or

(rC>B)xyz = {0.2j} m in

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# # (vC>B)xyz = (rC>B)xyz = (rC>B)x¿y¿z¿ + (v1 + v2) * (rC>B)xyz Dissemination

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= 3j + (4i + 5k) * (0.2j)

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$ $ # (aC>B)xyz = (rC>B)xyz = C (rC>B)x¿y¿z¿ + (v1 + v2) * (rC>B)x¿y¿z¿ D

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= {-1i + 3j + 0.8k} m>s

the

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# # # + C (v1 + v2) * (rC>B)xyz D + C (v1 + v2) * (rC>B)xyz D

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(aC>B)xyz = C 2j + (4i + 5k) * 3j D + C (1.5i - 6k) * 0.2j D + C (4i + 5k) * ( -1i + 3j + 0.8k) D

provided

integrity

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= {- 28.8i - 6.2j + 24.3k} m>s2

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vC = vB + Æ * rC>B + (vC>B)xyz their

destroy

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= 2j + 0 + ( -1i + 3j + 0.8k) sale

will

= {- 1.00i + 5.00j + 0.800k} m>s

Ans.

aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz = (0.75j + 8k) + 0 + 0 + 0 + (-28.8i - 6.2j + 24.3k) = { -28.8i - 5.45j + 32.3k} m>s2






Ans.

20–38. Solve Example 20–5 by fixing x, y, z axes to rod BD so that æ = V1 + V2. In this case the collar appears only to move radially outward along BD; hence æxyz = 0.

SOLUTION

# # Relative to XYZ, let x¿ y¿ z¿ be concident with XYZ and have Æ¿ = v1 and Æ ¿ = v1 # # # v = v1 + v2 = {4i + 5k} rad>s # # # # v = v1 + v2 = c a v1 b

+ v 1 * v1 d + c a v2 b x¿y¿z¿

+ v1 * v2 d x¿y¿z¿

= (1.5i + 0) + C - 6k + (4i) * (5k) D = {1.5i - 20j - 6k} rad>s2 rB = {- 0.5k} m # # vB = rB = arB b

x¿y¿z¿

# $ aB = rB = c a rB b

+ v1 * rB = 0 + (4i) * ( - 0.5k) = {2j} m>s # + v1 * arB b

x¿y¿z¿

x¿y¿z¿

# # d + v1 * rB + v1 * rB

= 0 + 0 + C (1.5i) * ( -0.5k) D + (4i * 2j) = {0.75j + 8k} m>s2

teaching

Web)

laws

# Æ x¿y¿z¿ = 0; in

Æ x¿y¿z¿ = 0;

or

Relative to x¿y¿z¿ , let xyz have

= {0.2j} m

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= 2j + C (4i + 5k) * (0.2j) D + 3j assessing

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aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz






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= (0.75j + 8k) + C (1.5i - 20j - 6k) * (0.2j) D + (4i + 5k) * C (4i + 5k) * (0.2j) D + 2 C (4i + 5k) * (3j) D + 2j Ans.

20–39. At the instant u = 60°, the telescopic boom AB of the construction lift is rotating with a constant angular velocity about the z axis of v1 = 0.5 rad>s and about the pin at A with a constant angular speed of v2 = 0.25 rad>s. Simultaneously, the boom is extending with a velocity of 1.5 ft>s, and it has an acceleration of 0.5 ft>s2, both measured relative to the construction lift. Determine the velocity and acceleration of point B located at the end of the boom at this instant.

z B v1, v1 15 ft

u

SOLUTION

v2, v2

O C

x

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are # # Æ = v1 = 0 Æ = v1 = {0.5k} rad>s

2 ft

A

Since point A rotates about a fixed axis (Z axis), its motion can determined from vA = v1 * rOA = (0.5k) * ( -2j) = {1i} ft>s

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# aA = v1 * rOA + v1 * (v1 * rOA)

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= 0 + (0.5k) * (0.5k) * ( -2j)

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= {0.5j} ft>s2

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In order to determine the motion of point B relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity of Æ¿ = v2 = {0.25i} rad>s, the direction of rB>A will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of rB>A, # # (vB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * rB>A D part

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= { - 2.4976j + 3.1740k} ft>s

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= (1.5 cos 60°j + 1.5 sin 60°k) + 0.25i * (15 cos 60°j + 15 sin 60°k)

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Since Æ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ = v2 = 0. Taking the time derivative of (rB>A)xyz, ## ## # # # (aB>A)xyz = (r B>A)xyz = C (r>A)x¿y¿z¿ + v2 * (rB>A)x¿y¿z¿ D + v2 * rB>A + v2 * (rB>A)xyz

= C (0.5 cos 60°j + 0.5 sin 60°k) + 0.25i * (1.5 cos 60°j + 1.5 sin 60°k) D + 0.25i * ( -2.4976j + 3.1740k)

= {- 0.8683j - 0.003886k} ft>s2






y

20–39. continued Thus, vB = vA + Æ * rB>A + (vB>A)xyz = (1i) + (0.5k) * (15 cos 60°j + 15 sin 60°k) + ( -2.4976j + 3.1740k) = { -2.75i - 2.50j + 3.17k} m>s

Ans.

and # aB = aA + Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz = 0.5j + 0 + 0.5k * C (0.5k) * (15 cos 60°j + 15 sin 60°k) D + 2(0.5k) * ( - 2.4976j + 3.1740k) + (-0.8683j - 0.003886k) = {2.50i - 2.24j - 0.00389k} ft>s2

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*20–40. At the instant u = 60°, the construction lift is rotating about the z axis with an angular velocity of v1 = 0.5 rad>s and an # angular acceleration of v1 = 0.25 rad>s2 while the telescopic boom AB rotates about the pin at A with an angular velocity of v2 = 0.25 rad>s and angular # acceleration of v2 = 0.1 rad>s2. Simultaneously, the boom is extending with a velocity of 1.5 ft> s, and it has an acceleration of 0.5 ft> s2, both measured relative to the frame. Determine the velocity and acceleration of point B located at the end of the boom at this instant.

z B v1, v1 15 ft

u v2, v2

SOLUTION The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are # # Æ = v1 = {0.25k} rad>s2 Æ = v1 = {0.5k} rad>s Since point A rotates about a fixed axis (Z axis), its motion can be determined from vA = v1 * rOA = (0.5k) * ( - 2j) = {1i} ft>s laws

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= {0.5i + 0.5j} ft>s2

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# # (vB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * rB>A D

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= { -2.4976j + 3.1740k} ft>s

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= (1.5 cos 60°j + 1.5 sin 60°k) + [0.25i * (15 cos 60°j + 15 sin 60°k)]

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Since Æ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ = v2 = [0.1i] rad>s2. Taking the time derivative of (rB>A)xyz, $ $ # # (aB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * (rB>A)x¿y¿z¿ D + v2 * rB>A # + v2 * (rB>A)xyz = (0.5 cos 60°j + 0.5 sin 60°k) + (0.25i) * (1.5 cos 60°j + 1.5 sin 60°k) + (0.1i) * (15 cos 60°j + 15 sin 60°k) + (0.25i) * (- 2.4976j + 3.1740k) = { - 2.1673j + 0.7461k} ft>s2






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In order to determine the motion of point B relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity of Æ¿ = v2 = [0.25i] rad>s, the direction of rB>A will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of rB>A,

O C

x

= (0.25k) * (- 2j) + (0 .5k) * [0.5k * ( -2j)]

2 ft

A

y

20–40. continued Thus, vB = vA + Æ * rB>A + (vB>A)xyz = [1i] + (0.5k) * (15 cos 60°j + 15 sin 60°k) + ( -2.4976j + 3.1740k) = { - 2.75i - 2.50j + 3.17k} ft>sAns. and # a B = aA + Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz = (0.5i + 0.5j) + (0.25k) * (15 cos 60°j + 15 sin 60°k) + (0.5k) * [(0.5k) * (15 cos 60°j + 15 sin 60°k)] + 2 (0.5k) * (-2.4976j + 3.1740k) + (- 2.1673j + 0.7461k) = {1.12i - 3.54j + 0.746k} ft>s2

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20–41. At a given instant, rod BD is rotating about the y axis with an angular velocity vBD = 2 rad>s and an angular # acceleration vBD = 5 rad>s2. Also, when u = 60° $ link AC is # rotating downward such that u = 2 rad>s and u = 8 rad>s2. Determine the velocity and acceleration of point A on the link at this instant.

B

vBD ⫽ 2 rad/s vBD ⫽ 5 rad/s2

z

C

x

D

u

SOLUTION

3 ft

Æ = - 2i - 2j

y

rA/C = 3 cos 60°j - 3 sin 60°k = 1.5j - 2.5980762k # Æ = - 5j - 8i + (- 2j) * (- 2i) = {- 8i - 5j - 4k} rad>s2

A

vA = vC + Æ * rA/C + (vA/C)xyz vA = (-2i - 2j) * (1.5j - 2.5980762k) + 0 vA = 5.19615i - 5.19615j - 3k + 0 vA = {5.20i - 5.20j - 3.00k} ft>s # aA = aC + Æ * rA/C + Æ * (Æ * rA/C)

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+ 2(Æ * (vA/C)xyz) + (aA/C)xyz

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= 0 + (-8i - 5j - 4k) * (1.5j - 2.5980762k)

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+ (-2i - 2j) * (5.19615i - 5.19615j - 3k) + 0 + 0

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aA = 24.9904i - 26.7846j + 8.7846k

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aA = {25i - 26.8j + 8.78k} ft>s2






Ans.

20–42. At the instant u = 30°, the frame of the crane and the boom AB rotate with a constant angular velocity of v1 = 1.5 rad>s and v2 = 0.5 rad>s, respectively. Determine the velocity and acceleration of point B at this instant.

z V1, V1 1.5 m

O

SOLUTION

12 m

u A

B

y

V2, V2

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. The angular velocity and angular acceleration of this frame with respect to the XYZ frame are # # Æ = v1 = [1.5k] rad>s Æ = v1 = 0 Since point A rotates about a fixed axis (Z axis), its motion can be determined from vA = v1 * rOA = (1.5k) * (1.5j) = [- 2.25i] m>s # aA = v1 * rOA + v1 * (v1 * rOA)

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= 0 + (1.5k) * [(1.5k) * (1.5j)]

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= [ -3.375j] m>s2

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In order to determine the motion of point B relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity relative to the xyz frame of Æ¿ = v2 = [0.5i] rad>s, the direction of (rB>A)xyz will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of (rB>A)xyz, the

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# # (vB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * (rB>A)xyz D assessing

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= 0 + (0.5i) * (12 cos 30° j + 12 sin 30°k)

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Since Æ ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ ¿ = v2 = 0. Taking the time derivative of (rA>B)xyz, sale

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$ $ # # # (aA>B)xyz = (rA>B)xyz = C (rA>B)x¿y¿z¿ + v2 * (rA>B)x¿y¿z¿ D + v2 * (rA>B)xyz + v2 * (rA>B)xyz = [0 + 0] + 0 + (0.5i) * (- 3j + 5.196k) = [ -2.598j - 1.5k] m>s2 Thus, vB = vA + Æ * rB>A + (vB>A)xyz = ( -2.25i) + 1.5k * (12 cos 30° j + 12 sin 30° k) + (-3j + 5.196k) = [ - 17.8i - 3j + 5.20k] m>s

Ans.

and # aB = aA + Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vAB)xyz + (aAB)xyz = ( -3.375j) + 0 + 1.5k * C (1.5k) * (12 cos 30° j + 12 sin 30° k) D + 2(1.5k) * ( -3j + 5.196k) + (-2.598j - 1.5k) = [9i - 29.4j - 1.5k] m>s2






Ans.

20–43. At the instant u = 30°, the frame of the crane is rotating with an angular velocity of v1 = 1.5 rad>s and angular # acceleration of v1 = 0.5 rad>s2, while the boom AB rotates with an angular velocity of v2 = 0.5 rad>s and angular # acceleration of v2 = 0.25 rad>s2. Determine the velocity and acceleration of point B at this instant.

z V1, V1 1.5 m

O

SOLUTION

u A

V2, V2

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are # Æ = [0.5k] rad>s2 Æ = v1 = [1.5k] rad>s Since point A rotates about a fixed axis (Z axis), its motion can be determined from vA = v1 * rOA = (1.5k) * (1.5j) = [-2.25i] m>s # aA = v1 * rOA + v1 * (v1 * rOA)

laws

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= (0.5k) * (1.5j) + (1.5k) * C (1.5k) * (1.5j) D

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= [ -0.75i - 3.375j] m>s2

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# # (vB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * (rB>A)xyz D

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In order to determine the motion of point B relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity relative to the xyz frame of Æ¿ = v2 = [0.5i] rad>s, the direction of (rB>A)xyz will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of (rB>A)xyz,

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= 0 + (0.5i) * (12 cos 30° j + 12 sin 30° k)

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= [-3j + 5.196k] m>s

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Since Æ ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ ¿ = v2 = [0.25i] m>s2. Taking the time derivative of (rB>A)xyz, # $ $ # # (aB>A) = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * (rB>A)x¿y¿z¿ D + Æ 2 * (rB>A)xyz + v2 * (rB>A)xyz = [0 + 0] + (0.25i) * (12 cos 30° j + 12 sin 30°k) + 0.5i * ( -3j + 5.196k) = [- 4.098j + 1.098k] m>s2






12 m

B

y

20–43. continued Thus, vB = vA + Æ * rB>A + (vB>A)xyz = ( -2.25i) + 1.5k * (12 cos 30° j + 12 sin 30°k) + ( -3j + 5.196k) = [- 17.8i - 3j + 5.20k] m>s

Ans.

and # aB = aA = Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz = (- 0.75i - 3.375j) + 0.5k * (12 cos 30°j + 12 sin 30° k) + (1.5k) * C (1.5k) * (12 cos 30° j + 12 sin 30° k) D + 2(1.5k) * ( -3j + 5.196k) + ( -4.098j + 1.098k) = [3.05i - 30.9j + 1.10k] m>s2

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*20–44. At the instant shown, the boom is rotating about the z axis with an angular velocity v1 = 2 rad>s and angular # acceleration v1 = 0.8 rad>s2. At this same instant the swivel # is rotating at v2 = 3 rad>s when v2 = 2 rad>s2, both measured relative to the boom. Determine the velocity and acceleration of point P on the pipe at this instant.

z

ω. 2 = 3 rad/s ω 2 = 2 rad/s2 ω. 1 = 2 rad/s ω 1 = 0.8 rad/s2

A 3m P

y 3m

4m

SOLUTION Relative to XYZ, let xyz have Æ = {2k} rad> s

4m

2m

x

Æ = {0.8k} rad> s ( Æ does not change direction relative to XYZ .)

relative to XYZ.) rA = { - 6j + 3k} m (rA changes direction 2

# # vA = rA = (rA)xyz + Æ * rA = 0 + (2k) * (- 6j + 3k) = {12i} m>s # $ $ # # aA = rA = [(rA)xyz + Æ * (rA)xyz] + Æ * rA + Æ * rA laws

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= 0 + 0 + (0.8k) * ( -6j + 3k) + (2k) * (12i)

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= {4.8i + 24j} m>s2 Dissemination

Relative to xyz, let x¿ , y¿ , z¿ have the origin at A and the

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Æ xyz = {3k} rad>s

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(rP>A)xyz = {4i + 2j} m ((rP>A)xyz changes direction relative to xyz.)

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# # (vP>A)xyz = (rP>A)xyz = [(rP>A)x¿y¿z¿ + Æ xyz * (rP>A)xyz ]

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= { - 6i + 12j} m>s

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# $ $ # # (aP>A)xyz = (rP>A)xyz = c (r P>A)x¿y¿z¿ + Æ xyz * (rP>A)x¿y¿z¿ d + c Æ xyz * (rP>A)xyz d + c Æ xyz * (rP>A)xyz d

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= 0 + 0 + [(2k) * (4i + 2j)] + [(3k) * ( - 6i + 12j)] = { - 40i - 10j} m>s2 Thus, vP = vA + Æ * rP>A + (vP>A)xyz = 12i + [(2k) * (4i + 2j)] + ( - 6i + 12j) = {2i + 20j} m>s

Ans.

aP = aA + Æ * rP>A + Æ * (Æ * rP>A) + 2Æ * (vP>A)xyz + (aP>A)xyz = (4.8i + 24j) + C (0.8k) * (4i + 2j) D + 2k * C 2k * (4i + 2j) D + 2(2k) * ( -6i + 12j) + ( - 40i - 10j) = { - 101i - 14.8j} m s2






Ans.

20–45. During the instant shown the frame of the X-ray camera is rotating about the vertical axis at vz = 5 rad>s and # vz = 2 rad>s2. Relative to the frame the arm is rotating at # vrel = 2 rad>s and vrel = 1 rad>s2. Determine the velocity and acceleration of the center of the camera C at this instant.

z 1.25 m vz . vz

1.75 m

5 rad/s 2 rad/s2 C vrel . vrel

SOLUTION A

Æ = {5k} rad>s # Æ = {2k} rad>s2

1.25 m

x

y

rB = { -1.25i} m vB = 0 + 5k * ( -1.25i) = - 6.25j aB = 0 + 2k * (-1.25i) + 0 + 5k * ( -6.25j) = 31.25i - 2.5j

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rC/B = {1.75j + 1k} m

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# (vC/B)xyz = rC/B = 0 + (2j) * (1.75j + 1k) = 2i

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$ (aC/B)xyz = rC/B = 0 + (1j) * (1.75j + 1k) + 0 + (2j) * (2i)

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= 1i - 4k

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vC = vB + Æ * rC/B + (vC/A)xyz assessing

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vC = -6.25j + 5k * (1.75j + 1k) + 2i

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vC = {-6.75i - 6.25j} m>s # aC = aB + Æ * rC/B + Æ * (Æ * rC/B) + 2(Æ * (vC/B)xyz) + (aC/B)xyz

aC = {28.75i - 26.25j - 4k} m>s2






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= (31.25i - 2.5j) + (2k) * (1.75j + 1k) + 5k * [(5k) * (1.75j + 1k)] + 2(5k) * (2i) + (1i - 4k)

2 rad/s 1 rad/s2

Ans.

1m

20–46. The boom AB of the crane is rotating about the z axis with an angular velocity vz = 0.75 rad>s, which is increasing at # vz = 2 rad>s2. At the same instant, u = 60° and # the boom is rotating upward at a constant rate u = 0.5 rad>s2. Determine the velocity and acceleration of the tip B of the boom at this instant.

z

B

· u

SOLUTION

40 ft

Motion of moving reference:

u

x

# # vO = rO = (rO)xyz + Æ * rO = 0 + (0.75k) * (5i) = {3.75j} ft>s laws

or

# $ $ # # aO = rO = [(rO)xyz + Æ * (rO)xyz] + Æ * rO + Æ * rO

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copyright

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= 0 + 0 + (2k) * (5i) + (0.75k) * (3.75j)

instructors

States

World

permitted.

Dissemination

= { - 2.8125i + 10j} ft>s2

on

learning.

is

of

the

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Motion of B with respect to moving reference:

the

(including

for

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student

the

by

and

use

United

Æ B>O = { - 0.5j} rad>s # Æ B>O = 0 assessing

is

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solely

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protected

rB>O = 40 cos 60°i + 40 sin 60°k = {20i + 34.64k} ft provided

integrity

of

and

work

this

# # (vB>O)xyz = rB>O = (rB>O)xyz + Æ B>O * rB>O

and

any

courses

part

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is

This

= 0 + (- 0.5j) * (20i + 34.64k) their

destroy

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= {- 17.32i + 10k} ft>s sale

will

# $ $ # # (aB>O)xyz = rB>O = [(rB>O)xyz + Æ B>O * (rB>O)xyz] + Æ B>O * rB>O + Æ B>O * rB>O = 0 + 0 + 0 + ( - 0.5j) * ( - 17.32i + 10k) = {- 5i - 8.66k} ft>s2 vB = vO + Æ * rB>O + (vB>O)xyz = (3.75j) + (0.75k) * (20i + 34.64k) + ( -17.32i + 10k) vB = { -17.3i + 18.8j + 10.0k} ft>s # aB = aO + Æ * rB>O + Æ * (Æ * rB>O) + 2Æ * (vB>O)xyz + (aB>O)xyz

Ans.

= ( - 2.8125i + 10j) + (2k) * (20i + 34.64k) + (0.75k) * [(0.75k) * (20i + 34.64k)] + 2(0.75k) * ( -17.32i + 10k) + ( -5i - 8.66j)





0.75 rad/s 2 rad/s2

60 5 ft

rO = {5i} ft


vz v· z

A

Æ = vz = {0.75k} rad>s # # Æ = (v)xyz = {2k} rad>s2

aB = { -19.1i + 24.0j - 8.66k} ft s2

0.5 rad/s2

Ans.

y

20–47. The boom AB of the crane is rotating about the z axis with an angular velocity of vz = 0.75 rad>s, which is increasing at # vz = 2 rad>s2. At the # same instant, u = 60° and the boom is rotating upward at u = 0.5 rad>s2, which is increasing at $ u = 0.75 rad>s2. Determine the velocity and acceleration of the tip B of the boom at this instant.

z

B

· θ = 0.5 rad/s 2

SOLUTION Motion of moving reference:

θ = 60° A

Æ = vz = {0.75k} rad>s # # Æ = (v)xyz = {2k} rad>s2

x

rO = {5i} ft # # vO = rO = (rO)xyz + Æ * rO = 0 + (0.75k) * (5i) = {3.75j} ft>s laws

or

$ $ # # aO = rO = [(rO)xyz + Æ * (rO)xyz] + v * rO + Æ * rO

Wide

copyright

in

teaching

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= 0 + 0 + (2k) * (5i) + (0.75k) * (3.75j)

instructors

States

World

permitted.

Dissemination

= { - 2.8125i + 10j} ft>s2

on

learning.

is

of

the

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Motion of B with respect to moving reference:

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for

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and

use

United

Æ B>O = { - 0.5j} rad>s # Æ B>O = { - 0.75j} rad>s2

integrity

of

and

work

provided

# # (vB>O)xyz = rB>O = (rB>O)xyz + Æ B>O * rB>O

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rB>O = 40 cos 60°i + 40 sin 60°k = {20i + 34.64k} ft

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= 0 + ( - 0.5j) * (20i + 34.64k) their

destroy

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= {- 17.32i + 10k} ft>s sale

will

# $ $ # # (aB>O)xyz = rB>O = [(rB>O)xyz + Æ B>O * (rB>O)xyz] + Æ B>O * rB>O + Æ B>O * rB>O = 0 + 0 + ( - 0.75j) * (20i + 34.64k) + ( - 0.5j) * ( - 17.32i + 10k) = { - 30.98i + 6.34k} ft>s2 vB = vO + Æ * rB>O + (vB>O)xyz = (3.75j) + (0.75k) * (20i + 34.64k) + ( - 17.32i + 10k) vB = { -17.3i + 18.8j + 10.0k} ft>s # aB = aO + Æ * rB>O + Æ * (Æ * rB>O) + 2Æ * (vB>O)xyz + (aB>O)xyz

Ans.

= (- 2.8125i + 10j) + (2k) * (20i + 34.64k) + (0.75k) * [(0.75k) * (20i + 34.64k)] + 2(0.75k) * (-17.32i + 10k) + ( - 30.98i + 6.34j) aB = { -45.0i + 24.0j + 6.34k} ft s2






ωz = 0.75 rad/s ω·z = 2 rad/s 2

40 ft

Ans.

5 ft

y

*20–48. At the instant shown, the motor rotates about the z axis with an angular velocity of v1 = 3 rad>s and angular acceleration # of v1 = 1.5 rad>s2. Simultaneously, shaft OA rotates with an angular velocity of v2 = 6 rad>s and angular acceleration of # v2 = 3 rad>s2, and collar C slides along rod AB with a velocity and acceleration of 6 m>s and 3 m>s2. Determine the velocity and acceleration of collar C at this instant.

z V1 V1 O A x

V2 V2

300 mm

SOLUTION

y

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are

300 mm

B

# v = [1.5k] rad>s2

Æ = v1 = [3k] rad>s

Since point A rotates about a fixed axis (Z axis), its motion can be determined from vA = v1 * rOA = (3k) * (0.3j) = [-0.9i] m>s # aA = v1 * rOA + v1 * (v * rOA)

or

= (1.5k) * (0.3j) + (3k) * (3k * 0.3j)

in

teaching

Web)

laws

= [-0.45i - 2.7j] m>s2

and

use

United

on

learning.

is

of

the

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instructors

States

World

permitted.

Dissemination

Wide

copyright

In order to determine the motion of point C relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the ¿x ¿y ¿z frame to have an angular velocity relative to the xyz frame of Æ¿ = v2 = [6j] rad>s , the direction of rC>A will remain # unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative A r C>A B xyz,

(including

for

work

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by

# # (vC>A)xyz = (rC>A)xyz = C (rC>A)x¿y¿z¿ + v2 * (rC>A)xyz D

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solely

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the

= ( -6k) + 6j * (-0.3k)

provided

integrity

of

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work

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assessing

= [- 1.8i - 6k] m>s

and

any

courses

part

the

is

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Since Æ ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ ¿ = v2 = [3j] rad>s2. Taking the time derivative of (rC>A)xyz, sale

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$ $ # # # (aC>A)xyz = (rC>A)xyz = C (rC>A)x¿y¿z¿ + v2 * (rC>A)x¿y¿z¿ D + v2 * rC>A + v2 * (rC>A)xyz = [( -3k) + 6j * ( -6k)] + (3j) * ( - 0.3k) + 6j * ( -1.8i - 6k) = [ -72.9i + 7.8k] m>s Thus, vC = vA + Æ * rC>A + (vC>A)xyz = (-0.9i) + 3k * (-0.3k) + ( -1.8i - 6k) = [- 2.7i - 6k] m>s

Ans.

and # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz = (-0.45i - 2.7j) + 1.5k * ( -0.3k) + (3k) * [(3k) * ( -0.3k)] + 2(3k) * ( -1.8i - 6k) + ( - 72.9i + 7.8k) = [- 73.35i - 13.5j + 7.8k] m>s






Ans.

6 m/s 3 m/s2

C

20–49. The motor rotates about the z axis with a constant angular velocity of v1 = 3 rad>s. Simultaneously, shaft OA rotates with a constant angular velocity of v2 = 6 rad>s. Also, collar C slides along rod AB with a velocity and acceleration of 6 m>s and 3 m>s2. Determine the velocity and acceleration of collar C at the instant shown.

z V1

O A

SOLUTION

V2

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are # # Æ = v1 = [3k] rad>s Æ = v1 = 0

x

300 mm

300 mm

y C

6 m/s 3 m/s2 B

Since point A rotates about a fixed axis (Z axis), its motion can be determined from vA = v1 * rOA = (3k) * (0.3j) = [ -0.9i] m>s # aA = v1 * rOA + v1 * (v1 * rOA)

or

= 0 + (3k) * C (3k) * (0.3j) D

in

teaching

Web)

laws

= [ -2.7j] m>s2

for

work

student

the

by

# # (vC>A)xyz = (rC>A)xyz = C (rC>A)x¿y¿z¿ + v2 * (rC>A)xyz D

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use

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on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

In order to determine the motion of the point C relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity relative to the xyz frame of Æ ¿ = v2 = [6j] rad>s, the direction of rC>A will remain unchanged with # respect to the x¿y¿z¿ frame. Taking the time derivative (rC>A)xyz,

is

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solely

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the

(including

= ( -6k) + 6j * ( -0.3k)

integrity

of

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= [- 1.8i - 6k] m>s

and

any

courses

part

the

is

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provided

Since Æ ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ ¿ = v2 = 0. Taking the time derivative of (rC>A)xyz,

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$ $ # # # (aC>A)xyz = (rC>A)xyz = C (rC>A)x¿y¿z¿ + v2 * (rC>A)x¿y¿z¿ D + v2 * rC>A + v2 * (rC>A)xyz = [(- 3k) + 6j * ( -6k)] + 0 + [6i * ( -1.8j - 6k)] = [ -72i + 7.8k] m>s2 Thus, vC = vA + Æ * rC>A + (vC>A)xyz = ( -0.9i) + 3k * ( -0.3k) + ( -1.8i - 6k) = [-2.7i - 6k] m>s

Ans.

and # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz = (-2.7j) + 0 + 3k * [(3k) * ( - 0.3k)] + 2(3k) * (- 1.8i - 6k) + ( - 72i + 7.8k) = [- 72i - 13.5j + 7.8k] m>s2






Ans.

20–50. At the instant shown, the arm OA of the conveyor belt is rotating about the z axis with a constant angular velocity v1 = 6 rad>s, while at the same instant the arm is rotating upward at a constant rate v2 = 4 rad>s. If the conveyor is # running at a constant rate r = 5 ft>s, determine the velocity and acceleration of the package P at the instant shown. Neglect the size of the package.

z

A P v1

SOLUTION

6 rad/s

Æ = v1 = {6k} rad>s # Æ = 0

r

u

O

rO = vO = aO = 0

v2

4 rad/s

x

Æ P/O = {4i} rad>s # Æ P/O = 0 rP/O = {4.243j + 4.243k} ft # (vP/O)xyz = (rP/O)xyz + Æ P/O * rP/O laws

or

= (5 cos 45°j + 5 sin 45°k) + (4i) * (4.243j + 4.243k) in

teaching

Web)

= { - 13.44j + 20.51k} ft>s

permitted.

Dissemination

Wide

copyright

# $ # # (aP/O) = (rP/O)xyz + Æ P/O * (rP/O)xyz + Æ P/O * rP/O + Æ P/O * rP/O

learning.

is

of

the

not

instructors

States

World

= 0 + (4i) * (3.536j + 3.536k) + 0 + (4i) * ( -13.44j + 20.51k)

the

by

and

use

United

on

= { -96.18j - 39.60k} ft>s2

the

(including

for

work

student

vP = vO + Æ * rP/O + (rP/O)xyz

assessing

is

work

solely

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= 0 + (6k) * (4.243j + 4.243k) + ( - 13.44j + 20.51k)

Ans.

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

vP = { -25.5i - 13.4j + 20.5k} ft>s # aP = aO + Æ * rP/O + Æ * (Æ * rP/O) + 2Æ * (vP/O)xyz + (aP/O)xyz






sale

will

their

destroy

of

= 0 + 0 + (6k) * [(6k) * (4.243j + 4.243k)] + 2(6k) * ( -13.44j + 20.51k) + ( -96.18j - 39.60k) aP = {161i - 249j - 39.6k} ft>s2

Ans.

6 ft

45

y

20–51. At the instant shown, the arm OA of the conveyor belt is rotating about the z axis with a constant angular velocity v1 = 6 rad>s, while at the same instant the arm is rotating upward at a constant rate v2 = 4 rad>s. If the conveyor is $ # running at a rate r = 5 ft>s, which is increasing at r = 8 ft>s2, determine the velocity and acceleration of the package P at the instant shown. Neglect the size of the package.

z

A P v1

SOLUTION

6 rad/s

Æ = v1 = {6k} rad>s # Æ = 0

r

u

O

rO = vO = aO = 0

v2

4 rad/s

x

Æ P/O = {4i} rad>s # Æ P/O = 0 rP/O = {4.243j + 4.243k} ft # (vP/O)xyz = (rP/O)xyz + Æ P/O * rP/O laws

or

= (5 cos 45°j + 5 sin 45°k) + (4i) * (4.243j + 4.243k)

Wide

copyright

in

teaching

Web)

= { -13.44j + 20.51k} ft>s

permitted.

Dissemination

(aP/O)xyz = 8 cos 45j + 8 sin 45°k - 96.18j - 39.60k

learning.

is

of

the

not

instructors

States

World

= { -90.52j - 33.945k} ft>s2

the

by

and

use

United

on

vP = vO + Æ * rP/O + (vP/O)xyz

the

(including

for

work

student

= 0 + (6k) * (4.243j + 4.243k) + ( - 13.44j + 20.51k)

Ans.

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

vP = {-25.5i - 13.4j + 20.5k} ft>s # aP = aO + Æ * rP/O + Æ * (Æ * rP/O) + 2Æ * (vP/O)xyz + (aP/O)xyz






sale

aP = {161i - 243j - 33.9k} ft>s2

will

their

destroy

of

and

any

courses

part

the

is

This

= 0 + 0 + (6k) * [(6k) * (4.243j + 4.243k)] + 2(6k) * ( - 13.44j + 20.51k) + ( -90.52j - 33.945k) = - 152.75j + 161.23i - 90.52j - 33.945k

Ans.

6 ft

45

y

*20–52. The boom AB of the locomotive crane is rotating about the Z axis with an angular velocity v1 = 0.5 rad>s, which is # increasing at v1 = 3 rad>s2. At this same instant, u = 30° and the boom is rotating upward at a constant rate of # u = 3 rad>s. Determine the velocity and acceleration of the tip B of the boom at this instant.

SOLUTION Æ = {0.5k} rad>s vA aA

# Æ = {3k} rad>s2

Z m v1

.5 rad/s

v· 1

rad/s2

· u

rad/s

2 m

rA = {3j} m

A

# # = rA = (rA)xyz + Æ * rA = 0 + (0.5k) * (3j) = {-1.5i} m>s # $ $ # # = rA = [(rA)xyz + Æ * (rA)xyz] + Æ * rA + Æ * rA = 0 + 0 + (3k) * (3j) + (0.5k) * ( -1.5i)

= { -9i - 0.75j} m>s2 # Æ xyz = 0 Æ xyz = {3i} rad>s rB>A = 20 cos 30°j + 20 sin 30° = {17.23j + 10k} m laws

or

# # (vB>A)xyz = rB>A = (rB>A)xyz + Æ xyz * rB>A

Wide

copyright

in

teaching

Web)

= 0 + (3i) * (17.32j + 10k)

the

by

and

use

United

(aB>A)xyz = 0 + 0 + 0 + [(3i) * ( -30j + 51.9k)] - { -155.8j - 90k} m>s2

on

learning.

is

of

the

the

(including

for

work

student

vB = vA + Æ * rB>A + (vB>A)xyz

assessing

is

work

solely

of

protected

= -1.5i + [(0.5k) * (17.32j + 10k)] + ( -30j + 51.96k)

Ans. integrity

of

and

work

this

= { -10.2i - 30j + 52.0k} m>s

part

the

is

and

any

courses

a B = aA

This

provided

# + v * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz

their

destroy

of

= ( -9i - 0.75j) + [(3k) * (17.32j + 10k)] + 0.5k * [0.5k * (17.32j + 10k)] sale

will

+ [2(0.5k) * ( - 30j + 51.96k)] + (-155.88j - 90k) = { -31.0i - 161j - 90k} m>s2






Ans.

not

instructors

States

World

$ $ # # # = rB>A = [(rB>A)xyz + Æ xyz * (rB>A)xyz] + [vxyz * rB>A] + [Æ xyz * rB>A]

permitted.

Dissemination

= {- 30j + 51.96k} m>s (aB>A)xyz

B

u

30 Y

20–53. The locomotive crane is traveling to the right at 2 m>s and has an acceleration of 1.5 m>s2, while the boom is rotating about the Z axis with an angular velocity v1 = 0.5 rad>s, # which is increasing at v1 = 3 rad>s2. At this same instant, u# = 30° and the boom is rotating upward at a constant rate u = 3 rad>s. Determine the velocity and acceleration of the tip B of the boom at this instant.

Z m v1

SOLUTION Æ = {0.5k} rad>s

# Æ = {3k} rad>s2

.5 rad/s

v· 1

rad/s2

· u

rad/s

2 m

rA = {3j} m

A

# # = rA = (rA)xyz + Æ * rA = 2j + (0.5k) * (3j) = {-1.5i + 2j} m>s $ $ $ # # = rA = [(rA)xyz + Æ * (rA)xyz] + Æ * rA + Æ * rA

vA aA

= 1.5j + (0.5k) * (2j) + (3k) * (3j) + (0.5k) * ( -1.5i + 2j) = { -11i + 0.75j} m>s2 # Æ xyz = 0 Æ xyz = {3i} rad>s rB/A = 20 cos 30°j + 20 sin 30° = {17.32j + 10k} m

teaching

Web)

laws

or

# # (vB/A)xyz = rB/A = (rB/A)xyz + Æ xyz * rB/A

Wide

copyright

in

= 0 + (3i) * (17.32j + 10k)

on

learning.

is

of

the

not

instructors

States

World

# $ $ # # (aB>A)xyz = rB>A = [(rB>A)xyz + Æ xyz * (rB>A)xyz] + [Æ xyz * rB>A] + [Æ xyz * rB>A] the

by

and

use

United

(aB>A)xyz = 0 + 0 + 0 + [(3i) * ( -30j + 51.96k)] = { -155.88j - 90k} m>s2 work

student

the

(including

for assessing

is

work

solely

of

protected

= -1.5i + 2j + [(0.5k) * (17.32j + 10k)] + ( -30j + 51.96k) integrity

of

and

work

this

= { -10.2i - 28j + 52.0k} m>s

Ans.

and

any

courses

part

the

is

This

provided

# aB = aA + Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz

will

their

destroy

of

= ( -11i + 0.75j) + [(3k) * (17.32j + 10k)] + 0.5k * [0.5k * (17.32j + 10k)] sale

+ [2(0.5k) * ( - 30j + 51.96k)] + ( - 155.88j - 90k) = { -33.0i - 159j - 90k} m>s2






Ans.

permitted.

Dissemination

= {-30j + 51.96k} m>s

vB = vA + Æ * rB>A + (vB>A)xyz

B

u

30 Y

20–54. The robot shown has four degrees of rotational freedom, namely, arm OA rotates about the x and z axes, arm AB rotates about the x axis, and CB rotates about the y axis. # At the instant shown, v2 = 1.5 rad>s, v2 = 1 rad>s2, v3 = # # 2 3 rad>s, v3 = 0.5 rad>s , v4 = 6 rad>s, v4 = 3 rad>s2, and # v1 = v1 = 0. If the robot does not translate, i.e., v = a = 0, determine the velocity and acceleration of point C at this instant.

z v1, v 1

O v2, v2 x

v3, v3

Here, arm OA rotates about a fixed axis (X axis), the motion of point A can be determined from vA = v2 * rOA = (1.5i) * (-1.5k) = [2.25j] m>s and

laws

or

# aA = v2 * rA>O + v2 * (v2 * rOA)

in

teaching

Web)

= (1i) * ( - 1.5k) + (1.5i) * [(1.5i) * ( -1.5k)]

permitted.

Dissemination

Wide

copyright

= [1.5j + 3.375k] m>s2

the

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for

work

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# $ (vC>A)xyz = (rC>A)xyz = [(rC>A)x¿y¿z¿ + Æ¿ xyz * rC>A]

provided

integrity

of

and

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this

= 0 + (3i + 6j) * (0.5j + 0.3k)

and

any

courses

part

the

is

This

= {1.8i - 0.9j + 1.5k} m>s

sale

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Here, the direction of Æ¿ changes with respect to the xyz frame. To account for this change a third x¿¿y¿¿z¿¿ rotating frame is set to coincide with the xyz frame, Fig. a. If the angular velocity of x¿¿y¿¿z¿¿ is Æ– xyz = v3 = {3i} rad>s, then the direction of v4 will remain unchanged with respect to the x¿¿y¿¿z¿¿ frame. Thus, # # (v4)xyz = (v4)x¿¿y¿¿z¿¿ + v3 * v4 = 3j + (3i * 6j) = {3j + 18k} rad>s2 Since v3 has a constant direction with respect to the xyz frame when Æ– = v3, then # # (v3)xyz = (v3)x¿¿y¿¿z¿¿ + v3 * v3 = {0.5i} rad>s2 Thus, # # # (Æ ¿)xyz = (v4)xyz + (v3)xyz = (3j + 18k) + (0.5i) = {0.5i + 3j + 18k} rad>s2 # Taking the time derivate of (rC>A)xyz, $ $ # (aC>A)xyz = (rC>A)xyz = [(rC>A)x¿y¿z¿ + Æ¿ xyz * (rC>A)x¿y¿z¿] # # + Æ ¿ xyz * rC>A + Æ¿ xyz * (rC>A)xyz = [0 + 0] + (0.5i + 3j + 18k) * (0.5j + 0.3k) + (3i + 6j) * (1.8i - 0.9j + 1.5k) = {0.9i - 4.65j - 13.25k} m>s2 or transmission in any form or by any means, electronic, mechanical,



not

instructors

States

World

In order to determine the motion of point C relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity of Æ¿ xyz = v3 + v4 = [3i + 6j] rad>s, the direction of rC>A will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of rC>A,

A

C

500 mm

B

y 300 mm

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a.Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are # # Æ = v2 = [1.5i] rad>s Æ = v2 = [1i] rad>s2


v

1500 mm

SOLUTION


a

v4, v4

20–54. continued Thus, vC = vA + Æ * rC>A + (vC>A)xyz = (2.25j) + (1.5i) * (0.5j + 0.3k) + (1.8i - 0.9j + 1.5k) = {1.8i + 0.9j + 2.25k} m>s

Ans.

and # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz = (1.5j + 3.375k) + 1i * (0.5j + 0.3k) + (1.5i) * [(1.5i) * (0.5j) + 0.3k)] + 2(1.5i) * (1.8i - 0.9j + 1.5k) + (0.9i - 4.65j - 13.25k) = {0.9i - 9.075j - 12.75k} m>s2

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21–1. Show that the sum of the moments of inertia of a body, Ixx + Iyy + Izz, is independent of the orientation of the x, y, z axes and thus depends only on the location of its origin.

SOLUTION Ixx + Iyy + Izz =

Lm

= 2

(y2 + z2)dm +

Lm

Lm

(x2 + z2)dm +

(x2 + y2)dm

Lm

(x2 + y2 + z2)dm

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However, x2 + y2 + z2 = r2, where r is the distance from the origin O to dm. Since ƒ r ƒ is constant, it does not depend on the orientation of the x, y, z axis. Consequently, Ixx + Iyy + Izz is also indepenent of the orientation of the x, y, z axis. Q.E.D.






21–2. Determine the moment of inertia of the cone with respect to a vertical y axis passing through the cone’s center of mass. What is the moment of inertia about a parallel axis y¿ that passes through the diameter of the base of the cone? The cone has a mass m.

–y

y

y¿

a x

SOLUTION rpa2

The mass of the differential element is dm = rdV = r(py2) dx =

=

L

r pa2 4h4

dIy =

(4h2 + a2) x4 dx

rpa2 4h4

h

(4h2 + a2)

x4dx =

r pa2h (4h2 + a2) 20

x2 dx =

r pa2h 3

L0

in

teaching

Web)

Iy =

r pa2 a 2 1 rpa2 2 B 2 x dx R a x b + ¢ 2 x2 ≤ x2 dx 4 h h h

or

=

1 dmy2 + dmx2 4

laws

dIy =

h

x2dx.

h2

World

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is

L0

the

h2

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h

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Lm

dm =

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m =

permitted.

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Iy =

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Iy = Iy + md2

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3h 2 3m (4h2 + a2) = Iy + m a b 20 4 Iy =

3m 2 (h + 4a2) 80

Ans.

Iy' = Iy + md2






=

h 2 3m 2 (h + 4a2) + m a b 80 4

=

m (2h2 + 3a2) 20

Ans.

21–3. Determine the moments of inertia Ix and Iy of the paraboloid of revolution. The mass of the paraboloid is m.

z z2

2

r — y a r y

SOLUTION

x a

m = r

L0

a

pz2 dy = rp

a

L0

r2 r2 b y dy = rpa b a a 2

a

Iy =

a

a

1 r4 1 1 r4 dm z2 = rp z4dy = rpa 2 b y2 dy = rpa ba 2 2 6 a L0 L0 Lm 2

Thus, 1 mr2 3

Ans.

a

a

laws

or

1 1 a dm z2 + dm y2 b = rp z4 dy + r pz2 y2 dy 4 L0 Lm 4 L0

a a rpr 2a 3 rpr 4a r4 1 r2 1 1 rpa 2 b + = mr 2 + ma 2 y2 dy + rpa b y3dy = a L0 4 12 4 6 2 a L0 Dissemination

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m 2 (r + 3a2) 6

of

Ix =

*21–4. y

Determine the radii of gyration kx and ky for the solid formed by revolving the shaded area about the y axis. The density of the material is r.

0.25 ft

xy ⫽ 1 4 ft 0.25 ft

SOLUTION For ky: The mass of the differential element is dm = rdV = r(px2) dy = rp dIy = 12 dmx2 = Iy =

L

dIy = 12rp

dy y2

x

.

dy 1 1 dy C rpy 2 D A y 2B = 2rpy 4

1 2

4

dy y4 L0.25

4 ft

1 C r(p)(4)2(0.25) D (4)2 2

+

= 134.03r 4

ky =

Lm

+ r C p(4)2(0.25) D = 24.35r

dy y2

L0.25

Iy 134.03r = = 2.35 ft Am A 24.35r

or

Hence,

dm = rp

Ans. in

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m =

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For kx: 0.25 ft 6 y … 4 ft

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1 I dy 2 c rpdy y d a 2 b + arp y by 4 y

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=

A 4y1 4 + 1 B dy = 28.53r

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C rp(4)2(0.25) D (4)2 + C rp(4)2(0.25) D (0.125)2 will sale

= 50.46r

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1 4

and

I¿¿ x =

Ix = I¿ x + I¿¿ x = 28.53r + 50.46r = 78.99r

Hence,






kx =

Ix = m

78.99r = 1.80 ft 24.35r

Ans.

21–5. Determine by direct integration the product of inertia Iyz for the homogeneous prism. The density of the material is r. Express the result in terms of the total mass m of the prism.

z a a

h

SOLUTION x

The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy. a

m =

Lm

dm = rh

L0

ra2h 2

(a - y)dy =

Using the parallel axis theorem: dIyz = (dIy¿z¿)G + dmyGzG h = 0 + (rhxdy) (y) a b 2 rh2 xydy 2

=

rh2 (ay - y2) dy 2

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a rh2 ra3h2 1 ra2h m = a b(ah) = ah (ay - y2) dy = 2 L0 12 6 2 6

the

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Iyz =






y

21–6. Determine by direct integration the product of inertia Ixy for the homogeneous prism. The density of the material is r. Express the result in terms of the total mass m of the prism.

z a a

h

SOLUTION x

The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy. a

m =

Lm

dm = rh

ra2h 2

(a - y)dy =

L0

Using the parallel axis theorem: dIxy = (dIx¿y¿)G + dmxGyG x = 0 + (rhxdy) a b(y) 2 rh2 2 x ydy 2

=

rh2 3 (y - 2ay2 + a 2 y) dy 2

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ra 4h 1 ra 2h 2 m 2 = ba = a a 24 12 2 12

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Ixy =

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y

21–7. Determine the product of inertia Ixy of the object formed by revolving the shaded area about the line x = 5 ft. Express the result in terms of the density of the material, r.

y 3 ft

2 ft

y2 ⫽ 3x

SOLUTION

x

3

L0

3

dm = r 2p

L0

3

(5 - x) 23x dx = 38.4rp

L0

3

~

L0

3

(5 - x)y dx = r 2p

y dm = r2p

y (5 - x)y dx L0 2 3

= rp

(5 - x)(3x) dx L0 = 40.5rp

or

40.5rp = 1.055 ft 38.4rp Web)

laws

Thus, y =

copyright

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The solid is symmetric about y, thus

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Ixy¿ = 0

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Ixy = Ixy¿ + x ym

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= 0 + 5(1.055)(38.4rp)

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Ixy = 636r






Ans.

*21–8. Determine the moment of inertia Iy of the object formed by revolving the shaded area about the line x = 5 ft. Express the result in terms of the density of the material, r.

y 3 ft

2 ft

y2 ⫽ 3x

SOLUTION

x

3

Iy¿ =

1 1 dm r2 - (m¿)(2)2 2 2 L0

3

3

1 1 dm r2 = r p(5 - x)4 dy 2 L0 L0 2 3 y2 4 1 rp b dy a5 2 3 L0

=

= 490.29 r p m¿ = r p (2)2(3) = 12 r p

or

1 (12 r p)(2)2 = 466.29 r p 2

teaching

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laws

Iy¿ = 490.29 r p -

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Mass of body; 3

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r p (5 - x)2 dy - m¿

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L0 = 38.4 r p

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Iy = 466.29 r p + (38.4 r p)(5)2

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Iy = 4.48(103) r

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L0

sale

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3

Iy¿ =

3

=

(5 - x)2 r (2p)(5 - x)y dx

L0

3

(5 - x)3 (3x)1/2 dx L0 = 466.29 r p = 2rp

3

m =

L0

dm 3

= 2rp

(5 - x)y dx

L0

3

= 2rp

L0 = 38.4 r p

(5 - x)(3x)1/2 dx

Iy = 466.29 r p + 38.4 r p(5)2 = 4.48(103)r © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





Ans.

21–9. Determine the elements of the inertia tensor for the cube with respect to the x, y, z coordinate system. The mass of the cube is m.

z, z¿ a a y¿

SOLUTION

a

dIzG =

x

1 (dm)(a2 + a2) 12

x¿

a

1 2 a r(a2) dz 6 L0 1 2 a (ra3) 6

=

1 ma2 6 1 a a 2 ma2 + m[( )2 + ( )2] = ma2 6 2 2 3

teaching

Web)

I x = Iy = Iz =

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IzG =

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1 a a Ixy = 0 + m(- )( ) = - ma2 2 2 4

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1 a a Iyz = 0 + m( )( ) = ma2 2 2 4

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1 a a Ixz = 0 + m(- )( ) = - ma2 2 2 4

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Ans.

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ma2

- 14 ma2 ¥ 2 2 3 ma

of

1 4

and

1 2 4 ma 2 2 3 ma - 14 ma2

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ma2 ma2 ma2

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Changing the signs of the produts of inertia as represented by the equation in the text, we have

y

21–10. Determine the mass moment of inertia of the homogeneous block with respect to its centroidal x¿ axis. The mass of the block is m.

z z¿

h y¿

SOLUTION

y b

The mass of the differential element is dm = rdV = rabdz. dIx¿ =

x¿ a

1 dma2 + dmz2 12

x

=

1 (rabdz)a2 + (rabdz)z2 12

=

rab 2 (a + 12z2) dz 12 h

dIx¿

or

L

teaching in

dz = rabh

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m 2 (a + h2) 12

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Ix¿ =

Lm

h 2

instructors

Hence,

dm = rab

States

m =

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However,

Web)

laws

Ix¿ =

rab 2 2 rabh 2 = (a + 12z2) dz = (a + h2) 12 L- h2 12






21–11. Determine the moment of inertia of the cylinder with respect to the a–a axis of the cylinder. The cylinder has a mass m.

a

a

SOLUTION

h

The mass of the differential element is dm = rdV = r A pa2 B dy. dIaa =

1 2 4 dma

C r A pa2 B dy D a2 + C r A pa2 B dy D y2

=

1 4

=

A 14 rpa4 + rpa2y2 B dy h

Iaa =

L

dIaa = =

a

+ dm A y2 B

L0

A 14rpa4 + rpa2y2 B dy

rpa2h (3a2 + 4h2) 12

r A pa2 B dy = rpa2h Lm L0 m = A 3a2 + 4h2 B 12

Web)

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Iaa

dm =

in

m =

copyright

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*21–12. Determine the moment of inertia Ix of the composite plate assembly. The plates have a specific weight of 6 lb>ft2.

z

0.25 ft 0.5 ft

0.5 ft x

Horizontial plate: Ixx =

0.5 ft

0.5 ft

SOLUTION 1 6(1)(1) ( )(1)2 = 0.0155 12 32.2

Vertical plates:

1 ( 3

Iy¿y¿ = (

32.2

6(14)(1 22) 32.2

lxz¿ = 0

1 )( )2 = 0.001372 4

)(

6(14)(122) 1 2 1 1 2 )[( ) + (1 22)2] + ( )( ) 12 4 32.2 8

or

Ix¿x¿ =

lxy¿ = 0.707,

6(14)(122)

laws

lxx¿ = 0.707,

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= 0.01235

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permitted.

Dissemination

Using Eq. 21–5,

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Ixx = (0.707)2(0.001372) + (0.707)2(0.01235)

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Ixx = 0.0155 + 2(0.00686) = 0.0292 slug # ft2






Ans.

y

21–13. Determine the product of inertia Iyz of the composite plate assembly. The plates have a weight of 6 lb>ft2.

z

0.25 ft 0.5 ft 0.5 ft

0.5 ft

SOLUTION

0.5 ft x

Due to symmetry, Iyz y = 0

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y

21–14. z

Determine the products of inertia Ixy, Iyz, and Ixz, of the thin plate. The material has a density per unit area of 50 kg>m2.

400 mm 200 mm

SOLUTION x

The masses of segments 1 and 2 shown in Fig. a are m1 = 50(0.4)(0.4) = 8 kg and m2 = 50(0.4)(0.2) = 4 kg. Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 for segment 1 and Ix–y– = Iy–z– = Ix–z– = 0 for segment 2 .

400 mm

Ixy = ©Ix¿y¿ + mxGyG = C 0 + 8(0.2)(0.2) D + C 0 + 4(0)(0.2) D = 0.32 kg # m2

Ans.

Iyz = ©Iy¿z¿ + myGzG or

= C 0 + 8(0.2)(0) D + C 0 + 4(0.2)(0.1) D laws

= 0.08 kg # m2

in

teaching

Web)

Ans. Wide

copyright

Ixz = ©Ix¿z¿ + mxGzG

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= C 0 + 8(0.2)(0) D + C 0 + 4(0)(0.1) D

the

= 0

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y

21–15. z

Determine the products of inertia Ixy, Iyz, and Ixz of the solid. The material is steel, which has a specific weight of 490 lb> ft3. 0.5 ft

0.5 ft 0.25 ft

0.25 ft

SOLUTION x

Consider the block to be made of a large one, 1, minus the slot 2. Thus, g g 490 490 m1 = V1 = (0.5)(0.75)(0.25) = 1.4266 slug and m2 = V2 = (0.25) g g 32.2 32.2 (0.25)(0.25) = 0.2378 slug. Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 and

0.125 ft y 0.125 ft

Ix–y– = Iy–z– = Ix–z– = 0. Since the slot 2 is a hole, it should be considered as a negative segment. Thus Ixy = ©(Ix¿y¿ + mxGyG) = [0 + 1.4266(0.25)(0.375)] - [0 + 0.2378(0.25)(0.625)] = 0.0966 slug # ft2

laws

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Iyz = ©(Iy¿z¿ + myGzG)

World

permitted.

Dissemination

= [0 + 1.4266(0.375)(0.125)] - [0 + 0.2378(0.625)(0.125)]

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= [0 + 1.4266(0.25)(0.125)] - [0 + 0.2378(0.25)(0.125)]

work

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Ixz = ©(Ix¿z¿ + mxGzG)

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= 0.0372 slug # ft2






Ans.

not

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= 0.0483 slug # ft2

*21–16. The bent rod has a mass of 4 kg> m. Determine the moment of inertia of the rod about the Oa axis.

z

y

a

0.4 m O

1.2 m

0.6 m

SOLUTION

x

Ixy = C 4(1.2) D (0)(0.6) + C 4(0.6) D (0.3)(1.2) + C 4(0.4) D (0.6)(1.2) = 2.016 kg # m2 Iyz = C 4(1.2) D (0.6)(0) + C 4(0.6) D (1.2)(0) + C 4(0.4) D (1.2)(0.2) = 0.384 kg # m2 Izx = C 4(1.2) D (0)(0) + C 4(0.6) D (0)(0.3) + C 4(0.4) D (0.2)(0.6) = 0.192 kg # m2 Ix =

1 1 C 4(1.2) D (1.2)2 + C 4(0.6) D (1.2)2 + c C 4(0.4) D (0.4)2 + C 4(0.4) D (1.22 + 0.22) d 3 12

= 8.1493 kg # m2 Iy = 0 +

1 1 C 4(0.6) D (0.6)2 + c C 4(0.4) D (0.4)2 + C 4(0.4) D A 0.62 + 0.22 B d 3 12

teaching

Web)

laws

or

= 0.9493 kg # m2 in

1 1 C 4(1.2) D (1.2)2 + c C 4(0.6) D (0.6)2 + C 4(0.6) D (0.32 + 1.22) d + C 4(0.4) D (1.22 + 0.62) 3 12 Dissemination

Wide

copyright

Iz =

is

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the

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instructors

States

World

permitted.

= 8.9280 kg # m2

learning.

0.6i + 1.2j + 0.4k 2

and

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on

6 2 3 i + j + k 7 6 7

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IOa = Ixu2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux

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20.6 + 1.2 + 0.4 2

the

vu

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= 1.21 kg # m2

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3 2 6 2 2 2 3 6 6 2 2 3 = 8.1493 a b + 0.9493 a b + 8.9280 a b - 2(2.016) a a b a b - 290.384) a b a b - 2(0.192) a b a b 7 7 7 7 7 7 7 7 7






21–17. z¿

The bent rod has a weight of 1.5 lb>ft. Locate the center of gravity G(x, y) and determine the principal moments of inertia Ix¿ , Iy¿ , and Iz¿ of the rod with respect to the x¿, y¿, z¿ axes.

z 1 ft

SOLUTION Due to symmetry

x

y = 0.5 ft

Ans.

1.5 1.5 1 b(0.5)2 d + a b (1)2 32.2 12 32.2

= 0.0272 slug # ft2

or

1 1.5 1.5 1.5 a b (1)2 + a b(0.667 - 0.5)2 d + a b(1 - 0.667)2 12 32.2 32.2 32.2

laws

Iy¿ = 2 c

Ans.

= 0.0155 slug # ft2

Ans. Wide

copyright

in

teaching

Web)

Ix¿ = 2c a

World

permitted.

Dissemination

1 1.5 1.5 a b(1)2 + a b(0.52 + 0.16672) d 12 32.2 32.2

of

the

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States

Iz¿ = 2c

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learning.

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1.5 1 1.5 a b(1)2 + a b (0.3333)2 12 32.2 32.2

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Ans.

y¿

_ x y

Ans.

( - 1)(1.5)(1) + 2 C ( - 0.5)(1.5)(1) D ©x W = = - 0.667 ft ©w 3 C 1.5(1) D

x =

G

1 ft A _ y x¿

21–18. z

Determine the moments of inertia about the x, y, z axes of the rod assembly.The rods have a mass of 0.75 kg/m.

D

30° B

2m A

SOLUTION

1 1 Ix = C 0.75(4) D (4)2 + C 0.75(2) D (2)2 = 4.50 kg # m2 12 12 Iy =

Ans.

1 1 C 0.75(4) D (4)2 + C 0.75(2) D (2 cos 30°)2 = 4.38 kg # m2 12 12

Ans.

1 0.75(2) (2 sin 30°)2 = 0.125 kg # m2 12

Ans.

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Dissemination

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in

teaching

Web)

laws

or

Iz = 0 +

x






2m

1m 1m C

y

21–19. Determine the moment of inertia of the composite body about the aa axis. The cylinder weighs 20 lb, and each hemisphere weighs 10 lb.

a

2 ft

a

SOLUTION

2 ft

uaz = 0.707 uax = 0 uay = 0.707 Izz =

1 20 2 10 ( )(1)2 + 2[ ( )(1)2] 2 32.2 5 32.2

= 0.5590 slug # ft2 Ixx = Iyy =

1 20 10 10 11 2 ( )[3(1)2 + (2)2] + 2[0.259( )(1)2 + ( )] 12 32.2 32.2 32.2 8

teaching

Web)

laws

or

Ixx = Iyy = 1.6975 slug # ft2

Wide

copyright

in

Iaa = 0 + (0.707)2(1.6975) + (0.707)2(0.559)

Dissemination

Iaa = 1.13 slug # ft2

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Ans.






*21–20. z

The assembly consists of a 15-lb plate A, 40-lb plate B, and four 7-lb rods. Determine the moments of inertia of the assembly with respect to the principal x, y, z axes.

1 ft A 4 ft y 4 ft

SOLUTION

x

Due to symmetry y = x = 0 z =

4(15) + 0(40) + 2(28) ©zw = ©w 15 + 40

= 1.3976 ft Iz = (Iz)upper + (Iz)lower + (Iz)rods 1 15 1 40 a b (1)2 + a b (4)2 2 32.2 2 32.2 or

=

teaching

Web)

laws

7 1 7 a b (3)2 + a b (2.5)2 d 12 32.2 32.2

Ans.

permitted.

Dissemination

copyright

Iz = 16.3 slug # ft2

Wide

in

+ 4c

learning.

is

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not

instructors

States

World

Ix = (Ix)upper + (Ix)lower + (Ix)rods - 1 + (Ix)rods - 2

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1 15 15 1 40 a b (1)2 + a b (4)2 d + a b (4)2 4 32.2 32.2 4 32.2

for

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= c

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Ix = 20.2 slug # ft2

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Iy = 20.2 slug # ft2

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And by symmetry,






Ans.

B

21–21. Determine the moment of inertia of the rod-and-thin-ring assembly about the z axis. The rods and ring have a mass per unit length of 2 kg>m.

z

A O

SOLUTION 500 mm

For the rod,

400 mm

ux¿ = 0.6,

uy¿ = 0,

D

uz¿ = 0.8 B

1 Ix = Iy = [(0.5)(2)](0.5)2 = 0.08333 kg # m2 3

x

Ix¿ = 0 Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 From Eq. 21–5, or

Iz = 0.08333(0.6)2 + 0 + 0 - 0 - 0 - 0

Wide

copyright

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teaching

Web)

laws

Iz = 0.03 kg # m2

World

permitted.

Dissemination

For the ring,

use

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The radius is r = 0.3 m by

and

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the

Iz = mR2 = [2 (2p)(0.3)](0.3)2 = 0.3393 kg # m2

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Iz = 3(0.03) + 0.339 = 0.429 kg # m2

this

assessing

is

work

Thus the moment of inertia of the assembly is






120 120

Ans.

y 120 C

21–22. z

If a body contains no planes of symmetry, the principal moments of inertia can be determined mathematically. To show how this is done, consider the rigid body which is spinning with an angular velocity V , directed along one of its principal axes of inertia. If the principal moment of inertia about this axis is I, the angular momentum can be expressed as H = IV = Ivx i + Ivy j + Ivz k. The components of H may also be expressed by Eqs. 21–10, where the inertia tensor is assumed to be known. Equate the i, j, and k components of both expressions for H and consider vx, vy, and vz to be unknown. The solution of these three equations is obtained provided the determinant of the coefficients is zero. Show that this determinant, when expanded, yields the cubic equation

V O

x

I 3 - (Ixx + Iyy + Izz)I 2 + (IxxIyy + IyyIzz + IzzIxx - I2xy - I2yz - I2zx)I - (IxxIyyIzz - 2IxyIyzIzx - IxxI 2yz - IyyI 2zx - IzzI 2xy) = 0

in

teaching

Web)

laws

or

The three positive roots of I, obtained from the solution of this equation, represent the principal moments of inertia Ix, Iy, and Iz.

Dissemination

Wide

copyright

SOLUTION

is

of

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not

instructors

States

World

permitted.

H = Iv = Ivx i + Ivy j + Ivz k

by

and

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learning.

Equating the i, j, k components to the scalar equations (Eq. 21–10) yields

(including

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work

student

the

(Ixx - I) vx - Ixy vy - Ixz vz = 0

is

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the

- Ixx vx + (Ixy - I) vy - Iyz vz = 0

any

- Ixz -Iyz 3 = 0 (Izz - I) destroy

will

courses

part

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is and

-Ixy (Iyy - I) - Izy their

(Ixx - I) - Iyx -Izx

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3

of

Solution for vx, vy, and vz requires

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work

this

assessing

-Izx vz - Izy vy + (Izz - I) vz = 0

Expanding I3 - (Ixx + Iyy + Izz)I2 + A Ixx Iyy + Iyy Izz + Izz Ixx - I2xy - I2yz - I2zx B I - A Ixx Iyy Izz - 2Ixy Iyz Izx - Ixx I2yz - IyyI2zx - Izz I2xy B = 0 Q.E.D.






y

21–23. Show that if the angular momentum of a body is determined with respect to an arbitrary point A, then H A can be expressed by Eq. 21–9. This requires substituting R A = R G + R G>A into Eq. 21–6 and expanding, noting that 1 R G dm = 0 by definition of the mass center and vG = vA + V : R G>A.

Z

z RG/A G

RG RA

P y

A Y

SOLUTION

= a

Lm Lm

rA dm b * vA +

(rG + rG>A) * C v * rG + rG>A) D dm Lm

rG dmb * vA + (rG>A * vA)

Lm

X

Lm

dm +

Lm

rG * (v * rG) dm

rGdmb * (v * rG>A) + rG>A * av *

rG dm = 0 and from Eq. 21–8 HG =

rG dmb + rG>A * (v * rG>A)

Lm

rG * (v * rG)dm

Lm

teaching

Web)

Lm

rA * (v * rA)dm

(rG + rG>A) dm b * vA +

+ a

Since

Lm

or

= a

Lm

laws

HA = a

x

Wide

copyright

in

HA = (rG>A * vA)m + HG + rG>A * (v * rG>A)m

States

World

permitted.

Dissemination

= rG>A * (vA + (v * rG>A))m + HG

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is

of

the

Q.E.D.






not

instructors

= (rG>A * mvG) + HG

Lm

dm

*21–24. z

The 15-kg circular disk spins about its axle with a constant angular velocity of v1 = 10 rad>s. Simultaneously, the yoke is rotating with a constant angular velocity of v2 = 5 rad>s. Determine the angular momentum of the disk about its center of mass O, and its kinetic energy.

150 mm O v1 ⫽ 10 rad/s x

SOLUTION The mass moments of inertia of the disk about the x, y, and z axes are Ix = Iz = Iy =

1 1 mr2 = (15)(0.152) = 0.084375 kg # m2 4 4

v2 ⫽ 5 rad/s

1 1 mr2 = (15)(0.152) = 0.16875 kg # m2 2 2

Due to symmetry, Ixy = Iyz = Ixz = 0

or

Here, the angular velocity of the disk can be determined from the vector addition of v1 and v2. Thus,

teaching

Web)

laws

v = v1 + v2 = [- 10j + 5k] rad>s

Wide

copyright

in

so that vz = 5 rad>s States

World

permitted.

vy = - 10 rad>s

Dissemination

vx = 0

learning.

is

of

the

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instructors

Since the disk rotates about a fixed point O, we can apply

and

use

United

on

Hx = Ixvx = 0.084375(0) = 0

work

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the

by

Hy = Iyvy = 0.16875(- 10) = - 1.6875 kg # m2>s is

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solely

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the

(including

for

Hz = Izvz = 0.084375(5) = 0.421875 kg # m2>s assessing

Thus,

Ans.

part

the

is

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provided

integrity

of

and

work

this

HO = [- 1.69j + 0.422k] kg # m2>s and

any

courses

The kinetic energy of the disk can be determined from

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will

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1 1 1 I v 2 + Iyvy2 + Izvz 2 2 x x 2 2 1 1 1 = (0.084375)(02) + (0.16875)(- 10)2 + (0.084375)(52) 2 2 2 = 9.49 J

T =






Ans.

y

21–25. The cone has a mass m and rolls without slipping on the conical surface so that it has an angular velocity about the vertical axis of V . Determine the kinetic energy of the cone due to this motion.

ω

r

h

SOLUTION vz r

=

vy h

h h vy = a b vz = a bv r r Iz = a

3 3 2 3 3 3 b m (4r2 + h2) + ma h b = a b mr2 + a b mh2 = a bm(r2 + 4h2) 80 4 20 5 20

1 1 1 I v2x + Iyv2y + Izv2z 2 x 2 2 or

1 3 h 2 1 3 mv2 3 a mr2 b a v b + c a b m(r2 + 4h2) d v2 = c3h2 + r2 + 6h2 d r 2 10 2 20 20 2 teaching

Web)

= 0 +

laws

T =

in

9mh2 r2 1 + v2 20 6h2

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permitted.

Dissemination

Wide

Ans.

copyright

T =






21–26. The circular disk has a weight of 15 lb and is mounted on the shaft AB at an angle of 45° with the horizontal. Determine the angular velocity of the shaft when t = 3 s if a constant torque M = 2 lb # ft is applied to the shaft. The shaft is originally spinning at v1 = 8 rad>s when the torque is applied.

0.8 ft 45 v1 A M

SOLUTION Due to symmetry Ixy = Iyz = Izx = 0 Iy = Iz = Ix =

A

1 15 2 32.2

A

1 15 2 32.2

B (0.8)2 = 0.07453 slug # ft2

B (0.8)2 = 0.1491 slug # ft2

For x' axis ux = cos 45° = 0.7071

uy = cos 45° = 0.7071

uz = cos 90° = 0 laws

or

Iz¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux in

teaching

Web)

= 0.1491(0.7071)2 + 0.07453(0.7071)2 + 0 - 0 - 0 - 0

instructors

States

World

permitted.

Dissemination

Wide

copyright

= 0.1118 slug # ft2

and

use

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on

learning.

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of

the

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Principle of impulse and momentum:

the

by

Mx¿ dt = (Hx¿)2

work

student

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for

(Hx¿)1 + ©

is

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the

0.1118(8) + 2(3) = 0.1118 v2

assessing

v2 = 61.7 rad/s

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this

Ans.






8 rad/s B

21–27. The circular disk has a weight of 15 lb and is mounted on the shaft AB at an angle of 45° with the horizontal. Determine the angular velocity of the shaft when t = 2 s if a torque M = 14e0.1t2 lb # ft, where t is in seconds, is applied to the shaft. The shaft is originally spinning at v1 = 8 rad>s when the torque is applied.

0.8 ft 45 v1 A M

SOLUTION Due to symmetry Ixy = Iyz = Izx = 0 Iy = Iz = Ix =

1 2

1 4

15 A 32.2 B (0.8)2 = 0.07453 slug # ft2

15 A 32.2 B (0.8)2 = 0.1491 slug # ft2

For x' axis ux = cos 45° = 0.7071

uy = cos 45° = 0.7071

uz = cos 90° = 0 laws

or

Iz¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux in

teaching

Web)

= 0.1491(0.7071)2 + 0.07453(0.7071)2 + 0 - 0 - 0 - 0

instructors

States

World

permitted.

Dissemination

Wide

copyright

= 0.1118 slug # ft2

and

use

United

on

learning.

is

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Principle of impulse and momentum:

the

by

Mx¿ dt = (Hx¿)2

work

student

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for

(Hx¿)1 + ©

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4e0.1 t dt = 0.1118v2 work

this

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is

L0

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2

0.1118(8) +

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v2 = 87.2 rad/s






Ans.

8 rad/s B

*21–28. The space capsule has a mass of 5 Mg and the radii of gyration are kx = kz = 1.30 m and ky = 0.45 m . If it travels with a velocity vG = 5400j + 200k6 m>s , compute its angular velocity just after it is struck by a meteoroid having a mass of 0.80 kg and a velocity vm = 5- 300i + 200j - 150k6 m>s . Assume that the meteoroid embeds itself into the capsule at point A and that the capsule initially has no angular velocity.

vm

vG

G

Conservation of Angular Momentum: The angular momentum is conserved about the center of mass of the space capsule G. Neglect the mass of the meteroid after the impact. (HG)1 = (HG)2 rGA * mm vm = IG v (0.8i + 3.2j + 0.9k) * 0.8( -300i + 200j - 150k)

laws

or

= 5000 A 1.302 B vx i + 5000 A 0.452 B vy j + 5000 A 1.302 B vz k

in

teaching

Web)

-528i - 120j + 896k = 8450vx i + 1012.5vy j + 8450 vz k

Dissemination

Wide

copyright

Equating i, j and k components, we have

is

of

the

not

instructors

States

World

permitted.

vx = - 0.06249 rad>s

-120 = 1012.5vy

by

and

use

United

on

learning.

vy = -0.11852 rad>s

896 = 8450vz

the

(including

for

work

student

the

vz = 0.1060 rad>s

assessing

is

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solely

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this

v = {-0.0625i - 0.119j + 0.106k} rad>s






A (0.8 m, 3.2 m, 0.9 m)

y

x

SOLUTION

-528 = 8450vx

z

21–29. z

The 2-kg gear A rolls on the fixed plate gear C. Determine the angular velocity of rod OB about the z axis after it rotates one revolution about the z axis, starting from rest. The rod is acted upon by the constant moment M = 5 N # m. Neglect the mass of rod OB. Assume that gear A is a uniform disk having a radius of 100 mm.

300 mm A O M = 5 N·m

SOLUTION vOB = vA tan 18.43° = 0.3333vA Iy =

1 2

(2)(0.1)2 = 0.01 kg # m2

Iz =

1 4

(2)(0.1)2 + 2(0.3)2 = 0.185 kg # m2

(1)

T1 + ©U1 - 2 = T2 0 + 5(2p)12 (0.01)v2A +

1 2

(0.185)v2OB

(2)

Solving Eqs. (1) and (2) yields: vOB = 15.1 rad s

or

Ans.

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the

(including

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by

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on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

vA = 45.35 rad s






y B C

100 mm

21–30. The rod weighs 3 lb>ft and is suspended from parallel cords at A and B. If the rod has an angular velocity of 2 rad>s about the z axis at the instant shown, determine how high the center of the rod rises at the instant the rod momentarily stops swinging.

z 3 ft

3 ft

v

SOLUTION

A

T1 + V1 = T2 + V2 1 1 W 2 2 c l dv + 0 = 0 + Wh 2 12 g h =

1 l2v2 1 (6)2(2)2 = 24 g 24 (32.2)

h = 0.1863 ft = 2.24 in.

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2 rad/s

21–31. Rod AB has a weight of 6 lb and is attached to two smooth collars at its ends by ball-and-socket joints. If collar A is moving downward with a speed of 8 ft>s when z = 3 ft, determine the speed of A at the instant z = 0. The spring has an unstretched length of 2 ft. Neglect the mass of the collars. Assume the angular velocity of rod AB is perpendicular to its axis.

z

vA ⫽ 8 ft/s A

z ⫽ 3 ft

7 ft

SOLUTION

y

vA = {- 8k} ft/s

k ⫽ 4 lb/ft

vB = v B i

B 6 ft

rB/A = {2i + 6j - 3k} ft

x

v = {vx i + vy j + vz k} rad/s vB = vA + v * rB/A j vy 6

k vz 3 -3 laws

or

i vB i = -8k + 3 vx 2

in

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expanding and equating components yields,

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(1)

copyright

vB = -3 vy - 6vz 0 = 3vx + 2vz

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0 = -8 + 6vx - 2vz by

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(vx i + vy j + vz k) # (2i + 6j - 3k) = 0

(4)

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2vx + 6vy - 3vz = 0 part

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Solving Eqs. (1)–(4) yields;

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vx = 0.9796 rad/s

sale

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vy = - 1.061 rad/s vz = -1.469 rad/s Thus,

v = 2(0.9796)2 + ( -1.061)2 + ( -1.469)2 = 2.06 rad/s vG = vA + v * rG/A where rG/A =

vG

1 1 = (2i + 6j - 3k)ft r 2 B/A 2

i 3 = -8k + 0.9796 1

j -1.061 3

k -1.469 3 - 1.5

vG = {5.9985i - 4k} ft/s vG = 2(5.9985)2 + ( -4)2 = 7.2097 ft/s






2 ft

21–31. continued Hence, since v is directed perpendicular to the axis of the rod, T1 = =

1 2 1 Iv + mv20 2 2 1 6 1 1 6 [ ( )(7)2](2.06)2 + ( )(7.2097)2 2 12 32.2 2 32.2

= 6.46 lb # ft

T1 + V1 = T2 + V2 6.46 + 1.5(6) = T2 + T2 = 10.304 lb # ft

1 (4)(3.6056 - 2)2 2

vA = vB + v * rA/B -vA k = vBi + 3

i vx 3.6056

j vy 6

k vz 3 0

0 = vB - 6vz or

0 = vz(0.36056) laws

-vA = 6vx - 3.6056 vy in

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3.6056 vx + 6vy = 0

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vG = vB + v * rG/B

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k 0 3 vy = {6.795 vy k} ft/s 0 will

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i vG = 0 + 3 -1.664 -1.803

1 6 1 1 6 ( )(6.795vy)2 + [ ( )(7)2][(-1.664vy)2 + v2y] 2 32.2 2 12 32.2

= 10.304 vy = -1.34 rad/s vA = 13.590(- 1.34) = 18.2 ft/s






T

Ans.

*21–32. z

The 5-kg circular disk spins about AB with a constant angular velocity of v1 = 15 rad>s. Simultaneously, the shaft to which arm OAB is rigidly attached, rotates with a constant angular velocity of v2 = 6 rad>s. Determine the angular momentum of the disk about point O, and its kinetic energy.

v2 ⫽ 6 rad/s O

A

SOLUTION B

x

The mass moments of inertia of the disk about the x¿ , y¿ , and z¿ axes are Ix¿ =

0.4 m

1 1 mr2 = (5)(0.12) = 0.025 kg # m2 2 2

Iy¿ = Iz¿ =

0.1 m

1 1 mr2 = (5)(0.12) = 0.0125 kg # m2 4 4

Due to symmetry, the products of inertia of the disk with respect to its centroidal planes are equal to zero. Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

laws

or

Here, the angular velocity of the disk can be determined from the vector addition of v1 and v2. Thus, in

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v = v1 + v2 = [15i + 6k] rad>s

permitted.

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The angular momentum of the disk about its mass center B can be obtained by applying

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(HB)x = Ix¿vx = 0.025(15) = 0.375 kg # m2>s by

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(HB)y = Iy¿vy = 0.0125(0) = 0

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(HB)z = Iz¿vz = 0.0125(6) = 0.075 kg # m2>s solely

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H B = [0.375i + 0.075k] kg # m2>s

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vB = v2 * rB>O = (6k) * (0.4i + 0.3j) = [- 1.8i + 2.4j] m>s sale

Since the disk does not rotate about a fixed point O, its angular momentum must be determined from H O = rB>O * mvB + H B = (0.4i + 0.3j) * 5( -1.8i + 2.4j) + (0.375i + 0.075k) = [0.375i + 7.575k] kg # m2>s

Ans.

The kinetic energy of the disk is therefore T =

1 # v HO 2

1 (15i + 6k) # (0.375i + 7.575k) 2 = 25.5 J =






0.3 m

Ans.

v1 ⫽ 15 rad/s

y

21–33. z

The 20-kg sphere rotates about the axle with a constant angular velocity of vs = 60 rad>s. If shaft AB is subjected to a torque of M = 50 N # m, causing it to rotate, determine the value of vp after the shaft has turned 90° from the position shown. Initially, vp = 0. Neglect the mass of arm CDE.

0.4 m D E

0.3 m

SOLUTION

C

The mass moments of inertia of the sphere about the x¿ , y¿ , and z¿ axes are Ix¿ = Iy¿ = Iz¿ =

2 2 mr2 = (20)(0.12) = 0.08 kg # m2 5 5

A

x

1 1 1 1 m(vG)12 + Ix¿(v1)x¿2 + Iy¿(v1)y¿2 + Iz¿(v1)z¿2 2 2 2 2

or

1 (0.08) (602) 2 laws

=0 + 0 + 0 +

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= 144 J

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(vG)22 = ( -0.4vp)2 + (- 0.3vp)2 = 0.25vp2. Also, its angular velocity at this position is v2 = vpi - 60j. Thus, its kinetic energy at this position is by

and

1 1 1 1 m(vG)22 + Ix¿(v2)x¿2 + Iy¿(v2)y¿2 + Iz¿(v2)z¿2 2 2 2 2 the

1 1 1 (20) A 0.25vp2 B + (0.08) A vp 2 B + (0.08)(-60)2 2 2 2 assessing

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T =

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When the sphere moves from position 1 to position 2 , its center of gravity raises vertically ¢z = 0.1 m. Thus, its weight W does negative work.

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UW = - W¢z = - 20(9.81)(0.1) = - 19.62 J Here, the couple moment M does positive work. p UW = Mu = 50 a b = 25pJ 2 Applying the principle of work and energy, T1 + ©U1 - 2 = T2 144 + 25p + ( -19.62) = 2.54vp2 + 144 vp = 4.82 rad>s






Ans.

permitted.

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in

When the sphere is at position 2 , Fig. a, vp = vpi. Then the velocity of its mass center is (vG)2 = vp * vG>C = (vpi) * (- 0.3j + 0.4k) = - 0.4vpj - 0.3vpk. Then

0.1 m

B vp

M ⫽ 50 N⭈m y

When the sphere is at position 1 , Fig. a, vp = 0. Thus, the velocity of its mass center is zero and its angular velocity is v1 = [60k] rad>s. Thus, its kinetic energy at this position is T =

vs ⫽ 60 rad/s

21–34. The 200-kg satellite has its center of mass at point G. Its radii of gyration about the z¿ , x¿ , y¿ axes are kz¿ = 300 mm, kx¿ = ky¿ = 500 mm, respectively. At the instant shown, the satellite rotates about the x¿, y¿ , and z¿ axes with the angular velocity shown, and its center of mass G has a velocity of vG = 5—250i + 200j + 120k6 m>s. Determine the angular momentum of the satellite about point A at this instant.

z, z¿ Vz¿

G

SOLUTION

Vx ¿

The mass moments of inertia of the satellite about the x¿ , y¿ , and z¿ axes are

600 rad/s x¿

Ix¿ = Iy¿ = 200 A 0.52 B = 50 kg # m2

x

Iz¿ = 200 A 0.32 B = 18 kg # m2 Due to symmetry, the products of inertia of the satellite with respect to the x¿ , y¿ , and z¿ coordinate system are equal to zero. Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

laws

or

The angular velocity of the satellite is

in

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v = [600i + 300j + 1250k] rad>s

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vz¿ = 1250 rad>s

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vy¿ = - 300 rad>s

permitted.

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Thus, vx¿ = 600 rad>s

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(HG)y¿ = Iy¿vy¿ = 50( - 300) = -15 000 kg # m2>s

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(HG)x¿ = Ix¿vx¿ = 50(600) = 30 000 kg # m2>s

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(HG)z¿ = Iz¿vz¿ = 18(1250) = 22 500 kg # m2>s and

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HG = [30 000i - 15 000j + 22 500k] kg # m2>s The angular momentum of the satellite about point A can be determined from HA = rG>A * mvG + HG = (0.8k) * 200( - 250i + 200j + 120k) + (30 000i - 15 000j + 22 500k) = [-2000i - 55 000j + 22 500k] kg # m2/s






800 mm

Ans.

A

1250 rad/s

vG Vy ¿

300 rad/s y¿ y

21–35. The 200-kg satellite has its center of mass at point G. Its radii of gyration about the z¿ , x¿ , y¿ axes are kz¿ = 300 mm, kx¿ = ky¿ = 500 mm, respectively. At the instant shown, the satellite rotates about the x¿ , y¿ , and z¿ axes with the angular velocity shown, and its center of mass G has a velocity of vG = 5—250i + 200j + 120k6 m>s. Determine the kinetic energy of the satellite at this instant.

z, z¿ Vz¿

SOLUTION

G Vx ¿

The mass moments of inertia of the satellite about the x¿ , y¿ , and z¿ axes are

600 rad/s

Ix¿ = Iy¿ = 200 A 0.52 B = 50 kg # m2

x¿

Iz¿ = 200 A 0.32 B = 18 kg # m2

x

Due to symmetry, the products of inertia of the satellite with respect to the x¿ , y¿ , and z coordinate system are equal to zero. ¿ Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 The angular velocity of the satellite is

teaching

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laws

or

v = [600i - 300j + 1250k] rad>s

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Thus, vz¿ = 1250 rad>s States

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vy¿ = - 300 rad>s

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vx¿ = 600 rad>s

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Since vG 2 = (- 250)2 + 2002 + 1202 = 116 900 m2>s2, the kinetic energy of the satellite can be determined from work

student

the

1 1 1 1 mvG 2 + Ix¿vx¿ 2 + Iy¿vy¿ 2 + Iz¿vz¿ 2 2 2 2 2 solely

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for

T =

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is

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1 1 1 1 (200)(116 900) + (50) A 6002 B + (50)(- 300)2 + (18) A 12502 B 2 2 2 2 part

the

is sale

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= 37.0025 A 106 B J = 37.0 MJ

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=






800 mm

Ans.

A

1250 rad/s

vG Vy ¿

300 rad/s y¿ y

*21–36. The 15-kg rectangular plate is free to rotate about the y axis because of the bearing supports at A and B.When the plate is balanced in the vertical plane, a 3-g bullet is fired into it, perpendicular to its surface, with a velocity v = 5-2000i6 m>s. Compute the angular velocity of the plate at the instant it has rotated 180°. If the bullet strikes corner D with the same velocity v, instead of at C, does the angular velocity remain the same? Why or why not?

z 150 mm 150 mm

D C A

150 mm

SOLUTION

v x

Consider the projectile and plate as an entire system. Angular momentum is conserved about the AB axis. (HAB)1 = - (0.003)(2000)(0.15)j = {- 0.9j} (HAB)1 = (HAB)2 -0.9j = Ixvx i + Iyvy j + Izvz k Equating components,

laws

or

vx = 0

in

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vz = 0

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permitted.

Dissemination

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copyright

-0.9 = - 8 rad/s 1 c (15)(0.15)2 + 15(0.075)2 d 12

is

of

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vy =

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T1 + V1 = T2 + V2

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1 1 c (15)(0.15)2 + 15(0.075)2 d(8)2 + 15(9.81)(0.15) 2 12

integrity

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this

assessing

1 1 c (15)(0.15)2 + 15(0.075)2 dv2AB 2 12

part

the

is

Ans.

and

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vAB = 21.4 rad/s

This

provided

=

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If the projectile strikes the plate at D, the angular velocity is the same, only the impulsive reactions at the bearing supports A and B will be different.






B

y

21–37. The circular plate has a weight of 19 lb and a diameter of 1.5 ft. If it is released from rest and falls horizontally 2.5 ft onto the hook at S, which provides a permanent connection, determine the velocity of the mass center of the plate just after the connection with the hook is made.

O

1.5 ft

2.5 ft

SOLUTION S

Conservation of energy: T1 + V1 = T2 + V2 0 + 19(2.5) =

A

1 19 2 32.2

B (vG)22 + 0

(vG)2 = 12.69 ft/s Conservation of momentum about point O: 19 (HO)1 = C - A 32.2 B (12.69)(0.75) D i = {- 5.6153i} slug # ft2/s 19 19 C 14 A 32.2 B (0.75)2 + A 32.2 B (0.75)2 D = 0.4149 slug # ft2

Iy =

1 19 a b (0.75)2 = 0.08298 slug # ft2 4 32.2

Iz =

19 19 C 12 A 32.2 B (0.75)2 + A 32.2 B (0.75)2 D = 0.4979 slug # ft2

Dissemination

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laws

or

Ix =

is

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permitted.

(HO)3 = Ix vxi + Iy vy j + Iz vz k

by

and

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learning.

= 0.4149vx i + 0.8298vy j + 0.4979vz k

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is

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-5.6153i = 0.4149vxi + 0.08298vyj + 0.4979vzk

the

(including

for

work

student

the

(HO)2 = (HO)3

provided

integrity

of

and

work

this

Equating i, j and k components

vx = - 13.54 rad/s

0 = 0.89298vy

vy = 0

0 = 0.4979vz

vz = 0

Hence

sale

will

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destroy

of

and

any

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part

the

is

This

-5.6153i = 0.4149vx

v = {-13.54i} rad/s

vG = v *rG/O = ( -13.54i) * (0.75j) = {- 10.2k} ft/s






Ans.

21–38. z

The 10-kg disk rolls on the horizontal plane without slipping. Determine the magnitude of its angular momentum when it is spinning about the y axis at 2 rad>s. M

0.3 m

O

SOLUTION

A

x

Here, the disk rotates about the fixed point O and its angular velocity is

0.1 m

v = - vS j + vP k Since the disk rolls without slipping on the horizontal plane, the instantaneous axis of zero velocity (IA) is shown in Fig. a. Thus, vP 1 1 = vP = vS vS 3 3 Then,

laws

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1 v . 3 S

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copyright

Thus, vx = 0, vy = - vS, and vz =

or

1 v k 3 S in

v = - vS j +

States

World

permitted.

Dissemination

The mass moments of inertia of the disk about the x, y, and z axes are

on

learning.

is

of

the

not

instructors

1 (10)(0.12) + 10(0.32) = 0.925 kg # m2 4

and

use

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Ix = Iz =

work

student

the

by

1 (10)(0.12) = 0.05 kg # m2 2

the

(including

for

Iy =

is

work

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Due to symmetry,

integrity

of

and

work

this

assessing

Ixy = Iyz = Ixz = 0

any

destroy

of

and

Hx = Ix¿vx = 0.925(0) = 0

courses

part

the

is

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provided

Thus, the components of angular momentum about point O are

sale

will

their

Hy = Iy¿vy = 0.05(-vS) = - 0.05vS 1 Hz = Iz¿vz = 0.925a vS b = 0.3083vS 3 At vS = 2 rad>s, H = 30 + [(0.05(2)]2 + [0.3083(2)]2 = 0.625 kg # m2>s






Ans.

y

21–39. If arm OA is subjected to a torque of M = 5 N # m, determine the spin angular velocity of the 10-kg disk after the arm has turned 2 rev, starting from rest. The disk rolls on the horizontal plane without slipping. Neglect the mass of the arm.

z

M

0.3 m

O

SOLUTION

A

x

Here, the disk rotates about the fixed point O and its angular velocity is

0.1 m

v = - vS j + vP k Since the disk rolls without slipping on the horizontal plane, the instantaneous axis of zero velocity (IA) is shown in Fig. a. Thus, vP 1 = vS 3

vP =

1 v 3 S

Then,

or

1 v k 3 S

teaching

Web)

laws

v = - vS j +

in

1 v . 3 S

World

permitted.

Dissemination

Wide

copyright

So that, vx = 0, vy = - vS, and vz =

learning.

is

of

the

not

instructors

States

The mass moment of inertia of the disk about the x, y, and z axes are

by

and

use

United

on

1 (10)(0.12) + 10 (0.32) = 0.925 kg # m2 4

work

student

the

Ix = Iz =

the

(including

for

1 (10)(0.12) = 0.05 kg # m2 2

this

assessing

is

work

solely

of

protected

Iy =

the

is

This

provided

integrity

of

and

work

Due to symmetry,

and

any

courses

part

Ixy = Iyz = Ixz = 0 their

destroy

of

Thus, the kinetic energy of the disk can be determined from sale

will

1 1 1 I v 2 + Iyvy2 + Izvz2 2 x x 2 2 2 1 1 1 1 = (0.925)(02) + (0.05)(- vS)2 + (0.925) a vS b 2 2 2 3 = 0.07639vS2

T =

Referring to the free-body diagram of the disk shown in Fig. b, W = 10(9.81) N, and N do no work. The frictional force F also does no work since the disk rolls without slipping. Only the couple moment does work. UM = Mu = 5[2(2p)] = 20p J Applying the principle of work and energy, T1 = ©U1-2 = T2 0 + 20p = 0.07639vS2 vS = 28.68 rad>s = 28.7 rad>s






Ans.

y

*21–40. Derive the scalar form of the rotational equation of motion about the x axis if æ Z V and the moments and products of inertia of the body are not constant with respect to time.

SOLUTION In general d (Hx i + Hy j + Hz k) dt # # # = A Hx i + Hy j + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)

M =

Substitute Æ = Æ x i + Æ y j + Æ z k and expanding the cross product yields # # M = a A Hx B xyz - Æ z Hy + Æ yHz b i + a A Hy B xyz - Æ x Hz + Æ zHx b j

or

# + a A Hz B xyz - Æ y Hx + Æ x Hy bk

teaching

Web)

laws

Subsitute Hx, Hy and Hz using Eq. 21–10. For the i component in

d (I v - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx) dt x x

Dissemination

Wide

copyright

©Mx =

+ Æ y (Iz vz - Izx vx - Izy vy)

is

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States

World

permitted.

Ans.

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and

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the

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This

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the

(including

for

work

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the

by

and

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United

on

learning.

One can obtain y and z components in a similar manner.






21–41. Derive the scalar form of the rotational equation of motion about the x axis if æ Z V and the moments and products of inertia of the body are constant with respect to time.

SOLUTION In general d (Hx i + Hy j + Hz k) dt # # # = A Hx i + Hyj + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)

M =


or


teaching

Web)

laws

Substitute Hx, Hy and Hz using Eq. 21–10. For the i component in

d (I v - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx) dt x x

permitted.

Dissemination

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copyright

©Mx =

on

learning.

is

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the

not

instructors

States

World

+ Æ y (Iz vz - Izx vx - Izy vy) by

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use

United

For constant inertia, expanding the time derivative of the above equation yields

the

(including

for

work

student

the

# # # ©Mx = (Ix vx - Ixy vy - Ixzvz) - Æ z (Iy vy - Iyz vz - Iyx vx)

this

assessing

is

work

solely

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+ Æ y (Iz vz - Izxvx - Izy vy)

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and

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part

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integrity

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work







Ans.

21–42. Derive the Euler equations of motion for æ Z V , i.e., Eqs. 21–26.

SOLUTION In general d (Hx i + Hy j + Hz k) dt # # # = A Hx i + Hyj + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)

M =


or


teaching

Web)

laws

Substitute Hx, Hy and Hz using Eq. 21–10. For the i component in

d (Ix vx - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx) dt + Æ y (Iz vz - Izx vx - Izy vy) not

instructors

States

World

permitted.

Dissemination

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©Mx =

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on

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is

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21–43. z

The 4-lb bar rests along the smooth corners of an open box. At the instant shown, the box has a velocity v = 53j6 ft>s and an acceleration a = 5 -6j6 ft>s2. Determine the x, y, z components of force which the corners exert on the bar.

A

SOLUTION

2 ft

©Fx = m(aG)x ;

Ax + Bx = 0

[1]

©Fy = m(aG)y ;

4 Ay + By = ¢ ≤ ( -6) 32.2

[2]

©Fz = m(aG)z ;

Bz - 4 = 0

B

x

Bz = 4 lb

1 ft

Ans.

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By (1) - Ay (1) + 4(0.5) = 0

[3]

# ©(MG)y = Iy vy - (Iz - Ix)vz vx ;

Ax (1) - Bx (1) + 4(1) = 0

[4] laws

By = - 1.37 lb

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# ©(MG)z = Iz vz - (Ix - Iy) vx vy ; States

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*21–44. The uniform rectangular plate has a mass of m = 2 kg and is given a rotation of v = 4 rad>s about its bearings at A and B. If a = 0.2 m and c = 0.3 m, determine the vertical reactions at the instant shown. Use the x, y, z axes shown mac c 2 - a 2 and note that Izx = - a b. ba 2 12 c + a2

x

A

V B z c

SOLUTION

y

vx = 0,

vy = 0,

vz = -4

# vx = 0,

# vy = 0,

# vz = 0

# # ©My = Iyy vy - (Izz - Ixx)vz vx - Iyz avz - vx vy b # - Izx A v2z - v2x B - Ixy avx + vy vz b 1

1

a 2 c 2 2 a 2 c 2 2 Bx B a b + a b R - Ax B a b + a b R = - Izx (v)2 2 2 2 2

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Bx = 9.98 N






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21–45. z

If the shaft AB is rotating with a constant angular velocity of v = 30 rad>s, determine the X, Y, Z components of reaction at the thrust bearing A and journal bearing B at the instant shown. The disk has a weight of 15 lb. Neglect the weight of the shaft AB.

1.5 ft A

30⬚ 1 ft

x

SOLUTION

0.5 ft B

The rotating xyz frame is set with its origin at the plate’s mass center, Fig. a. This frame will be fixed to the disk so that its angular velocity is Æ = v and the x, y, and z axes will always be the principle axes of inertia of the disk. Referring to Fig. b,

v ⫽ 30 rad/s y

v = [30 cos 30°j - 30 sin 30°k] rad>s = [25.98j -15k] rad>s Thus, vx = 0

vy = 25.98 rad>s

vz = - 15 rad>s

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A X + BX = 0

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AY = 0

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A Z + BZ - 15 = 0

(3) Ans. (4)

Solving Eqs. (1) through (4), A Z = 1.461 lb A X = BX = 0






BZ = 13.54 lb = 13.5 lb

Ans. Ans.

21–46. The 40-kg flywheel (disk) is mounted 20 mm off its true center at G. If the shaft is rotating at a constant speed v = 8 rad>s, determine the maximum reactions exerted on the journal bearings at A and B.

500 mm A

G

20 mm 0.75 m

SOLUTION vx = 0 vy = -8 rad>s vz = 0 # ©Mx = Ix vx - (Iy - Iz) vy vz ; Bz (1.25) - Az (0.75) = 0 - 0 # ©Mz = Iz vz - (Ix - Iy) vx vy ; - Bx (1.25) + Ax (0.75) = 0 - 0 Ax + B x = 0

©Fz = maz ;

Az + Bz - 40(9.81) = 40(8)2(0.020)

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1.25 m

v ⫽ 8 rad/s

B

21–47. The 40-kg flywheel (disk) is mounted 20 mm off its true center at G. If the shaft is rotating at a constant speed v = 8 rad>s, determine the minimum reactions exerted on the journal bearings at A and B during the motion.

500 mm A

G

20 mm 0.75 m

SOLUTION vx = 0 vy = -8 rad>s vz = 0 # ©Mx = Ix vx - (Iy - Iz) vy vz ; Bz (1.25) - Az (0.75) = 0 - 0 # ©Mz = Iz vz - (Ix - Iy) vx vy ; - Bx (1.25) + Ax (0.75) = 0 - 0 Ax + Bx = 0

©Fz = maz ;

Az + Bz - 40(9.81) = - 40(8)2(0.020)

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1.25 m

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*21–48. The man sits on a swivel chair which is rotating with a constant angular velocity of 3 rad>s. He holds the uniform 5-lb rod AB horizontal. He suddenly gives it an angular acceleration of 2 rad>s2, measured relative to him, as shown. Determine the required force and moment components at the grip, A, necessary to do this. Establish axes at the rod’s center of mass G, with +z upward, and + y directed along the axis of the rod towards A.

3 rad/s 3 ft 2 rad/s2 B

SOLUTION I x = Iz =

1 5 A B (3)2 = 0.1165 ft4 12 32.2

Iy = 0 Æ = v = 3k vx = v y = 0 vz = 3 rad>s # # Æ = (vxyz) + Æ * v = - 2i + 0 laws

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(aG)z = 2(1.5) = 3 ft>s2 Ax = 0

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# ©My = Iyvx - (Iz - Ix)vz vx; 0 + My = 0 - 0 My = 0

Ans.

# ©Mz = Izvz - (Ix - Iy)vx vy; Mz = 0 - 0 Mz = 0






2 ft

Ans.

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21–49. The 5-kg rod AB is supported by a rotating arm. The support at A is a journal bearing, which develops reactions normal to the rod. The support at B is a thrust bearing, which develops reactions both normal to the rod and along the axis of the rod. Neglecting friction, determine the x, y, z components of reaction at these supports when the frame rotates with a constant angular velocity of v = 10 rad>s.

z 0.5 m B 0.5 m G A

SOLUTION

x

I y = Iz =

1 (5)(1)2 = 0.4167 kg # m2 12

Ix = 0

# # # Applying Eq. 21–25 with vx = vy = 0 vz = 10 rad>s vx = vy = vz = 0 # ©Mx = Ix vx - (Iy - Iz)vy vz ;

0 = 0

# ©My = Iy vy - (Iz - Ix)vz vx ;

Bz (0.5) - Az(0.5) = 0

(1)

# ©Mz = Iz vz - (Ix - Iy)vx vy ;

Ay (0.5) - By(0.5) = 0

(2)

or

Also, Bx = -250N

©Fy = m(aG)y ;

Ay + B y = 0

©Fz = m(aG)z ;

Az + Bz - 5(9.81) = 0

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Bx = -5(10)2 (0.5)

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(3)

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v = 10 rad/s

21–50. z

The rod assembly is supported by a ball-and-socket joint at C and a journal bearing at D, which develops only x and y force reactions. The rods have a mass of 0.75 kg/m. Determine the angular acceleration of the rods and the components of reaction at the supports at the instant v = 8 rad>s as shown.

D

ω = 8 rad/s

2m

A

1m B

SOLUTION 2m

Æ = v = 8k vx = vy = 0,

vz = 8 rad>s

# # vx = vy = 0,

# # vz = vz

C x

Ixz = Ixy = 0 Iyz = 0.75(1)(2)(0.5) = 0.75 kg # m2 Izz =

1 (0.75)(1)(1)2 = 0.25 kg # m2 3

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-Dy (4) - 7.3575(0.5) = 0.75(8)2

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Dy = - 12.9 N

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# vz = 200 rad>s2

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Cy = - 11.1 N ©Fz = m(aG)z ; Cz = 36.8 N






Ans.

Cx - 37.5 = - 1(0.75)(200)(0.5)

Cx = - 37.5 N ©Fy = m(aG)y ;

50 N ⋅ m

Ans. Cz - 7.3575 - 29.43 = 0 Ans.

y

21–51. The uniform hatch door, having a mass of 15 kg and a mass center at G, is supported in the horizontal plane by bearings at A and B. If a vertical force F = 300 N is applied to the door as shown, determine the components of reaction at the bearings and the angular acceleration of the door.The bearing at A will resist a component of force in the y direction, whereas the bearing at B will not. For the calculation, assume the door to be a thin plate and neglect the size of each bearing. The door is originally at rest.

z 100 mm 200 mm

A

200 mm

G 30 mm

vx = vy = v z = 0 # # v x = vz = 0 Eqs. 21–25 reduce to 300(0.25 - 0.03) + Bz(0.15) - Az(0.15) = 0

Bz - Az = -440

(1) 1 # (15)(0.4)2 + 15(0.2)2]vy 12 laws

15(9.81)(0.2) - (300)(0.4 - 0.03) = [

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# ©My = Iyvy; # vy = -102 rad>s2

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-Bx(0.15) + Ax(0.15) = 0

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300 - 15(9.81) + Bz + Az = 15(101.96)(0.2)

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Bz + Az = 153.03

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Bz = - 143 N

F

x

SOLUTION

©Mx = 0;

150 mm 150 mm 100 mm B

Ans.

30 mm

y

*21–52. The conical pendulum consists of a bar of mass m and length L that is supported by the pin at its end A. If the pin is subjected to a rotation V, determine the angle u that the bar makes with the vertical as it rotates.

V A

L

SOLUTION Ix = I z =

1 mL2, 3

Iy = 0

vx = 0,

vy = - v cos u,

# vx = 0,

# vy = 0,

vz = v sin u

# vz = 0

# ©Mx = Ix vx - (Iy - Iz) vy vz -mg a

1 L sin u b = 0 - a0 - mL2 b( - v cos u)(v sin u) 2 3

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21–53. The car travels around the curved road of radius r such that its mass center has a constant speed vG. Write the equations of rotational motion with respect to the x, y, z axes. Assume that the car’s six moments and products of inertia with respect to these axes are known.

z y

G r

SOLUTION Applying Eq. 21–24 with vx = 0,

vy = 0,

vG , r

vz =

v x = vy = v z = 0 ©Mx = -Iyz B 0 - a ©My = -Izx B a

Iyz vG 2 b R = 2 v2G r r

Ans.

Izx vG 2 b - 0 R = - 2 v2G r r

Ans.

©Mz = 0

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Note: This result indicates the normal reactions of the tires on the ground are not all necessarily equal. Instead, they depend upon the speed of the car, radius of curvature, and the products of inertia Iyz and Izx. (See Example 13–6.)






x

21–54. The rod assembly is supported by journal bearings at A and B, which develops only x and y force reactions on the shaft. If the shaft AB is rotating in the direction shown at V = 5-5j6 rad>s, determine the reactions at the bearings when the assembly is in the position shown. Also, what is the shaft’s angular acceleration? The mass of each rod is 1.5 kg>m.

z

V

A x

300 mm 500 mm

400 mm B

300 mm

SOLUTION

500 mm

vx = vz = 0,

vy = - 5 rad>s

# # vx = v z = 0 Eqs. 21–24 become # ©Mx = -Ixy vy - Iy z v2y # ©My = Iyy vy # ©Mz = Ixy v2y - Iy z vy

or

1 1 (0.4)(1.5)(0.4)2 + (0.3)(1.5)(0.3)2 = 0.0455 kg # m2 3 3 laws

Iy y =

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Iy z = [0 + (1.5)(0.3)(0.15)(0.8)] = 0.0540 kg # m2

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Ixy = [0 + (1.5)(0.4)(0.2)(0.5)] = 0.0600 kg # m2

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Ax = -1.17 N ©Fz = m(aG)z ; Az = 12.3 N






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y

21–55. The 20-kg sphere is rotating with a constant angular speed of v1 = 150 rad>s about axle CD, which is mounted on the circular ring. The ring rotates about shaft AB with a constant angular speed of v2 = 50 rad>s. If shaft AB is supported by a thrust bearing at A and a journal bearing at B, determine the X, Y, Z components of reaction at these bearings at the instant shown. Neglect the mass of the ring and shaft.

z

A

D x

SOLUTION

0.5 m

C

The rotating xyz frame is established as shown in Fig. a. This frame will have an angular velocity of Æ = v2 = [50j] rad>s. Since the sphere is symmetric about its spinning axis, the x, y, and z axes will remain as the principal axes of inertia. Thus,

0.5 m

The angular velocity of the sphere is v = v1 + v2 = [150i + 50j] rad>s. Thus, vx = 150 rad>s

vy = 50 rad>s

vz = 0

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Since the directions of v1 and v2 do not change with respect to the xyz frame and # their magnitudes are constant, vxyz = 0. Thus, # # # vx = vy = vz = 0

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# ©My = Iyvy - Iz Æ xvz + Ix Æ z vx; # ©Mz = Izvz - Ix Æ yvx + Iy Æ x vy;

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A X - BX = - 7500

A X + BX = 0

©FY = m(aG)Y;

AY = 0

©FZ = m(aG)Z;

A Z + BZ - 20 (9.81) = 0

(3)

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Ans. (4)

Solving Eqs. (1) through (4), A Z = BZ = 98.1 N A X = - 3750 N = - 3.75 kN





BX = 3750 N = 3.75 kN

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Applying the equations of motion and referring to the free-body diagram shown in Fig. a, # ©Mx = Ixvx - Iy Æ zvy + Iz Æ y vz; BZ(0.5) - A Z(0.5) = 0 - 0 + 0

v2 ⫽ 50 rad/s B

v1 ⫽ 150 rad/s

2 2 2 mr = (20)(0.25)2 = 0.5 kg # m2 5 5

Ix = Iy = Iz =


0.25 m

y

*21–56. The rod assembly has a weight of 5 lb>ft. It is supported at B by a smooth journal bearing, which develops x and y force reactions, and at A by a smooth thrust bearing, which develops x, y, and z force reactions. If a 50-lb # ft torque is applied along rod AB, determine the components of reaction at the bearings when the assembly has an angular velocity v = 10 rad>s at the instant shown.

z

v ⫽ 10 rad/s B 50 lb ft

SOLUTION Iy =

4 ft

2(5) 2(5) 1 6(5) 1 2(5) c d (6)2 + c d (2)2 + c d (3)2 + c d (2)2 3 32.2 12 32.2 32.2 32.2

2 ft

2 ft

= 15.3209 slug # ft2

2 ft A

Ix =

2(5) 2(5) 1 6(5) 1 2(5) 1 2(5) c d (6)2 + c d (2)2 + c d (22 + 32) + c d (2)2 + c d (12 + 22) 3 32.2 12 32.2 32.2 12 32.2 32.2

x

Ix = 16.9772 slug # ft2 2(5) 1 2(5) c d (2)2 + c d (2)2 = 1.6563 slug # ft2 3 32.2 32.2 or

2(5) 2(5) d (1)(2) + c d (2)(3) = 2.4845 slug # ft2 32.2 32.2

Web)

laws

Ixy = Izx = 0 in

Iy z = c

teaching

Iz =

Dissemination

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copyright

# # Applying Eq. 21-24 with vx = vy = 0, vz = 10 rad>s, vx = vy = 0

is

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By = -46.4

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by

# Bx (6) = 0 - 0 - 2.4845vz - 0 - 0

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2(5) 2(5) d (1)(30.19) - c d (2)(30.19) 32.2 32.2 sale

©Fx = m(aG)x ;

Ax + ( -12.50) = - c Ax = -15.6 lb

©Fy = m(aG)y ;

Ay + ( -46.41) = - c

Ans. 2(5) 2(5) d (1)(10)2 - c d (2)(10)2 32.2 32.2

Ay = - 46.8 lb ©Fz = m(aG)z ;






Az - 2(5) - 2(5) - 6(5) = 0

Ans. Az = 50 lb

Ans.

y

21–57. The blades of a wind turbine spin about the shaft S with a constant angular speed of vs, while the frame precesses about the vertical axis with a constant angular speed of vp. Determine the x, y, and z components of moment that the shaft exerts on the blades as a function of u. Consider each blade as a slender rod of mass m and length l.

z

Vp

u

x S

SOLUTION

u Vs

The rotating xyz frame shown in Fig. a will be attached to the blade so that it rotates with an angular velocity of Æ = v, where v = vs + vp. Referring to Fig. b vp = vp sin ui + vp cos uk. Thus, v = vp sin ui + vs j + vp cos uk. Then vx = vp sin u

vy = vs vz = vp cos u

# The angular acceleration of the blade v with respect to the XYZ frame can be obtained by setting another x¿y¿z¿ frame having an angular velocity of Æ¿ = vp = vp sin ui + vp cos uk. Thus, # # v = A vx¿y¿z¿ B + Æ ¿ * v or

# # = (v1)x¿y¿z¿ + (v2)x¿y¿z¿ + Æ¿ * vS + Æ¿ * vP teaching

Web)

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= 0 + 0 + A vp sin ui + vp cos uk B * (vsj) + 0

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= -vsvp cos ui + vsvp sin uk

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# # Since Æ = v, vx¿y¿z¿ = v. Thus,

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Also, the x, y, and z axes will remain as principle axes of inertia for the blade. Thus, Iz = 0

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1 2 (2m)(2l)2 = ml2 12 3

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# ©Mx = Ixvx - A Iy - Iz B vyvz;

4 = - ml2 vsvp cos u 3 # ©My = Iyvy - A Iz - Ix B vzvx;

My = 0 - a0 =

# ©Mz = Izvz - (Ix - Iy)vxvy ;






2 2 ml b (vp cos u)(vp sin u) 3

1 2 2 ml vp sin 2u 3

Mz = 0 - 0 = 0

Ans.

Ans. Ans.

y

21–58. The cylinder has a mass of 30 kg and is mounted on an axle that is supported by bearings at A and B. If the axle is turning at V = 5-40j6 rad>s, determine the vertical components of force acting at the bearings at this instant.

z

0.5 m

1.5 m

A G

v B

SOLUTION

1m

vx = 0

x

vy = -40 sin 18.43° = - 12.65 rad>s vz = 40 cos 18.43° = 37.95 rad>s # vx = 0,

# vy = 0,

# vz = 0

(aG)x = (aG)y = (aG)z = 0 Ix = Iy =

or

1 (30)(0.25)2 = 0.9375 kg # m2 2

teaching

Web)

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Iz =

1 (30)[3(0.25)2 + (1.5)2] = 6.09375 kg # m2 12

copyright

in

Using the first of Eqs. 21–25,

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permitted.

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# ©Mx = Ix vx - (Iy - Iz)vy vz

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Bz (1) - Az (1) = 0 - (6.09375 - 0.9375)( -12.65)(37.95)

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Bz - Az = 2475

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Az = - 1.09 kN

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Bz = 1.38 kN sale

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Ans.






1m

21–59. The thin rod has a mass of 0.8 kg and a total length of 150 mm. It is rotating about its midpoint at a constant rate while the table to which its axle A is fastened is rotating at 2 rad>s. Determine the x, y, z moment components which the axle exerts on the rod when the rod is in any position u.

y 2 rad/s x

A z

SOLUTION The x,y,z axes are fixed as shown. vx = 2 sin u vy = 2 cos u # vz = u = 6 # # vx = 2u cos u = 12 cos u # # vy = -2u sin u = -12 sin u # vz = 0

Web)

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Ix = 0 in

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1 (0.8)(0.15)2 = 1.5(10-3) 12

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©My = 1.5(10-3)(- 12 sin u) - [1.5(10-3) -0](6)(2 sin u)

Ans.

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©Mz = 0.006 sin u cos u = (0.003 sin 2 u) N # m

Ans.

u

. u

*21–60. Show that the angular velocity of a body, in terms of Euler# angles f , # u , and c , # can be expressed as # v# = (f sin u #sin c + u cos c )i + (f sin u cos c - u sin c ) j + (f cos u + c)k, where i, j, and k are directed along the x, y, z axes as shown in Fig. 21–15d.

SOLUTION

# # From Fig. 21–15b. due to rotation f, the x, y, z components of f are simply f along z axis. # # # From Fig 21–15c,# due to rotation u, the x, y, z #components of f and u are f sin u in the y direction, f cos u in the z direction, and u in the x direction.

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Lastly, rotation c. Fig. 21–15d, produces the final components which yields # # # # # # v = A f sin u sin c + u cos c B i + A f sin u cos c - u sin c B j + A f cos u + c B k Q.E.D.






21–61. A thin rod is initially coincident with the Z axis when it is given three rotations defined by the Euler angles f = 30°, u = 45°, and c = 60°. If these rotations are given in the order stated, determine the coordinate direction angles a, b , g of the axis of the rod with respect to the X, Y, and Z axes. Are these directions the same for any order of the rotations? Why?

SOLUTION u = (1 sin 45°) sin 30° i - (1 sin 45°) cos 30°j + 1 cos 45° k u = 0.3536i - 0.6124j + 0.7071k a = cos - 1 0.3536 = 69.3°

Ans.

b = cos - 1( -0.6124) = 128°

Ans.

g = cos - 1(0.7071) = 45°

Ans.

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No, the orientation of the rod will not be the same for any order of rotation, because finite rotations are not vectors.






21–62. The turbine on a ship has a mass of 400 kg and is mounted on bearings A and B as shown. Its center of mass is at G, its radius of gyration is kz = 0.3 m, and kx = ky = 0.5 m. If it is spinning at 200 rad>s, determine the vertical reactions at the bearings when the ship undergoes each of the following motions: (a) rolling, v1 = 0.2 rad >s, (b) turning, v2 = 0.8 rad>s, (c) pitching, v3 = 1.4 rad>s.

y V2 G

B

200 rad/s A 0.8 m

SOLUTION

z

v1 = 0.2 + 200 = 200.2 rad>s

a)

©Fy = m(aG)y ;

1.3 m

V1

V3

Ay + By - 3924 = 0

# ©Mx = Ix vx - (Iy - Iz) vy vz ; By (0.8) - Ay (1.3) = 0 - 0 Thus, Ans.

By = 2.43 kN

Ans. or

Ay = 1.49 kN

teaching

Web)

laws

Æ = 0.8j

b)

in

v = 0.8j + 200k

States

World

permitted.

Dissemination

Wide

copyright

# v = 0 + 0.8j X (0.8j + 200k) = 160i

the

not

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Ay + By - 3924 = 0

is

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©Fy = m(aG)y ;

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# ©Mx = Ix vx - (Iy - Iz) vy vz;

the

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By (0.8) - Ay (1.3) = 400(0.5)2 (160) - [400(0.5)2 - 400(0.3)2](0.8)(200)

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By (0.8) - Ay (1.3) = 5760

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courses

Ay = -1.24 kN

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By = 5.17 kN Æ = 1.4i

c)

v = 1.4i + 200k # v = 1.4i * (1.4i + 200k) = -280j ©Fy = m(aG)y ;

Ay + By - 3924 = 0

# ©Mx = Ix vx - (Iy - Iz) vy vz ; By (0.8) - Ay (1.3) = 0 - 0 Thus,






Ay = 1.49 kN

Ans.

By = 2.43 kN

Ans.

x

21–63. The 10-kg disk spins about axle AB at a constant rate of vs = 100 rad>s . If the supporting arm precesses about the vertical axis at a constant rate of vp = 5 rad>s , determine the internal moment at O caused only by the gyroscopic action.

500 mm vp ⫽ 5 rad/s O 150 mm A

SOLUTION # # Here, u = 90°, f = vp = 5 rad>s, and c = - vs = - 100 rad>s are constants. This is the special case of precession. # # 1 Iz = (10)(0.152) = 0.1125 kg # m2, Æ z = f = 5 rad>s, and vz = c = - 100 rad>s. 2

vs ⫽ 100 rad/s

Thus, Mx = 0.1125(5)(- 100) = - 56.25 N # m

Ans.

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©Mx = Iz Æ yvz;






B

*21–64. The 10-kg disk spins about axle AB at a constant rate of vs = 250 rad>s, and u = 30°. Determine the rate of precession of arm OA. Neglect the mass of arm OA, axle AB, and the circular ring D.

B

D

C

vp A u

SOLUTION

600 mm

# # Since u = 30°, c = vs = 250 rad>s, and f = vp are constant, the disk undergoes 1 1 steady precession. Iz = (10)(0.152) = 0.1125 kg # m2 and I = Ix = Iy = (10) 2 4 (0.152) + 10(0.62) = 3.65625 kg # m2. Thus, # # # # ©Mx = - If 2 sin u cos u + Izf sin u(f cos u + c)

O

-10(9.81) sin 30°(0.6) = - 3.65625vp2 sin 30° cos 30° + 0.1125vp sin 30°(vp cos 30° + 250) 1.5345vp2 - 14.0625vp - 29.43 = 0 laws

or

Solving, vp = 10.9 rad>s or - 1.76 rad>s

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Ans.






vs

150 mm

21–65. When OA precesses at a constant rate of vp = 5 rad>s, when u = 90°, determine the required spin of the 10-kg disk C. Neglect the mass of arm OA, axle AB, and the circular ring D.

B

D

C

vp A u

SOLUTION

600 mm

# # Here, u = 90°, c = vs, and f = vp = 5 rad>s are constant. Thus, this is a special case of steady precession.

O

# # 1 (10)(0.152) = 0.1125 kg # m2, Æ y = - f = - 5 rad>s, and vz = c = vs. Then, 2 ©Mx = Iz Æ yvz; -10(9.81)(0.6) = 0.1125( - 5)(vs)

Iz =

vs = 104.64 rad>s = 105 rad>s

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vs

150 mm

21–66. The car travels at a constant speed of vC = 100 km>h around the horizontal curve having a radius of 80 m. If each wheel has a mass of 16 kg, a radius of gyration kG = 300 mm about its spinning axis, and a radius of 400 mm, determine the difference between the normal forces of the rear wheels, caused by the gyroscopic effect. The distance between the wheels is 1.30 m.

1.30 m vC ⫽ 100 km/h

SOLUTION

80 m

I = 2[16(0.3)2] = 2.88 kg # m2 vs =

100(1000) = 69.44 rad>s 3600(0.4)

vp =

100(1000) = 0.347 rad>s 80(3600)

M = I vs vp ¢F(1.30) = 2.88(69.44)(0.347) ¢F = 53.4 N

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Ans.






21–67. A wheel of mass m and radius r rolls with constant spin V about a circular path having a radius a. If the angle of inclination is u, determine the rate of precession. Treat the wheel as a thin ring. No slipping occurs.

. f u r

a V

SOLUTION Since no slipping occurs, # # (r) c = (a + r cos u) f or # a + r cos u # c = ( )f r

(1)

Also, # # v = f + c

World

permitted.

Dissemination

Iz = mr2

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# # # v = f sin u j + ( -c + f cos u)k

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# # # vy = f sin u, vz¿ = - c + f cos u vx = 0, # # # # # v = f * c = - f c sin u # # # # # vx = - f c sin u, vy = vz = 0 of

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F r sin u - N r cos u =

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# ©Mx = Ix vx + (Iz - Iy)vz vy

# # # # # m r2 mr2 (-f c sin u) + (m r2 )( - c + f cos u)(f sin u) 2 2

# Using Eqs. (1), (2) and (3), and eliminating c, we have # # # mr2 -f a # m r2 a + r cos u # (- f) sin u( ( )f + )f sin u m a f2 r sin u - m g r cos u = r r 2 2 # #2 m r2 # 2 m r2 - f2a ( (f sin u cos u) m a f sin u r - m g r cos u = ) sin u r 2 2 # # 2 g cos u = a f2 sin u + r f2 sin u cos u # 2 g cot u 1/2 f = ( ) a + r cos u






or Wide

mr2 , 2

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I x = Iy =

laws

N - mg = 0

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©Fz¿ = m(aG)z¿ ;

(2)

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# F = m(a f2)

copyright

©Fy¿ = m(aG)y¿ ;

Ans.

*21–68. The top consists of a thin disk that has a weight of 8 lb and a radius of 0.3 ft. The rod has a negligible mass and a length of 0.5 ft. If the top is spinning with an angular velocity vs = 300 rad>s, determine the steady-state precessional angular velocity vp of the rod when u = 40°.

0.3 ft

0.5 ft Vp u

SOLUTION # # # ©Mx = -If2 sin u cos u + Iz f sin u a f cos u + c b 8 8 1 b(0.3)2 + a b (0.5)2 dv2p sin 40° cos 40° 8(0.5 sin 40°) = - c a 4 32.2 32.2 8 1 b(0.3)2 d vp sin 40° (vp cos 40° + 300) +c a 2 32.2 0.02783v2p - 2.1559vp + 2.571 = 0 Ans. (Low precession)

vp = 76.3 rad>s

Ans. (High precession)

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vp = 1.21 rad>s






Vs

21–69. Solve Prob. 21–68 when u = 90°.

SOLUTION ©Mx = Iz Æ y vz

0.3 ft

0.5 ft Vp

8 1 b (0.3)2 d vp (300) 8(0.5) = c a 2 32.2

u

vp = 1.19 rad>s

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Vs

21–70. The top has a mass of 90 g, a center of mass at G, and a radius of gyration k = 18 mm about its axis of symmetry. About any transverse axis acting through point O the radius of gyration is kt = 35 mm. If the top is connected to a ball-andsocket joint at O and the precession is vp = 0.5 rad>s, determine the spin V s.

Vp Vs 45⬚

O

SOLUTION vp = 0.5 rad>s # # # # ©Mx = -If2 sin u cos u + Izf sin u af cos u + c b 0.090(9.81)(0.06) sin 45° = - 0.090(0.035)2 (0.5)2 (0.7071)2 # + 0.090(0.018)2(0.5)(0.7071) C 0.5(0.7071) + c D vs = c = 3.63 A 103 B rad>s

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Ans.






G

60 mm

21–71. The 1-lb top has a center of gravity at point G. If it spins about its axis of symmetry and precesses about the vertical axis at constant rates of vs = 60 rad>s and vp = 10 rad>s, respectively, determine the steady state angle u. The radius of gyration of the top about the z axis is kz = 1 in., and about the x and y axes it is kx = ky = 4 in.

z

u vp

10 rad/s

G

SOLUTION

3 in.

# # Since c = vs = 60 rad>s and f = vp = - 10 rad>s and u are constant, the top undergoes steady precession. Iz = a

1 1 b a b = 215.67 A 10 - 6 B slug # ft2 32.2 12 2

I = Ix = Iy = a

and

= 3.4507 10 - 3 slug # ft2.

A

1 4 ba b 32.2 12

O x y

2

B

Thus, # # # # ©Mx = -If2 sin u cos u + Izf sin u A f cos + c B

or

-1 sin u(0.25) = -3.4507 A 10 - 3 B ( -10)2 sin u cos u + 215.67 A 10 - 6 B (- 10) sin u[(-10) cos u + 60] u = 68.1°

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laws

Ans.






vs

60 rad/s

*21–72. While the rocket is in free flight, it has a spin of 3 rad>s and precesses about an axis measured 10° from the axis of spin. If the ratio of the axial to transverse moments of inertia of the rocket is 1>15, computed about axes which pass through the mass center G, determine the angle which the resultant angular velocity makes with the spin axis. Construct the body and space cones used to describe the motion. Is the precession regular or retrograde?

z

Vp

SOLUTION

vs ⫽ 3 rad/s

10

Determine the angle b from the result of Prob. 21-75 tan u = tan 10° =

G

I tan b Iz 15 tan b 1

b = 0.673°

Ans.

Thus,

laws

or

a = 10° - 0.673° = 9.33°

Wide

copyright

in

teaching

Web)

Hence,

Ans. States

World

permitted.

Dissemination

Regular Precession

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Since Iz 6 I.






21–73. The 0.2-kg football is thrown with a spin vz = 35 rad>s. If the angle u is measured as 60°, determine the precession about the Z axis. The radius of gyration about the spin axis is kz = 0.05 m, and about a transverse axis it is kt = 0.1 m.

Z

z

z=

SOLUTION

G

Gyroscopic Motion: Here, the spinning angular velocity c = vs = 35 rad>s. The moment inertia of the football about the z axis is Iz = 0.2(0.052) = 0.5(10-3) kg # m2 and the moment inertia of the football about its transverse axis is I = 0.2(0.12) = 2(10-3) kg # m2.Applying the third of Eq. 21–36 with u = 60°, we have I - Iz # HG cos u c = I Iz 35 = c

2(10-3) - 0.5(10-3)

d HG cos 60°

2(10-3)0.5(10-3)

or

HG = 0.04667 kg # m2>s

teaching

Web)

laws

Applying the second of Eq. 21–36, we have

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copyright

Ans.

Wide

in

# HG 004667 f = = 23.3 rad s = I 2(10-3)






35 rad/s

21–74. The projectile shown is subjected to torque-free motion. The transverse and axial moments of inertia are I and Iz, respectively. If u represents the angle between the precessional axis Z and the axis of symmetry z, and b is the angle between the angular velocity V and the z axis, show that b and u are related by the equation tan u = (I>Iz) tan b .

y Z V x

b G

SOLUTION From Eq. 21–34 vy =

vy Iz HG sin u HG cos u tan u and vz = Hence = I Iz vz I

However, vy = v sin b and vz = v cos b vy vz

= tan b =

I

tan u

I tan b Iz

Q.E.D.

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laws

or

tan u =

Iz






z

u

21–75. The 4-kg disk is thrown with a spin vz = 6 rad>s. If the angle u is measured as 160°, determine the precession about the Z axis.

Z

125 mm

u vz ⫽ 6 rad/s

SOLUTION I =

1 4

(4)(0.125)2 = 0.015625 kg # m2

(4)(0.125)2 = 0.03125 kg # m2

1 2

Iz =

z

# Applying Eq. 21–36 with u = 160° and c = 6 rad>s I - Iz # HO cos u c = IIz 6 =

0.015625 - 0.03125 HO cos 160° 0.015625(0.03125) HG = 0.1995 kg # m2>s HG 0.1995 = = 12.8 rad>s I 0.015625

or

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Web)

laws

f =

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Note that this is a case of retrograde precession since Iz 7 I.






*21–76. The rocket has a mass of 4 Mg and radii of gyration kz = 0.85 m and ky = 2.3 m. It is initially spinning about the z axis at vz = 0.05 rad>s when a meteoroid M strikes it at A and creates an impulse I = 5300i6 N # s. Determine the axis of precession after the impact.

y x

ωz

G

A 3m M

SOLUTION The impulse creates an angular momentum about the y axis of Hy = 300(3) = 900 kg # m2>s Since vz = 0.05 rad>s then HG = 900j + [4000(0.85)2](0.05)k = 900j + 144.5k

or

The axis of precession is defined by HG.

in

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laws

900j + 144.5k = 0.9874j + 0.159k 911.53

Wide

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uHG =

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Dissemination

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b = cos-1(0.9874) = 9.12°

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g = cos-1(0.159) = 80.9°

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not

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a = cos-1(0) = 90°

z

21–77. The football has a mass of 450 g and radii of gyration about its axis of symmetry (z axis) and its transverse axes (x or y axis) of k z = 30mm and k x = k y = 50mm, respectively. If the football has an angular momentum of HG = 0.02kg # m2>s, determine its precession f and spin c. Also, find the angle b that the angular velocity vector makes with the z axis.

HG y 45

x

SOLUTION

G

Since the weight is the only force acting on the football, it undergoes torque-free motion. = 1.125 A 10

Iz -3

= 0.45 A 0.032 B = 0.405 A 10 - 3 B kg # m2,

I = Ix = Iy = 0.45 A 0.052 B

B kg # m , and u = 45°. 2

Thus, # HG 0.02 f = = = 17.78 rad>s = 17.8 rad>s I 1.125 A 10 - 3 B

Ans.

1.125 A 10 - 3 B - 0.405 A 10 - 3 B I - Iz # HG cos u = (0.02) cos 45° c = IIz 1.125 A 10 - 3 B (0.405) A 10 - 3 B = 22.35 rad>s = 22.3 rad>s

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Ans.

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HG cos u 0.02 cos 45° = = 34.92 rad>s Iz 0.405 A 10 - 3 B

instructors

vz =

States

HG sin u 0.02 sin 45° = 12.57 rad>s = I 1.125 A 10 - 3 B

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Also, vy =

work

12.57 b = 19.8° 34.92 is

Ans.

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assessing

≤ = tan - 1 a

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Thus, b = tan - 1 ¢






0.02 kg m2/s z V

B

21–78. z

The #projectile precesses about the Z axis at a constant rate of f = 15 rad>s when it leaves the barrel of a gun. # Determine its spin c and the magnitude of its angular momentum H G. The projectile has a mass of 1.5 kg and radii of gyration about its axis of symmetry (z axis) and about its transverse axes (x and y axes) of kz = 65 mm and kx = ky = 125 mm, respectively.

Z

x 30 f

15 rad/s

G y

SOLUTION Since the only force that acts on the projectile is its own weight, the projectile undergoes

torque-free

Iz = 1.5 A 0.0652 B = 6.3375 A 10 - 3 B kg # m2,

motion.

I = Ix = Iy = 1.5 A 0.1252 B = 0.0234375 kg # m2, and u = 30°. Thus, # # HG ; HG = If = 0.0234375(15) = 0.352 kg # m2>s f = I

Ans.

I - Iz # # c = f cos u Iz 0.0234375 - 6.3375 A 10 - 3 B

(15) cos 30° or

6.3375 A 10 - 3 B

laws

=

= 35.1 rad>s

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Ans.






21–79. The space capsule has a mass of 3.2 Mg, and about axes passing through the mass center G the axial and transverse radii of gyration are kz = 0.90 m and kt = 1.85 m, respectively. If it spins at vs = 0.8 rev>s, determine its angular momentum. Precession occurs about the Z axis.

Vs

Z 6 G

SOLUTION Gyroscopic Motion: Here, the spinning angular velocity c = vs = 0.8(2p) = 1.6p rad>s. The moment of inertia of the satelite about the z axis is Iz = 3200 A 0.92 B = 2592 kg # m2 and the moment of inertia of the satelite about its transverse axis is I = 3200 A 1.852 B = 10 952 kg # m2. Applying the third of Eq. 21–36 with u = 6°, we have I - Iz # HG cos u c = IIz 10 952 - 2592 R HG cos 6° 10 952(2592)

HG = 17.16 A 103 B kg # m2>s = 17.2 Mg # m2>s

Ans.

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1.6p = B






z

22–1. A spring has a stiffness of 600 N>m. If a 4-kg block is attached to the spring, pushed 50 mm above its equilibrium position, and released from rest, determine the equation which describes the block’s motion. Assume that positive displacement is measured downward.

SOLUTION vn =

k 600 = = 12.25 rad>s Am A 4

y = 0,

x = - 0.05 m at t = 0

x = A sin vn t + B cos vn t -0.05 = 0 + B B = - 0.05 y = Avn cos vn t - B vn sin vn t

or

0 = A(12.25) - 0

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laws

A = 0 in

Thus, x = -0.05 cos (12.2t) m

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Dissemination

copyright

Ans.






22–2. When a 2-kg block is suspended from a spring, the spring is stretched a distance of 40 mm. Determine the frequency and the period of vibration for a 0.5-kg block attached to the same spring.

SOLUTION k =

2(9.81) F = 490.5 N>m = y 0.040

vn =

490.5 k = = 31.321 A 0.5 Am

f =

vn 31.321 = = 4.985 = 4.98 Hz 2p 2p 1 1 = = 0.201 s f 4.985

Ans.

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laws

or

t =

Ans.






22–3. A spring is stretched 200 mm by a 15-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 0.75 m>s, determine the equation which describes the motion. What is the phase angle? Assume that positive displacement is downward.

SOLUTION k =

15(9.81) F = 735.75 N>m = y 0.2

vn =

k 735.75 = = 7.00 Am A 15

y = A sin vn t + B cos vn t y = 0.1 m when t = 0, 0.1 = 0 + B;

B = 0.1

or

v = A vn cos vn t - Bvn sin vn t

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laws

v = 0.75 m>s when t = 0,

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0.75 = A(7.00)

Ans.

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y = 0.107 sin (7.00t) + 0.100 cos (7.00t) by

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B 0.100 b = tan - 1 a b = 43.0° A 0.107

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Ans.

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for

f = tan - 1 a






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Dissemination

A = 0.107

*22–4. When a 20-lb weight is suspended from a spring, the spring is stretched a distance of 4 in. Determine the natural frequency and the period of vibration for a 10-lb weight attached to the same spring.

SOLUTION k = vn =

4 12

= 60 lb>ft

k 60 = = 13.90 rad>s 10 Am A 32.2

Ans.

2p = 0.452 s vn

Ans.

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laws

or

t =

20






22–5. When a 3-kg block is suspended from a spring, the spring is stretched a distance of 60 mm. Determine the natural frequency and the period of vibration for a 0.2-kg block attached to the same spring.

SOLUTION k =

3(9.81) F = = 490.5 N>m ¢x 0.060 k 490.5 = = 49.52 = 49.5 rad>s Am A 0.2

Ans.

f =

vn 49.52 = = 7.88 Hz 2p 2p

Ans.

t =

1 1 = = 0.127 s f 7.88

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laws

or

vn =






22–6. An 8-kg block is suspended from a spring having a stiffness k = 80 N>m. If the block is given an upward velocity of 0.4 m>s when it is 90 mm above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the block measured from the equilibrium position. Assume that positive displacement is measured downward.

SOLUTION vn =

k 80 = = 3.162 rad>s Am A8

y = -0.4 m>s,

x = - 0.09 m at t = 0

x = A sin vn t + B cos vn t - 0.09 = 0 + B B = -0.09 y = Avn cos vn t - Bvn sin vn t

or

- 0.4 = A(3.162) - 0

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laws

A = -0.126

Ans.

C = 2A2 + B2 = 2(-0.126)2 + ( - 0.09) = 0.155 m

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States

World

Ans.






permitted.

Dissemination

Wide

in

x = -0.126 sin (3.16t) - 0.09 cos (3.16t) m

copyright

Thus,

22–7. A 2-lb weight is suspended from a spring having a stiffness k = 2 lb>in. If the weight is pushed 1 in. upward from its equilibrium position and then released from rest, determine the equation which describes the motion. What is the amplitude and the natural frequency of the vibration?

SOLUTION k = 2(12) = 24 lb>ft vn =

k 24 = = 19.66 = 19.7 rad>s 2 Am A 32.2

y = -

1 , 12

Ans.

y = 0 at t = 0

From Eqs. 22–3 and 22–4, -

1 = 0 + B 12 laws

or

B = -0.0833

Wide

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0 = Avn + 0

permitted.

Dissemination

A = 0

United

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Ans.

the

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Position equation,

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y = (0.0833 cos 19.7t) ft






Ans.

not

States

World

C = 2A2 + B2 = 0.0833 ft = 1 in.

*22–8. A 6-lb weight is suspended from a spring having a stiffness k = 3 lb>in. If the weight is given an upward velocity of 20 ft>s when it is 2 in. above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the weight, measured from the equilibrium position. Assume positive displacement is downward.

SOLUTION k = 3(12) = 36 lb>ft vn =

k 36 = = 13.90 rad>s Am A 6 32.2

t = 0,

1 y = - ft 6

y = - 20 ft>s,

From Eq. 22–3, -

1 = 0 + B 6

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laws

or

B = -0.167

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From Eq. 22–4,

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Dissemination

-20 = A(13.90) + 0

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use

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A = - 1.44

work

student

the

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Thus,

for

y = [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ft is

work

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the

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Ans.

work

this

assessing

From Eq. 22–10,

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provided

integrity

of

and

C = 2A2 + B2 = 2(1.44)2 + ( - 0.167)2 = 1.45 ft






Ans.

22–9. A 3-kg block is suspended from a spring having a stiffness of k = 200 N>m. If the block is pushed 50 mm upward from its equilibrium position and then released from rest, determine the equation that describes the motion. What are the amplitude and the frequency of the vibration? Assume that positive displacement is downward.

SOLUTION vn =

k 200 = = 8.165 Am A 3

Ans.

x = A sin vn t + B cos vn t x = - 0.05 m when t = 0, -0.05 = 0 + B;

B = - 0.05

v = Ap cos vn t - Bvn sin vn t v = 0 when t = 0, or

A = 0

teaching

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laws

0 = A(8.165) - 0;

Wide

copyright

in

Hence, x = - 0.05 cos (8.16t)

Ans.

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C = 2A2 + B2 = 2(0)2 + ( - 0.05) = 0.05 m = 50 mm






not

States

World

permitted.

Dissemination

Ans.

22–10. Determine the frequency of vibration for the block. The springs are originally compressed ¢.

k k

SOLUTION + : a x = max ;

$ -4kx = mx

4k $ x = 0 x + m vn 1 4k = 2p 2pA m

f =

1 k pA m

Ans.

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Dissemination

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laws

or

f =






k m

k

22–11. The semicircular disk weighs 20 lb. Determine the natural period of vibration if it is displaced a small amount and released.

O

1 ft

SOLUTION Moment of Inertia about O : IA = IG + md 2 where A is the center of the semicircle. 4(1) 2 1 20 20 a b (1)2 = IG + a b c d 2 32.2 32.2 3p IG = 0.1987 slug # ft2 IO = IG + md2 4(1) 2 20 b c1 d = 0.4045 slug # ft2 32.2 3p laws

or

= 0.1987 + a

in

teaching

Web)

Equation of Motion:

Wide

copyright

$ 4(1) d sin u = - 0.4045u 3p

permitted.

20 c 1 -

Dissemination

©MO = IOxs a;

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$ u + 28.462 sin u = 0

is

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the

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student

the

by

However, for small rotation u = u. Hence $ u + 28.462u = 0

integrity

of

and

work

this

assessing

From the above differential equation, vn = 228.462 = 5.335. provided

2p 2p = 1.18 s = vn 5.335

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t =






Ans.

*22–12. The uniform beam is supported at its ends by two springs A and B, each having the same stiffness k. When nothing is supported on the beam, it has a period of vertical vibration of 0.83 s. If a 50-kg mass is placed at its center, the period of vertical vibration is 1.52 s. Compute the stiffness of each spring and the mass of the beam.

A k

SOLUTION t = 2p

m Ak

m t2 = k (2p)2 (0.83)2 2

(2p)

(1.52)2 2

(2p)

=

mB 2k

(1)

=

mB + 50 2k

(2)

or

Eqs. (1) and (2) become

teaching

Web)

laws

mB = 0.03490k

Ans.

instructors

States

World

Dissemination

mB = 21.2 kg

not

k = 609 N m

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mB + 50 = 0.1170k


B k

22–13. The body of arbitrary shape has a mass m, mass center at G, and a radius of gyration about G of kG. If it is displaced a slight amount u from its equilibrium position and released, determine the natural period of vibration.

O d u G

SOLUTION a + ©MO = IO a;

$ -mgd sin u = C mk2G + md2 D u $ u +

gd k2G + d2

sin u = 0

However, for small rotation sin u Lu. Hence $ u +

gd 2 kG + d2

u = 0 gd . B k2G + d2 or

From the above differential equation, vn =

k2G + d2 C gd

Ans.

Web)

= 2p

laws

2p gd

teaching

2p = vn

in

t =

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A k2G + d2






22–14. The connecting rod is supported by a knife edge at A and the period of vibration is measured as tA = 3.38 s. It is then removed and rotated 180° so that it is supported by the knife edge at B. In this case the perod of vibration is measured as tB = 3.96 s. Determine the location d of the center of gravity G, and compute the radius of gyration kG.

A d

G

SOLUTION Free-body Diagram: In general, when an object of arbitary shape having a mass m is pinned at O and is displaced by an angular displacement of u, the tangential component of its weight will create the restoring moment about point O.

B

Equation of Motion: Sum monent about point O to eliminate Ox and Oy. a + ©MO = IO a;

-mg sin u(l) = IO a

(1)

$ d2u = u and sin u = u if u is small, then substitute these dt2 values into Eq. (1), we have Kinematics: Since a =

or

$ mgl u + u = 0 IO

(2)

in

teaching

Web)

or

laws

$ - mglu = IO u

mgl mgl , thus, vn = . Applying Eq. 22–12, we have IO B IO States

World

permitted.

Dissemination

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From Eq. (2), vn2 =

(3)

United

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not

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IO 2p = 2p vn B mgl

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t =

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IA = 0.2894mgd provided

integrity

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and

IA B mgd

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When the rod is rotating about A, t = tA = 3.38 s and l = d. Substitute these values into Eq. (3),we have

IB = 0.3972mg (0.25 - d) will

IB B mg (0.25 - d)

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When the rod is rotating about B, t = tB = 3.96 s and l = 0.25 - d. Substitute these values into Eq. (3), we have

However, the mass moment inertia of the rod about its mass center is IG = IA - md2 = IB - m(0.25 - d)2 Then, 0.2894mgd - md2 = 0.3972mg (0.25 - d) - m (0.25 - d)2 d = 0.1462 m = 146 mm

Ans.

Thus, the mass moment inertia of the rod about its mass center is IG = IA - md2 = 0.2894m (9.81)(0.1462) - m A 0.14622 B = 0.3937 m The radius of gyration is kG =






IG 0.3937m = = 0.627 m m Bm A

Ans.

250 mm

22–15. The thin hoop of mass m is supported by a knife-edge. Determine the natural period of vibration for small amplitudes of swing.

O

r

SOLUTION IO = mr 2 + mr 2 = 2mr 2 c + a MO = IOa;

$ - mgr u = (2mr 2)u

$ g u + a bu = 0 2r 2p = 2p vn

2r g

Ans.

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t =






*22–16. A block of mass m is suspended from two springs having a stiffness of k1 and k2, arranged a) parallel to each other, and b) as a series. Determine the equivalent stiffness of a single spring with the same oscillation characteristics and the period of oscillation for each case.

k1

k2

k1

k2

SOLUTION (a) When the springs are arranged in parallel, the equivalent spring stiffness is keq = k1 + k2

Ans.

The natural frequency of the system is vn =

keq =

Cm

(a)

k1 + k2 B m

Thus, the period of oscillation of the system is = 2p

m

Ans.

C k1 + k2

or

2p k1 + k2 B m

in

teaching

Web)

2p = vn

laws

t =

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F F F + = k1 k2 keq

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1 1 1 + = k1 k2 keq

part

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k2 + k1 1 = k1k2 keq

destroy

of

and

any

courses

k1k2 k1 + k2

Ans.

sale

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their

keq =

The natural frequency of the system is k1k2 b k2 + k1 vn = = m Cm S keq

a

Thus, the period of oscillation of the system is t =






2p = vn

2p k1k2 b a k2 + k1 m S

= 2p

C

m(k1 + k2) k1k2

Ans.

permitted.

Dissemination

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copyright

(b) When the springs are arranged in a series, the equivalent stiffness of the system can be determined by equating the stretch of both spring systems subjected to the same load F.

(b)

22–17. The 15-kg block is suspended from two springs having a different stiffness and arranged a) parallel to each other, and b) as a series. If the natural periods of oscillation of the parallel system and series system are observed to be 0.5 s and 1.5 s, respectively, determine the spring stiffnesses k1 and k2.

k1

k2

k1

k2

SOLUTION The equivalent spring stiffness of the spring system arranged in parallel is A keq B P = k1 + k2 and the equivalent stiffness of the spring system arranged in a series can be determined by equating the stretch of the system to a single equivalent spring when they are subjected to the same load. F F F + = k1 k2 (keq)S

(a)

k2 + k1 1 = k1k2 k A eq B S or

k1k2 k1 + k2 in

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laws

A keq B S =

Wide

B

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Dissemination

k1 + k2 15

not

=

is

of

the

D m

instructors

A keq B P

States

(vn)P =

copyright

Thus the natural frequencies of the parallel and series spring system are

on

learning. United

and

use by

work

student

the

k1k2 D 15 A k1 + k2 B

=

15

(including

U

the

=

k1k2 ≤ k1 + k2

for

D m

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is

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(vn)S =

A keq B S

provided

integrity

of

and

work

this

Thus, the natural periods of oscillation are

(1)

15 A k1 + k2 B 2p = 2p = 1.5 (vn)S D k1k2

any

courses

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the

is

This

2p 15 = 2p = 0.5 (vn)P B k1 + k2

will

tS =

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their

destroy

of

and

tP =

(2)

Solving Eqs. (1) and (2),






k1 = 2067 N>m or 302 N>m

Ans.

k2 = 302 N>m or 2067 N>m

Ans.

(b)

22–18. The pointer on a metronome supports a 0.4-lb slider A, which is positioned at a fixed distance from the pivot O of the pointer. When the pointer is displaced, a torsional spring at O exerts a restoring torque on the pointer having a magnitude M = 11.2u2 lb # ft, where u represents the angle of displacement from the vertical, measured in radians. Determine the natural period of vibration when the pointer is displaced a small amount u and released. Neglect the mass of the pointer.

A

0.25 ft

SOLUTION

O

0.4 IO = (0.25)2 = 0.7764(10 -3) slug # ft2 32.2

c + a MO = IO a; For small u,

k ⫽ 1.2 lb · ft/rad

$ -1.2u + 0.4(0.25) sin u = 0.7764(10-3)u

sin u = u

So that $ u + 1417.5u = 0

laws

or

vn = 21417.5 = 37.64 rad>s 2p = 0.167 s 37.64

in

teaching

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Ans.

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t =






22–19. The 50-kg block is suspended from the 10-kg pulley that has a radius of gyration about its center of mass of 125 mm. If the block is given a small vertical displacement and then released, determine the natural frequency of oscillation.

k ⫽ 1500 N/m

O

SOLUTION

150 mm

Equation of Motion: When the system is in the equilibrium position, the moment equation of equilibrium written about the IC using the free-body diagram of the system shown in Fig. a gives a + ©MIC = 0;

A Fsp B st (0.3) - 10(9.81)(0.15) - 50(9.81)(0.15) = 0 A Fsp B st = 294.3 N

A Fsp B st

294.3 = 0.1962 m. Referring k 1500 the pulley shown in Fig. a, the spring stretches further sA = rA>ICu = 0.3u when the pulley rotates through a small angle u. Thus, Fsp = k(s0 + s1) = $ $ 1500(0.1962 + 0.3u) = 294.3 + 450u. Also, aG = urG>IC = u(0.15). The mass moment of inertia of the pulley about its mass center is IG = mk2G = 10(0.1252) = 0.15625 kg # m2. Referring to the free-body and kinetic diagrams of the pulley shown in Fig. b,

or

=

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Thus, the initial stretch of the spring is s0 =

10(9.81)(0.15) + 50(9.81)(0.15) - (294.3 + 450u)(0.3) $ $ $ = 10 C u(0.15) D (0.15) + 50 C u(0.15) D (0.15) + 0.15625u the

by

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not

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©MIC = ©(Mk)IC;

$

the

(including

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u + 89.63u = 0

integrity

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and

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is

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solely

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protected

Comparing this equation to that of the standard form, the natural frequency of the system is Ans.

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provided

vn = 289.63 rad>s = 9.47 rad>s






*22–20. A uniform board is supported on two wheels which rotate in opposite directions at a constant angular speed. If the coefficient of kinetic friction between the wheels and board is m, determine the frequency of vibration of the board if it is displaced slightly, a distance x from the midpoint between the wheels, and released.

x A

B

d

SOLUTION Freebody Diagram: When the board is being displaced x to the right, the restoring force is due to the unbalance friction force at A and B C (Ff)B 7 (Ff)A D . Equation of Motion: a + ©MA = ©(MA)k ;

NB (2d) - mg(d + x) = 0 mg(d + x) 2d

NB = + c ©Fy = m(aG)y ;

mg(d + x) - mg = 0 2d laws

or

NA +

Wide

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mg(d - x) 2d

NA =

the

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permitted.

Dissemination

mg(d-x) mg(d + x) d - mc d = ma 2d 2d mg a + x = 0 d

is learning.

on

(1)

the

by

and

use

United

mc

of

+ ©F = m(a ) ; : x G x

the

(including

for

work

student

d 2x ## = x, then substitute this value into Eq.(1), we have dt2

integrity

of

and

work

this

(2)

part

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is

This

mg x = 0 d

provided

##

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assessing

is

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Kinematics: Since a =

any

courses

mg mg , thus, vn = . Applying Eq. 22–4, we have d A d f =






vn 1 = 2p 2p

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will

their

destroy

of

and

From Eq.(2), vn 2 =

mg d

Ans.

d

22–21. If the 20-kg block is given a downward velocity of 6 m>s at its equilibrium position, determine the equation that describes the amplitude of the block’s oscillation.

k1 ⫽ 1500 N/m

k2 ⫽ 1000 N/m

SOLUTION

y

The equivalent stiffness of the springs in a series can be obtained by equating the stretch of the spring system to an equivalent single spring when they are subjected to the same load. Thus F F F + = k1 k2 keq k2 + k1 1 = k1k2 keq 1500(1000) k1k2 = = 600 N>m k1 + k2 1500 + 1000 laws

or

keq =

in

teaching

Web)

The natural frequency of the system is

Wide

permitted.

Dissemination

600 = 5.4772 rad>s D 20

World

=

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Dm

copyright

keq

States

vn =

for

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by

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The equation that describes the oscillation of the system becomes

(1)

is

work

solely

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the

(including

y = C sin (5.4772t + f) m

and

work

this

assessing

Since y = 0 when t = 0, Eq. (1) gives

part

the

is

This

provided

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of

0 = C sin f

will sale

y = C sin (5.4772t) m

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Since C Z 0, sin f = 0. Then f = 0. Thus, Eq. (1) becomes (2)

Taking the time derivative of Eq. (2), # y = v = 5.4772C cos (5.4772t) m>s

(3)

Here, v = 6 m>s when t = 0. Thus, Eq. (3) gives 6 = 5.4772C cos 0 C = 1.095 m = 1.10 m Then y = 1.10 sin (5.48t) m






Ans.

22–22. The bar has a length l and mass m. It is supported at its ends by rollers of negligible mass. If it is given a small displacement and released, determine the natural frequency of vibration.

R A

B

l

SOLUTION Moment of inertia about point O: IO =

1 2 l2 2 1 ml + ma R2 b = maR2 - l2 b 12 4 6 B

c + ©MO = IOa;

mg a

B

R2 -

l2 1 $ b u = -m a R2 - l2 bu 4 6 1

$ 3g(4R 2 - l2)2 u + u = 0 6R 2 - l2 1

3g(4R2 - l2)2

Ans.

.

6R2 - l2

or

D

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From the above differential equation, vn =






22–23. The 50-lb spool is attached to two springs. If the spool is displaced a small amount and released, determine the natural period of vibration. The radius of gyration of the spool is kG = 1.5 ft. The spool rolls without slipping.

k

2 ft G

SOLUTION IIC =

50 50 (1.5)2 + (1)2 = 5.047 slug # ft2 32.2 32.2

c + a MIC = IICa; $ u + 5.5483u = 0

k

1 lb/ft

$ -3(3u)(3) - 1(u) = 5.047u

vn = 25.5483 2p

t =

= 2.67 s

Ans.

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laws

or

25.5483






1 ft

3 lb/ft

*22–24. The cart has a mass of m and is attached to two springs, each having a stiffness of k1 = k2 = k, unstretched length of l0, and a stretched length of l when the cart is in the equilibrium position. If the cart is displaced a distance of x = x0 such that both springs remain in tension (x0 6 l - l0), determine the natural frequency of oscillation.

D

SOLUTION Equation of Motion: When the cart is displaced x to the right, the stretch of springs AB and CD are sAB = (l - l0) - x0 and sAC = (l - l0) + x. Thus, FAB = ksAB = k[(l - l0) - x] and FAC = ksAC = k[(l - l0) + x]. Referring to the free-body diagram of the cart shown in Fig. a, + ©F = ma ; : x x

k[(l - l0) - x] - k[(l - l0) + x] = mx -2kx = mx x +

2k x = 0 m

teaching

Web)

laws

or

Simple Harmonic Motion: Comparing this equation with that of the standard form, the natural circular frequency of the system is Ans.

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Dissemination

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2k Am

copyright

vn =






k2

C

x

A

k1

B

22–25. The cart has a mass of m and is attached to two springs, each having a stiffness of k1 and k2, respectively. If both springs are unstretched when the cart is in the equilibrium position shown, determine the natural frequency of oscillation.

D

SOLUTION Equation of Motion: When the cart is displaced x to the right, spring CD stretches sCD = x and spring AB compresses sAB = x. Thus, FCD = k2sCD = k2x and FAB = k1sAB = k1x. Referring to the free-body diagram of the cart shown in Fig. a, + ©F = ma ; : x x

- k1x - k2x = mx

-(k1 + k2)x = mx x + a

k1 + k2 bx = 0 m

laws

or

Simple Harmonic Motion: Comparing this equation with that of the standard equation, the natural circular frequency of the system is Web)

k1 + k2 m

teaching

Ans. in

B

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vn =






k2

C

x

A

k1

B

22–26. A flywheel of mass m, which has a radius of gyration about its center of mass of kO, is suspended from a circular shaft that has a torsional resistance of M = Cu. If the flywheel is given a small angular displacement of u and released, determine the natural period of oscillation.

L

SOLUTION Equation of Motion: The mass moment of inertia of the wheel about point O is IO = mkO 2. Referring to Fig. a, $ a+ ©MO = IO a; -Cu = mkO 2u $ u +

u

C u = 0 mkO 2

Comparing this equation to the standard equation, the natural circular frequency of the wheel is

or

1 C C = kO A m A mkO 2 laws

vn =

in

teaching

Web)

Thus, the natural period of the oscillation is

Ans. permitted.

Dissemination

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2p m = 2pkO vn AC

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t =






22–27. If a block D of negligible size and of mass m is attached at C, and the bell crank of mass M is given a small angular displacement of u, the natural period of oscillation is t1. When D is removed, the natural period of oscillation is t2. Determine the bell crank’s radius of gyration about its center of mass, pin B, and the spring’s stiffness k. The spring is unstrectched at u = 0°, and the motion occurs in the horizontal plane.

k a

B

SOLUTION

u

Equation of Motion: When the bell crank rotates through a small angle u, the spring stretches s = au. Thus, the force in the spring is Fsp = ks = k(au). The mass

$ u +

ka2 (cos u)u = 0 MkB2

(1)

laws

or

Since u is very small, cos u ⬵ 1. Then Eq.(1) becomes $ ka2 u + u = 0 MkB2

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$ Since the bell crank rotates about point B, aC = arBC = u(a). Referring to the freebody diagram shown in Fig. b, $ $ -k(au) cos u(a) = MkB2u + m C u(a) D (a) c + ©MB = ©(Mk)B; is

of

the

not

ka2 (cos u)u = 0 MkB2 + ma2

on

learning. United

and work

student

(including

for

Again, cos u ⬵ 1, since u is very small. Thus, Eq. (2) becomes

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the

ka2 u = 0 MkB2 + ma2






(2)

use by

the

$ u +

D C

a

moment of inertia of the bell crank about its mass center B is IB = MkB2. Referring to the free-body diagram of the bell crank shown in Fig. a, $ c + ©MB = IBa; - k(au) cos u(a) = MkB2uB

$ u +

A

22–27. continued Thus, the natural frequencies of the two oscillations are (vn)2 =

ka2 B MkB2

(vn)1 =

ka2 B MkB2 + ma2

The natural periods of the two oscillations are

t2 =

MkB2 2p = 2p (vn)2 B ka2

t1 =

MkB2 + ma2 2p = 2p (vn)1 B ka2

Solving, Ans.

4p2 m t1 - t22

Ans.

2

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k =

t22 m ¢ 2 ≤ D M t1 - t22

or

kB = a






*22–28. The platform AB when empty has a mass of 400 kg, center of mass at G1, and natural period of oscillation t1 = 2.38 s. If a car, having a mass of 1.2 Mg and center of mass at G2, is placed on the platform, the natural period of oscillation becomes t2 = 3.16 s. Determine the moment of inertia of the car about an axis passing through G2.

O 1.83 m

A

B

SOLUTION Free-body Diagram: When an object arbitrary shape having a mass m is pinned at O and being displaced by an angular displacement of u, the tangential component of its weight will create the restoring moment about point O. Equation of Motion: Sum moment about point O to eliminate Ox and Oy. - mg sin u(l) = IOa

a + ©MO = IOa :

(1)

$ d2u = u and sin u = u if u is small, then substituting these dt2 values into Eq. (1), we have Kinematics: Since a =

laws

or

mgl $ u + u = 0 IO

Web)

(2) teaching

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permitted.

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mgl mgl , thus, vn = , Applying Eq. 22–12, we have IO B IO

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From Eq. (2), v2n =

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IO 2p = 2p vn B mgl

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the

t =

(IO)p = 1407.55 kg # m2

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When the platform is empty, t = t1 = 2.38 s, m = 400 kg and l = 2.50 m. Substituting these values into Eq. (3), we have 2.38 = 2p

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When the car is on the platform, t = t2 = 3.16 s, m = 400 kg + 1200 kg = 1600 kg. 2.50(400) + 1.83(1200) and l = = 1.9975 m IO = (IO)C + (IO)p = (IO)C + 1600 1407.55. Substituting these values into Eq. (3), we have

3.16 = 2p

(IO)C + 1407.55 (I ) = 6522.76 kg # m2 D 1600(9.81)(1.9975) O C

Thus, the mass moment inertia of the car about its mass center is (IG)C = (IO)C - mCd2 = 6522.76 - 1200(1.832) = 2.50(103) kg # m2






2.50 m

G2 G1

Ans.

22–29. A wheel of mass m is suspended from three equal-length cords. When it is given a small angular displacement of u about the z axis and released, it is observed that the period of oscillation is t. Determine the radius of gyration of the wheel about the z axis. z

SOLUTION L

Equation of Motion: Due to symmetry, the force in each cord is the same. The mass moment of inertia of the wheel about is z axis is Iz = mkz 2. Referring to the freebody diagram of the wheel shown in Fig. a, + c ©Fz = maz;

3T cos f - mg = 0

120

mg 3 cos f

T =

u

Then, $ mg ≤ sin f(r) = mkz 2u 3 cos f

-3 ¢

$ gr u + tan f = 0 kz 2

(1) teaching

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laws

or

c + ©Mz = Iz a;

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Since u is very small, from the geometry of Fig. b,

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ru = Lf

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$ gr2 u + u = 0 kz2L

This

r ub L

and

Since u is very small, tan a

Comparing this equation to that of the standard form, the natural circular frequency of the wheel is vn =

gr2

C kz L 2

=

g r kz A L

Thus, the natural period of oscillation is t =

2p vn

t = 2p ¢ kz =






kz

L ≤ r Ag

g tr 2p A L

120

120

Ans.

r

22–30. Determine the differential equation of motion of the 3-kg block when it is displaced slightly and released. The surface is smooth and the springs are originally unstretched.

k

T + V = const. 1 # (3)x2 2

V =

1 1 (500)x 2 + (500)x2 2 2

# T + V = 1.5x 2 + 500x 2 # $ # 1.5(2x) x + 1000xx = 0 $ 3x + 1000x = 0 or

$ x + 333x = 0

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Ans.






k 3 kg

SOLUTION

T =

500 N/m

500 N/m

22–31. Determine the natural period of vibration of the pendulum. Consider the two rods to be slender, each having a weight of 8 lb>ft.

O

2 ft

SOLUTION y =

1(8)(2) + 2(8)(2) = 1.5 ft 8(2) + 8(2) 1 ft

1 1 IO = c (2)(8)(2)2 + 2(8)(1)2 d 32.2 12 +

1 1 c (2)(8)(2)2 + 2(8)(2)2 d = 2.8157 slug # ft2 32.2 12

h = y (1 - cos u) T + V = const or

# # 1 (2.8157)(u)2 = 1.4079 u2 2

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V = 8(4)(1.5)(1 - cos u) = 48(1 - cos u) # T + V = 1.4079u2 + 48(1 - cos u) # $ # 1.4079 (2u)u + 48(sin u)u = 0

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Ans.

assessing

= 1.52 s

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1 ft

*22–32. The uniform rod of mass m is supported by a pin at A and a spring at B. If the end B is given a small downward displacement and released, determine the natural period of vibration.

A B k l

SOLUTION T = V = =

1 1 2 $2 a ml b u 2 3

1 k(yeq + y2)2 - mgy1 2 1 1 k(lueq + lu)2 - mg a b u 2 2

T + V =

1 2#2 1 lu ml u + k(lueq + lu)2 - mg a b 6 2 2

Time derivative

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# # u 1 2#$ ml uu + kl(ueq + u)u - mgl 3 2

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Ans.

22–33. The 7-kg disk is pin-connected at its midpoint. Determine the natural period of vibration of the disk if the springs have sufficient tension in them to prevent the cord from slipping on the disk as it oscillates. Hint: Assume that the initial stretch in each spring is dO . This term will cancel out after taking the time derivative of the energy equation.

100 mm O

SOLUTION

k ⫽ 600 N/m

E = T + V 1 1 1 # k(ur + dO)2 + k(ur - dO)2 + lO(u)2 2 2 2 # # #$ E = k(ur + dO)ur + k(ur - dO)ur + IOu u = 0 =

Thus, $ 2kr 2 u + u = 0 IO

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IO 2p = 2p vn B 2kr 2

Dissemination

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t =

2kr 2 B IO

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vn =

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1 (7)(0.1)2 2 t = 2p = 0.339 s Q 2(600)(0.1)2

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Ans.

k ⫽ 600 N/m

22–34. The machine has a mass m and is uniformly supported by four springs, each having a stiffness k. Determine the natural period of vertical vibration. G

SOLUTION

k

T + V = const. 1 # m(y)2 2

T = V = mgy + T + V =

d — 2

1 (4k)(¢s - y)2 2

1 1 # m(y)2 + m g y + (4k)(¢s - y)2 2 2

# $ # # m y y + m g y - 4k(¢s-y )y = 0

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laws

or

$ m y + m g + 4ky - 4k¢s = 0 in

mg 4k

permitted.

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Since ¢s =

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k

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22–35. Determine the natural period of vibration of the 3-kg sphere. Neglect the mass of the rod and the size of the sphere.

k

500 N/m O

300 mm

SOLUTION E = T + V # 1 1 (3)(0.3u)2 + (500)(dst + 0.3u)2 - 3(9.81)(0.3u) 2 2 $ # E = u[(3(0.3)2u + 500(dst + 0.3u)(0.3) - 3(9.81)(0.3)] = 0 =

By statics, T(0.3) = 3(9.81)(0.3) T = 3(9.81) N or

3(9.81) 500 in

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dst =

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$ 3(0.3)2u + 500(0.3)2u = 0 $ u + 166.67u = 0

for

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vn = 2166.67 = 12.91 rad>s

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2p 2p = 0.487 s = vn 12.91

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Ans.

300 mm

*22–36. The slender rod has a mass m and is pinned at its end O. When it is vertical, the springs are unstretched. Determine the natural period of vibration.

O

a

k

SOLUTION

a

# 1 1 1 1 T + V = c m(2a)2 d u2 + k(2ua)2 + k(ua)2 + mga(1 - cos u) 2 3 2 2 0 =

# # # 4 2# $ ma u u + 4ka 2uu + ka 2uu + mga sin uu 3

k

sin u = u $ 4 ma 2u + 5ka 2u + mgau = 0 3 $ 15ka + 3mg u + a bu = 0 4ma laws

or

1 2

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Ans. teaching

4p 2p ma = 2 vn 3 5ka + mg

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t =






22–37. Determine the natural frequency of vibration of the 20-lb disk. Assume the disk does not slip on the inclined surface.

10 lb/in.

1 ft

SOLUTION u is the displacement of the disk. The disk rolls a distance s = ru

30

¢k = ru sin 30° E = T + V # 1 1 IIC(u)2 + k[dst + ur]2 - W(ru sin 30°) 2 2 # $ E = u(IICu + kdstr + kur2 - Wr sin 30°) = 0 =

laws

or

Since kdst = Wr sin 30°

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$ kr 2 u + u = 0 IIC

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1 2 3 mr + mr2 = mr2 2 2

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kr2 B IIC

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vn =

this

10(12)(1)2

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= 11.3 rad/s provided

3 20 a b (1)2 T 2 32.2

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Ans.

22–38. If the disk has a mass of 8 kg, determine the natural frequency of vibration. The springs are originally unstretched.

k

400 N/m

100 mm O k

SOLUTION lO =

1 (8)(0.1)2 = 0.04 2

Tmax = Vmax 1 1 I (v u )2 = 2 c k(rumax)2 d 2 O n max 2 Thus,

2(400)(0.1)2 = 14.1 rad/s 0.04 B

or

Ans.

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=

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vn =






400 N/m

22–39. The semicircular disk has a mass m and radius r, and it rolls without slipping in the semicircular trough. Determine the natural period of vibration of the disk if it is displaced slightly and released. Hint: IO = 12mr2.

2r O r

SOLUTION AB = (2r - r) cos f = r cos f, 4r cos u, 3p

AC = r cos f +

4r cos u 3p

BC =

DE = 2rf = r(u + f)

f = u AC = ra 1 +

4 b cos u 3p

or

Thus, the change in elevation of G is

teaching

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laws

4r 4 b - AC = r a 1 + b (1 - cosu) 3p 3p

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h = 2r - a r -

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Since no slipping occurs,

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# 4r vG = u ar b 3p

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the

4 2 4r 2 1 b = ¢ - a b ≤ mr2 3p 2 3p

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for

IG = IO - ma

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# 4 2 1 1 1 3 4 2 8 #2 1 #2 2 mu r a1 b + ¢ - a b ≤ mr2 u2 = mr2 a bu 2 3p 2 2 3p 2 2 3p provided

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and

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this

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is

T =

8 #2 4 1 2 3 mr a b u + mgr a 1 + b(1 - cos u) 2 2 3p 3p

0 = mr2 a

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T + V =

sin u L u $ u +

4 b 3p

ga1 +

8 3 b ra 2 3p

u = 0

g Ar

vn = 1.479 t =






2p = 4.25 vn

r g

Ans.

*22–40. The gear of mass m has a radius of gyration about its center of mass O of kO. The springs have stiffnesses of k1 and k2, respectively, and both springs are unstretched when the gear is in an equilibrium position. If the gear is given a small angular displacement of u and released, determine its natural period of oscillation.

u

A

Potential and Kinetic Energy: Since the gear rolls on the gear rack, springs AO and BO stretch and compress sO = rO>ICu = ru. When the gear rotates a small angle u, Fig. a, the elastic potential energy of the system is 1 1 k s 2 + k2sO 2 2 1 O 2

=

1 1 k1 (ru)2 + k2 (ru)2 2 2

=

1 2 r (k1 + k2)u2 2

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# # Also, from Fig. a, vO = urO>IC = ur. The mass moment of inertia of the gear about its mass center is IO = mkO 2.

permitted.

Dissemination


the

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1 1 mvO 2 + IO v2 2 2

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# # 1 1 m(ur)2 + A mkO 2 B u2 2 2

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The energy function of the system is therefore

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# 1 1 m A r 2 + kO 2 B u2 + r 2(k1 + k2)u2 = constant 2 2 Taking the time derivative of this equation, # $ # m A r 2 + kO 2 B u u + r 2(k1 + k2)uu = 0 # $ u c m A r 2 + kO 2 B u + r 2(k1 + k2)u d = 0 # Since u is not always equal to zero, then $ m A r 2 + kO 2)u + r 2(k1 + k2)u = 0 $ r 2(k1 + k2) u + u = 0 m A r 2 + kO 2 B






k2

r O

SOLUTION

V = Ve =

k1

B

*22–40. continued Comparing this equation to that of the standard form, the natural circular frequency of the system is vn =

r 2(k1 + k2)

C m A r 2 + kO 2 B

Thus, the natural period of the oscillation is m A r + kO B 2p = 2p vn D r 2(k1 + k2) 2

2

Ans.

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22–41. If the block is subjected to the periodic force F = F0 cos v t, show that the differential equation of motion $ is y + (k>m)y = (F0>m) cos vt, where y is measured from the equilibrium position of the block. What is the general solution of this equation?

k

y m

SOLUTION + T ©Fy = may ;

$ F0 cos vt + W - kdst - ky = my

Since W = kdst,

F

F0 k $ y + a by = cos vt m m

(1) (Q.E.D.)

yc = A sin vn y + B cos v n y yp = C cos vt

(complementary solution)

(particular solution)

Substitute yp into Eq. (1).

teaching

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laws

or

F0 k cos vt ≤ cos vt = m m in

C ¢ - v2 +

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C =

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y = A sin vn + B cos vn + ¢

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y = y c + yp






Ans.

F0 cos vt

22–42. The block shown in Fig. 22–16 has a mass of 20 kg, and the spring has a stiffness k = 600 N>m. When the block is displaced and released, two successive amplitudes are measured as x1 = 150 mm and x2 = 87 mm. Determine the coefficient of viscous damping, c.

SOLUTION Assuming that the system is underdamped. x1 = De - A2m Bt 1

(1)

x2 = De - A Bt 2

(2)

c

c 2m

c

x1 e - 12 m2t1 = - 1 c 2t x2 e 2m 2

or Web)

c b cc

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t2 - t1 =

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Cc 2m

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2

laws

2p

copyright

2p = vd

(3)

instructors

However, t2 - t1 = tc =

x1 c b(t - t1) b = a x2 2m 2

States

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Divide Eq. (1) by Eq. (2)

work

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by

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Substitute Eq. (4) into Eq. (3) yields:

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x1 ln a b = x2

From Eq. (5) x1 = 0.15 m x2 = 0.087m vn =

k 600 = = 5.477 rad>s Am A 20

Cc = 2mvn = 2(20)(5.477) = 219.09 N # s>m

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2pa

0.15 b = 0.087 B

c b 219.09

1 - a

2 c b 219.09

c = 18.9 N # s>m Since C 6 Cc, the system is underdamped. Therefore, the assumption is OK!






Ans.

22–43. A 4-lb weight is attached to a spring having a stiffness k = 10 lb>ft. The weight is drawn downward a distance of 4 in. and released from rest. If the support moves with a vertical displacement d = 10.5 sin 4t2 in., where t is in seconds, determine the equation which describes the position of the weight as a function of time.

SOLUTION d0

y = A sin vnt + B cos vnt +

v

1 - a v0 b

sin v0t

2

n

# v = y = Avn cos vnt - Bvn sin vnt +

d0v0 2

v0

1 - av b

cos v0t

n

The initial condition when t = 0, y = y0 , and v = v0 is B = y0

v0

d0v0 v0 2

vn -

vn in

teaching

Web)

1 - a vn b

v0 vn

A =

2

or

d 0 v0

v0 = Avn - 0 +

laws

y0 = 0 + B + 0

Dissemination

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d0

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sin vnt + y0 cos vnt +

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v02 r vn

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k 10 = = 8.972 Am A 4>32.2

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(0.5>12)4 d0 v 0 v0 = 0 = - 0.0232 v02 42 vn vn 8.972 - 8.972

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vn =

y = (-0.0232 sin 8.97t + 0.333 cos 8.97t + 0.0520 sin 4t) ft






Ans.

*22–44. A 4-kg block is suspended from a spring that has a stiffness of k = 600 N>m. The block is drawn downward 50 mm from the equilibrium position and released from rest when t = 0. If the support moves with an impressed displacement of d = 110 sin 4t2 mm, where t is in seconds, determine the equation that describes the vertical motion of the block. Assume positive displacement is downward.

SOLUTION vn =

k 600 = = 12.25 Am A 4

The general solution is defined by Eq. 22–23 with kd0 substituted for F0.

d0

y = A sin vnt + B cos vnt + ±

≤ sin vt

v 2 b d vn

c1 - a

or

d = (0.01 sin 4t)m, hence d0 = 0.01, v = 4, so that

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laws

y = A sin 12.25t + B cos 12.25t + 0.0112 sin 4t

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y = 0.05 when t = 0 B = 0.05 m

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0.05 = 0 + B + 0;

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# y = A(12.25) cos 12.25t - B(12.25) sin 12.25t + 0.0112(4) cos 4t the

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v = y = 0 when t = 0

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A = - 0.00366 m solely

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0 = A(12.25) - 0 + 0.0112(4);

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y = ( - 3.66 sin 12.25t + 50 cos 12.25t + 11.2 sin 4t) mm

Ans.

22–45. Use a block-and-spring model like that shown in Fig. 22–14a, but suspended from a vertical position and subjected to a periodic support displacement d = d0 sin v0t, determine the equation of motion for the system, and obtain its general solution. Define the displacement y measured from the static equilibrium position of the block when t = 0.

SOLUTION $ k(y - d0 sin v0t + yst) - mg = -my

+ c ©Fx = max;

$ my + ky + kyst - mg = kd0 sin v0t However, from equilibrium kyst - mg = 0, therefore kd0 k $ y = sin v t y + m m

where vn =

k Am

kd0 $ sin v t y + vn 2 y = m

laws

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Ans. (1)

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The general solution of the above differential equation is of the form of y = yc + yp, where

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yc = A sin vnt + B cos vnt

not

yp = C sin v0t

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$ yp = -Cv02 sin v0t

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kd0 sin v0t m

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-Cv2 sin v0 t + vn2(C sin v0 t) =

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Substitute Eqs. (2) and (3) into (1) yields:

The general solution is therefore y = A sin vnt + B cos vnt +

d0 sin v t v0 2 1 - a b vn

The constants A and B can be found from the initial conditions.






Ans.

22–46. A 5-kg block is suspended from a spring having a stiffness of 300 N>m. If the block is acted upon by a vertical force F = 17 sin 8t2 N, where t is in seconds, determine the equation which describes the motion of the block when it is pulled down 100 mm from the equilibrium position and released from rest at t = 0. Assume that positive displacement is downward.

k

300 N/m

SOLUTION The general solution is defined by: F0 k y = A sin vnt + B cos vnt + § ¥ sin v0t v0 2 1 - a b vn

F

Since F = 7 sin 8t,

v0 = 8 rad>s,

k = 300 N>m

or

k 300 = = 7.746 rad>s Am A 5 laws

vn =

F0 = 7 N,

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7 300 ¥ sin 8t y = A sin 7.746t + B cos 7.746t + § 2 8 1 - a b 7.746

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y = 0.1 m when t = 0,

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# y = A(7.746) - 2.8 = 0;

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# y = y = 0 when t = 0,

y = (361 sin 7.75t + 100 cos 7.75t - 350 sin 8t) mm






Ans.

7 sin 8t

22–47. The electric motor has a mass of 50 kg and is supported by four springs, each spring having a stiffness of 100 N>m. If the motor turns a disk D which is mounted eccentrically, 20 mm from the disk’s center, determine the angular velocity v at which resonance occurs. Assume that the motor only vibrates in the vertical direction.

20 mm

V

D

k ⫽ 100 N/m k ⫽ 100 N/m

SOLUTION vn =

4(100) k = = 2.83 rad>s Cm C 50 vn = v = 2.83 rad>s

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Ans.






*22–48. The 20-lb block is attached to a spring having a stiffness of 20 lb>ft. A force F = 16 cos 2t2 lb, where t is in seconds, is applied to the block. Determine the maximum speed of the block after frictional forces cause the free vibrations to dampen out.

k

F

SOLUTION F0 k C = v0 2 1 - a b vn vn =

C =

k 20 = = 5.6745 rad>s Am A 20 32.2 6 20

1 -

2

Q 5.6745R

2

= 0.343 ft

laws

or

xp = C cos 2t

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# xp = -C(2) sin 2t

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Maximum velocity is instructors

vmax = C(2) = 0.343(2) = 0.685 ft>s United

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20 lb/ft 6 cos 2t

22–49. The light elastic rod supports a 4-kg sphere. When an 18-N vertical force is applied to the sphere, the rod deflects 14 mm. If the wall oscillates with harmonic frequency of 2 Hz and has an amplitude of 15 mm, determine the amplitude of vibration for the sphere.

0.75 m

SOLUTION k =

18 F = = 1285.71 N>m ¢y 0.014

v0 = 2 Hz = 2(2p) = 12.57 rad>s d0 = 0.015 m vn =

k 1285.71 = = 17.93 4 Am A

Using Eq. 22–22, the amplitude is

in

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laws

or

d0 0.015 3 = 3 3 12.57 2 v0 2 1 R Q 1 - a b 17.93 vn

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copyright

(xp)max = 3

(xp)max = 0.0295 m = 29.5 mm

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Dissemination

Ans.






22–50. The instrument is centered uniformly on a platform P, which in turn is supported by four springs, each spring having a stiffness k = 130 N>m. If the floor is subjected to a vibration v0 = 7 Hz, having a vertical displacement amplitude d0 = 0.17 ft, determine the vertical displacement amplitude of the platform and instrument. The instrument and the platform have a total weight of 18 lb.

P

k

SOLUTION k = 4(130) = 520 lb>ft d0 = 0.17 ft v0 = 7 Hz = 7(2p) = 43.98 rad>s Using Eq. 22–22, the amplitude is (xp)max = 3

teaching

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laws

or

k 520 = = 30.50 rad>s Am A 18 32.2 in

Since vn =

d0 3 v0 2 1 - a b vn

Dissemination

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Then, 0.17 3 = 0.157 ft 1 - Q 43.98 R 2 30.50 = 1.89 in.

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(xp)max

permitted.

(xp)max = 3






Ans.

k

22–51. The uniform rod has a mass of m. If it is acted upon by a periodic force of F = F0 sin vt, determine the amplitude of the steady-state vibration.

A L 2 k

L 2

SOLUTION Equation of Motion: When the rod rotates through a small angle u, the springs L u. Thus, the force in each spring is compress and stretch s = r AGu = 2 kL Fsp = ks = u. The mass moment of inertia of the rod about point A is 2 1 IA = mL2. Referring to the free-body diagram of the rod shown in Fig. a, 3 FO sin vt cos u(L) - mg sin u a

L kL L b - 2a u bcos u a b 2 2 2 or

# 1 mL2u 3

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=

F ⫽ F0 sin vt

laws

+ ©MA = IAa;

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Since u is small, sin u ⬵ 0 and cos u ⬵ 1. Thus, this equation becomes

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$ 1 1 mLu + (mg + kL)u = FO sin vt 3 2

(1)

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$ 3FO 3 g k u + ¢ + ≤u = sin vt m 2 L mL

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the

The particular solution of this differential equation is assumed to be in the form of (2)

provided

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up = C sin vt

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is

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Taking the time derivative of Eq. (2) twice, $ up = -Cv2 sin vt

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(3)

-Cv2 sin vt +

C =





3FO 3 g k a + ≤ (C sin vt) = sin vt m 2 L mL

3FO 3 g k a + ≤ - v2 R sin vt = sin vt m 2 L mL

C =


sale

Substituting Eqs. (2) and (3) into Eq. (1),

CB

k

3FO > mL 3 g k a + b - v2 m 2 L 3FO 3 (mg + Lk) - mLv2 2

Ans.

*22–52. Use a block-and-spring model like that shown in Fig. 22–14a but suspended from a vertical position and subjected to a periodic support displacement of d = d0 cos v0t, determine the equation of motion for the system, and obtain its general solution. Define the displacement y measured from the static equilibrium position of the block when t = 0.

SOLUTION $ + T ©Fy = may; kd0 cos v0t + W - kdst - ky = my Since W = kdst, kd0 k $ y + y = cos v0t m m

(1)

yC = A sin vny + B cos vny (General sol.) yP = C cos v0t (Particular sol.) Substitute yp into Eq. (1)

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kd0 k ) cos v0t = cos v0t m m in

C(- v02 +

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k 2 m - v0 R

the

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of

C =

kd0 m

and by

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k - v02 R m

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Q

the

kd0 m

is

y = A sin vnt + B cos vnt +

use

Thus, y = yC + yP






Ans.

22–53. V

The fan has a mass of 25 kg and is fixed to the end of a horizontal beam that has a negligible mass. The fan blade is mounted eccentrically on the shaft such that it is equivalent to an unbalanced 3.5-kg mass located 100 mm from the axis of rotation. If the static deflection of the beam is 50 mm as a result of the weight of the fan, determine the angular velocity of the fan blade at which resonance will occur. Hint: See the first part of Example 22.8.

SOLUTION k = vn =

25(9.81) F = = 4905 N>m ¢y 0.05 k

Cm

=

4905

C 25

= 14.01 rad>s

Resonance occurs when v = vn = 14.0 rad>s

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Ans.






22–54. V

In Prob. 22–53, determine the amplitude of steady-state vibration of the fan if its angular velocity is 10 rad>s .

SOLUTION k = vn =

25(9.81) F = = 4905 N>m ¢y 0.05 k

Cm

=

4905

= 14.01 rad>s

C 25

The force caused by the unbalanced rotor is F0 = mr v2 = 3.5(0.1)(10)2 = 35 N

laws

or

Using Eq. 22–22, the amplitude is

(xp)max

35 4905 4 = 0.0146 m = 4 10 2 1 - a b 14.01 solely

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(xp)max

F0 k 4 = 4 v 2 1 -a b p

is

work

(xp)max = 14.6 mm

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Ans.






22–55. V

What will be the amplitude of steady-state vibration of the fan in Prob. 22–53 if the angular velocity of the fan blade is 18 rad>s? Hint: See the first part of Example 22.8.

SOLUTION k = vn =

25(9.81) F = = 4905 N>m ¢y 0.05 k

Cm

=

4905

= 14.01 rad>s

C 25

The force caused by the unbalanced rotor is F0 = mrv2 = 3.5(0.1)(18)2 = 113.4 N

laws

or

Using Eq. 22–22, the amplitude is teaching

Web)

F0 in

4

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v 2 1 - a b p

Dissemination

k

copyright

(xp)max = 4

the on

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the

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for

(xp)max

113.4 4905 4 = 0.0355 m = 4 18 2 b 1 - a 14.01

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(xp)max = 35.5 mm






Ans.

*22–56. The small block at A has a mass of 4 kg and is mounted on the bent rod having negligible mass. If the rotor at B causes a harmonic movement dB = (0.1 cos 15t) m, where t is in seconds, determine the steady-state amplitude of vibration of the block.

0.6 m O

A

1.2 m

SOLUTION $ 4(9.81)(0.6) - Fs (1.2) = 4(0.6)2u

+ ©MO = IO a;

V k

Fs = kx = 15(x + xst - 0.1 cos 15t) xst =

4(9.81)(0.6) 1.2(15)

B

Thus, $ -15(x - 0.1 cos 15t)(1.2) = 4(0.6)2u x = 1.2u laws

or

u + 15u = 1.25 cos 15t

Wide

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Set xp = C cos 15t

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Dissemination

-C(15)2 cos 15t + 15(C cos 15t) = 1.25 cos 15t

United

on

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is

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the

not

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1.25 = -0.00595 m 15 - (15)2 by

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C =

for

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the

umax = C = 0.00595 rad

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the

(including

ymax = (0.6 m)(0.00595 rad) = 0.00357 rad






15 N/m

Ans.

22–57. V

The electric motor turns an eccentric flywheel which is equivalent to an unbalanced 0.25-lb weight located 10 in. from the axis of rotation. If the static deflection of the beam is 1 in. due to the weight of the motor, determine the angular velocity of the flywheel at which resonance will occur. The motor weights 150 lb. Neglect the mass of the beam.

SOLUTION F 150 = = 1800 lb>ft d 1>12

k =

k 1800 = = 19.66 Am A 150>32.2

v0 = vn = 19.7 rad>s

Ans.

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Resonance occurs when

vn =






22–58. V

What will be the amplitude of steady-state vibration of the motor in Prob. 22–57 if the angular velocity of the flywheel is 20 rad>s?

SOLUTION The constant value FO of the periodic force is due to the centrifugal force of the unbalanced mass. FO = ma n = mrv2 = a

0.25 10 b a b (20)2 = 2.588 lb 32.2 12

Hence F = 2.588 sin 20t k =

150 F = = 1800 lb>ft d 1>12

vn =

k 1800 = = 19.657 Am A 150>32.2

laws

or

From Eq. 22–21, the amplitude of the steady state motion is

teaching

Web)

F0>k 2.588>1800 4 = 4 4 = 0.04085 ft = 0.490 in. 2 v0 2 20 b 1 - a b 1 - a vn 19.657

in

Ans.

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C = 4






22–59. V

Determine the angular velocity of the flywheel in Prob. 22–57 which will produce an amplitude of vibration of 0.25 in.

SOLUTION The constant value FO of the periodic force is due to the centrifugal force of the unbalanced mass. FO = man = mrv2 = a

0.25 10 b a bv2 = 0.006470v2 32.2 12

F = 0.006470v2 sin vt k =

150 F = = 1800 lb>ft d 1>12

k 1800 = = 19.657 Am A 150>32.2

vn =

laws

or

From Eq. 22.21, the amplitude of the steady-state motion is

in

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F0>k 4 v 2 1 - a b vn

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C = 4

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v2 b 1800 0.25 4 = 4 2 v 12 b 1 - a 19.657

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v = 19.7 rad>s






Ans.

*22–60. The engine is mounted on a foundation block which is springsupported. Describe the steady-state vibration of the system if the block and engine have a total weight of 1500 lb and the engine, when running, creates an impressed force F = 150 sin 2t2 lb, where t is in seconds. Assume that the system vibrates only in the vertical direction, with the positive displacement measured downward, and that the total stiffness of the springs can be represented as k = 2000 lb>ft.

SOLUTION The steady-state vibration is defined by Eq. 22–22. F0 k sin v0t xp = v0 2 1 - a b vn Since F = 50 sin 2t Then F0 = 50 lb, v0 = 0 rad>s laws

or

k = 2000 lb>ft

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2000 1500 k vn = = = 6.55 rad>s m A Q 32.2

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xp = (0.0276 sin 2t) ft

protected

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50 2000 Hence, xp = sin 2t 2 2 b 1 - a 6.55






Ans.

22–61. Determine the rotational speed v of the engine in Prob. 22–60 which will cause resonance.

SOLUTION Resonance occurs when 2000

v = vn =






k 1500 = = 6.55 rad>s C 32.2 Am

Ans.

22–62. The motor of mass M is supported by a simply supported beam of negligible mass. If block A of mass m is clipped onto the rotor, which is turning at constant angular velocity of v, determine the amplitude of the steady-state vibration. Hint: When the beam is subjected to a concentrated force of P at its mid-span, it deflects d = PL3>48EI at this point. Here E is Young’s modulus of elasticity, a property of the material, and I is the moment of inertia of the beam’s crosssectional area.

r A

L 2

SOLUTION In this case, P = keqd. Then, keq = frequency of the system is

48EI P P = = . Thus, the natural d PL3>48EI L3

48EI 48EI L3 vn = = = Cm R M B ML3 keq

Here, FO = man = m(v2r). Thus, Y =

FO>keq 1 - a

v 2 b vn

m(v2r) 48EI>L3

Y = 1 -

Y =






v2 48EI>ML3

mrv2L3 48EI - Mv2L3

Ans.

L 2

22–63. A block having a mass of 0.8 kg is suspended from a spring having a stiffness of 120 N>m. If a dashpot provides a damping force of 2.5 N when the speed of the block is 0.2 m s, determine the period of free vibration.

SOLUTION F = cy vn =

c = k

Cm

=

F 2.5 = 12.5 N # s >m = y 0.2 120

C 0.8

= 12.247 rad>s

Cc = 2m vn = 2(0.8)(12.247) = 19.60 N # s>m vd = vn td =






C

1 - a

c 2 12.5 2 b = 9.432 rad>s b = 12.247 1 - a cc 19.6 C

2p 2p = 0.666 s = vd 9.432

Ans.

*22–64. The block, having a weight of 15 lb, is immersed in a liquid such that the damping force acting on the block has a magnitude of F = (0.8 ƒ v ƒ ) lb, where v is the velocity of the block in ft>s. If the block is pulled down 0.8 ft and released from rest, determine the position of the block as a function of time. The spring has a stiffness of k = 40 lb>ft. Consider positive displacement to be downward.

k

SOLUTION Viscous Damped Free Vibration: Here c = 0.8 lb # s>ft, vn =

k 40 = Am A 15>32.2

15 b (9.266) = 8.633 lb # s>ft. Since c 6 cc, the 32.2 system is underdamped and the solution of the differential equation is in the form of = 9.266 rad>s and cc = 2m vn = 2 a

y = D C e - (c>2m)t sin (vd t + f) D

(1)

Taking the time derivative of Eq. (1),we have c # b e - (c>2m)t sin (vdt + f) + vde - (c>2m)t cos ( vdt + f) d y = y = Dc - a 2m = De - (c>2m)t c - a

c b sin (vdt + f) + vd cos (vdt + f) d 2m

(2)

Applying the initial condition y = 0 at t = 0 to Eq. (2),we have 0 = De - 0 c - a

c b sin (0 + f) + vd cos (0 + f) d 2m

0 = Dc - a

c b sin f + vd cos f d 2m

(3)

c 2 0.8 2 c 0.8 = = 0.8587 and vd = vn 1 - a b = 9.266 1 - a b cc B B 2m 2(15>32.2) 8.633 = 9.227 rad>s. Substituting these values into Eq. (3) yields

Here,

0 = D[ - 0.8587 sin f + 9.227 cos f]

(4)

Applying the initial condition y = 0.8 ft at t = 0 to Eq. (1),we have 0.8 = D C e - 0 sin (0 + f) D 0.8 = D sin f

(5)

Solving Eqs. (4) and (5) yields f = 84.68° = 1.50 rad

D = 0.8035 ft

Substituting these values into Eq. (1) yields y = 0.803 C e - 0.8597 sin (9.23t + 1.48) D






Ans.

40 lb/ft

22–65. A 7-lb block is suspended from a spring having a stiffness of k = 75 lb>ft. The support to which the spring is attached is given simple harmonic motion which can be expressed as d = (0.15 sin 2t) ft, where t is in seconds. If the damping factor is c>cc = 0.8, determine the phase angle f of forced vibration.

SOLUTION vn =

k

Cm

=

75 = 18.57 7 b a S 32.2

d = 0.15 sin 2t d0 = 0.15, v = 2 2a f¿ = tan-1 §

c v ba b cc vn

v 2 1 - a b vn

2(0.8)a ¥ = tan-1 §

f¿ = 9.89°






2 b 18.57

2 2 b 1 - a 18.57

¥

Ans.

22–66. Determine the magnification factor of the block, spring, and dashpot combination in Prob. 22–65.

SOLUTION vn =

k = Am

75

= 18.57

7 Q ¢ 32.2 ≤

d = 0.15 sin 2t d0 = 0.15,

v = 2 1

MF =

C

B1 - a

MF = 0.997






v 2 2 v 2 c b R + c2a b a b d cn vn vn

1

=

C

B1 - a

2 2 2 2 2 b R + c2(0.8) a bR 18.57 18.57

Ans.

22–67. A block having a mass of 7 kg is suspended from a spring that has a stiffness k = 600 N>m. If the block is given an upward velocity of 0.6 m>s from its equilibrium position at t = 0, determine its position as a function of time. Assume that positive displacement of the block is downward and that motion takes place in a medium which furnishes a damping force F = 150 ƒ v ƒ 2 N, where v is the velocity of the block in m>s.

SOLUTION c = 50 N s>m k = 600 N>m vn =

m = 7 kg

k 600 = = 9.258 rad>s Am A 7

cc = 2mvn = 2(7)(9.258) = 129.6 N # s>m Since c 6 cz, the system is underdamped, vd = vn

B

1 - a

c 2 50 2 b = 8.542 rad>s b = 9.258 1 - a cc 129.6 B 50 c = = 3.751 2m 2(7)

From Eq. 22-32 y = D C e- A2m Bt sin (vdt + f) S c

c # v = y = D C e- A2m Bt vd cos (vdt + f) +

A-

y = De- A2m Bt C vd cos (vdt + f) c

c c B e- A2m Bt sin (vdt + f) S 2m

c sin (vdt + f) D 2m

Applying the initial condition at t = 0, y = 0 and y = -0.6 m>s. 0 = D[e - 0 sin (0 + f)] sin f = 0

since D Z 0 f = 0°

-0.6 = De - 0 [8.542 cos 0° - 0] D = - 0.0702 m y = [- 0.0702 e - 3.57t sin (8.540)] m






Ans.

*22–68. The 4-kg circular disk is attached to three springs, each spring having a stiffness k = 180 N>m. If the disk is immersed in a fluid and given a downward velocity of 0.3 m>s at the equilibrium position, determine the equation which describes the motion. Consider positive displacement to be measured downward, and that fluid resistance acting on the disk furnishes a damping force having a magnitude F = 160 ƒ v ƒ 2 N, where v is the velocity of the block in m>s.

120

SOLUTION k = 540 N>m vn =

k 540 = = 11.62 rad>s Am A 4

cc = 2mvn = 2(4)(11.62) = 92.95 F = 60y, so that c = 60 Since c 6 cc, system is underdamped. vd = vn

A

c 1 - a b2 cc

60 2 = 11.62 1 - a b A 92.95 = 8.87 rad>s c

y = A[e - (2m)t sin (vdt + f)]

(1)

y = 0, v = 0.3 at t = 0 0 = A sin f A Z 0 (trivial solution) so that f = 0 v = y = A[-

c c - ( c )t e 2m sin (vdt + f) + e - (2m)t cos (vdt + f)(vd)] 2m

Since f = 0 0.3 = A[0 + 1(8.87)] A = 0.0338 Substituting into Eq. (1) 60

y = 0.0338[e - (2(4))t sin (8.87)t] Expressing the result in mm y = [33.8 e - 7.5t sin (8.87t)] mm






120

120

Ans.

22–69. If the 12-kg rod is subjected to a periodic force of F = (30 sin 6t) N, where t is in seconds, determine the steady-state vibration amplitude umax of the rod about the pin B. Assume u is small.

F ⫽ (30 sin 6 t) N

k ⫽ 3 kN/m 0.2 m B

0.2 m C

D

A

SOLUTION

c ⫽ 200 N⭈s/m

# Equation of Motion: When the rod is in equilibrium, u = 0°, F = 0, Fc = cyc = 0, $ and u = 0. Writing the moment equation of equilibrium about point B by referring to the free-body diagram of the rod, Fig. a, + g MB = 0;

FA(0.2) - 12(9.81)(0.1) = 0

FA = 58.86 N

FA 58.86 = = 0.01962 m. When k 3000 the rod rotates about point B through a small angle u, the spring compresses further by s1 = 0.2u. Thus, the force in the spring is FA = k(s0 + s1) = 3000(0.01962 + 0.2u) = 58.85 + 600u. Also, the velocity of point C on the rod is # # # # # vc = yc = 0.2u. Thus, Fc = cyc = 200(0.2u) = 40u. The mass moment of inertia of 1 (12)(0.6)2 + 12(0.1)2 = 0.48 kg # m2. Again, referring to the rod about B is IB = 12 Fig. a, # + g MB = IBa; (58.86 + 600u) cos u(0.2) + 40u cos u(0.2) - (30 sin 6t) cos u(0.4) $ - 12(9.81) cos u(0.1) = - 0.48u $ # u + 16.67 cos uu + 250(cos u)u = 25 sin 6t cos u Thus, the initial compression of the spring is sO =

Since u is small, cos u ⬵ 1. Thus, this equation becomes $ # u + 16.67u + 250u = 25 sin 6t Comparing this equation to that of the standard form, keq m ceq m

= 250

keq = 250(12) = 3000 N>m

= 16.667

ceq = 16.667(12) = 200 N # s>m

FO = 25 m

FO = 25(12) = 300 N

vn = 33000>12 = 3250 Thus, cc = 2mvn = 2(12)3250 = 379.47 N # s>m Then, ceq cc

=

200 = 0.5270 379.47 300>3000

umax = D © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





B1 - ¢

= 0.106 rad

6 3250

2 2

≤ R + B

2(0.5270)(6) 3250

R

2

Ans.

0.2 m

22–70. The damping factor, c>cc , may be determined experimentally by measuring the successive amplitudes of vibrating motion of a system. If two of these maximum displacements can be approximated by x1 and x2 , as shown in Fig. 22–17, show that the ratio ln x1>x2 = 2p1c>cc2> 21 - 1c>ce22. The quantity ln x1>x2 is called the logarithmic decrement.

SOLUTION Using Eq. 22–32, x = Dc e - A2m B t sin (vdt + f) d c

The maximum displacement is xmax = De - A2m B t c

At t = t1, and t = t2 x1 = De - A2m B t1 c

x2 = De - A2m B t2 c

Hence, x1 De - A2m Bt1 = c c x2 De - A2m Bt2 = e - A2m B(t1 - t2) c

Since vdt2 - vdt1 = 2p then t2 - t1 =

so that ln a

2p vd

x1 cp b = x2 mvd

Using Eq. 22–33, cc = 2mvn vd = vn

B

1 - a

cc c 2 c 2 1 - a b b = cc c 2mA r

So that,

x1 ln a b = x2






2p a

c b cc

c 2 1- a b cc B

Q.E.D.

22–71. v

If the amplitude of the 50-lb cylinder’s steady-vibration is 6 in., determine the wheel’s angular velocity v.

9 in.

c ⫽ 25 lb⭈s/ft

k ⫽ 200 lb/ft

SOLUTION In this case, Y =

6 9 = 0.5 ft, dO = = 0.75 ft, and keq = 2k = 2(200) = 400 lb>ft. 12 12

Then vn =

keq

Cm

=

400 = 16.05 rad>s B (50>32.2)

cc = 2mvn = 2 ¢

50 ≤ (16.05) = 49.84 lb # s>ft 32.2

c 25 = = 0.5016 cc 49.84 dO

Y =

2

2(c>cc)v 2 v 2 B1 - ¢ ≤ R + ¢ ≤ D vn vn 0.75

0.5 = D

B1 - ¢

2

2

2(0.5016)v 2 v ≤ R + ¢ ≤ 16.05 16.05

15.07(10 - 6)v4 - 3.858(10 - 3)v2 - 1.25 = 0 Solving for the positive root of this equation, v2 = 443.16 v = 21.1 rad>s






Ans.

k ⫽ 200 lb/ft

*22–72. The 10-kg block-spring-damper system is continually damped. If the block is displaced to x = 50 mm and released from rest, determine the time required for it to return to the position x = 2 mm.

x k ⫽ 60 N/m

SOLUTION c = 80,

m = 10 kg,

k = 60

cc = 2 2km = 2 2600 = 49.0 6 c

(Overdamped)

x = Ael1 t + Bel2 t l1,2 =

-c c 2 k - 80 80 2 60 ; a b = ; a b m 2m B 2m 20 B 20 10

l1 = -0.8377,

l2 = - 7.1623

# At t = 0, x = 0.05, x = 0 0.05 = A + B A = 0.05 - B # x = Al1el1t + Bl2el2 t 0 = Al1 + Bl2 0 = (0.05 - B)l1 + Bl2 B =

0.05l1 = - 6.6228(10-3) l1 - l2

A = 0.056623 x = 0.056623e -0.8377t - 6.6228(10-3)e -7.1623t Set x = 0.0002 m and solve, t = 3.99 s






Ans.

c ⫽ 80 N ⭈ s/m

22–73. The 20-kg block is subjected to the action of the harmonic force F = 190 cos 6t2 N, where t is in seconds. Write the equation which describes the steady-state motion.

k ⫽ 400 N/m

F ⫽ 90 cos 6t 20 kg

k ⫽ 400 N/m

SOLUTION F = 90 cos 6t F0 = 90 N, vn =

v0 = 6 rad>s

k 800 = = 6.32 rad>s Am A 20

From Eq. 22–29, cc = 2mvn = 2(20)(6.32) = 253.0 Using Eqs. 22–39, x = C¿ cos (v0 t - f) Fn k

C¿ =

C =

B1 - a

v0 2 2 v0 2 b R + c 2 a CCr b a b d vr vn 90 800

2[1 -

6 2 2 125 6 A 6.32 B ] + [2 A 253.0 B A 6.32 B ]2

= 0.119 cv0 k T f = tan-1 D v0 2 1 - a b vn = tan-1 C

125(6) 800

1 -

6 2 A 6.32 B

S

f = 83.9° Thus, x = 0.119 cos (6t - 83.9°) m






Ans.

c ⫽ 125 N ⭈ s/m

22–74. A bullet of mass m has a velocity of v0 just before it strikes the target of mass M. If the bullet embeds in the target, and the vibration is to be critically damped, determine the dashpot’s critical damping coefficient, and the springs’ maximum compression. The target is free to move along the two horizontal guides that are “nested” in the springs.

k

v0 c

k

SOLUTION Since the springs are arranged in parallel, the equivalent stiffness of the single spring system is keq = 2k. Also, when the bullet becomes embedded in the target, mT = m + M. Thus, the natural frequency of the system is vn =

keq =

B mT

2k Bm + M

When the system is critically damped c = cc = 2mTvn = 2(m + M)

2k = 28 (m + M)k Bm + M

Ans.

The equation that describes the critically dampened system is x = (A + Bt)e-vn t When t = 0, x = 0. Thus, A = 0 Then, x = Bte-vn t

(1)

Taking the time derivative, # v = x = Be - vn t - Bvn te-vn t v = Be-vn t(1 - vn t)

(2)

Since linear momentum is conserved along the horizontal during the impact, then + B A;

mv0 = (m + M)v v = a

m bv m + M 0

Here, when t = 0, v = a B = a

m bv . Thus, Eq. (2) gives m + M 0

m bv m + M 0

And Eqs. (1) and (2) become






x = ca

m b v d te-vn t m + M 0

(3)

v = ca

m b v d e-vn t(1 - vn t) m + M 0

(4)

22–74. continued The maximum compression of the spring occurs when the block stops. Thus, Eq. (4) gives 0 = ca Since a

m b v d (1 - vn t) m + M 0

m b v Z 0, then m + M 0 1 - vn t = 0 t =

1 m + M = vn B 2k

Substituting this result into Eq. (3) xmax = c a

m 1 dv e A 2k(m + M) 0

Ans.

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or

= c

m m + M -1 b v0 d a be m + M A 2k






22–75. A bullet of mass m has a velocity v0 just before it strikes the target of mass M. If the bullet embeds in the target, and the dashpot’s damping coefficient is 0 6 c V cc, determine the springs’ maximum compression. The target is free to move along the two horizontal guides that are“nested” in the springs.

k

v0 c

k

SOLUTION Since the springs are arranged in parallel, the equivalent stiffness of the single spring system is keq = 2k. Also, when the bullet becomes embedded in the target, mT = m + M. Thus, the natural circular frequency of the system vn =

keq

C mT

=

2k Bm + M

The equation that describes the underdamped system is

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copyright

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When t = 0, x = 0. Thus, Eq. (1) gives

or

(1)

laws

x = Ce - (c>2mT)t sin (vdt + f)

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permitted.

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0 = C sin f

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on

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of

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not

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Since C Z 0, sin f = 0. Then f = 0. Thus, Eq. (1) becomes

(2)

for

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the

by

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x = Ce - (c>2mT)t sin vd t

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Taking the time derivative of Eq. (2), is

work

c - (c>2mT)t e sin vdt R 2mT the

is

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# v = x = C B vde - (c>2mT)t cos vdt -

any

courses

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c sin vdt R 2mT

(3)

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and

v = Ce - (c>2mT)t B vd cos vdt -

sale

will

Since linear momentum is conserved along the horizontal during the impact, then + B A;

mv0 = (m + M)v v = a

When t = 0, v = a a

m bv m + M 0

m b v . Thus, Eq. (3) gives m + M 0

m b v = Cvd m + M 0

C = a

v0 m b m + M vd

And Eqs. (2) becomes x = ca






v0 m b d e-(c>2mT) t sin vdt m + M vd

(4)

22–75. continued The maximum compression of the spring occurs when sin vdt = 1 p 2

vdt = t =

p 2vd

Substituting this result into Eq. (4), xmax = c a

However, vd =

keq

C mT

p v0 m b d e -[c>2(m + M)] A 2vd B m + M vd

- a

c 2 2k c2 1 b = = 2 2mT C m + M 4(m + M) 2(m + M)

28k(m + M) - c2. Substituting this result into Eq. (5), 2mv0 28k(m + M) - c2

-c

e

d

pc 2 28k(m + M) - c2)

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xmax =






*22–76. Determine the differential equation of motion for the damped vibratory system shown. What type of motion occurs? Take k = 100 N>m, c = 200 N # s>m, m = 25 kg. k

SOLUTION

k

k

m

Free-body Diagram: When the block is being displaced by an amount y vertically downward, the restoring force is developed by the three springs attached the block. c

Equation of Motion: + c ©Fx = 0;

# $ 3ky + mg + 2cy - mg = -my $ # my + 2cy + 3ky = 0

(1)

Here, m = 25 kg, c = 200 N # s>m and k = 100 N>m. Substituting these values into Eq. (1) yields or

$ # 25y + 400y + 300y = 0 laws

$ # y + 16y + 12y = 0

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cc = 2mvn = 2(1)(3.464) = 6.928 N # s>m for

Since c 7 cc , the system will not vibrate. Therefore it is overdamped.

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protected © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by





the

(including

Ans.

permitted.

Dissemination

Comparing the above differential equation with Eq. 22–27, we have m = 1kg, k 12 c = 16 N # s>m and k = 12 N>m. Thus, vn = = = 3.464 rad>s. Am A1

c

22–77. Draw the electrical circuit that is equivalent to the mechanical system shown. Determine the differential equation which describes the charge q in the circuit.

k F ⫽ F0 cos vt m c

SOLUTION For the block, mx + cx + kx = F0 cos vt Using Table 22–1, 1 Lq + Rq + ( )q = E0 cos vt C

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Ans.






22–78. Draw the electrical circuit that is equivalent to the mechanical system shown. What is the differential equation which describes the charge q in the circuit?

k

c m

k

SOLUTION Electrical Circuit Analogs: The differential equation that deseribes the motion of the given mechanical system is $ # mx + cx + 2kx = F0 cos vt From Table 22–1 of the text, the differential equation of the analog electrical circuit is 2 $ # Lq + Rq + a bq = E0 cos vt C

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F

F0 cos vt

22–79. Draw the electrical circuit that is equivalent to the mechanical system shown. Determine the differential equation which describes the charge q in the circuit. k

SOLUTION

m

For the block $ # my + cy + ky = 0

c

Using Table 22–1 1 $ # Lq + Rq + q = 0 C

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R1–1. An automobile is traveling with a constant speed along a horizontal circular curve that has a radius r = 750 ft. If the magnitude of acceleration is a = 8 ft>s2, determine the speed at which the automobile is traveling.

SOLUTION a = an = 8 =

v2 r

v2 750

8 =

v = 77.5 ft>s

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R1–2. Block B rests on a smooth surface. If the coefficients of friction between A and B are ms = 0.4 and mk = 0.3, determine the acceleration of each block if (a) F = 6 lb, and (b) F = 50 lb.

20 lb

B

SOLUTION a) The maximum friction force between blocks A and B is Fmax = 0.4(20) = 8 lb 7 6 lb Thus, both blocks move together. + : a Fx = max;

6 =

70 a 32.2

aB = aA = a = 2.76 ft>s2

Ans.

b) In this case 8 lb 6 F = 50 lb

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Block A: in

20 a 32.2 A

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Ans. is

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aA = 70.8 ft>s2

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Block B:

work

student

50 a 32.2 B

the

(including

for

20(0.3) =

solely

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+ : a Fx = max;

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aB = 3.86 ft>s2






Ans.

permitted.

Dissemination

20(0.3) =

copyright

+ : a Fx = max;

F

A

50 lb

R1–3. z

The small 2-lb collar starting from rest at A slides down along the smooth rod. During the motion, the collar is acted upon by a force F = 510i + 6yj + 2zk6 lb, where x, y, z are in feet. Determine the collar’s speed when it strikes the wall at B.

4 ft A F B

10 ft

SOLUTION rAB = rB - rA = - 4i + 8j - 9k T1 + ©

L

Fds = T2

x

0

0 + 2(10-1) +

L4

8

10dx +

L0

1

6y dy +

L10

1 2 a bv2B 2 32.2

2z dz =

vB = 47.8 ft>s

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1 ft 8 ft

y

*R1–4. The automobile travels from a parking deck down along a cylindrical spiral ramp at a constant speed of v = 1.5 m>s. If the ramp descends a distance of 12 m for every full revolution, u = 2p rad, determine the magnitude of the car’s acceleration as it moves along the ramp, r = 10 m. Hint: For part of the solution, note that the tangent to the ramp at any point is at an angle of f = tan - 1 (12>32p(10)4) = 10.81° from the horizontal. Use this to determine the velocity# components vu and vz, which in turn are used to determine u # and z.

10 m 12 m

SOLUTION f = tan-1 a

12 b = 10.81° 2p(10)

v = 1.5 m>s vr = 0 vu = 1.5 cos 10.81° = 1.473 m>s

laws

or

vz = -1.5 sin 10.81° = - 0.2814 m>s

in

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Wide World

1.473 = 0.1473 10

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u =

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# vu = r u = 1.473

permitted.

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# r = 0

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# ar = r - r u2 = 0 - 10(0.1473)2 = -0.217 #$ au = r u + 2 r u = 10(0) + 2(0)(0.1473) = 0

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$ az = z = 0

Ans.

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a = 2(- 0.217)2 + (0)2 + (0)2 = 0.217 m>s2






R1–5. A rifle has a mass of 2.5 kg. If it is loosely gripped and a 1.5-g bullet is fired from it with a horizontal muzzle velocity of 1400 m/s, determine the recoil velocity of the rifle just after firing.

SOLUTION + :

©mv1 = ©mv2 0 + 0 = 0.0015(1400) - 2.5(vR)2 (vR)2 = 0.840 m>s

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R1–6. If a 150-lb crate is released from rest at A, determine its speed after it slides 30 ft down the plane. The coefficient of kinetic friction between the crate and plane is mk = 0.3.

A 30 ft

SOLUTION +a©Fy = 0;

NC - 150 cos 30° = 0

30°

NC = 129.9 lb T1 + ©U1 - 2 = T2 1 150 2 a bv 2 32.2 2

0 + 150 sin 30°(30) - (0.3)129.9(30) = v2 = 21.5 ft>s

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R1–7. The van is traveling at 20 km> h when the coupling of the trailer at A fails. If the trailer has a mass of 250 kg and coasts 45 m before coming to rest, determine the constant horizontal force F created by rolling friction which causes the trailer to stop.

20 km/h

A

F

SOLUTION 20 km>h = + b a;

20(103) = 5.556 m>s 3600 y2 = y20 + 2ac (s - s0) 0 = 5.5562 + 2(a)(45 - 0) a = - 0.3429 m>s2 = 0.3429 m>s2 : F = 250(0.3429) = 85.7 N

Ans.

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laws

or

+ ©F = ma ; : x x






*R1–8. The position of a particle along a straight line is given by s = 1t3 - 9t2 + 15t2 ft, where t is in seconds. Determine its maximum acceleration and maximum velocity during the time interval 0 … t … 10 s.

SOLUTION s = t3 - 9t2 + 15t y =

ds = 3t2 - 18t + 15 dt

a =

dy = 6t - 18 dt

amax occurs at t = 10 s, amax = 6(10) - 18 = 42 ft>s2

Ans.

or

ymax occurs when t = 10 s ymax = 3(10)2 - 18(10) + 15 = 135 ft>s

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Ans.






R1–9. The spool, which has a mass of 4 kg, slides along the rotating rod. At the # instant shown, the angular rate of rotation of the rod is u = 6 rad >s and this rotation is increasing at $ u = 2 rad>s2. At this same instant, the spool has a velocity of 3 m/s and an acceleration of 1 m>s2, both measured relative to the rod and directed away from the center O when r = 0.5 m. Determine the radial frictional force and the normal force, both exerted by the rod on the spool at this instant.

r = 0.5 m ••

θ = 2 rad/s2 O •

θ = 6 rad/s

vs = 3 m/s as = 1 m/s2

SOLUTION r = 0.5 m # r = 3 m>s $ r = 1 m>s2

# u = 6 rad>s $ u = 2 rad>s

©Fz = maz ;

Nz - 4(9.81) = 0

Web)

Nu = 4(37) = 148 N

laws

©Fu = mau ;

teaching

Fr = 4( - 17) = - 68 N

permitted.

Dissemination

Wide

copyright

in

©Fr = mar ;

or

# $ ar = r - ru2 = 1 - 0.5(6)2 = - 17 $ ## au = ru + 2ru = 0.5(2) + 2(3)(6) = 37

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Nz = 39.24 N

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Fr = - 68 N

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N = 2(148)2 + (39.24)2 = 153 N






Ans.

R1–10. Packages having a mass of 6 kg slide down a smooth chute and land horizontally with a speed of 3 m/ s on the surface of a conveyor belt. If the coefficient of kinetic friction between the belt and a package is mk = 0.2, determine the time needed to bring the package to rest on the belt if the belt is moving in the same direction as the package with a speed v = 1 m/s.

3 m/s 1 m/s

SOLUTION (+c )

m(v1)y + ©

L

Fy dt = m(v2)y

0 + Np(t) - 58.86(t) = 0 Np = 58.86 N +) (:

m(v1)x + ©

L

Fx dt = m(v2)x

6(3) - 0.2(58.86)(t) = 6(1) t = 1.02 s

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Ans.






R1–11. A 20-kg block is originally at rest on a horizontal surface for which the coefficient of static friction is ms = 0.6 and the coefficient of kinetic friction is mk = 0.5. If a horizontal force F is applied such that it varies with time as shown, determine the speed of the block in 10 s. Hint: First determine the time needed to overcome friction and start the block moving.

F

F (N)

200

SOLUTION The crate starts moving when 5

F = Fr = 0.6(196.2) = 117.72 N From the graph since F =

200 t, 5

0 … t … 5s

The time needed for the crate to start moving is 5 (117.72) = 2.943 s 200 laws

or

t =

Wide

copyright

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Hence, the impulse due to F is equal to the area under the curve from 2.943 s … t … 10 s Dissemination

+ :

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permitted.

Fx dt = m(vx)2

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m(vx)1 + ©

10

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200 t dt + 200 dt - (0.5)196.2(10 - 2.943) = 20v2 L2.943 5 L5 the assessing

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5 1 40( t2) 2 + 200(10 - 5) - 692.292 = 20v2 2 2.943

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634.483 = 20v2

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v2 = 31.7 m>s






Ans.

10

t (s)

*R1–12. The 6-lb ball is fired from a tube by a spring having a stiffness k = 20 lb>in. Determine how far the spring must be compressed to fire the ball from the compressed position to a height of 8 ft, at which point it has a velocity of 6 ft/s.

v = 6 ft/s

8 ft


1 6 1 (20)(12)(x2) = a b(6)2 + 8(6) 2 2 32.2

k = 20 lb/in.

x = 0.654 ft = 7.85 in.

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R1–13. A train car, having a mass of 25 Mg, travels up a 10° incline with a constant speed of 80 km/h. Determine the power required to overcome the force of gravity.

SOLUTION v = 80 km>h = 22.22 m>s P = F

#

v = 25(103)(9.81)(22.22)(sin 10°)

P = 946 kW

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R1–14. v

The rocket sled has a mass of 4 Mg and travels from rest along the smooth horizontal track such that it maintains a constant power output of 450 kW. Neglect the loss of fuel mass and air resistance, and determine how far it must travel to reach a speed of v = 60 m>s.

T

SOLUTION F = m a = ma

+ ©F = m a ; : x x P = Fv = ma

v2 dv b ds

P ds = m

v2 dv

L

L

s

P

v

ds = m

L0

L0

v2 dv

m v3 3 laws

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Ps =

v dv b ds

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m v3 3P

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permitted.

Dissemination

= 640 m instructors

3(450)(103)

World

4 (103)(60)3

States

s =

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s =

■R1–15.

A projectile, initially at the origin, moves vertically downward along a straight-line path through a fluid medium such that its velocity is defined as v = 318e-t + t21>2 m>s, where t is in seconds. Plot the position s of the projectile during the first 2 s. Use the Runge-Kutta method to evaluate s with incremental values of h = 0.25 s.

SOLUTION n = 3 (8 e - 1 + t)1>2 s0 = 0 at t = 0

Web)

laws

teaching

Wide

0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

in

0 2.01 3.83 5.49 7.03 8.48 9.87 11.2 12.5

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Using the Runge–Kutta method:






*R1–16. The chain has a mass of 3 kg/m. If the coefficient of kinetic friction between the chain and the plane is mk = 0.2, determine the velocity at which the end A will pass point B when the chain is released from rest.

A

2m

SOLUTION + a©Fy = may ;

40°

- 2(3)(9.81) cos 40° + NC = 0

B

NC = 45.09 N + b©Fx = max ;

2(3)(9.81) sin 40° - 0.2(45.09) = 2(3) a a = 4.80 m>s2

+ b v22 = v21 + 2as v22 = 0 + 2(4.80)(2) v2 = 4.38 m>s

or

Ans.

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Also,

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T1 + ©U1-2 = T2

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Dissemination

1 (2)(3)(v2) 2

0 + 2(3)(9.81)(2 sin 40°) - 0.2(45.09)(2) =

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v = 4.38 m>s






Ans.

R1–17. The motor M pulls in its attached rope with an acceleration ap = 6 m/s2. Determine the towing force exerted by M on the rope in order to move the 50-kg crate up the inclined plane. The coefficient of kinetic friction between the crate and the plane is mk = 0.3. Neglect the mass of the pulleys and rope.

aP = 6 m/s2

M

P

SOLUTION a+ ©Fy = may ;

NC - 50(9.81) cos 30° = 0

θ = 30°

NC = 424.79 Q+ ©Fx = max ;

3T - 0.3(424.79) - 50(9.81) sin 30° = 50aC

(1)

Kinematics, 2sC + (sC - sp) = l Taking two time derivatives, yields 3aC = ap or

6 = 2 3

teaching

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laws

Thus, aC =

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in

Substituting into Eq. (1) and solving,

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permitted.

Dissemination

T = 158 N

R1–18. The drinking fountain is designed such that the nozzle is located from the edge of the basin as shown. Determine the maximum and minimum speed at which water can be ejected from the nozzle so that it does not splash over the sides of the basin at B and C.

vA 40° A 50 mm

B

C

250 mm 100 mm

SOLUTION + bs = v t a: x x R = vA sin 40°t (+c ) sy = A sy B 0 + vy t +

t = 1 2

R vA sin 40°

(1)

ac t2

-0.05 = 0 + vA cos 40°t +

1 2

( - 9.81)t2

(2)

Substituting Eq. (1) into (2) yields: 2 R R 1 ≤ + ( - 9.81) ¢ ≤ vA sin 40° 2 vA sin 40° laws

or

-0.05 = vA cos 40° ¢

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4.905 sin 40°R2 A sin2 40°(R cos 40° + 0.05 sin 40°)

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vA =

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Dissemination

At point B, R = 0.1 m. of

the

not

4.905 sin 40°(0.1)2 = 0.838 m>s A sin 40°(0.1 cos 40° + 0.05 sin 40°)

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At point C, R = 0.35 m.

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4.905 sin 40°(0.35)2 = 1.76 m>s A sin 40°(0.35 cos 40° + 0.05 sin 40°) is

vA =

Ans.

R1–19. The 100-kg crate is subjected to the action of two forces, F1 = 800 N and F2 = 1.5 kN, as shown. If it is originally at rest, determine the distance it slides in order to attain a speed of 6 m/s. The coefficient of kinetic friction between the crate and the surface is mk = 0.2.

F1 30°

SOLUTION + c a Fy = 0;

NC - 800 sin 30° - 100(9.81) + 1500 sin 20° = 0 NC = 867.97 N T1 + a U1 - 2 = T2 1 (100)(6)2 2

0 + 800 cos 30°(s) - 0.2(867.97)(s) + 1500 cos 20° (s) = s(1928.7) = 1800 s = 0.933 m

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F2 20°

*R1–20. If a particle has an initial velocity v0 = 12 ft>s to the right, and a constant acceleration of 2 ft>s2 to the left, determine the particle’s displacement in 10 s. Originally s0 = 0.

SOLUTION +) (:

s = s0 + v0 t +

1 2 at 2 c

s = 0 + 12(10) +

1 ( - 2)(10)2 2

s = 20.0 ft

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R1–21. The ping-pong ball has a mass of 2 g. If it is struck with the velocity shown, determine how high h it rises above the end of the smooth table after the rebound. Take e = 0.8.

30⬚ 18 m/s h 2.25 m

SOLUTION + ) (:

s = s0 + v0 t 2.25 = 0 + 18 cos 30°t t = 0.14434 s

(vx)1 = (vx)2 = 18 cos 30° = 15.5885 m>s ( + T)

v = v0 + ac t

(vy)1 = 18 sin 30° + 9.81(0.14434) (vy)1 = 10.4160 m>s laws

or

(vy)2

Web)

10.4160

teaching

e = 0.8 =

in

(+ c)

Dissemination

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copyright

(vy)2 = 8.3328 m>s

World

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s = s0 + v0 t

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0.75 = 0 + 15.5885t

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t = 0.048112 s

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s = s0 + v0 t +

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h = 0 + 8.3328(0.048112) -

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h = 0.390 m






Ans.

0.75 m

R1–22. A sports car can accelerate at 6 m/s 2 and decelerate at 8 m/s 2. If the maximum speed it can attain is 60 m/s, determine the shortest time it takes to travel 900 m starting from rest and then stopping when s = 900 m.

SOLUTION Time to accelerate to 60 m/s, +) (:

v = v0 + ac t 60 = 0 + 6t t = 10 s

+) (:

s = s0 + v0 t +

1 (6)(102) 2 or

s = 0 + 0 +

1 2 a t 2 c

in

teaching

Web)

laws

s = 300 m

Wide Dissemination

World

permitted.

v = v0 + ac t

not

instructors

States

+) (:

copyright

Time to decelerate to a stop,

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of

the

0 = 60 - 8t

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t = 7.5 s

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any sale

Time to travel at 60 m/s,

will

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s = 225 m

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1 (8)(7.52) 2

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s = 0 + 60(7.5) -

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this

assessing

s = s0 + v0 t +

is

+) (:

900 - 300 - 225 = 375 m +) (:

s = s0 + v0 t 375 = 0 + 60t t = 6.25 s

Total time t = 10 + 7.5 + 6.25 = 23.8 s






Ans.

R1–23. A 2-kg particle rests on a smooth horizontal plane and is acted upon by forces Fx = 0 and Fy = 3 N. If x = 0, y = 0, vx = 6 m/s, and vy = 2 m/s when t = 0, determine the equation y = f1x2 which describes the path.

SOLUTION + c ©Fy = may;

3 = 2ay

ay = 1.5 m>s2

(1)

+ ©F = ma ; : x x

0 = 2ax

ax = 0

(2)

dvy dt

= 1.5

vy

t

dvy = 1.5

vy =

L0

dt

dy = 1.5t + 2 dt y

t

(1.5t + 2) dt

Web)

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in

L0

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dy =

or

L2

laws

ay =

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(3)

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y = 0.75t2 + 2t

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dx = 6 dt

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(4)

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x = 6t Eliminating t from Eq. (3) and (4) yields: y = 0.0208x2 + 0.333x (Parabola)






Ans.

*R1–24. A skier starts from rest at A (30 ft, 0) and descends the smooth slope, which may be approximated by a parabola. If she has a weight of 120 lb, determine the normal force she exerts on the ground at the instant she arrives at point B.

y 1 2 y = __ 60 x – 15

30 ft

A 15 ft B

SOLUTION TA + VA = TB + VB 0 + (120)(15) =

1 120 2 a b vB + 0 2 32.2

vB = 31.08 ft>s y =

1 2 x - 15 60

dy 1 2 2 x = = 0 dx x = 0 30 x = 0 or

1 30

Web)

laws

=

in

dx2

teaching

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permitted.

Dissemination

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copyright

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the

dx2

(1 + 0)3>2 = 30 ft 1 30

World

dy

3 =

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120 (31.08)2 c d 32.2 30

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Ns - 120 =

the

+ c ©Fn = man ;

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Ns = 240 lb






Ans.

x

R1–25. The 20-lb block B rests on the surface of a table for which the coefficient of kinetic friction is mk = 0.1. Determine the speed of the 10-lb block A after it has moved downward 2 ft from rest. Neglect the mass of the pulleys and cords.

20 lb B

A

C 6 lb

10 lb

SOLUTION Block A: + T ©Fy = may ;

- T1 + 10 =

10 a 32.2

(1)

Block B: + ©F = ma ; ; x x

-T2 + T1 - 0.1NB =

+ c ©Fy = may ;

NB - 20 = 0

20 a 32.2

(2) (3)

laws

or

Block C: 6 a 32.2

teaching

Web)

(4) in

T2 - 6 =

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copyright

+ c ©Fy = may ;

instructors

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Dissemination

Solving Eqs. (1)–(4) for a,

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(+ T )v2 = v20 + 2ac(s - s0)

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v = 2.68 ft>s






Ans.

R1–26. At a given instant the 10-lb block A is moving downward with a speed of 6 ft/s. Determine its speed 2 s later. Block B has a weight of 4 lb, and the coefficient of kinetic friction between it and the horizontal plane is mk = 0.2. Neglect the mass of the pulleys and cord.

B

SOLUTION

A

Block A: + T ©Fy = may ;

10 - 2T =

10 a 32.2 A

Block B: + ©F = ma ; ; x x

-T + 0.2(4) =

4 a 32.2 B

Kinematics:

laws

or

2sA + sB = l

in

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Web)

2aA = - aB

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vA = 26.8 ft>s

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vA = 6 + 10.40(2)






Ans.

R1–27. Two smooth billiard balls A and B have an equal mass of m = 200 g. If A strikes B with a velocity of 1vA21 = 2 m/s as shown, determine their final velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e = 0.75.

y A (vA)1 40°

SOLUTION

B

(vA)x1 = - 2 cos 40° = - 1.532 m>s (vA)y1 = - 2 sin 40° = - 1.285 m>s + ) (:

mA(vA)x1 + mB (vB)x1 = mA(vA)x2 + mB (vB)x2 -2(1.532) + 0 = 0.2(vA)x2 + 0.2(vB)x2

+ ) (:

e =

(1)

(vrel)2 (vrel)1 (vA)x2 - (vB)x2

(2)

1.532

laws

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0.75 =

in

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(vA)x2 = - 0.1915 m>s

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For B:

Ans.

(vA)2 = 2(- 0.1915)2 + (1.285)2 = 1.30 m>s

Ans.

0.1915 (uA)2 = tan - 1 a b = 8.47° e 1.285

Ans.






x

*R1–28. A crate has a weight of 1500 lb. If it is pulled along the ground at a constant speed for a distance of 20 ft, and the towing cable makes an angle of 15° with the horizontal, determine the tension in the cable and the work done by the towing force. The coefficient of kinetic friction between the crate and the ground is mk = 0.55.


NC - 1500 + T sin 15° = 0

+ ©F = 0; : x

T cos 15° - 0.55NC = 0 T = 744.4 lb = 744 lb

Ans.

NC = 1307.3 lb U1 - 2 = (744.4 cos 15°)(20) = 14 380.7 ft # lb U1-2 = 14.4 ft # kip

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Ans.






R1–29. The collar of negligible size has a mass of 0.25 kg and is attached to a spring having an unstretched length of 100 mm. If the collar is released from rest at A and travels along the smooth guide, determine its speed just before it strikes B.

A

400 mm

k ⫽ 150 N/m 200 mm

SOLUTION

B

TA + VA = TB + VB 0 + (0.25)(9.81)(0.6) +

1 1 1 (150)(0.6 - 0.1)2 = (0.25)(vB)2 + (150)(0.4 - 0.1)2 2 2 2

vB = 10.4 m>s

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laws

or

Ans.






R1–30. Determine the tension developed in the two cords and the acceleration of each block. Neglect the mass of the pulleys and cords. Hint: Since the system consists of two cords, relate the motion of block A to C, and of block B to C. Then, by elimination, relate the motion of A to B.

10 kg A

C

SOLUTION

3m

Block A: + T ©Fy = may ;

B 4 kg

10(9.81) - TA = 10aA

(1)

TB - 4(9.81) = 4aB

(2)

Block B: + c ©Fy = may ; Pulley C: + c ©Fy = 0;

TA - 2TB = 0

(3)

laws

or

Kinematics:

in

teaching

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sA + sC = l

Dissemination

Wide

copyright

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sC ¿ + (sC ¿ - sB) = l¿

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(4)

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aB = 2 aA

aA = 0.755 m>s2

Ans.

aB = 1.51 m>s2

Ans.

TA = 90.6 N

Ans.

TB = 45.3 N

Ans.






R1–31. The baggage truck A has a mass of 800 kg and is used to pull each of the 300-kg cars. Determine the tension in the couplings at B and C if the tractive force F on the truck is F = 480 N. What is the speed of the truck when t = 2 s, starting from the rest? The car wheels are free to roll. Neglect the mass of the wheels.

A C

F


480 = 3800 + 2130024a a = 0.3429 m>s2

+ 2 1:

n = n0 + act n = 0 + 0.3429122 = 0.686 m>s

+ ©F = ma ; : x x

Ans.

TB = 21300210.34292 TB = 205.71 = 206 N

+ ©F = ma ; : x x

Ans.

TC = 1300210.34292 or

TC = 102.86 = 103 N

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laws

Ans.






B

*R1–32. The baggage truck A has a mass of 800 kg and is used to pull each of the 300-kg cars. If the tractive force F on the truck is F = 480 N, determine the initial acceleration of the truck. What is the acceleration of the truck if the coupling at C suddenly fails? The car wheels are free to roll. Neglect the mass of the wheels.

A C

F


480 = 3800 + 2130024a a = 0.3429 = 0.343 m>s2

+ ©F = ma ; : x x

Ans.

480 = 1800 + 3002a a = 0.436 m>s2

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Ans.






B

■ R1–33.

Packages having a mass of 2.5 kg ride on the surface of the conveyor belt. If the belt starts from rest and with constant acceleration increases to a speed of 0.75 m/s in 2 s, determine the maximum angle of tilt, u, so that none of the packages slip on the inclined surface AB of the belt. The coefficient of static friction between the belt and each package is ms = 0.3. At what angle f do the packages first begin to slip off the surface of the belt if the belt is moving at a constant speed of 0.75 m/s?

φ B 350 mm

θ

SOLUTION

A

+ ©Fy = may ;

NP - 2.5(9.81) cos u = 0

(1)

+ ©Fx = max ;

0.3NP - 2.5(9.81) sin u = 2.5aP

(2)

Kinematics: v = v0 + ac t 0.75 = 0 + aP (2)

laws

or

aP = 0.375 m>s2

teaching

Web)

Combining Eqs. (1) and (2), using ap, Wide

copyright

in

0.3 cos u - sin u = 0.0382 u = 14.6° instructors

States

World

permitted.

Dissemination

Ans. not the on

learning.

is

of

and

2.5(9.81) sin f - 0.3NP = 0

United

Q+ ©Ft = mat ;

(0.75)2 d 0.350

use

2.5(9.81) cos f - NP = 2.5 c

the

(including

for

work

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the

by

a+ ©Fn = man ;

assessing

is

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Combining these equations,

provided

integrity

of

and

work

this

cos f - 3.33 sin f = 0.164

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f = 14.0°






Ans.

R1–34. A particle travels in a straight line such that for a short time 2 s … t … 6 s its motion is described by v = 14>a2 ft>s where a is in ft>s2. If v = 6 ft>s when t = 2 s, determine the particle’s acceleration when t = 3 s.

SOLUTION 4 dn = v dt

a = n

L6

t

n dn =

L2

4 dt

1 2 n - 18 = 4 t - 8 2 n2 = 8 t + 20 At t = 3 s, choosing the positive root laws

or

n = 6.63 ft>s in

teaching

Web)

4 = 0.603 ft>s2 6.63

Ans.

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Wide

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a =






R1–35. The blocks A and B weigh 10 and 30 lb, respectively. They are connected together by a light cord and ride in the frictionless grooves. Determine the speed of each block after block A moves 6 ft up along the plane. The blocks are released from rest.

z 1 ft

1 ft

B

SOLUTION A

15 ft

2152 + 22 6 = z 15

3 ft

z = 5.95 ft

3 ft

x

T1 + V1 = T2 + V2 0 + 0 =

1 30 2 1 10 2 ( )v2 + ( )v2 + 10(5.95) - 30(5.95) 2 32.2 2 32.2

v2 = 13.8 ft>s

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Dissemination

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Web)

laws

or

Ans.






y

*R1–36. A motorcycle starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft>s2 until it reaches a speed of 50 ft>s. Afterwards it maintains this speed. Also, when t = 0, a car located 6000 ft down the road is traveling toward the motocycle at a constant speed of 30 ft>s. Determine the time and the distance traveled by the motorcycle when they pass each other.

SOLUTION Motorcycle: + B A:

v = v0 + act¿ 50 = 0 + 6t t¿ = 8.33 s

v2 = v02 + 2ac(s - s0) (50)2 = 0 + 2(6)(s¿ - 0) s¿ = 208.33 ft laws

or

In t¿ = 8.33 s car travels

Wide

copyright

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s¿¿ = v0t¿ = 30(8.33) = 250 ft

States

World

permitted.

Dissemination

Distance between motorcycle and car: the

not

instructors

6000 - 250 - 208.33 = 5541.67 ft use

United

on

learning.

is

of

s = v0t;

work

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by

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When passing occurs for motorcycle: x = 50(t¿¿)

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s = v0t;

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For car: provided

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5541.67 - x = 30(t¿¿)

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s = v0t;

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x = 3463.54 ft

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destroy

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and

any

courses

Solving,

t¿ = 69.27 s Thus, for the motorcycle, t = 69.27 + 8.33 = 77.6 s

Ans.

sm = 208.33 + 3463.54 = 3.67(10)3 ft

Ans.






R1–37. The 5-lb ball, attached to the cord, is struck by the boy. Determine the smallest speed he must impart to the ball so that it will swing around in a vertical circle, without causing the cord to become slack. 4 ft

SOLUTION

v

+ T ©Fn = man ;

5 =

5 32.2

v22 4

A B

v22 = 128.8 ft2>s2

T1 + V1 = T2 + V2 1 2

5 5 A 32.2 B v2 + 0 = 12 A 32.2 B (128.8) + 5(8)

v = 25.4 ft>s

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in

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Web)

laws

or

Ans.






R1–38. A projectile, initially at the origin, moves along a straightline path through a fluid medium such that its velocity is v = 180011 - e-0.3t2 mm>s, where t is in seconds. Determine the displacement of the projectile during the first 3 s.

SOLUTION v =

ds = 1800(1 - e-0.3t) dt

s

L0

ds =

t

L0

1800(1 - e-0.3t) dt 1 -0.3t b - 6000 e 0.3

s = 1800 at + Thus, in t = 3 s

laws

or

1 -0.3(3) b - 6000 e 0.3

s = 1800 a3 +

s = 1839.4 mm = 1.84 m

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Web)

Ans.






R1–39. A particle travels along a straight line with a velocity v = (12 - 3t2) m>s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t = 4 s, the displacement from t = 0 to t = 10 s, and the distance the particle travels during this time period.

SOLUTION v = 12 - 3t2

(1)

dv = -6t 2 = -24 m>s2 dt t=4

a = s

t

L-10

ds =

L1

t

v dt =

L1

Ans.

A 12 - 3t2 B dt

s + 10 = 12t - t3 - 11 s = 12t - t3 - 21

or

s| t = 0 = -21

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laws

s|t = 10 = - 901 ¢s = -901 - ( -21) = - 880 m

Dissemination

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Ans.

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From Eq. (1):

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v = 0 when t = 2s

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s|t = 2 = 12(2) - (2)3 - 21 = -5

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sT = (21 - 5) + (901 - 5) = 912 m






Ans.

*R1–40. A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by u = 15t22 rad, where t is in seconds. Determine the magnitude of the particle’s acceleration when u = 30°. The particle starts from rest when u = 0°.

SOLUTION r = 2m # r = 0 $ r = 0

u = 5t2 # u = 10t $ u = 10

# $ $ ## a = (r - ru 2)ur + (ru + 2ru)uu = [0 - 2(10t)2]ur + [2(10) + 0]uu = { - 200t2ur + 20uu} m>s2

teaching

Web)

laws

or

p When u = 30° = 30 a b = 0.524 rad 180

Wide

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0.524 = 5t2

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t = 0.324 s

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a = [ - 200(0.324)2]ur + 20uu assessing

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= { - 20.9ur + 20uu} m>s2

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a = 2( - 20.9)2 + (20)2 = 29.0 m>s2






Ans.

R1–41. If the end of the cable at A is pulled down with a speed of 2 m/s, determine the speed at which block B rises.

C A 2 m/s

SOLUTION Two cords: sA + 2sC = l

B

sB + (sB - sC) = l¿ Thus, vA = - 2vC 2vB = vC 4vB = - vA -2 = - 0.5 m>s = 0.5 m>s c 4

or

Ans.

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laws

vB =






R1–42. The bottle rests at a distance of 3 ft from the center of the horizontal platform. If the coefficient of static friction between the bottle and the platform is ms = 0.3, determine the maximum speed that the bottle can attain before slipping. Assume the angular motion of the platform is slowly increasing.

3 ft

Motion

SOLUTION ©Fz = maz;

Nz - mg = 0

©Fx = man;

0.3(mg) = m a

Nz = mg v2 b r

v = 20.3gr = 20.3(32.2)(3) = 5.38 ft>s

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Ans.






R1–43. Work Prob. R1–42 assuming that the platform starts rotating from rest so that the speed of the bottle is increased at 2 ft>s2.

3 ft

Motion

SOLUTION ©Fz = maz;

Nz - mg = 0

Nz = mg

©Ft = mat;

- 0.3(mg) cos u = - m(2)

©Fn = man;

0.3(mg) sin 78.05° = - ma

u = 78.05° v2 b 3

v = 5.32 ft>s

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*R1–44. A 3-lb block, initially at rest at point A, slides along the smooth parabolic surface. Determine the normal force acting on the block when it reaches B. Neglect the size of the block.

y y = x2 – 4 2 ft

x

A

SOLUTION

4 ft

T1 + V1 = T2 + V2 B

1 3 a bv22 + 0 2 32.2

0 + 3(4) =

v2 = 16.05 ft>s [1 + r =

dy

A dx B 2]3>2 d2y

=

[1 + (2x)2]3>2 2

dx2 or

r = 0.5 ft laws

At x = 0 ,

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3 (16.05)2 c d 32.2 0.5

in

N-3 =

Ans.

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N = 51.0 lb






permitted.

Wide

copyright

+ c ©Fn = man ;

R1–45. A car starts from rest and moves along a straight line with an acceleration of a = 13s-1>32 m>s2, where s is in meters. Determine the car’s acceleration when t = 4 s.

SOLUTION 1

a = 3s- 3 a ds = n dn s

L0

v

1

3s - 3 ds =

L0

n dn

2 3 1 (3)s3 = n2 2 2 1

n = 3s3

laws

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3 dt

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L0

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L0

t

1

s - 3 ds =

copyright

s

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1 ds = 3s3 dt

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3 2 s3 = 3t 2 3

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by

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s = (2t)2 3

the

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s t = 4 = (2(4))2 = 22.62 = 22.6 m 1

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a t = 4 = 3(22.62)- 3 = 1.06 m s2






Ans.

R1–46. A particle travels along a curve defined by the equation s = (t3 - 3t2 + 2t) m. where t is in seconds. Draw the s- t, v - t, and a - t graphs for the particle for 0 … t … 3 s .

SOLUTION s = t3 - 3t2 + 2t v =

ds = 3t2 - 6t + 2 dt

Ans.

a =

dv = 6t - 6 dt

Ans.

v = 0 at 0 = 3t2 - 6t + 2 t = 1.577 s, and t = 0.4226 s,

laws

or

s|t = 1.577 = -0.386 m

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s|t = 0.4226 = 0.385 m






R1–47. The crate, having a weight of 50 lb, is hoisted by the pulley system and motor M. If the crate starts from rest and, by constant acceleration, attains a speed of 12 ft>s after rising 10 ft, determine the power that must be supplied to the motor at the instant s = 10 ft. The motor has an efficiency P = 0.74.

M

SOLUTION (+ c )

v2 = v20 + 2ac(s - s0) (12)2 = 0 + 2ac(10 - 0) ac = 7.20 ft>s2

+ c ©Fy = may;

2T- 50 =

s

50 (7.20) 32.2

T = 30.6 lb sC + (sC - sM) = l laws

or

vM = 2vC teaching in

v = 30.6(24) = 734.2 lb # ft>s

Wide Dissemination

#

copyright

P0 = T

Web)

vM = 2(12) = 24 ft>s

the

instructors

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permitted.

734.2 = 992.1 lb # ft>s = 1.80 hp 0.74

of

Pi =

*R1–48. The block has a mass of 0.5 kg and moves within the smooth vertical slot. If the block starts from rest when the attached spring is in the unstretched position at A, determine the constant vertical force F which must be applied to the cord so that the block attains a speed vB = 2.5 m/s when it reaches B; sB = 0.15 m. Neglect the mass of the cord and pulley.

0.3 m

C

0.3 m

SOLUTION

B F

sB

The work done by F depends upon the difference in the cord length AC–BC. A

TA + ©UA - B = TB 0 + F[2(0.3)2 + (0.3)2 - 2(0.3)2 + (0.3 - 0.15)2] - 0.5(9.81)(0.15)

k = 100N/m

1 1 - (100)(0.15)2 = (0.5)(2.5)2 2 2 F(0.0889) = 3.423 F = 38.5 N

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R1–49. A ball having a mass of 200 g is released from rest at a height of 400 mm above a very large fixed metal surface. If the ball rebounds to a height of 325 mm above the surface, determine the coefficient of restitution between the ball and the surface.

SOLUTION Just before impact T1 + T2 = T2 + V2 0 + 0.2(9.81)(0.4) =

1 (0.2)(v2)2 + 0 2

v2 = 2.80 m>s Just after impact T3 + V3 = T4 + V4

teaching

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laws

or

1 (0.2)(v3)2 + 0 = 0 + 0.2(9.81)(0.325) 2

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permitted.

Dissemination

(vrel)2 2.53 = = 0.901 (vrel)1 2.80

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Ans.

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is

of

e =

in

v3 = 2.53 m>s






R1–50. Determine the speed of block B if the end of the cable at C is pulled downward with a speed of 10 ft>s. What is the relative velocity of the block with respect to C? C 10 ft/s

SOLUTION 3sB + sC = l 3nB = - nC

B

3nB = - 1102 nB = - 3.33 ft>s = 3.33 ft>s c (+ T )

Ans.

nB = nC + nB>C - 3.33 = 10 + nB>C vB>C = - 13.3 ft>s = 13.3 ft>s c

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