engineering mechanics 13e chap 04 solution

4–1.

If A, B, and D are give given n vect vectors ors,, pro prove ve the dist di stri ribu buti tive ve la law w fo forr the the vec vecto torr cro cross ss pr prod oduc uct, t, i. i.ee., A : (B + D) = (A : B) + (A : D).

SOLUTION

Consider the three three vectors; with A vertical. Note obd is perpendicular to A. od = ƒ A * (B + D) ƒ = ƒ A ƒ ƒ B + D ƒ sin u3 ob = ƒ A * B ƒ = ƒ A ƒ ƒ B ƒ sin u1 bd = ƒ A * D ƒ = ƒ A ƒ ƒ D ƒ sin u2

Also, these three three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product product is proportional to the length of each side of the triangle. The three vector cross products also form a closed triangle o ¿ b ¿ d ¿ which is similar to triangle obd. Thus from the figure, figure, A

* (B + D) = (A * B) + (A * D)

(QED)

Note also, A

= A x i + A y j + A z k

B = Bx i + By j + Bz k D = Dx i + Dy j + Dz k

A

3

i Ax * (B + D) = Bx + Dx

j Ay By + Dy

k Az Bz + Dz

3

= [A y (Bz + Dz) - A z(By + Dy)]i j - [A x(Bz + Dz) - A z(Bx + Dx)] j

+ [A x(By + Dy) - A y(Bx + Dx)]k j + (A x By - A y Bx)k = [(A y Bz - A zBy)i - (A x Bz - A z Bx)] j

+ [(A y Dz - A z Dy)i - (A x Dz - A z Dx) j + (A x Dy - A y Dx)k

3

i = Ax Bx

j Ay By

3 3

k i Az + Ax Bz Dx

= (A * B) + (A * D)

j Ay Dy

k Az Dz

3 (QED)

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4–2.

Prove A

# (B

the triple scalar : C) = (A : B) # C.

product

identity

SOLUTION

As shown in the figure Area = B(C sin u) = |B * C| Thus, Volume of parallelepiped is | B * C||h| But, |h| = |A # u(B * C)| =

` # a A A

B * C |B * C|

b`

Thus, Volume = |A # (B * C)| Since |(A * B) # C| represents this same volume then A

# (B

: C) = (A : B) # C

(QED)

Also, LHS =

A

# (B

: C)

= (A x i + A y j + A z k) #

3

i Bx Cx

j By Cy

k Bz Cz

3

= A x (ByCz - BzCy) - A y (BxCz - BzCx) + A z (BxCy - ByCx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx RHS = (A : B) # C

3

i = Ax Bx

j Ay By

k Az Bz

3

# (Cx i + Cy j + Cz k)

= Cx(A y Bz - A zBy) - Cy(A xBz - A zBx) + Cz(AxBy - A yBx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx

Thus, LHS = RHS A

# (B

: C) = (A : B) # C

(QED)


4–2.

Prove A

# (B

the triple scalar : C) = (A : B) # C.

product

identity

SOLUTION

As shown in the figure Area = B(C sin u) = |B * C| Thus, Volume of parallelepiped is | B * C||h| But, |h| = |A # u(B * C)| =

` # a A A

B * C |B * C|

b`

Thus, Volume = |A # (B * C)| Since |(A * B) # C| represents this same volume then A

# (B

: C) = (A : B) # C

(QED)

Also, LHS =

A

# (B

: C)

= (A x i + A y j + A z k) #

3

i Bx Cx

j By Cy

k Bz Cz

3

= A x (ByCz - BzCy) - A y (BxCz - BzCx) + A z (BxCy - ByCx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx RHS = (A : B) # C

3

i = Ax Bx

j Ay By

k Az Bz

3

# (Cx i + Cy j + Cz k)

= Cx(A y Bz - A zBy) - Cy(A xBz - A zBx) + Cz(AxBy - A yBx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx

Thus, LHS = RHS A

# (B

: C) = (A : B) # C

(QED)


4–3.

Given the three nonzero vectors A, B, an and d C, sho show w that that if A # (B : C) = 0, the three vect vectors ors must lie in the same plane.

SOLUTION

Consider, |A # (B * C)| = |A| |B * C | co coss u coss u)|B * C| = (|A| co = |h| |B * C| = BC |h| si sin nf

parallelepiped. ped. = volume of parallelepi If A # (B * C) = 0, then the volume equals zero, so that A, B, and C are coplanar.


*4–4.

Determine the moment about point A of each of the three forces acting on the beam.

F 2 = 500 lb

F 1 = 375 lb 5

A

3

8 ft

SOLUTION

1 2

a + MF1

A

1 2

a + MF2

A

1 2

a + MF3

A

4

6 ft

B

0.5 ft

5 ft

30˚

F 3 = 160 lb

12

= - 375 8

= - 3000 lb # ft = 3.00 kip # ft (Clockwise) = - 500

Ans.

a 45 b1142

= - 5600 lb # ft = 5.60 kip # ft (Clockwise)

1

21 2

Ans.

1 2

= - 160 cos 30° 19 + 160 sin 30° 0.5

= - 2593 lb # ft = 2.59 kip # ft (Clockwise)

Ans.


4–5.

Determine the moment about point B of each of the three forces acting on the beam.

F 2 = 500 lb

F 1 = 375 lb 5

A

3

8 ft

SOLUTION

1 2

a + MF1

B

1 2

a + MF2

B

1 2

a + MF3

B

4

6 ft

B

0.5 ft

5 ft

30˚

F 3 = 160 lb

1 2

= 375 11

= 4125 lb # ft = 4.125 kip # ft (Counterclockwise) = 500

Ans.

a 45 b152

= 2000 lb # ft = 2.00 kip # ft (Counterclockwise)

1 2

Ans.

12

= 160 sin 30° 0.5 - 160 cos 30° 0 = 40.0 lb # ft (Counterclockwise)

Ans.


4–6.

The crane can be adjusted for any angle 0° … u … 90° and any extension 0 … x … 5 m. For a suspended mass of 120 kg, determine the moment developed at A as a function of x and u. What values of both x and u develop the maximum possible moment at A? Compute this moment. Neglect the size of the pulley at B.

x 9m

B

1.5 m θ

A

SOLUTION

1 21

2

a + MA = - 120 9.81 7.5 + x cos u

5 - 1177.2 cos 17.5 + 26 N # m = 51.18 cos 17.5 + 26 kN # m =

x

u

u

x

(Clockwise)

The maximum moment at A occurs when u = 0° and x = 5 m.

Ans.

Ans.

1 2 = 5 - 1177.2 cos 0°17.5 + 526 N # m

a + MA

max

= - 14 715 N # m

= 14.7 kN # m (Clockwise)

Ans.


4–7.

Determine the moment of each of the three forces about point A.

F 1

F 2

250 N 30

300 N

60

A

2m

3m

4m

SOLUTION

he moment arm measured perpendicular to each force from point A is d1 = 2 sin 60° = 1.732 m

B

4

5 3

d2 = 5 sin 60° = 4.330 m

F 3

500 N

d3 = 2 sin 53.13° = 1.60 m

Using each force where MA = Fd, we have

1 2

a + MF1

1 2

a + MF2

1 2

a + MF3

A

1 2

= - 250 1.732

= - 433 N # m = 433 N # m (Clockwise) A

1 2

= - 300 4.330

= - 1299 N # m = 1.30 kN # m (Clockwise) A

Ans.

Ans.

1 2

= - 500 1.60

= - 800 N # m = 800 N # m (Clockwise)

Ans.


*4–8.

Determine the moment of each of the three force s about point B.

F 1

 250

F 2

N 30

 300

N

60

A

2m

3m

SOLUTION

4m

The forces are resolved into horizontal and vertical component as shown in Fig. a. For F1, a + MB = 250 cos 30°(3) - 250 sin 30°(4)

= 149.51 N # m = 150 N # m d

Ans.

B

4

5 3

For F2,

F 3  500 N

a + MB = 300 sin 60°(0) + 300 cos 60°(4)

= 600 N # m d

Ans.

Since the line of action of F3 passes through B, its moment arm about point B is zero.Thus MB

=0

Ans.


4–9.

Determine the moment of each force about the bolt located at A.Take FB = 40 lb, FC = 50 lb.

0.75 ft B

2.5 ft

30 FC

20

A

C

25 FB

SOLUTION

#

a + MB = 40 cos 25°(2.5) = 90.6 lb ft d

#

a + MC = 50 cos 30°(3.25) = 141 lb ftd

Ans. Ans.


4–10.

If FB = 30 lb and FC = 45 lb, determine the resultant moment about the bolt located at A.

0.75 ft B

2.5 ft

A

C

30 FC

20

25 FB

SOLUTION a + MA = 30 cos 25°(2.5) + 45 cos 30°(3.25)

= 195 lb # ft d


4–11.

The railway cro ssing gate consists of the 100-kg gate arm having a center of mass at Ga and the 250-kg counterweight having a center of ma ss at GW . Determine the magnitude and directional sense of the resultant moment produced by the weights about point A.

A

2.5 m

Ga

GW

0.5 m

0.75 m

1m

B

SOLUTION

+ (MR)A =

g

Fd;

0.25 m

(MR)A = 100(9.81)(2.5 + 0.25) - 250(9.81)(0.5 - 0.25)

= 2084.625 N # m = 2.08 kN # m (Counterclockwi se )

Ans.


*4–12.

The railway cro ssing gate consists of the 100-kg gate arm having a center of mass at Ga and the 250-kg counterweight having a center of ma ss at GW . Determine the magnitude and directional sense of the resultant moment produced by the weights about point B.

A

2.5 m

SOLUTION a + (MR)B =

g

Fd;

Ga

(MR)B = 100(9.81)(2.5) - 250(9.81)(0.5)

= 1226.25 N # m = 1.23 kN # m (Counterclockwi se )

GW

0.5 m

0.75 m B

Ans. 0.25 m


1m

*4–13.

The two boys push on the gate with forces of FA = 30 lb , and FB = 50 lb , as shown. Determine the moment of each force about C . Which way will the gate rotate, clockwise or counterclockwise? Neglect the thickness of the gate.

6 ft

3 ft 4

F A

3

A

C B

5



60

FB

SOLUTION a + (MFA)C = - 30

a 35 b (9)

= - 162 lb # ft = 162 lb # ft (Clockwi se )

Ans.

a + (MFB)C = 50(sin 60°)(6)

= 260 lb # ft (Counterclockwi se )

Since (MFB)C

7 (MFA)C, the gate will rotate Counterclockwi se .

Ans.

Ans.


4–14.

Two boys push on the gate as shown. If the boy at B exerts a force of FB = 30 lb , determine the magnitude of the force FA the boy at A must exert in order to prevent the gate from turning. Neglect the thickness of the gate.

6 ft

3 ft 4

F A

3

A

C B

5



60

FB

SOLUTION

In order to prevent the gate from turning, the resultant moment about point C must be equal to zero.

+ MRC = ©Fd;

MRC

= 0 = 30 sin 60°(6) - FA FA

= 28.9 lb

a 35 b (9) Ans.


4–15.

The Achilles tendon force of Ft = 650 N is mobilized when the man tries to stand on his toes.As this is done, each of his feet is subjected to a reactive force of N f = 400 N. Determine the resultant moment of Ft and N f about the ankle joint A.

Ft

5

A

200 mm

SOLUTION

Referring to Fig. a, a + (MR)A = © Fd;

(MR)A = 400(0.1) - 650(0.065) cos 5°

= - 2.09 N # m = 2.09 N # m (Clockwise )

Ans.

65 mm

100 mm

N f

400 N


*4–16.

The Achilles tendon force Ft is mobilized when the man tries to stand on his toes. As this is done, each of his feet is subjected to a reactive force of N t = 400 N. If the resultant moment produced by forces Ft and N f about the ankle joint A is required to be zero, determine the magnitude of F f.

Ft

5 A

200 mm

SOLUTION

Referring to Fig. a, a + (MR)A = © Fd;

0 = 400(0.1) - F cos 5°(0.065) F

= 618 N

Ans.

65 mm

100 mm

N f

400 N


4–17.

The total hip replacement is subjected to a force of F = 120 N. Determine the moment of this force about the neck at A and the stem at B.

120 N

15°

40 mm

A

SOLUTION

15 mm

150°

Moment About Point A: The angle between the line of action of the load and the

neck axis is 20° - 15° = 5°.

1 2

a + MA = 120 sin 5° 5° 0.04

= 0. 0.41 4188 N # m (Counterclockwise)

10°

Ans.

B

Moment About Point B: The dimension l can be determined using the law of sines. l

sin 150°

=

55 sin 10°

l

= 15 158.4 8.4 mm = 0.1584 m

Then,

1

a + MB = - 120 sin 15° 0.1584

2

= - 4. 4.92 92 N # m = 4. 4.92 92 N # m (Clockwise)

Ans.


4–18. 4m

The tower crane i s used to hoist the 2-Mg load upward at constant velocity elocity.. The 1.5-Mg jib BD, 0. 0.55-M Mg jib BC , and 6-Mg counterweight C have centers of mass at G1, G2, and G3, respectively ely.. Determine the resultant moment produced by the load and the wei ghts of the tower crane jib s about point A and about point B.

G2

9.5m

B

D

C G3

7.5 m

12.5 m

G1

23 m

SOLUTION Since the moment arms of the weights and the load mea sured to points A and B are the same ame,, the resultant moments produced by the load and the wei ght about points A and B are the same.

a + (MR)A = (MR)B = © Fd;

A

(MR)A = (MR)B = 6000(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5)

- 2000(9.81)(12.5) = 76 027.5 N # m = 76.0 kN # m (Counterclockwi se )

Ans.


4–19.

The tower crane i s used to hoist a 2-Mg load upward at conelocity.. The 1.5-Mg jib BD and 0.5-Mg jib BC have stant velocity centers of mass at G1 and G2, respectively ely.. Dete Determin rminee the required mass of the counterweight C s s o that the resultant moment produced by the load and the wei ght of the tower crane jibs about point A is zero. The center center of mass for the counterweight is located at G3.

4m

G2

9.5m

B

D

C G3

7.5 m

12.5 m

G1

23 m

SOLUTION Fd;; a + (MR)A = © Fd

A

0 = MC(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5) MC

= 4966.67 k g = 4.97 Mg

Ans.


*4–20.

The handle of the hammer is subjected to the force of F = 20 lb. Determine the moment of this force about the point A.

F

30

5 in. 18 in.

SOLUTION

Resolving the 20-lb force into components parallel and perpendicular to the hammer, Fig. a, and applying the principle of moments,

A B

a + MA = - 20 cos 30°(18) - 20 sin 30°(5)

= - 361.77 lb # in = 362 lb # in (Clockwise )

Ans.


4–21.

In order to pull out the nail at B, the force F exerted on the handle of the hammer must produce a clockwise moment of 500 lb in. about point A. Determine the required magnitude of force F.

F

#

30

5 in. 18 in.

SOLUTION

Resolving force F into components parallel and perpendicular to the hammer, Fig. a, and applying the principle of moments,

A B

a + MA = - 500 = - F cos 30°(18) - F sin 30°(5) F

= 27.6 lb

Ans.


4–22.

The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench. If a force of 50 N is applied to the wrench at B in the direction shown, determine the moment it creates about the nut at C . What is the magnitude of force F at A so that it creates the opposite moment about C ?

13 12

F

5

400 mm

50 N

C

SOLUTION B

c + M A = 50 sin 60°(0.3) MA

F

300 mm

= 12.99 = 13.0 N # m

a + M A = 0;

= 35.2 N

- 12.99 + F

60

Ans.

a 1213 b (0.4) = 0 Ans.


4–23.

The towline exerts a force of P = 4 kN at the end of the 20-m-long crane boom. If u = 30°, determine the placement x of the hook at A so that this force creates a maximum moment about point O. What is this moment?

B

P

4 kN

20 m O

u

1.5 m

SOLUTION

Maximum moment, OB

A x

BA

#

a + (MO)max = - 4kN(20) = 80 kN m b

Ans.

#

4 kN sin 60°(x) - 4 kN cos 60°(1.5) = 80 kN m x

= 24.0 m

Ans.


*4–24.

The towline exerts a force of P = 4 k N at the end of the 20-m-long crane boom. If x = 25 m, determine the position u of the boom so that this force creates a maximum moment about point O. What is this moment?

B

P

4 kN

20 m O

u

1.5 m

SOLUTION

Maximum moment, OB

A x

BA

#

#

c + (MO)max = 4000(20) = 80 000 N m = 80.0 kN m

Ans.

4000 sin f(25) - 4000 cos f(1.5) = 80 000 25 sin f - 1.5 cos f = 20 f

= 56.43°

u

= 90° - 56.43° = 33.6°

Ans.

Also, (1.5)2 + z2 = y 2 2.25 + z2 = y2 Similar triangles 20 + y z

=

25 + z y

20y + y2 = 25z + z2 20( 2 2.25 + z2) + 2.25 + z2 = 25z + z2 z

= 2.260 m

y

= 2.712 m

u

= cos -1

a 2.260 b = 33.6° 2.712

Ans.


4–25.

If the 1500-lb boom AB, the 200-lb cage BCD, and the 175-lb man have centers of gravity located at points G1, G2 and G3, respectively, determine the re sultant moment produced by each weight about point A.

G3 D G2

B

C

2.5 ft 1.75 ft 20 ft

G1

SOLUTION

10 ft 75

Moment of the we i ght of boom AB about poi nt A :

#

A

#

a + MA = - 1500(10 cos 75°) = - 3882.29 lb ft = 3.88 kip ft (Clockwi se )

Ans.

Moment of the we i ght of cage BCD about poi nt A :

#

#

a + MA = - 200(30 cos 75° + 2.5) = - 2052.91 lb ft = 2.05 kip ft (Clockwi se )

Ans.

Moment of the we i ght of the man about poi nt A :

#

#

a + MA = - 175(30 cos 75° + 4.25) = - 2102.55 lb ft = 2.10 kip ft (Clockwi se ) Ans.


4–26.

If the 1500-lb boom AB, the 200-lb cage BCD, and the 175-lb man have centers of gravity located at point s G1, G2 and G3, respectively, determine the resultant moment produced by all the wei ghts about point A.

G3 D G2

B

C

2.5 ft 1.75 ft 20 ft

G1

SOLUTION

10 ft

Referring to Fig. a, the resultant moment of the wei ght about point A is given by a + (MR)A = © Fd;

75 A

(MR)A = - 1500(10 cos 75°) - 200(30 cos 75° + 2.5) - 175(30 cos 75° + 4.25)

= - 8037.75 lb # ft = 8.04 kip # ft (Clockwi se )

Ans.


4–27.

The connected bar BC is u sed to increase the lever arm of the crescent wrench as shown. If the applied force is F = 200 N and d = 300 mm, determine the moment produced by this force about the bolt at A.

C

d

15 30

300 mm B

SOLUTION

By resolving the 200-N force into components parallel and perpendicular to the bo x wrench BC , Fig. a, the moment can be obtained by adding algebraically the moment s of these two components about point A in accordance with the principle of moments.

a + (MR)A = ©Fd;

MA

A

= 200 sin 15°(0.3 sin 30°) - 200 cos 15°(0.3 cos 30° + 0.3) = - 100.38 N # m = 100 N # m (Clockwi se )

Ans.


F

*4–28.

The connected bar BC is used to increase the lever arm of the crescent wrench as shown. If a clockwise moment of MA = 120 N m is needed to tighten the bolt at A and the force F = 200 N, determine the required extension d in order to develop this moment.

C

#

d

15 30

300 mm B

SOLUTION

A

By resolving the 200-N force into components parallel and perpendicular to the box wrench BC , Fig. a, the moment can be obtained by adding algebraically the moment s of these two components about point A in accordance with the principle of moments.

a + (MR)A = ©Fd; - 120 = 200 sin 15°(0.3 sin 30°) - 200 cos 15°(0.3 cos 30° + d) d

= 0.4016 m = 402 mm

Ans.


F

4–29.

The connected bar BC is used to increase the lever arm of the crescent wrench as shown. If a clockwise moment of MA = 120 N m is needed to tighten the nut at A and the extension d = 300 mm, determine the required force F in order to develop this moment.

C

#

d

15 30

300 mm B

SOLUTION A

By resolving force F into components parallel and perpendicular to the bo x wrench BC , Fig. a, the moment of F can be obtained by addin g al gebraically the moment s of these two components about point A in accordance with the principle of moments.

a + (MR)A = ©Fd;

- 120 = F sin 15°(0.3 sin 30°) - F cos 15°(0.3 cos 30° + 0.3) F

= 239 N

Ans.


F

4–30.

A force F having a magnitude of F = 100 N acts along the diagonal of the parallelepiped. Determine the moment of F about point A, using M A = rB : F and M A = rC : F.

z

C

F

200 mm rC

SOLUTION F = 100 F =

a

400 mm

- 0.4 i + 0.6 j + 0.2 k

5 - 53.5

0.7483

B

b

r B

600 mm

F

6

A

x

i + 80.2 j + 26.7 k N

M A = rB * F =

3

i

j

0 - 53.5

- 0.6 80.2

k

3

0 = 26.7

5 - 16.0

6

#

i - 32.1 k N m

Ans.

Also, M A = rC * F =

i

j

- 0.4 - 53.5

0 80.2

k

0.2 = 26.7

- 16.0 i - 32.1 k N # m

Ans.


y

4–31.

5

6

The force F = 600i + 300 j - 600k N acts at the end of the beam. Determine the moment of the f orce about point A.

z

A

x

O

F

SOLUTION 1.2 m

r = {0.2i + 1.2 j } m MO = r * F =

3

B

0.4 m

i

j

k

0.2 600

1.2 300

0 - 600

#

3

M O = { - 720i + 120 j - 660k} N m

0.2 m

y

Ans.


*4–32.

Determine the moment produced by force FB about point O. Express the result as a Cartesian vector.

z

A

6m F C

420 N F B

780 N

SOLUTION Position Vector and Force Force Vectors: Vectors: Either position vector rOA or rOB can be used to determine the moment of FB about point O. rOA = [6k] m

rOB = [2.5 j] m

B

2.5 m

C

3m

The force vector FB is given by FB = FB u FB = 780

2m

x

(0 - 0)i + (2.5 - 0) j + (0 - 6)k

2 (0 (0 - 0)2 + (2.5 - 0)2 + (0 - 6)2

R

O B

y

= [300 j - 720k] N

Vector Cross Product: The moment of FB about point O is given by

3

i M O = rOA * FB = 0

0

j

0 300

3

k

#

#

Ans.

#

#

Ans.

= [ - 1800i] N m = [ - 1.80i] kN m 6 - 720

or

3

i M O = rOB * FB = 0

0

j

2.5 300

k

3

= [ - 1800i] N m = [ - 1.80i] kN m 0 - 720


4–33.

Determine the moment produced by force FC about point O. .Express the result as a Cartesian vector

z

A

6m F C

420 N F B

780 N

SOLUTION Position Vector and Force Force Vectors: Vectors: Either position vector rOA or rOC can be used to determine the moment of FC about point O. rOA

= {6k} m

rOC

= (2 - 0)i + ( - 3 - 0) j + (0 - 0)k = [2i - 3 j] m

2m 2.5 m

C

3m x

O B

y

The force vector FC is given by FC

= FCuFC = 420 B

(2 - 0)i + ( - 3 - 0) j + (0 - 6)k

2 (2 (2 - 0)2 + ( - 3 - 0)2 + (0 - 6)2

R=

[120i - 180 j - 360k] N

Vector Cross Product: The moment of FC about point O is given by

MO

= rOA * FC =

3

= rOC * FC =

3

i

j

0 120

0 - 180

i

j

2 120

-3 - 180

k

3

#

Ans.

3

#

Ans.

6 = [1080i + 720 j] N m - 360

or

MO

k

0 = [1080i + 720 j] N m - 360


4–34.

Determine the resultant moment produced by forces FB and FC about point O. Express the result as a Cartesian Cartesian vector.

z

A

6m F C

420 N F B

780 N

SOLUTION Position Vector and Force Vectors: Vectors: The position vector rOA and force vectors FB and FC, Fig. a, must be determined first. rOA = {6k} m

2m 2.5 m

C

3m

B B

FB = FB uFB = 780 FC = FCuFC = 420

(0 - 0)i + (2.5 - 0) j + (0 - 6)k

R R

= [300 j - 720k] N

(0 - 0)2 + (2.5 - 0)2 + (0 - 6)2 2 (0 (2 - 0)i + ( - 3 - 0) j + (0 - 6)k

(2 - 0)2 + ( - 3 - 0)2 + (0 - 6)2 2 (2

x

O B

y

= [120i - 180 j - 360k] N

Resultant Moment: The resultant moment of FB and FC about point O is given by MO = rOA * FB + rOA * FC

3

i

= 0 0

j

0 300

k

3 3

i

+ 0 6 - 720 120

j

k

0 - 180

6 - 360

= [ - 720i + 720 j] N # m

3 Ans.


4–35.

■

Using a ring collar the 75-N force can act in the vertical plane at various angles u. Determine the magnitude of the moment it produces about point A, plot the result of M (ordinate) versus u (abscissa) for 0° … u … 180°, and specify the angles that give the maximum and minimum moment.

z

A 2m

1.5 m

SOLUTION

3

i MA = 2

0

j

k

1.5 75 cos u

0 75 sin u

3

y

x 75 N θ

= 112.5 sin u i - 150 sin u j + 150 cos u k MA dMA du

1 2 + 1 - 150 sin 2 + 1150 cos 2 = 2 12 656.25 sin 1 = 112 656.25 sin + 22 5002 112 656.25 212 sin cos 2 = 0 2

= 2 112.5 sin u

sin u cos u = 0; Mmax Mmin

2

2

u

u

u

= 0°, 90°, 180°

- 12

2

u

2

u

2

u

+ 22 500

u

Ans.

= 187.5 N # m at u = 90° = 150 N # m at u = 0°, 180°


*4–36.

The curved rod lies in the x– y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point O.

z

O y B

3m 45

3m

SOLUTION 1m

rAC = {1i - 3 j - 2k} m

A F

rAC = 2 (1)2 + ( - 3)2 + ( - 2)2 = 3.742 m M O = rOC * F =

3

i

j

k

4 1 3.742(80)

0

-2

3 - 3.742 (80)

2 - 3.742 (80)

#

M O = { - 128i + 128 j - 257k} N m

3

80 N

2m x

C

Ans.


4–37.

The curved rod lies in the x– y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point B.

z

O y B

3m 45

SOLUTION rAC = {1i - 3 j - 2k} m

1m

A

rAC = 2 (1) + ( - 3) + ( - 2) = 3.742 m 2

2

3

3m

2

i

M B = rBA * F = 3 cos 45° 1 3.742(80)

j

F

k

(3 - 3 sin 45°) 0 3 2 - 3.742(80) - 3.742(80)

#

M B = { - 37.6i + 90.7 j - 155k} N m

3

80 N

2m x

C

Ans.


4–38.

Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point A. Express the result as a Cartesian vector.

z

3m

A F

400 N

3m B

SOLUTION

x

4m

C y

Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF is equal to the unit vector of the cross product, b = rAC * rBC , Fig. a. Here rAC = (0 - 0)i + (4 - 0) j + (0 - 3)k = [4 j - 3k] m rBC = (0 - 3)i + (4 - 0) j + (0 - 0)k = [ - 3i + 4 j] m

Thus, b = rCA * rCB =

3

i

j

k

0 -3

4 4

-3 0

3

= [12i + 9 j + 12k] m2 Then, uF =

12i + 9 j + 12k b = = 0.6247 i + 0.4685 j + 0.6247 k b 2 122 + 92 + 12 2

And finally F = FuF = 400(0.6247 i + 0.4685 j + 0.6247 k)

= [249.88i + 187.41 j + 249.88 k] N Vector Cross Product: The moment of F about point A is M A = rAC * F =

3

i

j

k

0 249.88

4 187.41

-3 249.88

3

= [1.56i - 0.750 j - 1.00k] kN # m

Ans.


4–39.

Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point B. Express the result as a Cartesian vector.

z

3m

A F

400 N

3m B

SOLUTION

x

C

4m

Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF is equal to the unit vector of the cross product, b = rAC * rBC , Fig. a. Here rAC = (0 - 0)i + (4 - 0) j + (0 - 3)k = [4 j - 3k] m rBC = (0 - 3)i + (4 - 0) j + (0 - 0)k = [ - 3k + 4 j] m

Thus, b = rCA * rCB =

3

3

i

j

k

0 -3

4 4

- 3 = [12i + 9 j + 12k] m2 0

Then, uF =

12i + 9 j + 12k b = = 0.6247 i + 0.4685 j + 0.6247 k b 2 122 + 92 + 122

And finally F = FuF = 400(0.6247 i + 0.4685 j + 0.6247 k)

= [249.88 i + 187.41 j + 249.88 k] N Vector Cross Product: The moment of F about point B is

MB = rBC * F =

3

i

j

k

-3

4 187.41

0 249.88

249.88

3

= [1.00i + 0.750 j - 1.56k] kN # m

Ans.


y

*4–40.

The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point A.

z

A

400 mm B x

SOLUTION

300 mm

Position Vector And Force Vector: 200 mm

rAC = {(0.55 - 0)i + (0.4 - 0) j + ( - 0.2 - 0)k} m

200 mm

C

250 mm

= {0.55i + 0.4 j - 0.2k} m 40

F = 80(cos 30° sin 40° i + cos 30° cos 40° j - sin 30°k) N

30

= (44.53 i + 53.07 j - 40.0k} N

F

80 N

Moment of Force F About Point A: Applying Eq. 4–7, we have MA = rAC * F

=

3

i

j

k

0.55 44.53

0.4 53.07

- 0.2 - 40.0

3

= { - 5.39i + 13.1 j + 11.4k} N # m

Ans.


y

4–41.

The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point B.

z

A

400 mm B x

SOLUTION

300 mm

y

Position Vector And Force Vector: 200 mm

rBC = {(0.55 - 0) i + (0.4 - 0.4) j + ( - 0.2 - 0)k} m

200 mm

C

250 mm

= {0.55i - 0.2k} m 40

F = 80 (cos 30° sin 40°i + cos 30° cos 40° j - sin 30°k) N

30

= (44.53 i + 53.07 j - 40.0k} N

F

80 N

Moment of Force F About Point B: Applying Eq. 4–7, we have MB = rBC * F

=

3

i

j

k

0.55 44.53

0 53.07

- 0.2 - 40.0

3

= {10.6i + 13.1 j + 29.2k} N # m

Ans.


4–42.

Strut AB of the 1-m-diameter hatch door exerts a force of 450 N on point B. Determine the moment of this force about point O.

z

B 30° 0.5 m

F = 450 N O

SOLUTION

y 0.5 m

Position Vector And Force Vector: i

j

j

k

x

k

i

i

F

A

510 - 02 + 11 cos 30° - 02 + 11 sin 30° - 02 6 m = 50.8660 + 0.5 6 m = 510.5 sin 30° - 02 + 10.5 + 0.5 cos 30° - 02 + 10 - 02 6 m = 50.250 + 0.9330 6 m 10 - 0.5 sin 30°2 + 31 cos 30° - 10.5 + 0.5 cos 30°24 + 11 sin 30° - 02 ≤ N = 450 ¢ 2 10 - 0.5 sin 30°2 + 31 cos 30° - 10.5 + 0.5 cos 30°24 + 11 sin 30° - 02 = 5 - 199.82 - 53.54 + 399.63 6 N

rOB =

rOA

30°

j

k

j

i

j

2

i

j

k

2

2

k

Moment of Force F About Point O: Applying Eq. 4–7, we have MO = rOB * F

=

3

i

j

k

0 - 199.82

0.8660 - 53.54

0.5 399.63

= 373i - 99.9 j + 173k N # m

5

6

3 Ans.

Or M O = rOA * F

=

i

j

k

0.250 - 199.82

0.9330 - 53.54

0 399.63

= 373i - 99.9 j + 173k N # m

Ans.


4–43.

The curved rod has a radius of 5 ft. If a force of 60 lb acts at its end as shown, determine the moment of this force about point C .

z

C

5 ft

60°

A

5 ft

y

SOLUTION

60 lb 6 ft

Position Vector and Force Vector:

515 sin 60° - 02 + 15 cos 60° - 52 6 m = 54.330 - 2.50 6 m 16 - 02 + 17 - 5 sin 60°2 + 10 - 5 cos 60°2 ≤ lb = 60 ¢ 2 16 - 02 + 17 - 5 sin 60°2 + 10 - 5 cos 60°2 = 551.231 + 22.797 - 21.346 6 lb

rCA =

j

j

FAB

B

k

x

7 ft

k

i

j

2

i

k

2

j

2

k

Moment of Force F AB About Point C: Applying Eq. 4–7, we have M C = rCA * FAB

=

i

j

k

0 51.231

4.330 22.797

- 2.50 - 21.346

= - 35.4i - 128 j - 222k lb # ft

Ans.


*4–44.

Determine the smallest force F that must be applied alon g the rope in order to cause the curved rod, which has a radius of 5 ft, to fail at the support C . This requires a moment of M = 80 lb ft to be developed at C .

z

#

C

5 ft

60

A

5 ft

y

60 lb

SOLUTION

6 ft

B

Posi ti o n Vector and Force Vector : x

7 ft

rCA = {(5 sin 60° - 0) j + (5 cos 60° - 5)k} m

= {4.330 j - 2.50 k} m FAB = F

a 2 (6(6--0)0) + +(7(7- -5 5inin60°)60°) + +(0(0- -5 5coco 60°) b lb 60°) i

j

s

2

2

s

k

s

s

2

= 0.8539 Fi + 0.3799 F j - 0.3558 Fk Moment of Force F AB About Poi nt C : MC = rCA * FAB

=

3

i

j

k

0 0.8539 F

4.330 0.3799 F

- 2.50 - 0.3558 F

3

= - 0.5909 Fi - 2.135 j - 3.697k Require 80 = 2 (0.5909) 2 + ( - 2.135) 2 + ( - 3.697) 2 F F

= 18.6 lb.

Ans.


4–45.

5

6

A force of F = 6i - 2 j + 1k kN produces a moment of M O = 4i + 5 j - 14k kN m about the origin of coordinates, point O. If the force acts at a point having an x coordinate of x = 1 m, determine the y and z coordinates.

5

6

#

z

F

P

MO z

d

y

O

1m

SOLUTION y

MO = r * F

3 3 i

4i + 5 j - 14k = 1 6

j y

k z

-2

1

x

4 = y + 2z

5 = - 1 + 6z

- 14 = - 2 - 6y y

= 2m

Ans.

z

= 1m

Ans.


4–46.

5

6

The force F = 6i + 8 j + 10k N creates a moment about point O of M O = - 14i + 8 j + 2k N m. If the force passes through a point having an x coordinate of 1 m, determine the y and z coordinates of the point.Also, realizing that MO = Fd, determine the perpendicular distance d from point O to the line of action of F.

5

#

6

z

F

P

MO z

d

y

O

1m

SOLUTION

3 3 i

- 14i + 8 j + 2k = 1

j y

k z

6

8

10

y x

- 14 = 10y - 8z

8 = -10 + 6z 2 = 8 - 6y y

= 1m

Ans.

z

= 3m

Ans.

F

2 ( - 14) + (8) + (2) = 16.25 N # m = 2 (6) + (8) + (10) = 14.14 N

d

=

MO

=

2

2

2

2

16.25 = 1.15 m 14.14

2

2

Ans.


4–47.

Determine the magnitude of the moment of each of the three forces about the axis AB. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach.

z

F 1 = 60 N

F 2 = 85 N

F 3 = 45 N

SOLUTION A

a) Vector Analysis

B

Position Vector and Force Vector:

y

x 1.5 m

5- 1.5 6 m = 5 - 60 6 N

r1 =

j

r2 = r3 = 0

F1

k

F2 =

585 6 N i

F3 =

2m

545 6 N j

Unit Vector Alon g AB Axis:

12 - 02 + 10 - 1.52 2 12 - 02 + 10 - 1.52 i

uAB =

j

2

2

= 0.8i - 0.6 j

Moment of Each Force About AB Axis: Applying Eq. 4–11, we have

1 2 MAB

1

= uAB # r1 * F1 =

3 3 3

1

2

- 0.6 - 1.5

0.8 0 0

0 0 - 60

0

3

= 0.8 - 1.5 - 60 - 0 - 0 + 0 = 72.0 N # m

1 2 MAB

2

= uAB r2 =

1 2 MAB

3

31 21 2 4 #1 * 2 F2

- 0.6

0.8 0 85

0 0

= uAB # r3 * F3 =

1

0.8 0 0

2

- 0.6 0 45

3 3

Ans.

0 0 = 0 0

Ans.

0 0 = 0 0

Ans.

b) Scalar Analysis: Since moment arm from force F2 and F3 is equal to zero,

1 2 = 1 2 MAB

2

MAB

3

=0

Ans.

Moment arm d from force F1 to axis AB is d = 1.5 sin 53.13° = 1.20 m, MAB

1

= F1d = 60 1.20 = 72.0 N # m

Ans.


*4–48.

The flex-headed ratchet wrench is subjected to a force of P = 16 lb, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A.

P



60

10 in.

SOLUTION M =

16(0.75

M =

151 lb # in.

+

A

10 sin 60°)

0.75 in.

Ans.


4–49.

If a torque or moment of 80 lb # in. is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench.

P



60

10 in.

SOLUTION 80

= P(0.75 +

P =

80 9.41

=

A

10 sin 60°)

8.50 lb

0.75 in.

Ans.


4–50.

The chain AB exerts a force of 20 lb on the door at B. Determine the magnitude of the moment of this force along the hinged axis x of the door.

z

3 ft

2 ft

A F = 20 lb

SOLUTION Position Vector and Force Vector: i

rOB

i

k

i

k

j

j

k

20˚ 3 ft

k

i

j

2

j

x

k

2

i

y

4 ft

513 - 02 + 14 - 02 6 ft = 53 + 4 6 ft = 510 - 02 + 13 cos 20° - 02 + 13 sin 20° - 02 6 ft = 52.8191 + 1.0261 6 ft 13 - 02 + 10 - 3 cos 20°2 + 14 - 3 sin 20°2 ≤ lb = 20 ¢ 2 13 - 02 + 10 - 3 cos 20°2 + 14 - 3 sin 20°2 = 511.814 - 11.102 + 11.712 6 lb

rOA =

F

B

O

2

k

Moment of Force F About x Axis: The unit vector along the x axis is i. Applying

Eq. 4–11, we have Mx

= i # rOA * F

3

1

1 = 3 11.814

2

0 0 - 11.102

31

0 4 11.712

2 1

21 24 - 0 + 0

= 1 0 11.712 - - 11.102 4 = 44.4 lb # ft

3

Ans.

Or Mx

= i # rOB * F

3

1

1 = 0 11.814

3

2

0 2.8191 - 11.102

1

0 1.0261 11.712

2 1

3

21

= 1 2.8191 11.712 - - 11.102 1.0261 = 44.4 lb # ft

24 - 0 + 0 Ans.


4–51.

The hood of the automobileis supported by the strut AB,which exerts a force of F = 24lb on the hood. Determine the moment of this force about the hinged axis y.

z

B

F

4 ft A x

SOLUTION

2 ft

2 ft

4 ft

y

r = {4i} m F = 24

a 2 ( --2)2 ++ 2(2)+ 4+ (4) b i

2

j

k

2

2

= { - 9.80 i + 9.80 j + 19.60k} lb

3

0 My = 4 - 9.80

1 0 9.80

#

My = { - 78.4 j} lb ft

3

0 = - 78.4 lb ft 0 19.60

#

Ans.


*4–52.

Determine the magnitude of the moments of the force F about the x, y , and z axes. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach.

z

A

4 ft y

SOLUTION x

a) Vector Analysis

3 ft C

PositionVector: rAB = {(4 - 0) i + (3 - 0) j + ( - 2 - 0)k} ft = {4 i + 3 j - 2k} ft

2 ft

Moment of Force F About x,y, and z Axes: The unit vectors along x, y, and z axes are i, j, and k respectively. Applying Eq. 4–11, we have Mx

= i # (r

AB

0 3 12

12 j

3k} lb

Ans.

= j # (rAB * F) 0 = 4 4

1 3 12

0 -2 -3

= 0 - 1[4( - 3) - (4)( - 2)] + 0 = 4.00 lb # ft Mz

{4i

0 -2 -3

= 1[3( - 3) - (12)( - 2)] - 0 + 0 = 15.0 lb # ft My

F

* F)

3 3 3 3 3 3

1 = 4 4

B

Ans.

= k # (rAB * F) 0 = 4 4

0 3 12

1 -2 -3

= 0 - 0 + 1[4(12) - (4)(3)] = 36.0 lb # ft

Ans.

b) ScalarAnalysis Mx

= © Mx ;

Mx

= 12(2) - 3(3) = 15.0 lb # ft

Ans.

My

= © My ;

My

= - 4(2) + 3(4) = 4.00 lb # ft

Ans.

Mz

= © Mz ;

Mz

= - 4(3) + 12(4) = 36.0 lb # ft

Ans.


4–53.

Determine the moment of the force F about an axis extending between A and C . Express the result as a Cartesian vector.

z

A

4 ft y

SOLUTION x

PositionVector:

3 ft C

rCB = { - 2k} ft rAB = {(4 - 0)i + (3 - 0) j + ( - 2 - 0)k} ft = {4i + 3 j - 2k} ft

2 ft

Unit Vector Alon g AC Axis: uAC =

B

(4 - 0)i + (3 - 0) j

2 (4 - 0)

2

+ (3 - 0)2

F

= 0.8i + 0.6 j

{4i

12 j

3k} lb

Moment of Force F About AC Axis: With F = {4i + 12 j - 3k} lb, applying Eq. 4–7,

we have MAC

= uAC # (rCB * F)

3 3

0.8 = 0 4

0.6 0 12

0 -2 -3

= 0.8[(0)( - 3) - 12( - 2)] - 0.6[0( - 3) - 4( - 2)] + 0 = 14.4 lb # ft Or MAC

= uAC # (rAB * F)

3 3

0.8 = 4 4

0.6 3 12

0 -2 -3

= 0.8[(3)( - 3) - 12( - 2)] - 0.6[4( - 3) - 4( - 2)] + 0 = 14.4 lb # ft Expressing M AC as a Cartesian vector yields M AC = MAC uAC

= 14.4(0.8 i + 0.6 j) = {11.5i + 8.64 j} lb # ft

Ans.


4–54.

The board is used to hold the end of a four-way lu g wrench in the position shown when the man applies a force of F = 100 N. Determine the magnitude of the moment produced by this force about the x axis. Force F lies in a vertical plane.

z

F 60

250 mm y

SOLUTION

x

250 mm

Vector Analysi s Moment About the x A xi s : The position vector r AB, Fig. a, will be used to determine the moment of F about the x axis. rAB = (0.25 - 0.25) i + (0.25 - 0) j + (0 - 0)k = {0.25 j} m

The force vector F, Fig. a, can be written as F = 100(cos 60° j - sin 60°k) = {50 j - 86.60k} N

Knowing that the unit vector of the x axis i s i, the magnitude of the moment of F about the x axis is given by Mx

3

1

= i # rAB * F = 0 0

0 0.25 50

0 0 - 86.60

3

= 1[0.25( - 86.60) - 50(0)] + 0 + 0 = - 21.7 N # m

Ans.

The negative sign indicates that M x is directed towards the negative x axis. Scalar Analysi s This problem can be solved by summing the moment about the x axis Mx

= ©Mx;

Mx

= - 100 sin 60°(0.25) + 100 cos 60°(0) = - 21.7 N # m Ans.


4–55.

The board is used to hold the end of a four-way lug wrench in position. If a torque of 30 N m about the x axis is required to tighten the nut, determine the required magnitude of the force F that the man’s foot must apply on the end of the wrench in order to turn it. Force F lies in a vertical plane.

#

z

F 60

250 mm y

x

250 mm

SOLUTION Vector Analysi s Moment About the x A xi s : The position vector r AB, Fig. a, will be used to determine the moment of F about the x axis. rAB = (0.25 - 0.25) i + (0.25 - 0) j + (0 - 0)k = {0.25 j} m

The force vector F, Fig. a, can be written as F = F(cos 60° j - sin 60°k) = 0.5F j - 0.8660 Fk

Knowing that the unit vector of the x axis i s i, the magnitude of the moment of F about the x axis is given by

Mx

= i#r

AB

1 *F = 0 0 3

0 0.25 0.5F

0 3 0 - 0.8660 F

= 1[0.25( - 0.8660 F) - 0.5F(0)] + 0 + 0 = - 0.2165 F

Ans.

The negative sign indicates that M x is directed towards the negative x axis. The magnitude of F required to produce Mx = 30 N m can be determined from

#

30 = 0.2165 F F = 139 N

Ans.

Scalar Analysi s This problem can be solved by summing the moment about the x axis Mx

= ©Mx;

- 30 = - F sin 60°(0.25) + F cos 60°(0) F = 139 N

Ans.


*4–56.

The cutting tool on the lathe exerts a force F on the shaft as shown. Determine the moment of this force about the y axis of the shaft.

z

F

{6i

4 j

7k} kN

SOLUTION My

= uy # (r * F) =

My

3

0 30 cos 40° 6

1 0 -4

0 30 sin 40° -7

= 276.57 N # mm = 0.277 N # m

3

30 mm 40 y

Ans.

x


4–57. z

The cutting tool on the lathe exerts a force F on the shaft as shown. Determine the moment of this force about the x and z axes.

F  {6i  4 j  7k} kN

30 mm 40

SOLUTION y

Moment Abo u t x and y A x es : Pos ition v ectors r x and r z s hown in Fig. a can be conveniently u s ed in computing the moment of F about x and z ax es re spectiv ely. rx = {0.03 sin 40° k} m

x

rz = {0.03 cos 40° i} m

Knowing that the unit vectors for x and z axes are i and k respectively. Thus, the magnitudes of moment of F about x and z axes are given by

Mx

3

1

= i # rx * F = 0 6

0 0 -4

0 0.03 sin 40° -7

3

= 1[0( - 07) - ( - 4)(0.03 sin 40°)] - 0 + 0 = 0.07713 kN # m = 77.1 N # m

Mz

= k#r

z

3

0 * F = 0.03 cos 40° 6

0 0 -4

1 0 7

3

= 0 - 0 + 1[0.03 cos 40°( - 4) - 6(0)] = - 0.09193 kN # m = - 91.9 N # m Thus, Mx

= Mxi = {77.1i} N # m

Mz

= Mzk = { - 91.9 k} N # m

Ans.


4–58.

If the tension in the cable is F = 140 lb, determine the agnitude of the moment produced by this force about the inged axis, CD, of the panel.

z

4 ft

4 ft B

SOLUTION oment About the CD Axis: Either position vector rCA or rDB , Fig. a, can be used o determine the moment of F about the CD axis.

6 ft

C

6 ft A

rCA = (6 - 0)i + (0 - 0) j + (0 - 0)k = [6i] ft

D

F

6 ft

x

rDB = (0 - 0)i + (4 - 8) j + (12 - 6)k = [ - 4 j + 6k] ft

Referring to Fig. a, the force vector F can be written as F = FuAB = 140

C

(0 - 6)i + (4 - 0) j + (12 - 0)k

2 (0 - 6)

2

+ (4 - 0)2 + (12 - 0)2

S

= [ - 60i + 40 j + 120k] lb

he unit vector uCD, Fig. a, that specifies the direction of the CD axis is given by uCD =

(0 - 0)i + (8 - 0) j + (6 - 0)k

2 (0 - 0)

2

+ (8 - 0) + (6 - 0) 2

2

4 5

3 5

= j + k

hus, the magnitude of the moment of F about the CD axis is given by

MCD

= uCD # rCA * F =

5

0 6 - 60

=0-

4 5 0 40

3 5 0 120

5

4 3 [6(120) - ( - 60)(0)] + [6(40) - ( - 60)(0)] 5 5

= - 432 lb # ft

Ans.

or

MCD

= uCD # rDB * F =

5

0 0 - 60

=0-

4 5 -4 40

3 5 6 120

5

4 3 [0(120) - ( - 60)(6)] + [0(40) - ( - 60)( - 4)] 5 5

= - 432 lb # ft he negative sign indicates that M CD acts in the opposite sense to that of uCD. hus,

#

MCD = 432 lb ft

Ans.


y

4–59.

Determine the magnitude of force F in cable AB in order to produce a moment of 500 lb ft about the hinged axis CD, which is needed to hold the panel in the position shown.

#

z

4 ft

4 ft B

SOLUTION

D

F

Moment About the CD Axis: Either position vector rCA or rCB, Fig. a, can be used to determine the moment of F about the CD axis. rCA = (6 - 0)i + (0 - 0) j + (0 - 0)k = [6i]ft

6 ft

C

6 ft A

6 ft

x

rCB = (0 - 0)i + (4 - 0) j + (12 - 0)k = [4 j + 12k]ft

Referring to Fig. a, the force vector F can be written as F = FuAB = F

C

(0 - 6)i + (4 - 0) j + (12 - 0)k

2 (0 - 6)

2

+ (4 - 0)2 + (12 - 0)2

S

= -

3 2 6 F i + F j + Fk 7 7 7

The unit vector uCD, Fig. a, that specifies the direction of the CD axis is given by uCD =

(0 - 0)i + (8 - 0) j + (6 - 0)k

2 (0 - 0)

+ (8 - 0)2 + (6 - 0)2

2

4 5

3 5

= j + k

Thus, the magnitude of the moment of F about the CD axis is required to be M CD = 500 lb ft. Thus,

#

MCD

= uCD # rCA * F

5

500 =

0 6 3 - F 7

- 500 = 0 F

4 5 0 2 F 7

5

3 5 0 6 F 7

B ¢ ≤ ¢ ≤ R B ¢ ≤ ¢ ≤ R

4 6 6 F 5 7

-

-

3 F (0) 7

+

3 2 6 F 5 7

-

-

3 F (0) 7

= 162 lb

Ans.

or MCD

= uCD # rCB * F

500 =

5

0 0 3 - F 7

- 500 = 0 F

4 5 4 2 F 7

5

B ¢ ≤ ¢ ≤ R B ¢ ≤ ¢ ≤ R

4 6 0 5 7

= 162 lb

3 5 12 6 F 7 F

-

-

3 F (12) 7

+

3 2 0 F 5 7

-

-

3 F (4) 7

Ans.


y

*4–60.

The force of F = 30 N acts on the bracket as shown. Determine the moment of the force about the a - a axis of the pipe.Also, determine the coordinate direction angles of F in order to produce the maximum moment about the a - a axis. What is this moment?

z

F = 30 N y

45

°

60

°

50 mm

60

°

x

100 mm

100 mm

SOLUTION

1

2

F = 30 cos 60° i + cos 60° j + cos 45° k

5 = 5 - 0.1

6

a

= 15 i + 15 j + 21.21 k N

r

6

i + 0.15 k m

u = j

Ma

3

0 = - 0.1 15

1 0 15

3

a

0 0.15 = 4.37 N m 21.21

#

Ans.

F must be perpendicular to u and r. uF =

0.15 0.1 i + k 0.1803 0.1803

= 0.8321 i + 0.5547 k a

= cos-1 0.8321 = 33.7°

Ans.

b

= cos-1 0 = 90°

Ans.

g

= cos-1 0.5547 = 56.3°

Ans.

= 30 0.1803 = 5.41 N # m

Ans.

M


4–61. z

The pipe a ssembly is secured on the wall by the two brackets. If the flower pot has a wei ght of 50 lb, determine the magnitude of the moment produced by the weight about the x, y, and z axes.

4 ft A O

60



3 ft 4 f t 3 f t

SOLUTION

30

x



B

Moment About x , y, and z A x es : Position vectors r x, r y, and rz s hown in Fig. a can be conveniently used in computing the moment of W about x, y, and z axes. r x = {(4 + 3 cos 30°) sin 60° j + 3 sin 30°k} ft

= {5.7141 j + 1.5k} ft r y = {(4 + 3 cos 30°) cos 60°i + 3 sin 30°k} ft

= {3.2990 i + 1.5k} ft rz = {(4 + 3 cos 30°) cos 60°i + (4 + 3 cos 30°) sin 60° j} ft

= {3.2990 i + 5.7141 j} ft The Force vector is given by W = W( - k) = { - 50 k} lb

Knowing that the unit vectors for x, y, and z axes are i, j, and k respectively. Thus, the magnitudes of the moment of W about x, y, and z axes are given by

M x

= i#r

x

3

1 *W= 0 0

0 5.7141 0

0 1.5 - 50

3

= 1[5.7141( - 50) - 0(1.5)] - 0 + 0 = - 285.70 lb # ft = - 286 lb # ft

My

= j#r

y

3

0 * W = 3.2990 0

1 0 0

0 1.5 - 50

Ans.

3

= 0 - 1[3.2990( - 50) - 0(1.5)] + 0 = 164.95 lb # ft = 165 lb # ft

Mz

= k#r

z

3

0 * W = 3.2990 0

0 5.7141 0

1 0 -5

Ans.

3

= 0 - 0 + 1[3.2990(0) - 0(5.7141)] = 0Ans. The negative sign indicates that M x is directed towards the negative x axis.


y

4–62.

The pipe assembly is secured on the wall by the two brackets. If the flower pot has a weight of 50 lb, determine the magnitude of the moment produced by the weight about the OA axis.

4 ft A O

60

3 ft 4 f t 3 f t

SOLUTION

30

x

B

About the OA Axis: The coordinates of point B are [(4 + 3 cos 30°) cos 60°, (4 + 3 cos 30°) sin 60°, 3 sin 30°] ft = (3.299, 5.714, 1.5) ft. Either position vector rOB or r AB can be used to determine the moment of W about the OA axis. Moment

rOB = (3.299 - 0)i + (5.714 - 0) j + (1.5 - 0)k = [3.299 i + 5.714 j + 1.5k] ft rAB = (3.299 - 0)i + (5.714 - 4) j + (1.5 - 3)k = [3.299i + 1.714 j - 1.5k] ft

Since W is directed towards the negative z axis, we can write W = [ - 50k] lb The unit vector uOA, Fig. a, that specifies the direction of the OA axis is given by uOA =

(0 - 0)i + (4 - 0) j + (3 - 0)k

2 (0 - 0)

2

+ (4 - 0) + (3 - 0) 2

2

4 5

3 5

= j + k

The magnitude of the moment of W about the OA axis is given by

MOA

5

0

= uOA # rOB * W = 3.299

=0-

0

4 5 5.714 0

3 5 1.5 - 50

5

4 3 [3.299( - 50) - 0(1.5)] + [3.299(0) - 0(5.714)] 5 5

= 132 lb # ft

Ans.

or

MOA

5

0

= uOA # rAB * W = 3.299

=0-

0

4 5 1.714 0

3 5 - 1.5 - 50

5

4 3 [3.299( - 50) - 0( - 1.5)] + [3.299(0) - 0(1.714)] 5 5

= 132 lb # ft

Ans.


y

4–63.

The pipe assembly is secured on the wall by the two brackets. If the frictional force of both brackets can resist a maximum moment of 150 lb ft, determine the largest weight of the flower pot that can be supported by the assembly without causing it to rotate about the OA axis.

z

#

4 ft A O

60

SOLUTION

3 ft 4 f t 3 f t

Moment About the OA Axis: The coordinates of point B are

x

30

B

[(4 + 3 cos 30°) cos 60°, (4 + 3 cos 30°) sin 60°, 3 sin 30°]ft = (3.299, 5.174, 1.5) ft . Either position vector rOB or rOC can be used to determine the moment of W about the OA axis. rOA = (3.299 - 0)i + (5.714 - 0) j + (1.5 - 0)k = [3.299 i + 5.714 j + 1.5k] ft rAB = (3.299 - 0)i + (5.714 - 4) j + (1.5 - 3)k = [3.299 i + 1.714 j - 1.5k] ft

Since W is directed towards the negative z axis, we can write W = - Wk The unit vector uOA, Fig. a, that specifies the direction of the OA axis is given by (0 - 0)i + (4 - 0) j + (3 - 0)k

2 (0 - 0)

uOA =

2

+ (4 - 0)2 + (3 - 0)2

4 5

3 5

= j + k

Since it is required that the magnitude of the moment of W about the OA axis not exceed 150 ft lb, we can write

#

MOA

= uOA # rOB * W

5

0

150 = 3.299 0

150 = 0 W

4 5 5.714 0

3 5 1.5

-W

5

4 3 [3.299( - W) - 0(1.5)] + [3.299(0) - 0(5.714)] 5 5

= 56.8 lb

Ans.

or MOA

= uOA # rAB * W

5

0

150 = 3.299 0

150 = 0 -

W

4 5 5.714 0

3 5 0

-W

5

4 3 [3.299( - W) - 0(0)] + [3.299(0) - 0(5.714)] 5 5

= 56.8 lb

Ans.


y

*4–64

The wrench A is used to hold the pipe in a stationary position while wrench B is used to tighten the elbow fitting. If F B = 150 N, determine the magnitude of the moment produced by this force about the y axis. Also, what is the magnitude of force F A in order to counteract this moment?

z

50 mm 50 mm

y

SOLUTION Vector Analysi s Moment of F B About the y A xi s : The position vector rCB, Fig. a, will be used to determine the moment of FB about the y axis. rCB = ( - 0.15 - 0) j + (0.05 - 0.05) j + ( - 0.2598 - 0)k = { - 0.15i - 0.2598 k} m

x

300 mm

300 mm 30



30



F A 135



120



A

Referring to Fig. a, the force vector FB can be written a s

B

FB

FB = 150(co s 60°i - sin 60°k) = {75i - 129.90 k} N

Knowing that the unit vector of the y axis is j, the magnitude of the moment of FB about the y axis is given by My

3

0

1 0 0

= j # rCB * FB = - 0.15 75

0

- 0.2598 -129.90

3

= 0 - 1[ - 0.15( - 129.90) - 75( - 0.2598)] + 0

= - 38.97 N # m = 39.0 N # m

Ans.

The negative sign indicates that M y is directed towards the negative y axis. Moment of F A About the y A xi s : The position vector rDA, Fig. a, will be used to determine the moment of F A about the y axis. rDA = (0.15 - 0)i + [ - 0.05 - ( - 0.05)] j + ( - 0.2598 - 0)k = {0.15i - 0.2598 k} m

Referring to Fig. a, the force vector F A can be written a s FA = F A( - cos 15°i + sin 15°k) = - 0.9659 F Ai + 0.2588 F Ak Since the moment of F A about the y axis is required to counter that of FB about the same axis, F A must produce a moment of equal ma gnitude but in the opposite sense to that of F A. Mx

= j # rDA * FB

+ 0.38.97 =

3

0 0.15 - 0.9659 FA

1 0 0

0 - 0.2598 0.2588 FA

3

+ 0.38.97 = 0 - 1[0.15(0.2588 FA) - ( - 0.9659 FA)( - 0.2598)] + 0 FA

= 184 N

Ans.

Scalar Analysi s This problem can be solved by first taking the moments of FB and then F A about the y axis. For FB we can write My

= ©My;

My

= - 150 cos 60°(0.3 co s 30°) - 150 sin 60°(0.3 sin 30°) = - 38.97 N # m

Ans.

The moment of F A, about the y axis also must be equal in ma gnitude but opposite in sense to that of FB about the same axis

©MPearson = FA Inc., -F My©=2013 38.97 cos 15°(0.3 co s 30°) sin 30°) This publication is protected by y; A sin Education, Upper Saddle River, NJ. All15°(0.3 rights reserved. Copyright and written prohibited reproduction, storage in a retrieval system, FA permission = 184 N should be obtained from the publisher prior to any Ans. or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4–65.

The wrench A is used to hold the pipe in a stationary position while wrench B is used to tighten the elbow fitting. Determine the magnitude of force FB in order to develop a torque of 50 N m about the y axis. Also, what is the required magnitude of force FA in order to counteract this moment?

z

50 mm 50 mm

#

y

x

SOLUTION Vector Analysi s Moment of F B About the y A xi s : The position vector rCB, Fig. a , will be used to determine the moment of FB about the y axis. rCB = ( - 0.15 - 0)i + (0.05 - 0.05) j + ( - 0.2598 - 0)k = { - 0.15i - 0.2598 k} m

300 mm

300 mm 30



F A

30



135



120



A

B

FB

Referring to Fig. a, the force vector FB can be written a s FB = FB(cos 60°i - sin 60°k) = 0.5FBi - 0.8660 FBk

Knowing that the unit vector of the y axis is j, the moment of FB about the y axis is required to be equal to - 50 N m, which is given by

#

My

= j # rCB * FB

†

0

- 50i = - 0.15 0.5FB

1 0 0

†

0

- 0.2598 - 0.8660 FB

- 50 = 0 - 1[ - 0.15( - 0.8660 FB) - 0.5FB( - 0.2598)] + 0 FB = 192 N

Ans.

Moment of F A About the y A xi s : The po s ition v ector rDA , Fi g. a, will be u s ed to determine the moment of FA about the y a x i s. rDA = (0.15 - 0)i + [ - 0.05 - ( - 0.05)] j + ( - 0.2598 - 0)k = { - 0.15i - 0.2598 k} m

Referring to Fig. a, the force vector F A can be written a s FA = FA( - cos 15°i + sin 15°k) = - 0.9659 FAi + 0.2588 FAk Since the moment of F A about the y axis i s required to produce a countermoment of 50 N m about the y axis, we can write

#

My

= j # rDA * FA

†

0 50 = 0.15 - 0.9659 FA

1 0 0

0 - 0.2598 0.2588FA

†

50 = 0 - 1[0.15(0.2588 FA) - ( - 0.9659 FA)( - 0.2598)] + 0 FA = 236 N m

#

Ans.

Scalar Analysi s This problem can be solved by first taking the moments of FB and then F A about the y axis. For FB we can write My

= ©My;

- 50 = - FB cos 60°(0.3 co s 30°) - FB sin 60°(0.3 sin 30°) FB = 192 N

Ans.

For F A, we can write My

= ©My;

50 = FA cos 15°(0.3 co s 30°) - FA sin 15°(0.3 sin 30°) FA = 236 N

Ans.


4–66. z

The A-frame is being hoisted into an upright position by the vertical force of F = 80 lb. Determine the moment of this force about the y axis when the frame is in the position shown.

F C

SOLUTION

A

Using x ¿ , y ¿ , z:

x¿

6 ft

F = 80 k

3

- sin 30° My = - 6 cos 15° My

0

cos 30° 3 0

x

3

6 ft

30

uy = - sin 30° i ¿ + cos 30° j ¿ rAC = - 6 cos 15°i ¿ + 3 j ¿ + 6 sin 15° k

15 y B y¿

0 6 sin 15° = - 120 + 401.53 + 0 80

= 282 lb # ft

Ans.

Also, using x, y, z: Coordinates of point C : x

= 3 sin 30° - 6 cos 15° cos 30° = -3.52 ft

y

= 3 cos 30° + 6 cos 15° sin 30° = 5.50 ft

z

= 6 sin 15° = 1.55 ft

rAC = - 3.52 i + 5.50 j + 1.55 k F = 80 k

3

0 My = - 3.52 0

1 5.50 0

3

0 1.55 = 282 lb ft 80

#

Ans.


4–67.

#

A twist of 4 N m is applied to the handle of the screwdriver. Resolve this couple moment into a pair of couple forces F exerted on the handle and P exerted on the blade.

–F –P

P

5 mm

4 N·m

30 mm

F

SOLUTION

For the handle MC

= © Mx ;

1 2=4

F 0.03

F

= 133 N

Ans.

For the blade, MC

= © Mx ;

1 2=4

P 0.005

P

= 800 N

Ans.


*4–68.

The ends of the triangular plate are subjected to three couples. Determine the plate dimension d so that the resultant couple is 350 N m clockwise.

100 N

#

600 N

d 100 N 30

°

600 N

SOLUTION a + MR = © MA ;

1

2

1

2

200 N

- 350 = 200 d cos 30° - 600 d sin 30° - 100d d

= 1.54 m

200 N

Ans.


4–69.

The caster wheel is subjected to the two couples. Determine the forces F that the bearings exert on the shaft so that the resultant couple moment on the caster is zero.

500 N

F

A

40 mm B

F

100 mm

SOLUTION a + © MA = 0;

45 mm

500(50) - F (40) = 0 F

= 625 N

Ans.

50 mm

500 N


4–70.

Two couples act on the beam. If F = 125 lb , determine the resultant couple moment.

200 lb

F

30



1.5 ft

1.25 ft

SOLUTION

F

30



200 lb

125 lb couple i s resolved in to their horizontal and vertical components as shown in Fig. a.

2 ft

a + (MR)C = 200(1.5) + 125 cos 30° (1.25)

= 435.32 lb # ft = 435 lb # ftd

Ans.


4–71.

Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is 450 lb # ft, counterclockwise. Where on the beam does the resultant couple moment act?

200 lb

F 

30 1.5 ft

1.25 ft

F



30

200 lb 2 ft

SOLUTION a + MR

= ©M

;

450

=

F =

200(1.5) 139 lb

+ Fcos

30°(1.25) Ans.

The resultant couple moment is a free vector. It can act at any point on the beam.


*4–72.

Friction on the concrete s urface create s a couple moment of MO = 100 N m on the blade s of the trowel. Determine the ma g nitude of the couple force s so that the re s ultant couple moment on the trowel i s zero. The forces lie in the horizontal plane and act perpendicular to the handle of the trowel.

#

–F

750 mm

F

MO

SOLUTION Co u ple Moment : The couple moment of F about the vertical axis is MC = F(0.75) = 0.75F. Since the resultant couple moment about the vertical axis is

required to be zero, we can write (Mc)R = © Mz;

0 = 100 - 0.75F

F

= 133 N

1.25 mm

Ans.


4–73.

The man tries to open the valve by applyin g the couple forces of F = 75 N to the wheel. Determine the couple moment produced.

150 mm

150 mm

F

SOLUTION a + Mc = © M;

Mc

= - 75(0.15 + 0.15) = - 22.5 N # m = 22.5 N # m b

Ans. F


4–74.

#

If the valve can be opened with a couple moment of 25 N m, determine the required magnitude of each couple force which mu st be applied to the wheel.

150 mm

150 mm

F

SOLUTION a + Mc = © M;

- 25 = - F(0.15 + 0.15) F = 83.3 N

Ans. F


4–75.

When the engine of the plane is running, the vertical reaction that the ground exerts on the wheel at A is measured as 650 lb. When the engine is turned off, however, the vertical reactions at A and B are 575 lb each. The difference in readings at A is caused by a couple acting on the propeller when the engine is running. This couple tends to overturn the plane counterclockwise, which is opposite to the propeller’s clockwise rotation. Determine the magnitude of this couple and the magnitude of the reaction force exerted at B when the engine is running.

A

B 12 ft

SOLUTION

When the engine of the plane is turned on, the resulting couple moment exerts an additional force of F = 650 - 575 = 75.0 lb on wheel A and a lesser the reactive force on wheel B of F = 75.0 lb as well. Hence, M

= 75.0 12 = 900 lb # ft

1 2

Ans.

The reactive force at wheel B is RB

= 575 - 75.0 = 500 lb

Ans.


*4–76.

Determine the magnitude of the couple force F so that the resultant couple moment on the crank is zero.

150 lb

–F

5 in. 30

30

30 5 in.

150 lb

30

45

4 in. 45 4 in.

SOLUTION

F

By resolving F and the 150-lb couple into components parallel and perpendicular to the lever arm of the crank, Fig. a, and summing the moment of these two force components about point A, we have a + (MC)R = © MA;

0 = 150 cos 15°(10) - F cos 15°(5) - F sin 15°(4) - 150 sin 15°(8) Ans. F = 194 lb

Note: S ince the line of action of the force component parallel to the le ver arm of the crank passes through point A, no moment is produced about this point.


4–77.

Two couples act on the beam a s shown. If F = 150 lb, determine the resultant couple moment.

–F

5

3 4

200 lb 1.5 ft 200 lb 5

SOLUTION

F

150 lb couple i s resolved into their horizontal and vertical components as shown in Fig. a a + (MR)c = 150

3

4

4 ft

a 45 b (1.5) + 150 a 35 b (4) - 200(1.5)

= 240 lb # ftd

Ans.


4–78.

Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is 300 lb ft counterclockwise. Where on the beam does the resultant couple act?

–F

5

#

3 4

200 lb 1.5 ft 200 lb 5

F

SOLUTION a + (MC)R =

4 ft

3 4 F(4) + F(1.5) - 200(1.5) = 300 5 5 F

= 167 lb

Resultant couple can act anywhere.

3

4

Ans. Ans.


4–79.

If F = 200 lb, determine the resultant couple moment.

2 ft

2 ft

F

5

4 3

B

30 2 ft

150 lb 150 lb

2 ft

SOLUTION

30

a) By resolving the 150-lb and 200-lb couples into their x and y components, Fig. a, the couple moments (MC)1 and (MC)2 produced by the 150-lb and 200-lb couples, respectively, are given by

#

5

4 3

F

2 ft A

#

a + (MC)1 = - 150 cos 30°(4) - 150 sin 30°(4) = - 819.62 lb ft = 819.62 lb ftb 4 3 (2) + 200 (2) = 560 lb ft a + (MC)2 = 200 5 5

ab

ab

#

Thus, the resultant couple moment can be determined from a + (MC)R = (MC)1 + (MC)2

= - 819.62 + 560 = - 259.62 lb # ft = 260 lb # ft (Clockwise )

Ans.

b) By resolving the 150-lb and 200-lb couples into their x and y components, Fig. a, and summing the moments of these force components algebraically about point A, a + (MC)R = © MA ; (MC)R = - 150 sin 30°(4) - 150 cos 30°(6) + 200

- 200

a 45 b (2) + 200 a 35 b (6)

a 35 b (4) + 200 a 45 b (0) + 150 cos 30°(2) + 150 sin 30°(0) = - 259.62 lb # ft = 260 lb # ft (Clockwise )

Ans.


*4–80.

Determine the required magnitude of force F if the resultant couple moment on the frame is 200 lb ft, clockwise.

2 ft

#

2 ft

F

5

4 3

B

30 2 ft

150 lb 150 lb

2 ft

SOLUTION

30

By resolving F and the 150-lb couple into their x and y components, Fig. a, the couple moments (MC)1 and (MC)2 produced by F and the 150-lb couple, respectively, are given by a + (MC)1 = F

5

4 3

F

2 ft A

a 45 b (2) + a 35 b (2) = 2.8 F

F

#

#

a + (MC)2 = - 150 cos 30°(4) - 150 sin 30°(4) = - 819.62 lb ft = 819.62 lb ftb

#

The resultant couple moment acting on the beam is required to be 200 lb ft, clockwise. Thus, a + (MC)R = (MC)1 + (MC)2

- 200 = 2.8F - 819.62 F

= 221 lb

Ans.


4–81.

Two couples act on the cantilever beam. If F = 6 kN, determine the resultant couple moment.

3m

3m

5 kN 3

A

SOLUTION

30

F

5

4

B

0.5 m

30

0.5 m 5

4 F

3

5 kN

a) By resolving the 6-kN and 5-kN couples into their x and y components, Fig. a,the couple moments (Mc)1 and (M c)2 produced by the 6-kN and 5-kN couples, respectively, are given by

#

a + (MC)1 = 6 sin 30°(3) - 6 cos 30°(0.5 + 0.5) = 3.804 kN m a + (MC)2 = 5

a 35 b (0.5 + 0.5) - 5 a 45 b (3) = - 9 kN # m

Thus, the resultant couple moment can be determined from (MC)R = (MC)1 + (MC)2

= 3.804 - 9 = - 5.196 kN # m = 5.20 kN # m (Clockwise )

Ans.

b) By resolving the 6-kN and 5-kN couples into their x and y components, Fig. a, and summing the moments of these force components about point A, we can write

a 35 b (0.5) + 5 a 45 b (3) - 6 cos 30°(0.5) - 6 sin 30°(3) 3 4 + 6 sin 30°(6) - 6 cos 30°(0.5) + 5 a b (0.5) - 5 a b (6) 5 5

a + (MC)R = © MA ;

(MC)R = 5

= - 5.196 kN # m = 5.20 kN # m (Clockwise )

Ans.


4–82.

Determine the required magnitude of force F, if the resultant couple moment on the beam is to be zero.

3m

3m

5 kN 3

A

SOLUTION

30

F

5

4

B

0.5 m

30

0.5 m 5

4 F

3

5 kN

By resolving F and the 5-kN couple into their x and y components, Fig. a, the couple moments (M c)1 and (M c)2 produced by F and the 5-kN couple, respectively, are given by a + (MC)1 = F sin 30°(3) - F cos 30°(1) = 0.6340 F a + (MC)2 = 5

a 35 b (1) - 5 a 45 b (3) = - 9 kN # m

The resultant couple moment acting on the beam is required to be zero.Thus, (MC)R = (MC)1 + (MC)2 0 = 0.6340 F - 9 F

= 14.2 kN # m

Ans.


4–83.

Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4–13, and (b) summing the moment of each force about point O. Take F = 25k N.

z

5 6

O y

300 mm

200 mm F

150 mm

SOLUTION B

(a) MC = rAB * (25k)

3 3 3 33 i

= - 0.35 0

MC

j

k

- 0.2

0 25

0

x

400 mm

Ans.

(b) MC = rOB * (25k) + rOA * ( - 25k)

= 0.3 0

j

0.2 0

k

i

0 + 0.65 25 0

j

k

0.4 0

0 - 25

MC = (5 - 10)i + (-7.5 + 16.25) j

#

MC = { - 5i + 8.75 j} N m

200 mm A

= { - 5i + 8.75 j} N # m

i

F

3 Ans.


*4–84.

If the couple moment acting on the pipe has a magnitude of 400 N m, determine the magnitude F of the vertical force applied to each wrench.

#

z

O y

300 mm

200 mm F

150 mm

SOLUTION B

M C = rAB * (Fk)

3

i

= - 0.35

j

k

- 0.2

0

0

F

0

3

x

400 mm

#

F

2 ( - 0.2 )

=

F

2 ( - 0.2)

2

+ (0.35F)2 = 400

400 2

200 mm A

MC = { - 0.2Fi + 0.35F j} N m MC =

F

+ (0.35) 2

= 992 N

Ans.


4–85.

The gear reducer is subjected to the couple moments shown. Determine the resultant couple moment and specify its magnitude and coordinate direction angles.

z

M 2 = 60 N·m M 1 = 50 N·m 30

SOLUTION

°

x

Ex press Each Cou ple Moment as a Cartesian Vector:

550 6 N # m = 601cos 30° + sin 30° 2 N # m = 551.96 + 30.0 6 N # m

M1 = M2

j

i

k

i

k

Resultant Cou ple Moment: M R = © M;

MR = M1 + M2

= 51.96i + 50.0 j + 30.0k N # m

5 = 552.0

6

6 #

i + 50 j + 30k N m

Ans.

The magnitude of the resultant couple moment is MR

=

2 51.96

2

+ 50.0 2 + 30.0 2

= 78.102 N # m = 78.1 N # m

Ans.

The coordinate direction angles are a

= cos-1

51.96 a 78.102 b = 48.3°

Ans.

b

= cos-1

50.0 a 78.102 b = 50.2°

Ans.

g

= cos-1

30.0 78.102

= 67.4°

Ans.


y

4–86.

The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles.

z M 2 = 20 N· m

20 30

°

°

SOLUTION

550 6 N # m = 201 - cos 20° sin 30° - cos 20° cos 30° = 5 - 9.397 - 16.276 + 6.840 6 N # m

M1 = M2

y

k

i

2

#

j + sin 20°k N m

i

j

x

k

M 1 = 50 N · m


MR = M1 + M2

= - 9.397 i - 16.276 j + 50 + 6.840 k N # m

5 = 5 - 9.397

1

26

#

6

i - 16.276 j + 56.840 k N m


=

2 1 - 9.3972 + 1- 16.2762 + 156.8402 2

2

= 59.867 N # m = 59.9 N # m

2 Ans.


= cos-1

9.397 a -59.867 b = 99.0°

Ans.

b

= cos-1

16.276 a -59.867 b = 106°

Ans.

g

= cos-1

56.840 59.867

= 18.3°

Ans.


4–87.

The gear reducer is subject to the couple moments shown. Determine the resultant couple moment and specify its magnitude and coordinate direction angles.

z M 2 = 80 lb·ft

M 1 = 60 lb·ft

30

°

45

°

SOLUTION

x

Ex press Each:

560 6 lb # ft = 801 - cos 30° sin 45° - cos 30° cos 45° = 5 - 48.99 - 48.99 - 40.0 6 lb # ft

M1 = M2

i

i

2

#

j - sin 30°k lb ft

i

j

k


MR = M 1 + M 2

5160 - 48.992 - 48.99 - 40.0 6 lb # ft = 511.01 - 48.99 - 40.0 6 lb # ft = 511.0 - 49.0 - 40.0 6 lb # ft =

i

i

j

j

i

k

k

j

k

Ans.


=

2 11.01 + 1- 48.992 + 1- 40.02 2

2

= 64.20 lb # ft = 64.2 lb # ft

2 Ans.


= cos-1

a 11.01 b = 80.1° 64.20

Ans.

b

= cos-1

48.99 a -64.20 b = 140°

Ans.

g

= cos-1

- 40.0 64.20

= 129°

Ans.


y

*4–88.

A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment.

z

35 N 25 N 60

SOLUTION Mx My

y

= - 35(0.35) - 25(0.35) cos 60° = - 16.625 = - 25(0.35) sin 60° = - 7.5777 N # m

|M| =

2 ( - 16.625)

a

= cos-1

b

= cos-1

g

= cos-1

2

+ ( - 7.5777)

16.625 a -18.2705 b = 155° 7.5777 a -18.2705 b = 115° 0 a 18.2705 b = 90°

2

= 18.2705 = 18.3 N # m

175 mm

175 mm

x

Ans.

35 N

25 N

Ans.

Ans.

Ans.


4–89.

Determine the resultant couple moment of the two couples that act on the pipe assembly. The distance from A to B is d = 400 mm. Express the result as a Cartesian vector.

z

{35k} N B

250 mm d

{ 50i} N C

30

{ 35k} N

SOLUTION

350 mm

Vector Analysis

x

A {50i} N

Position Vector: rAB = {(0.35 - 0.35) i + ( - 0.4 cos 30° - 0) j + (0.4 sin 30° - 0)k} m

= { - 0.3464 j + 0.20k} m Cou ple Moments: With F1 = {35k} N and F2 = { - 50i} N, applying Eq. 4–15, we have

(M C)1 = rAB * F1

3 3

i

j

= 0

k

- 0.3464

0

0

(M C)2 = rAB * F2

=

i

j

0 - 50

- 0.3464 0

3

#

0.20 = { - 12.12i} N m 35

k

3

#

0.20 = { - 10.0 j - 17.32k} N m 0


M R = (M C)1 + (M C)2

= { - 12.1i - 10.0 j - 17.3k}N # m

Ans.

Scalar Analysis: Summing moments about x, y, and z axes, we have

#

(MR)x = © Mx ;

(MR)x = - 35(0.4 cos 30°) = - 12.12 N m

(MR)y = © My ;

(MR)y = - 50(0.4 sin 30°) = - 10.0 N m

(MR)z = © Mz ;

(MR)z = - 50(0.4 cos 30°) = - 17.32 N m

#

#

Express M R as a Cartesian vector, we have

#

M R = { - 12.1i - 10.0 j - 17.3k} N m


y

4–90.

Determine the distance d between A and B so that the resultant couple moment has a magnitude of MR = 20 N m.

#

z

{35k} N B

250 mm d

{ 50i} N C

30

{ 35k} N

SOLUTION

350 mm

Position Vector:

x

A {50i} N

rAB = {(0.35 - 0.35) i + ( - d cos 30° - 0) j + (d sin 30° - 0)k} m

= { - 0.8660 d j + 0.50d k} m Cou ple Moments: With F1 = {35k} N and F2 = { - 50i} N, applying Eq. 4–15, we have

(M C)1 = rAB * F1

3 3

i

j

= 0

k

- 0.8660 d

0

0

(M C)2 = rAB * F2

=

3

#

0.50d = { - 30.31d i} N m 35

i

j

0 - 50

- 0.8660 d

3

k

0

#

0.50d = { - 25.0d j - 43.30d k} N m 0


M R = (M C)1 + (MC)2

= { - 30.31d i - 25.0d j - 43.30d k} N # m

#

The magnitude of M R is 20 N m, thus 20 =

2 ( - 30.31 ) d

d

2

+ ( - 25.0d)2 + (43.30 d)2

= 0.3421 m = 342 mm

Ans.


y

4–91.

If F = 80 N, determine the magnitude and coordinate direction angles of the couple moment. The pipe assembly lies in the x–y plane.

z

F

300 mm 300 mm F

x

200 mm

SOLUTION

It is easiest to find the couple moment of F by taking the moment of F or –F about point A or B , respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

rAB = (0.3 - 0.2)i + (0.8 - 0.3) j + (0 - 0)k = [0.1i + 0.5 j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8) j + (0 - 0)k = [ - 0.1i - 0.5 j] m

The force vectors F and –F can be written as F = {80 k} N and - F = [ - 80 k] N

Thus, the couple moment of F can be determined from

3 3 i

M c = rAB * F = 0.1

0

or

3

Mc = rBA * - F =

=

2

Mx 2

k

j

- 0.1

- 0.5

0

0

+ My 2 + Mz 2 =

2 40

2

#

0 = [40i - 8 j] N m 80

0.5 0

i

The magnitude of M c is given by Mc

j

k

3

#

0 = [40i - 8 j] N m - 80

+ ( - 8)2 + 02 = 40.79 N # m = 40.8 N # m

Ans.

The coordinate angles of Mc are

¢ ≤ ¢ ≤ ¢ ≤

a

= cos - 1

b

= cos - 1

g

= cos - 1

Mx M

My M

Mz M

= cos = cos = cos

¢ ≤ ¢ ≤ ¢ ≤ 40 40.79

-8

40.79

0 40.79

= 11.3°

Ans.

= 101°

Ans.

= 90°

Ans.


y

*4–92.

If the magnitude of the couple moment acting on the pipe assembly is 50 N m, determine the magnitude of the couple forces applied to each wrench. The pipe assembly lies in the x–y plane.

#

z

F

300 mm 300 mm F

x

200 mm

SOLUTION

It is easiest to find the couple moment of F by taking the moment of either F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

rAB = (0.3 - 0.2) i + (0.8 - 0.3) j + (0 - 0)k = [0.1i + 0.5 j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8) j + (0 - 0)k = [ - 0.1i - 0.5 j] m

The force vectors F and –F can be written as F = {Fk} N and - F = [ - Fk]N Thus, the couple moment of F can be determined from

3 3 i

j

M c = rAB * F = 0.1

0.5 0

0

The magnitude of Mc is given by Mc

=

2

Mx 2

+ My 2 + Mz 2 =

k

0 = 0.5Fi - 0.1F j

F

2 (0.5 ) F

2

+ (0.1F)2 + 02 = 0.5099 F

#

Since Mc is required to equal 50 N m, 50 = 0.5099 F F

= 98.1 N

Ans.


y

4–93.

5 6

If F = 100k N, determine the couple moment that acts on the assembly. Express the result as a Cartesian vector. Member BA lies in the x–y plane.

z O O

B

60

F

SOLUTION f

= tan-1

a 23 b - 30° = 3.69°

x

200 mm 300 mm –F

150 mm

r1 = { - 360.6 sin 3.69°i + 360.6 cos 3.69° j}

200 mm

= { - 23.21i + 359.8 j} mm u

= tan-1

y

A

a 4.52 b + 30° = 53.96°

r2 = {492.4 sin 53.96° i + 492.4 cos 53.96° j}

= {398.2i + 289.7 j} mm Mc = (r1 - r2) * F

3

i

= - 421.4 0

j

k

70.10 0

0 100

#

Mc = {7.01i + 42.1 j} N m

3

Ans.


4–94.

If the magnitude of the resultant couple moment is 15 N m, determine the magnitude F of the forces applied to the wrenches.

z

#

O O

B

60

SOLUTION f

= tan-1

x

a 23 b - 30° = 3.69°

200 mm 300 mm 150 mm 200 mm

= { - 23.21i + 359.8 j} mm = tan-1

y

–F

r1 = { - 360.6 sin 3.69° i + 360.6 cos 3.69° j}

u

F

A

a 4.52 b + 30° = 53.96°

r2 = {492.4 sin 53.96° i + 492.4 cos 53.96° j}

= {398.2 i + 289.7 j} mm Mc = (r1 - r2) * F

3

i

= - 421.4 0

j

k

70.10 0

0 F

3

#

Mc = {0.0701 Fi + 0.421 F j} N m Mc = F

=

2 (0.07

01F)2

2 (0.0701)

+ (0.421 F)2 = 15

15 2

+ (0.421) 2

= 35.1 N

Ans.

Also, align y ¿ axis along BA. Mc = - F(0.15) i ¿ + F(0.4) j ¿

15 = F

2 ( (-0.15)) F

= 35.1 N

2

+ (F(0.4))2 Ans.


4–95.

If F1 = 100 N, F2 = 120 N and F3 = 80 N, determine the magnitude and coordinate direction angles of the resultant couple moment.

z

–F4  [150 k] N

0.3 m

0.2 m

0.2 m

F1 0.2 m

x

30

0.3 m

0.2 m – F1

F4  [150 k] N y

SOLUTION Cou ple Moment : The position vectors r1, r2, r3, and r4, Fig. a, must be determined first. r1 = {0.2i} m

r2 = {0.2 j} m

– F2 0.2 m F2 – F3

r3 = {0.2 j} m

0.2 m

From the geometry of Figs. b and c, we obtain

F3

r4 = 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45° j - 0.3 sin 30°k

= {0.1837 i + 0.1837 j - 0.15k} m The force vectors F1, F2, and F3 are given by F1 = {100k} N

F2 = {120k} N

F3 = {80i} N

Thus,

#

M 1 = r1 * F1 = (0.2i) * (100 k) = { - 20 j} N m

#

M 2 = r2 * F2 = (0.2 j) * (120k) = {24i} N m

#

M 3 = r3 * F3 = (0.2 j) * (80i) = { - 16k} N m

#

M 4 = r4 * F4 = (0.1837 i + 0.1837 j - 0.15k) * (150k) = {27.56i - 27.56 j} N m Resultant Moment : The resultant couple moment is given by

(M c)R = ∑M c;

(M c)R = M1 + M2 + M 3 + M 4

= ( - 20 j) + (24i) + ( - 16k) + (27.56 i - 27.56 j) = {51.56 i - 47.56 j - 16k} N # m The magnitude of the couple moment i s (M c)R = 2 [(M c)R]x2 + [(M c)R]y 2 + [(Mc)R]z 2

= 2 (51.56) 2 + ( - 47.56) 2 + ( - 16)2 = 71.94 N # m = 71.9 N # m

Ans.

The coordinate angles of (Mc)R are a

b

g

= cos -1

a [(( a [(( a [((

= cos -1 = cos -1

Mc)R]x

51.56 = co a b b = 44.2° ) 71.94 ) ] - 47.56 = co a b b = 131° ) 71.94 ) ] - 16 = co a b b = 103° ) 71.94

Mc

R

Mc

R y

Mc

R

Mc

R z

Mc

R

s

Ans.

s

Ans.

s

Ans.


*4–96.

Determine the required ma gnitude of F1, F2, and F3 so that the resultant couple moment is (Mc)R = [50 i - 45 j - 20 k] N m.

z

–F4  [150 k] N

#

0.3 m

0.2 m

0.2 m

F1 0.2 m – F1

x

Cou ple Moment : The position vectors r1, r2, r3, and r4, Fig. a, must be determined first. r1 = {0.2i} m

0.3 m

0.2 m

SOLUTION

r2 = {0.2 j} m

30

F4  [150 k] N y

– F2 0.2 m

r3 = {0.2 j} m

F2 – F3

From the geometry of Figs. b and c, we obtain

0.2 m

r4 = 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45° j - 0.3 sin 30°k

F3

= {0.1837 i + 0.1837 j - 0.15k} m The force vectors F1, F2, and F3 are given by F1 = F1k

F2 = F2k

F3 = F3i

Thus, M 1 = r1 * F1 = (0.2i) * (F1k) = - 0.2 F1 j M 2 = r2 * F2 = (0.2 j) * (F2k) = 0.2 F2i M 3 = r3 * F3 = (0.2 j) * (F3i) = - 0.2 F3k

#

M 4 = r4 * F4 = (0.1837 i + 0.1837 j - 0.15k) * (150k) = {27.56 i - 27.56 j} N m Res u ltant Moment : The res ultant couple moment required to equal (M c)R = {50i - 45 j - 20k} N m. Thus,

#

(M c)R = © M c;

(M c)R = M 1 + M 2 + M 3 + M 4 50i - 45 j - 20k = ( - 0.2F1 j) + (0.2F2i) + ( - 0.2F3k) + (27.56 i - 27.56 j) 50i - 45 j - 20k = (0.2F2 + 27.56) i + ( - 0.2F1 - 27.56) j - 0.2F3k

Equating the i, j, and k components yields 50 = 0.2F2 + 27.56

F2

= 112 N

Ans.

- 45 = - 0.2F1 - 27.56

F1

= 87.2 N

Ans.

- 20 = - 0.2F3

F3

= 100 N

Ans.


4–97.

Replace the force and couple system by an equivalent force and couple moment at point O.

y

3m 8 kN m

P

O

3m 4 kN

6 kN 4m

5m 12

13

SOLUTION

60

5

+ © FRx = © Fx ;

FRx

:

=6

a 135 b - 4 cos 60°

A

4m

= 0.30769 kN + c © FRy = © Fy ;

FR u

=

2 (0.30769)

= tan-1

FRy

=6

= 2.0744 kN 2

+ (2.0744) 2 = 2.10 kN

2.0744 c 0.30769 d = 81.6°

a + M O = © MO ;

a 1213 b - 4 sin 60° Ans.

Ans.

a

a 1213 b (4) + 6 a 135 b (5) - 4 cos 60°(4)

MO

= 8 - 6

MO

= - 10.62 kN # m = 10.6 kN # m b

Ans.


x

4–98.

Replace the force and couple system by an equivalent force and couple moment at point P .

y

3m 8 kN m

P

O

3m 4 kN

6 kN 4m

5m 12

13

SOLUTION

60

5

+ © FRx = © Fx ;

FRx

:

=6

a 135 b - 4 cos 60°

A

4m

= 0.30769 kN + c © FRy = © Fy ;

FR = u

2 (0.30769)

= tan-1

FRy

=6

= 2.0744 kN 2

+ (2.0744) 2 = 2.10 kN

2.0744 c 0.30769 d = 81.6°

a + M P = © MP ;

a 1213 b - 4 sin 60° Ans.

Ans.

a

a 1213 b (7) + 6 a 135 b (5) - 4 cos 60°(4) + 4 sin 60°(3)

MP

=8-6

MP

= - 16.8 kN # m = 16.8 kN # m b

Ans.


x

4–99.

Replace the force system acting on the beam by an equivalent force and couple moment at point A.

3 kN 2.5 kN 1.5 kN 30 5

3

4 B

A

2m

4m

2m

SOLUTION

+ F = ©F ; Rx x

:

FRx

= 1.5 sin 30° - 2.5

a 45 b

= - 1.25 kN = 1.25 kN + c FRy = © Fy ;

FRy

= - 1.5 cos 30° - 2.5

;

a 35 b - 3

= - 5.799 kN = 5.799 kN T Thus, FR

=

2

F 2Rx

+ F 2Ry =

2 1.25

2

+ 5.7992 = 5.93 kN

Ans.

and u

= tan - 1

a + MRA = © MA ;

¢ ≤ FRy

FRx

MRA

= tan - 1

= - 2.5

a 5.799 b = 77.8° 1.25

d

Ans.

a 35 b (2) - 1.5 cos 30°(6) - 3(8)

= - 34.8 kN # m = 34.8 kN # m (Clockwise )

Ans.


*4–100.

Replace the force system acting on the beam by an equivalent force and couple moment at point B.

3 kN 2.5 kN 1.5 kN 30 5

3

4 B

A

2m

4m

2m

SOLUTION

+ FR = © Fx ; x

FRx

:

= 1.5 sin 30° 30° - 2.5

a 45 b

= - 1. 1.25 25 kN = 1.2 1.255 kN + c FRy = © Fy ;

FRy

= - 1.5 cos 30° 30° - 2.5

;

a 35 b - 3

= - 5.7 5.799 99 kN = 5.799 kN T Thus, FR

=

2

F 2Rx

+ F 2Ry =

1.25 2 1.25

2

+ 5.7992 = 5. 5.93 93 kN

Ans.

and u

= tan - 1

¢ ≤ FRy

FRx

a + M RB = © MRB ;

= tan - 1

MB

a 5.799 b = 77.8° 1.25

= 1.5cos 30°(2) + 2.5

d

Ans.

a 35 b (6)

= 11 11.6 .6 kN # m (Counterclockwise )

Ans.


4–101.

Replace the force system actin g on the po st by a re sultant force and couple moment at point O.

300 lb 30 150 lb 3

2 ft

5 4

SOLUTION Equi valent valent Resultant Force : Forces F1 and F2 are resolved into their x and y components, Fig. a. S umming these force components algebraically alon g the x and y axes, we have

+ © (FR)x = © Fx;

a 45 b + 200 = 339.81 lb 3 = 300 in 30° + 150 a b = 240 lb c 5

(FR)x = 300 cos 30° - 150

:

+ c (FR)y = © Fy;

(FR)y

2 ft 200 lb

2 ft

:

O

s

The magnitude of the resultant force FR is given by F R

= 2 (FR)x2 + (FR)y2 = 2 339.81 339.812 + 2402 = 416.02 lb = 416 lb

Ans.

The angle u of FR is u

= tan-1

240 c (( )) d = tan c 339.81 d = 35.23° = 35.2° FR

y

FR

x

-1

a

Ans.

Equi valent valent Resultant Cou ple Moment : Applying the principle of moment s, Figs. a and b, and summing the moments of the force components algebraically about point A, we can can write

a + (MR)A = © M A;

(MR)A = 150

a 45 b (4) - 200(2) - 300 co 30°(6) s

= - 1478.85 lb # ft = 1.48 kip # ft (Clockwi se )Ans. ) Ans.


4–102.

Replace the two forces by an equivalent resultant force and couple moment at point O. Se Sett F = 20 lb.

y

20 lb 30

F

5 4

6 in.

40

O

SOLUTION

1.5 in.

x

4 (20) - 20 sin 30° = 6 lb 5

+ F = ©F ; Rx x

FRx

=

+ c FRy = © Fy;

FRy

= 20 cos 30° 30° +

:

FR u

a + MRO = © MO ;

=

2

F 2Rx

= tan - 1

MRO

4 5

FRx

2 6

= tan - 1

2

+ (29.32) 2 = 29 29.9 .9 lb

a 29.32 b = 78.4° 6

a

Ans.

Ans.

= 20 sin 30°(6 30°(6 sin 40°) 40°) + 20 cos 30°(3. 30°(3.55 + 6 cos 40 40°) °)

- (20)(6 sin 40°) + = 214 lb # in.d

3 (20) = 29 29.32 .32 lb 5

+ F 2Ry =

FRy

2 in.

3 (20)(3.5 + 6 cos 40°) 40°) 5 Ans.


3

4–103.

Replace the two forces by an equivalent resultant force and couple moment at point O. Set F = 15 lb .

y

20 lb 30

F

5 4

6 in.

x

4 (15) - 20 sin 30° = 2 lb 5

+ F = ©F ; Rx x

FRx

=

+ c FRy = © Fy;

FRy

= 20 cos 30° +

:

FR u

a + M RO = © MO;

1.5 in. 40

O

SOLUTION

3

MRO

=

2

F 2Rx

= tan - 1

2 in.

3 (15) = 26.32 lb 5

+ F 2Ry =

FRy FRx

2 2

= tan - 1

2

+ 26.32 2 = 26.4 lb

a 26.32 b = 85.7° 2

a

Ans.

Ans.

= 20 sin 30°(6 sin 40°) + 20 cos 30°(3.5 + 6 cos 40°) 4 5

- (15)(6 sin 40°) + = 205 lb # in.d

3 (15)(3.5 + 6 cos 40°) 5 Ans.


*4–104.

Replace the force system acting on the crank by a resultant force, and specify where it s line of action intersects BA measured from the pin at B.

60 lb

A

12 in. 10 lb

20 lb

SOLUTION B

Equi valent Resultant Force : S umming the forces, Fig. a, algebraically alon g the x and y axes, we have

+ © (FR)x = © Fx;

(FR)x = - 60 lb = 60 lb

:

+ c (FR)y = ©Fy;

C

4.5 in.

4.5 in.

3 in.

;

(FR)y = - 10 - 20 = - 30 lb = 30 lb T

The magnitude of the resultant force FR is given by FR

= 2 (FR)x2 + (FR)y2 = 2 602 + 302 = 67.08 lb = 67.1 lb

Ans.


= tan-1

c (( )) d = tan c 3060 d = 26.57° = 26.6° FR

y

FR

x

-1

d

Ans.

Locat io n of Resultant Force : S umming the moments of the forces shown in Fig. a and the force components shown in Fig. b algebraically about point B, we can write

a + (MR)B = © M B;

60(d) = 60(12) - 10(4.5) - 20(9) d

= 8.25 in.

Ans.


4–105.

Replace the force system acting on the frame by a re sultant force and couple moment at point A.

5 kN 3

3 kN

2 kN

5

1m

4m

4

B

C

1m

13

12

5 D

5m

SOLUTION Equi valent Resultant Force : Resolving F1, F2, and F3 into their x and y components, Fig. a, and summing these force components al gebraically alon g the x and y axes, we have

+ © (FR)x = © Fx;

a 45 b - 3 a 135 b = 2.846 kN 3 12 = - 5 a b - 2 - 3 a b = - 7.769 kN = 7.769 kN T 5 13

(FR)x = 5

:

+ c (FR)y = © Fy;

(FR)y

A

:

The magnitude of the resultant force FR is given by F R

= 2 (FR)x2 + (FR)y2 = 2 2.8462 + 7.7692 = 8.274 kN = 8.27 kN

Ans.


= tan-1

c (( )) d = tan c 7.769 d = 69.88° = 69.9° 2.846 FR

y

FR

x

-1

c

Ans.

Equi valent Cou ple Moment : Applying the principle of moment s and summing the moments of the force components algebraically about point A, we can write

a + (MR)A = © M A;

(MR)A = 5

a 35 b (4) - 5 a 45 b (5) - 2(1) - 3 a 1312 b (2) + 3 a 135 b (5)

= - 9.768 kN # m = 9.77 kN # m (Clockwi se )

Ans.


4–106.

Replace the force system acting on the bracket by a resultant force and couple moment at point A.

450 N 600 N B

30

45

0.3 m

SOLUTION A

Equi valent Resultant Force : Forces F1 and F2 are resolved into their x and y components, Fig. a. S umming the se force components al gebraically alon g the x and y axes, we have

+ © (FR)x = © Fx; (FR)x = 450 cos 45° - 600 cos 30° = - 201.42 N = 201.42 N

:

+ c (FR)y = ©Fy;

0.6 m

;

c

(FR)y = 450 sin 45° + 600 sin 30° = 618.20 N


= 2 (FR)x2 + (FR)y2 = 2 201.4 2 + 618.20 2 = 650.18 kN = 650 N

Ans.


= tan - 1

c (( )) d = tan c 618.20 d = 71.95° = 72.0° 201.4 FR

y

FR

x

-1

b

Ans.

Equi valent Resultant Cou ple Moment : Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point A, we can write

a + (MR)A = © MA; (MR)A = 600 sin 30°(0.6) + 600 cos 30°(0.3) + 450 sin 45°(0.6) - 450 cos 45°(0.3)

= 431.36 N # m = 431 N # m (Counterclockwi se )

Ans.


4–107.

A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of FR = 35 N for the rectus, FO = 45 N for the oblique, FL = 23 N for the lumbar latissimus dorsi, and FE = 32 N for the erector spinae. These loadings are symmetric with respect to the y–z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form.

z

FO FR

FO

MROx

;

= © MOx

FL

15 mm 45 mm

50 mm

30 mm 40 mm

y

x

FR

;

FL

O

SOLUTION = © Fz

FE FE

75 mm

FR

FR

=

{2(35

+

45

+

23

+

32)k }

+

2(32)(0.015)

MR O

=

[ - 2(35)(0.075)

MR O

=

{ - 2.22i} N # m

=

{270k} N +

Ans.

2(23)(0.045)]i Ans.


*4–108.

Replace the two forces acting on the post by a resultant force and couple moment at point O. Express the results in Cartesian vector form.

z A C F D

SOLUTION

vector form can be written as

C C

(0 - 0)i + (6 - 0) j + (0 - 8)k =5 (0 - 0)2 + (6 - 0)2 + (0 - 8)2

FD = FDuCD = 7

S

(2 - 0)i + ( - 3 - 0) j + (0 - 6)k (2 - 0)2 + ( - 3 - 0)2 + (0 - 6)2

The resultant force FR is given by

5 kN

7 kN

6m

Equivalent Resultant Force: The forces FB and FD, Fig. a, expressed in Cartesian

FB = FBuAB

F B

8m 2m

D

O

3m

= [3 j - 4k] kN

S

x

6m

B y

= [2i - 3 j - 6k] kN

FR = πF; FR = FB + FD

= (3 j - 4k) + (2i - 3 j - 6k) = [2i - 10k] kN

Ans.

Equivalent Resultant Force: The position vectors rOB and rOC are rOB = {6 j} m

rOC = [6k] m

Thus, the resultant couple moment about point O is given by (M R)O = © M O;

(M R)O = rOB * FB + rOC * FD

3 33 3 i

j

k

= 0

6 3

0 + 0 -4 2

0

i

= [ - 6i + 12 j] kN # m

j

k

0 -3

6 -6

Ans.


4–109.

Replace the force system by an equivalent force and couple moment at point A.

z

F2

{100i

100 j

50k} N

F1

{300i

400 j

100k} N

4m

F3

SOLUTION FR = © F;

{ 500k} N 8m

FR = F1 + F2 + F3

1m

1 2 1 2 1 = 5400 + 300 - 650 6 N = 512 6 m and = 5 - 1 6 m. The position vectors are

2

= 300 + 100 i + 400 - 100 j + - 100 - 50 - 500 k i

j

rAB

M RA = © M A ;

k

k

rAE

j

M RA = rAB * F1 + rAB * F2 + rAE * F3

=

3

i

j

0 300

0 400

i

+ 0 0

k

i

+ 0 12 - 100 100

j

k

-1

0 - 500

0

33

= - 3100i + 4800 j N # m

j

k

0 - 100

12 - 50

3 Ans.


4–110.

The belt passing over the pulley is subjected to forces F1 and F2, eachhaving a magnitude of 40 N. F1 actsinthe - k direction. Replace these forces by an equivalent force and couple momentat point A. Express the result in Cartesianvector form. Set u = 0° so that F2 acts in the - j direction.

z



r



80 mm

y

300 mm

A

SOLUTION FR

=

F1

FR

=

{ - 40 j

M RA

MRA

+

x

F2

= © (r *

=

3

=

{ - 12 j

-

40 k} N

Ans.

F)

i - 0.3 0

j 0 - 40 +

F2

F1

k 0.08 0

12k} N # m

3 3 +

i - 0.3 0

j 0.08 0

k 0 - 40

3 Ans.


4–111.

The belt passing over the pulley is subjected to two forces F1 and F2, ea each ch hav having ing a mag magnit nitude ude of 40 N. F1 ac acts ts in the - k direction. Replace these forces by an equivalent force and and couple moment at point A. Express the result in Cartesian Cartesian vector form. Take u = 45°.

z

r

80 mm

y

300 mm A

SOLUTION

x

FR = F1 + F2

= - 40 cos 45° j j + ( - 40 - 40 sin 45°) k

F2

FR = { - 28.3 j - 68.3k} N

Ans.

F1

rAF1 = { - 0.3i + 0.08 j} m rAF2 = - 0.3i - 0.08 sin 45° j + 0.08 cos 45°k

= {- 0.3i - 0.0566 j + 0.0566 k} m MRA = (rAF1 * F 1) + (rAF2 * F 2)

3

i

= - 0.3 0

j

0.08 0

k

33

i

+ - 0.3 0 - 40 0

j

k

- 0.0566

0.0566 - 40 sin 45°

- 40 cos 45° 45°

#

MRA = { - 20.5 j + 8.49k} N m

3

Ans.

Also, MRAx

= © MAx

MRAx

= 28.28(0.0566) + 28.28(0.0566) - 40(0.08)

MRAx

= 0

MRAy

= © MAy

MRAy

= - 28.28(0.3) - 40(0.3)

MRAy

= - 20.5 N # m

MRAz

= © MAz

MRAz

= 28.28(0.3)

MRAz

= 8. 8.49 49 N # m

#

MRA = - 20.5 + 8.49k N m

Ans.


*4–112.

Handle forces F1 and F2 are applied to the electric drill. Replace this force system by an equi valent resultant force and couple moment acting at point O. Express the results in Cartesian vector form.

F2

{2 j



4k} N



z

0.15 m

F1

{6i



3 j



10k} N



0.25 m

0.3 m

SOLUTION FR = © F;

O

FR = 6i - 3 j - 10k + 2 j - 4k

x

= {6i - 1 j - 14k} N

y

Ans.

MRO = © M O ;

3

i MRO = 0.15

6

j

k

3 3

i

0.3 + 0 - 10 0

0 -3

j

k

- 0.25

0.3 -4

2

3

= 0.9i + 3.30 j - 0.450k + 0.4i

= {1.30i + 3.30 j - 0.450k} N # m

Ans.

Note that FRz = - 14 N pushes the drill bit down into the stock.

#

#

(MRO)x = 1.30 N m and (MRO)y = 3.30 N m cause the drill bit to bend.

#

(MRO)z = - 0.450 N m causes the drill case and the spinning drill bit to rotate about the z-axis.


4–113.

The weights of the various components of the truck are shown. Replace this system system of forces by an equivalent equivalent resultant force and specify its location measured from B.

3500 lb

B

SOLUTION

+ c FR = © Fy;

FR

= - 1750 - 5500 - 3500

= - 10 75 7500 lb = 10 10.75 .75 kip T

5500 lb 14 ft

3 ft

A

1750 lb

6 ft 2 ft

Ans.

a + MRA = © MA ; -10 750d = - 3500(3) - 5500(17) - 1750(25) d

= 13 13.7 .7 ft

Ans.


4–114.

The weights of the various components of the truck are shown. Replace this system of forces by an equivalent resultant force and specify its location measured from point A.

5500 lb 14 ft

Equivalent Force:

+ c FR = © Fy ;

3500 lb

B

SOLUTION

3 ft

FR

A

1750 lb

6 ft 2 ft

= - 1750 - 5500 - 3500 = - 10 750 lb = 10.75 kip T

Ans.

Location of Resultant Force From Point A:

a + MRA = © MA ;

10 750( d) = 3500(20) + 5500(6) - 1750(2) d

= 9.26 ft

Ans.


4–115.

Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts,measured from end A.

5 ft

3 ft

2 ft

4 ft

A

B 5

12

3 4

500 lb

13 5

200 lb

260 lb

SOLUTION

+ FRx = © Fx ;

FRx

+ c FRy = © Fy ;

FRy

:

F u

=

2 ( - 300)

= tan-1

2

a 45 b + 260 a 135 b = - 300 lb = 300 lb 3 12 = - 500 a b - 200 - 260 a b = - 740 lb = 740 lb 5 13

= - 500

;

+ ( - 740)2 = 798 lb

a 740 b = 67.9° 300

c + M RA = © MA;

Ans.

Ans.

d

740(x) = 500

T

a 35 b (5) + 200(8) + 260 a 1213 b (10)

740(x) = 5500 x

= 7.43 ft

Ans.


*4–116.

Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts,measured from end B.

5 ft

3 ft

2 ft

4 ft

A

B 5

12

3 4

500 lb

13 5

200 lb

260 lb

SOLUTION

+ FRx = © Fx ;

FRx

+ c FRy = © Fy ;

FRy

:

F u

=

2 ( - 300)

= tan-1

2

a 45 b + 260 a 135 b = - 300 lb = 300 lb 3 12 = - 500 a b - 200 - 260 a b = - 740 lb = 740 lb 5 13

= - 500

;

+ ( - 740)2 = 798 lb

a 740 b = 67.9° 300

a + M RB = © MB;

x

Ans.

Ans.

d

740(x) = 500

= 6.57 ft

T

a 35 b (9) + 200(6) + 260 a 1213 b (4) Ans.


4–117.

Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from end A.

700 N 450 N

30

300 N

60 B A

2m

4m

3m

1500 N m

SOLUTION

+ FRx = © Fx ;

FRx

= 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = © Fy ;

FRy

= - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N T

:

F u

=

2 ( - 125)

= tan-1

2

;

+ ( - 1296) 2 = 1302 N

a 1296 b = 84.5° 125

c + M RA = © MA ;

d

Ans.

Ans.

1296(x) = 450 sin 60°(2) + 300(6) + 700 cos 30°(9) + 1500 x

= 7.36 m

Ans.


4–118.

Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from B.

700 N 450 N

30

300 N

60 B A

2m

4m

3m

1500 N m

SOLUTION

+ FRx = © Fx ;

FRx

= 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = © Fy ;

FRy

= - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N T

:

F u

=

2 ( - 125)

= tan-1

2

+ ( - 1296) 2 = 1302 N

a 1296 b = 84.5° 125

c + M RB = © MB ;

d

;

Ans.

Ans.

1296(x) = - 450 sin 60°(4) + 700 cos 30°(3) + 1500 x

= 1.36 m (to the right)

Ans.


4–119.

Replace the force system acting on the frame by a re sultant force, and specify where its line of action inter sects member AB, measured from point A.

2.5 ft

B

3 ft 45

2 ft 300 lb

200 lb 5

3

4

250 lb

4 ft

SOLUTION A

Equi valent Resultant Force : Resolving F1 and F3 into their x and y components, Fig. a , and summing these force components algebraically alon g the x and y axes, we have

+ © (FR)x = © Fx;

:

+ c (FR)y = ©Fy;

a 45 b - 300 = - 358.58 lb = 358.58 lb 3 = - 200 in 45° - 250 a b = - 291.42 lb = 291.42 lb T 5

(FR)x = 200 cos 45° - 250 (FR)y

;

s


= 2 (F R)x2 + (F R)y2 = 2 358.58 2 + 291.42 2 = 462.07 lb = 462 lb

Ans.


= tan -1

c (( )) d = tan c 291.42 d = 39.1° 358.58 FR

y

FR

x

-1

d

Ans.

Locat io n of Resultant Force : Applying the principle of moments to Figs. a and b, and summing the moments of the force components algebraically about point A, we can write

a + (MR)A = © MA;

358.58( d) = 250

a 35 b (2.5) + 250 a 45 b (4) + 300(4) - 200 co 45°(6) s

- 200 sin 45°(3) d

= 3.07 ft

Ans.


*4–120.

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB , measured from A.

300 N 250 N 1m C

2m

3m

5

B

4

D

2m

400 N m 60

SOLUTION

3m

500 N

+ © Fx = FRx ;

+ c © Fy = © Fy ; FR u

=

2 ( - 450)

= tan-1

2

a 45 b - 500(cos 60°) = - 450N = 450 N 3 = - 300 - 250 a b - 500 sin 60° = - 883.0127 N = 883.0127 N T 5

= - 250

FRx

:

FRy

+ ( - 883.0127) 2 = 991 N

a 883.0127 b = 63.0° 450

a + MRA = © MA ;

=

A

Ans.

d

450y = 400 + (500 cos 60°)(3) + 250

y

;

800 = 1.78 m 450

a 45 b (5) - 300(2) - 250 a 35 b (5) Ans.


3

4–121.

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD , measured from end C .

300 N 250 N 1m C

2m

3m

5

B

3

4

D

2m

400 N m 60

SOLUTION

3m

500 N

+ © Fx = FRx ;

+ c © Fy = © Fy ; FR u

=

2 ( - 450)

= tan-1

2

a 45 b - 500(cos 60°) = - 450 N = 450 N 3 = - 300 - 250 a b - 500 sin 60° = - 883.0127 N = 883.0127 N T 5

= - 250

FRx

:

FRy

+ ( - 883.0127) 2 = 991 N

a 883.0127 b = 63.0° 450

c + MRA = © MC ;

=

A

Ans.

d

883.0127 x = - 400 + 300(3) + 250

x

;

2333 = 2.64 m 883.0127

a 35 b (6) + 500 cos 60°(2) + (500 sin 60°)(1) Ans.


4–122.

Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant ’s line of action intersects member AB, measured from point A.

35 lb

20 lb

30 4 ft

A

B

2 ft 3 ft 25 lb

SOLUTION

+ FRx = © Fx ;

FRx

:

+ T FRy = © Fy ; FR

=

2

= tan - 1

2 ft

= 35 cos 30° + 20 = 50.31 lb

FRy

2 (42.5)

u

c + MRA = © MA ;

= 35 sin 30° + 25 = 42.5 lb

+ (50.31) = 65.9 lb 2

a 50.31 b = 49.8° 42.5

c

C Ans.

Ans.

50.31 (d) = 35 cos 30°(2) + 20(6) - 25(3) d

= 2.10 ft

Ans.


4–123.

Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant ’s line of action intersects member BC , measured from point B .

35 lb

20 lb

30 4 ft

A

B

2 ft 3 ft 25 lb

SOLUTION

+ FRx = © Fx;

FRx

= 35 sin 30° + 25 = 42.5 lb

+ T FRy = © Fy ;

FRy

= 35 cos 30° + 20 = 50.31 lb

:

FR

=

2 (42.5)

u

c + MRA = © MA ;

2

= tan - 1

2 ft

+ (50.31) = 65.9 lb 2

a 50.31 b = 49.8° 42.5

c

C Ans.

Ans.

50.31 (6) - 42.5(d) = 35 cos 30° (2) + 20(6) - 25 (3) d

= 4.62 ft

Ans.


*4–124.

Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point A.

0.5 m B

1m

500 N

5

3

0.2 m

4

250 N

30 1m

SOLUTION

300 N

Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes,

+ (FR)x = © Fx;

a 45 b - 500cos 30° - 300 = - 533.01 N = 533.01 N 3 = 500 sin 30° - 250 a b = 100 N c 5

(FR)x = 250

:

+ c (FR)y = © Fy;

(FR)y

1m A ;


=

2 (

FR)x 2

+ (FR)y 2 =

2 533.01 + 100 2

2

= 542.31 N = 542 N

Ans.


= tan - 1

B R (FR)y

(FR)x

= tan - 1

100 c 533.01 d = 10.63° = 10.6°

b

Ans.

Location of the Resultant Force: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point A,

a + (MR)A = © MA;

533.01( d) = 500 cos 30°(2) - 500 sin 30°(0.2) - 250 d

= 0.8274 mm = 827 mm

a 35 b (0.5) - 250 a 45 b (3) + 300(1) Ans.


4–125.

Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point B.

0.5 m B

1m

500 N

5

3

0.2 m

4

250 N

30

1m

SOLUTION

300 N

Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y

components, Fig. a. Summing these force components algebraically along the x and y axes,

+ © (FR)x = © Fx;

a 45 b - 500cos 30° - 300 = - 533.01N = 533.01 N 3 = 500 sin 30° - 250 a b = 100 N c 5

(FR)x = 250

:

+ c (FR)y = © Fy;

(FR)y

1m A ;


=

2 (

FR)x 2

+ (FR)y 2 =

2 533.01 + 100 2

2

= 542.31 N = 542 N

Ans.


= tan - 1

B R (FR)y

(FR)x

= tan - 1

100 c 533.01 d = 10.63° = 10.6°

b

Ans.

Location of the Resultant Force: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point B ,

a + (MR)B = © Mb;

- 533.01( d) = - 500cos 30°(1) - 500sin 30°(0.2) - 250 d

= 2.17 m

a 35 b (0.5) - 300(2)

Ans.


4–126.

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB , measured from A.

300 lb

200 lb 3 ft

400 lb 4 ft B

A

2 ft

600 lb ft 200 lb

SOLUTION

+ FRx = © Fx ;

FRx

= - 200 lb = 200lb

+ c FRy = © Fy ;

FRy

= - 300 - 200 - 400 = - 900 lb = 900 lb T

:

F u

=

2 ( - 200)

= tan-1

2

;

+ ( - 900)2 = 922 lb

a 900 b = 77.5° 200

c + MRA = © MA ;

d

900(x) = 200(3) + 400(7) + 200(2) - 600 = 0 3200 = 3.56 ft x = 900

7 ft

Ans.

C

Ans.

Ans.


4–127.

The tube supports the four parallel forces. Determine the magnitudes of forces FC and FD acting at C and D so that the equivalent resultant force of the force system acts through the midpoint O of the tube.

z FD

600 N D FC

A

400 mm

SOLUTION

Since the resultant force passes through point O, the resultant moment components about x and y axes are both zero.

© Mx = 0;

FD(0.4)

500 N C

400 mm x

z B

200 mm 200 mm y

+ 600(0.4) - FC(0.4) - 500(0.4) = 0 FC

© My = 0;

O

- FD = 100

(1)

500(0.2) + 600(0.2) - FC(0.2) - FD(0.2) = 0 FC

+ FD = 1100

(2)

Solving Eqs. (1) and (2) yields: FC

= 600 N

F D

= 500 N

Ans.


*4–128.

Three parallel bolting forces act on the circular plate. Determine the resultant force, and specify its location ( x, z) on the plate. FA = 200 lb, FB = 100 lb, and FC = 400 lb.

z

C

FC

1.5 ft 45

SOLUTION x

30 B

Equivalent Force: FR

= © Fy;

A

FB

F A

y

- FR = - 400 - 200 - 100 FR

= 700 lb

Ans.

Location of Resultant Force: MRx

= © Mx;

700(z) = 400(1.5) - 200(1.5 sin 45°) - 100(1.5 sin 30°) z

MRz

= © Mz;

= 0.447 ft

Ans.

- 700(x) = 200(1.5 cos 45°) - 100(1.5 cos 30°) x

= - 0.117 ft

Ans.


4–129.

The three parallel bolting forces act on the circular plate. If the force at A has a magnitude of FA = 200 lb, determine the magnitudes of FB and FC so that the resultant force FR of the system has a line of action that coincides with the y axis. Hint: This requires © Mx = 0 and © Mz = 0.

z

C

FC

1.5 ft 45

SOLUTION x

Since FR coincides with y axis, MRx = MRy = 0. MRz

= © Mz; FB

30 B

A

FB

F A

y

0 = 200(1.5 cos 45°) - FB (1.5 cos 30°)

= 163.30 lb = 163 lb

Ans.

Using the result FB = 163.30 lb , MRx

= © Mx;

0 = FC (1.5) - 200(1.5 sin 45°) - 163.30(1.5 sin 30°) FC

= 223 lb

Ans.


4–130.

The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location ( x, y) on the slab. Take F1 = 30 kN, F2 = 40 kN.

z

20 kN

F1

50 kN

F2

SOLUTION

+ c FR = © Fz;

x

FR

= - 20 - 50 - 30 - 40 = - 140kN = 140 kN T

(MR)x = © Mx;

8m Ans.

6m 2m

- 140y = - 50(3) - 30(11) - 40(13) y

(MR)y = © My;

4m

3m

= 7.14 m

Ans.

140x = 50(4) + 20(10) + 40(10) x

= 5.71 m

Ans.


y

4–131.

The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location ( x, y) on the slab. Take F1 = 20 kN, F2 = 50 kN.

z

20 kN

F1

50 kN

F2

SOLUTION

+ T FR = © Fz;

x

FR

MR y

= 20 + 50 + 20 + 50 = 140 kN

= © My;

= © Mx;

8m Ans.

y

6m 2m

140(x) = (50)(4) + 20(10) + 50(10) x

MR x

4m

3m

= 6.43 m

Ans.

- 140(y) = - (50)(3) - 20(11) - 50(13) y

= 7.29 m

Ans.


*4–132.

If FA = 40 kN and FB = 35kN, determine the magnitude of the resultant force and specify the location of its point of application ( x, y) on the slab.

z 30 kN FB

0.75 m 2.5 m

90 kN 20 kN

2.5 m 0.75 m F A

0.75 m

SOLUTION

x

3m

Equivalent Resultant Force: By equating the sum of the forces along the z axis to the resultant force FR, Fig. b,

+ c FR = © Fz;

3m 0.75 m

- FR = - 30 - 20 - 90 - 35 - 40 FR

= 215 kN

Ans.

Point of A pplication: By equating the moment of the forces and FR, about the x and y axes,

(MR)x = © Mx;

- 215(y) = - 35(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - 40(6.75) y

(MR)y = © My;

= 3.68 m

Ans.

215(x) = 30(0.75) + 20(0.75) + 90(3.25) + 35(5.75) + 40(5.75) x

= 3.54 m

Ans.


y

4–133.

If the resultant force is required to act at the center of the slab, determine the magnitude of the column loadings FA and FB and the magnitude of the resultant force.

z 30 kN FB

0.75 m 2.5 m

90 kN 20 kN

2.5 m 0.75 m F A

0.75 m x

SOLUTION

3m 3m

Equivalent Resultant Force: By equating the sum of the forces along the z axis to the resultant force FR,

+ c FR = © Fz;

0.75 m

- FR = - 30 - 20 - 90 - FA - FB FR

= 140 + FA + FB

(1)

Point of A pplication: By equating the moment of the forces and FR, about the x and y axes,

(MR)x = © Mx;

- FR(3.75) = - FB(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - FA(6.75) FR

(MR)y = © My;

= 0.2FB + 1.8FA + 132

FR(3.25) FR

(2)

= 30(0.75) + 20(0.75) + 90(3.25) + FA(5.75) + FB(5.75)

= 1.769FA + 1.769FB + 101.54

(3)

Solving Eqs.(1) through (3) yields FA

= 30kN

FB

= 20 kN

FR

= 190kN

Ans.


y

4–134.

Replace the two wrenches and the force, acting on the pipe assembly, by an equivalent resultant force and couple moment at point O.

100 N · m

300 N

z

C

SOLUTION

O 0.5 m

Force And Moment Vectors:

F2

k

F3

i

M2

0.6 m 45

j

k

i

M1

B

100 N

0.8 m

x

= 5100 6 N 5300 6 N = 2005cos 45° - sin 45° 6 N = 5141.42 - 141.42 6 N = 5100 6 N # m = 1805cos 45° - sin 45° 6 N # m = 5127.28 - 127.28 6 N # m

F1 =

A

°

200 N

180 N · m

k

k

i

k

i

k

Equivalent Force and Cou ple Moment At Point O: FR = © F;

FR = F1 + F2 + F3

1

2

= 141.42 i + 100.0 j + 300 - 141.42 k

5

6

= 141i + 100 j + 159k N

5 6

Ans.

5 6

The position vectors are r1 = 0.5 j m and r2 = 1.1 j m. M RO = © M O ;

MRO = r1 * F1 + r2 * F2 + M 1 + M 2

3 3 i

= 0 0

+

j

k

0.5 0

0 300

i

j

k

0 141.42

1.1 0

0 - 141.42

+ 100k + 127.28 i - 127.28 k = 122i - 183k N # m

Ans.


y

4–135.

The three forces acting on the block each have a magnitude of 10 lb. Replace this system by a wrench and specify the point where the wrench intersects the z axis, measured from point O.

z

F2

2 ft O

y

F3 F1

6 ft

SOLUTION 6 ft

FR = { - 10 j} lb

x

MO = (6 j + 2k) * ( - 10 j) + 2(10)( - 0.707i - 0.707 j)

= { 5.858i - 14.14 j} lb # ft Require z

=

5.858 = 0.586 ft 10

FW = { - 10 j} lb

Ans.

Ans.

#

MW = { - 14.1 j} lb ft

Ans.


*4–136.

Replace the force and couple moment system acting on the rectangular block by a wrench.Specify the magnitude of the force and couple moment of the wrench and where its line of action intersects the x–y plane.

z

4 ft 600 lb ft

450 lb

600 lb

2 ft y x

SOLUTION

3 ft

Equivalent Resultant Force: The resultant forces F1, F2, and F3 expressed in Cartesian vector form can be written as F1 = [600 j] lb, F2 = [-450i] lb, and F3 = [300k] lb. The force of the wrench can be determined from

300 lb

FR = © F; FR = F1 + F2 + F3 = 600 j - 450i + 300k = [ - 450i + 600 j + 300k] lb

Thus, the magnitude of the wrench force is given by FR =

2 (

FR)x 2

+ (FR)y 2 + (FR)z 2 =

2 ( - 450)

+ 6002 + 3002 = 807.77 lb = 808 lb

2

Ans.

Equivalent Cou ple Moment: Here,we will assume that the axis of the wrench passes through point P , Figs. a and b. Since MW is collinear with F R, M W = MW uFR = MW

C

- 450i + 600 j + 300k

2 ( - 450)

2

+ 6002 + 3002

= - 0.5571 Mwi + 0.7428 Mw j + 0.3714 Mwk

S

The position vectors rPA, rPB, and rPC are rPA = (0 - x)i + (4 - y) j + (2 - 0)k = - xi + (4 - y) j + 2k rPB = (3 - x)i + (4 - y) j + (0 - 0)k = (3 - x)i + (4 - y) j rPC = (3 - x)i + (4 - y) j + (2 - 0)k = (3 - x)i + (4 - y) j + 2k

The couple moment M expressed in Cartesian vector form is written as M = [600i] lb ft. Summing the moments of F1, F2, and F3 about point P and including M,

#

MW = © MP;

MW = rPA * F1 + rPC * F2 + rPB * F3 + M

- 0.5571 Mw i + 0.7428 Mw j + 0.3714 Mw k =

3

i -x

0

j

(4 - y) 600

k

33

i

2 + (3 - x) - 450 0

j

(4 - y) 0

k

33

i

2 + (3 - x) 0 0

- 0.5571 Mw i + 0.7428 Mw j + 0.3714 Mw k = (600 - 300y)i + (300x - 1800) j + (1800 - 600x - 450y)k

j

(4 - y) 0

k

3

0 + 600i 300

Equating the i, j, and k components,

- 0.5571 Mw = 600 - 300y

(1)

0.7428 Mw = 300x - 1800

(2)

0.3714 Mw = 1800 - 600x - 450y

(3)

Solving Eqs. (1),(2),and (3) yields x

= 3.52 ft

y

= 0.138 ft

MW

= - 1003 lb # ft

Ans.

The negative sign indicates that MW acts in the opposite sense to that of FR. © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4–137.

Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P ( x, y) where its line of action intersects the plate.

z

FB

F A

{800k} N

{500i} N

A

y

P

B

x

y

4m

x

SOLUTION

C

FR = {500 i + 300 j + 800k} N FR

6m

=

2 (500)

2

+ (300)2 + (800)2 = 990 N

FC

{300 j} N

Ans.

uFR = {0.5051 i + 0.3030 j + 0.8081 k} MRx¿

= © Mx¿;

MRx¿

= 800(4 - y)

MRy¿

= © My¿;

MRy¿

= 800x

MRz¿

= © Mz¿;

MRz¿

= 500y + 300(6 - x)

Since MR also acts in the direction of uFR, MR (0.5051)

= 800(4 - y)

MR (0.3030)

= 800x

MR (0.8081)

= 500y + 300(6 - x) = 3.07 kN # m

Ans.

x

= 1.16 m

Ans.

y

= 2.06 m

Ans.

MR


4–138.

The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. 2 lb/ ft

3.5 lb/ ft

O

A

2.75 ft 4 ft

SOLUTION

+ T FR O = © F; c + M R O = © MO ;

FRO

= 8 + 5.25 = 13.25 = 13.2 lb T

1.5 ft

Ans.

13.25x = 5.25(0.75 + 1.25) - 8(2 - 1.25) x

= 0.340 ft

Ans.


4–139.

Replace the distributed loading with an equivalent resultant force, and specify its location on the beam mea sured from point O.

3 kN/ m

O

3m

SOLUTION

1.5 m

Load in g : The distributed loading can be divided into two part s a s shown in Fig. a. Equat io ns of Equi li b r ium : Equating the forces along the y axis of Figs. a and b, we have

+ T FR = ©F;

FR

1 2

= (3)(3) +

1 (3)(1.5) = 6.75 kN T 2

Ans.

If we equate the moment of F R, Fig. b , to the sum of the moment of the forces in Fig. a about point O, we have a + (MR)O = © MO;

1 2 x = 2.5 m

1 2

- 6.75( x) = - (3)(3)(2) - (3)(1.5)(3.5) Ans.


*4–140.

Replace the loading by an equivalent force and couple moment acting at point O.

200 N/m

O

4m

3m

SOLUTION Equivalent Force and Cou ple Moment At Point O:

+ c FR = © Fy ;

FR

= - 800 - 300 = - 1100 N = 1.10 kN T

a+ MRO = © MO ;

MRO

12

Ans.

12

= - 800 2 - 300 5 = - 3100 N # m

= 3.10 kN # m (Clockwise)

Ans.


4–141.

The column is used to support the floor which exerts a force of 3000 lb on the top of the column.The effect of soil pressure along its side is distributed as shown. Replace this loading by an equivalent resultant force and specify where it acts along the column, measured from its base A.

3000 lb

80 lb/ ft

9 ft

SOLUTION

+ © FRx = © Fx ;

:

+ T FRy = © Fy ;

FRx

= 720 + 540 = 1260 lb

FRy

= 3000 lb =

FR

= 3.25 kip

u

a + M RA = © MA ;

2 (1260)

FR

= tan - 1

200 lb/ ft A

2

+ (3000) = 3254 lb

B R 3000 1260

2

Ans.

= 67.2°

d

Ans.

1260x = 540(3) + 720(4.5) x

= 3.86 ft

Ans.


4–142.

Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point B.

800 lb/ ft 500 lb/ ft

A

B

12 ft

SOLUTION

+ T FR = © F;

= 4800 + 1350 + 4500 = 10 650 lb

FR FR

c + MRB = © MB ;

9 ft

= 10.6 kip T

Ans.

10 650x = - 4800(4) + 1350(3) + 4500(4.5) x

= 0.479 ft

Ans.


4–143.

The masonry support creates the loading distribution acting on the end of the beam. Simplify this load to a single resultant force and specify its location measured from point O.

0.3 m O

1 kN/m 2.5 kN/m

SOLUTION Equivalent Resultant Force:

+ c FR = © Fy ;

FR

= 1(0.3) + 12 (2.5 - 1)(0.3) = 0.525 kN c

Ans.

Location of Equivalent Resultant Force:

1 2

a + MR

O

= © MO ;

12

1 2

1 2

0.525 d = 0.300 0.15 + 0.225 0.2 d

= 0.171 m

Ans.


*4–144.

The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O.

O

50 lb/ ft

100 lb/ ft

12 ft

SOLUTION

300 lb/ ft 9 ft

+ c FR = © Fy; FR = 50(12) + 12 (250)(12) + 12 (200)(9) + 100(9) = 3900 lb = 3.90 kip c

Ans.

1 1 a + MRo = © MO; 3900(d) = 50(12)(6) + 2 (250)(12)(8) + 2 (200)(9)(15) + 100(9)(16.5)

d

= 11.3 ft

Ans.


4–145.

Replace the distributed loading by an equivalent resultant force,and specify its location on the beam,measured from the pin at C. 30

A C

SOLUTION

+ T FR = © F;

c + MRC = © MC;

800 lb/ ft

FR

= 12 000 + 6000 = 18 000 lb

FR

= 18.0 kip T

15 ft

15 ft

Ans.

18 000x = 12 000(7.5) + 6000(20) x

= 11.7 ft

Ans.


B

4–146.

Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A.

w0

w0

A

B

L –– 2

L –– 2

SOLUTION Loadin g : The distributed loading can be divided into two parts as shown in Fig. a.

The magnitude and location of the resultant force of each part acting on the beam are also shown in Fig. a. Resultants: Equating the sum of the forces along the y axis of Figs. a and b,

+ T FR = © F;

FR

=

a b + 12 a 2 b = 12

1 L w0 2 2

w0

L

w0L T

Ans.

If we equate the moments of FR, Fig. b, to the sum of the moment of the forces in Fig. a about point A, a + (MR)A = © MA;

-

a b a 6 b - 12 a 2 b a 23 b

L 1 1 w 0L(x) = w 0 2 2 2 x

=

5 L 12

L

L

w 0

L

Ans.


4–147.

The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero.

b

40 lb/ ft a

SOLUTION

60 lb/ ft 10 ft

Require FR = 0.

6 ft

+ c FR = © Fy; 0 = 180 - 40b b

= 4.50 ft

Ans.

Require MRA = 0. Using the result b = 4.50 ft, we have

a

a + MRA = © MA; 0 = 180(12) - 40(4.50) a + a

= 9.75 ft

4.50 2

b Ans.


*4–148.

If the soil exerts a trapezoidal di stribution of load on the bottom of the footing, determine the intensities w1 and w2 of this distribution needed to support the column loadings.

80 kN

60 kN 1m

50 kN 2.5 m

3.5 m

1m

SOLUTION w2

Load in g : The trapezoidal reacti ve di stributed load can be divided into two parts as shown on the free-body diagram of the footing, Fig. a. The magnitude and location measured from point A of the resultant force of each part are al so indicated in Fig. a.

w1

Equat io ns of Equi li b r ium : Writing the moment equation of equilibrium about point B, we have

¢ ≤ ¢ ≤ ¢

a + © MB = 0; w2(8) 4 w2

8 3

+ 60

>

8 -1 3

- 80 3.5 -

8 3

≤ ¢ ≤ - 50 7 -

>

= 17.1875 kN m = 17.2 kN m

8 3

=0 Ans.

Using the result of w2 and writing the force equation of equilibrium along the y axis, we obtain

+ c © Fy = 0;

1 (w - 17.1875)8 + 17.1875(8) - 60 - 80 - 50 = 0 2 1 w1 = 30.3125 kN m = 30.3 kN m

>

>

Ans.


4–149.

The post is embedded into a concrete footing so that it is fixed supported. If the reaction of the concrete on the post can be approximated by the di stributed loading shown, determine the intensity of w1 and w2 s o that the resultant force and couple moment on the post due to the loadin gs are both zero.

30 lb/ ft

3 ft

3 ft w2

SOLUTION

1.5 ft

Load in g : The magnitude and location of the re sultant forces of each triangular distributed load are indicated in Fig. a. Resultants : The resultant force FR of the triangular distributed load is required to be zero. Referring to Fig. a and summing the forces along the x axis, we have

+ FR = 0 = ©Fx;

0 =

:

w1

1 1 1 (30)(3) + (w1)(1.5) - (w2)(1.5) 2 2 2

0.75w2 - 0.75w1 = 45

(1)

Also, the resultant couple moment MR of the triangular distributed load is required to be zero. Here, the moment will be summed about point A, as in Fig. a. a + (MR)A = 0 = © MA;

0 =

1 1 1 (w2)(1.5)(1) - (w1)(1.5)(0.5) - (30)(3)(6.5) 2 2 2

0.75w2 - 0.375w1 = 292.5

(2)

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

w1

>

= 660 lb ft

w2

>

= 720 lb ft

Ans.


4–150.


6 kN/ m

15 kN

500 kN m O

7.5 m

4.5 m

SOLUTION

+ c FR = © Fy ;

FR

= - 22.5 - 13.5 - 15.0

= - 51.0 kN = 51.0 kN T a + MRo = © Mo ;

MRo

Ans.

= - 500 - 22.5(5) - 13.5(9) - 15(12)

= - 914 kN # m = 914 kN # m (Clockwise )

Ans.


4–151.

Replace the loading by a single resultant force, and specify the location of the force measured from point O.

6 kN/ m

15 kN

500 kN m O

7.5 m

4.5 m


+ c FR = © Fy ;

- FR = - 22.5 - 13.5 - 15 FR

= 51.0 kN T

Ans.

Location of Equivalent Resultant Force:

a + (MR)O = © MO ;

- 51.0( d) = - 500 - 22.5(5) - 13.5(9) - 15(12) d

= 17.9 m

Ans.


*4–152.

Replace the loading by an equivalent resultant force and couple moment at point A.

50 lb/ ft 50 lb/ ft B

4 ft

SOLUTION

6 ft

=

F1

1 (6) (50) = 150 lb 2

100 lb/ ft 60

F2

= (6) (50) = 300 lb

F3

= (4) (50) = 200 lb

+ FRx = © Fx ;

FRx

+ T FRy = © Fy ;

FRy

:

FR

= u

c + MRA = © MA ;

A

= 150 sin 60° + 300 sin 60° = 389.71 lb = 150 cos 60° + 300 cos 60° + 200 = 425 lb

2 (389.71) = tan - 1 MRA

2

+ (425)2 = 577 lb

425 a 389.71 b = 47.5°

c

Ans.

Ans.

= 150 (2) + 300(3) + 200(6 cos 60° + 2)

= 2200 lb # ft = 2.20 kip # ft b

Ans.


4–153.

Replace the loading by an equivalent resultant force and couple moment acting at point B.

50 lb/ ft 50 lb/ ft B

4 ft

SOLUTION

6 ft

1 (6) (50) = 150 lb 2 F2 = (6) (50) = 300 lb

F1

F2

=

100 lb/ ft 60 A

= (4) (50) = 200 lb

+ FRx = © Fx ;

FRx

= 150 sin 60° + 300 sin 60° = 389.71 lb

+ T FRy = © Fy ;

FRy

= 150 cos 60° + 300 cos 60° + 200 = 425 lb

:

FR

=

2 (389.71)

u

a + MRB = © MB ;

= tan - 1

MRB

2

+ (425)2 = 577 lb

425 a 389.71 b = 47.5°

c

Ans.

Ans.

= 150 cos 60° (4 cos 60° + 4) + 150 sin 60° (4 sin 60°)

+ 300 cos 60° (3 cos 60° + 4) + 300 sin 60° (3 sin 60°) + 200 (2) MRB

= 2800 lb # ft = 2.80 kip # ftd

Ans.


4–154.

Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A.

200 N/ m 100 N/ m B C

6m 5m

SOLUTION

+ © FRx = © Fx ;

;

FRx

= 1000 N

+ T FRy = © Fy ;

FRy

= 900 N

2 (1000)

FR

=

FR

= 1.35 kN

u

= tan-1

2

+ (900)2 = 1345 N

A

Ans.

900 c 1000 d = 42.0°

a + M RA = © MA ;

200 N/ m

d

Ans.

1000y = 1000(2.5) - 300(2) - 600(3) y

= 0.1m

Ans.


4–155.

Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C.

200 N/ m 100 N/ m B C

6m 5m

SOLUTION

+ © FRx = © Fx ;

;

+ T FRy = © Fy ;

2 (1000)

FR

=

FR

= 1.35 kN

u

= tan-1

2

FRx

= 1000 N

FRy

= 900 N

+ (900)2 = 1345 N

A

Ans.

900 c 1000 d = 42.0°

a + M RC = © MC ;

200 N/ m

d

Ans.

900x = 600(3) + 300(4) - 1000(2.5) x

= 0.556 m

Ans.


*4–156.

Replace the distributed loading with an equi valent resultant force, and specify its location on the beam measured from point A.

w p w  w0 sin ( –– x) 2L

w0

x

A

SOLUTION

L

Resultant : The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a.Thus, dFR

¢

= w dx = w0 sin

p

2L

≤

x dx

Integrating dFR over the entire len gth of the beam gives the resultant force FR.

+T

FR

=

L

L¢ L

dFR

=

0

L

w0 sin

p

2L

≤ ¢

x dx

= -

2w0L p

cos

p

2L

x

≤`

L

0

=

2w0L p

T

Ans.

Locat io n : The location of dFR on the beam is xc = x measured from point A.Thus, the location x of FR measured from point A is given by

L L

xcdFR

x

=

L

L

L ¢ L

= dFR

0

x w0 sin

p

2L

2w0L p

≤

4w0L2

x dx

=

p2

2w0L

=

2L p

Ans.

p


4–157.

Replace the di stributed loading with an equi valent resultant force, and specify its location on the beam measured from point A.

w

10 kN/ m

w

1 ( x2  4 x  60) kN/ m –– 6

A

B

6m

SOLUTION Resultant : The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a.Thus,

1 6

= w dx = ( - x2 - 4x + 60)dx

dFR


+T

FR

=

L

=

dFR

L 0

L

6m

B

x3 1 1 - - 2x2 + 60x ( - x2 - 4x + 60)dx = 6 6 3

= 36 kN T

R`

6m 0

Ans.

Locat io n : The location of dFR on the beam is xc = x, measured from point A.Thus the location x of FR measured from point A is

= L L

xcdFR

x

B L =

R L

6m

L

dFR

0

1 x (- x2 - 4x + 60) dx 6 36

=

0

6m

¢

1 x4 4x3 1 3 2 - + 30x2 ( - x - 4x + 60x)dx 6 4 3 6 = 36 36

≤`

6m 0

L

= 2.17 m

Ans.


x

4–158.

Replace the distributed loading with an equi valent resultant force, and specify its location on the beam measured from point A.

w

2 w  ( x  3 x  100)

lb/ ft

370 lb/ ft

100 lb/ ft A

x B

SOLUTION 15 ft

Resultant : The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a.Thus,

a

b

= w dx = x2 + 3x + 100 dx

dFR


+T

FR

=

L

L

dFR

L

=

La

x2

b ¢ 3 + 32

+ 3x + 100 dx =

0

x3

x2

+ 100x

≤`

15 ft 0

= 2962.5 lb = 2.96 kip

Ans.

Locat io n : The location of dFR on the beam is xc = x measured from point A.Thus, the location x of FR measured from point A is given by

= L L

15 ft

xcdFR

x

L

dFR

a L =

x x

2

b

+ 3x + 100 dx

0

2962.5

=

¢

x4

4

3

+ x + 50x 2962.5

2

≤`

15 ft 0

= 9.21 ft

Ans.

L


4–159.

Wet concrete exerts a pressure distribution along the wall of the form. Determine the resultant force of this distribution and specify the height h where the bracing strut should be placed so that it lies through the line of action of the resultant force.The wall has a width of 5 m.

p

4m

p

1

(4 z /2) kPa


+ FR = © Fx;

- FR = - LdA = -

:

4m

L

h

z

8 kPa

dz

w

0

L a 20z b A 10 B dz

FR =

1 2

3

0

= 106.67 A 103 B N = 107 kN

z

;

Ans.

Location of Equivalent Resultant Force: z

z dz L L = z = L dA L dz z c A 20z B (10 ) d dz L = L A 20z B (10 )dz c A 20z B (10 ) d dz L = L A 20z B (10 )dz zdA

A

w

0

z

w

A

0

4m

1 2

3

0

4m

1 2

3

0

4m

3 2

3

1 2

3

0

4m

0

= 2.40 m Thus,

h = 4 - z = 4 - 2.40 = 1.60 m

Ans.


*4–160.


w

1 ––

w = (200 x 2 ) N/m

600 N/m

x

O 9m

SOLUTION Equivalent Resultant Force And Moment At Point O:

+ c FR = © Fy ;

L = L

FR = FR

L

dA = -

x

wdx

0

A

9m

1 2

A 200x B dx

0

= - 3600 N = 3.60 kN T

Ans.

x

a + MRO = © MO ;

MRO = -

L x dx w

0

L = L

9m

0

0

1

x A 200x2 B dx

= -

9m

3 2

A 200x B dx

= - 19 440 N # m = 19.4 kN # m (Clockwise)

Ans.


4–161.

Determine the magnitude of the equivalent resultant force of the distributed load and specify its location on the beam measured from point A.

w 420 lb/ft

w = (5 ( x – 8) 2 +100) lb/ft 100 lb/ft

120 lb/ft

A

x

SOLUTION 8 ft

Equivalent Resultant Force:

+ c FR = © Fy ;

2 ft

x

L L dx = L 351x - 82 + 1004dx

- FR = -

dA = -

w

0

A

10 ft

FR

2

0

= 1866.67 lb = 1.87 kip T

Ans.

Location of Equivalent Resultant Force: x

xdA x dx L L = x = L dA L dx x351x - 82 + 1004dx L = L 351x - 82 + 1004dx 15x - 80x + 420x2dx L = L 351x - 82 + 1004dx

'

w

0

A

x

w

A

0

10 ft

2

0

10 ft

2

0

10 ft

3

2

0

10 ft

2

0

= 3.66 ft

Ans.


■4–162.

Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integral using Simpson ’s rule.

w w

5 x

(16

x

2 1/2

)

kN/ m 5.07 kN/ m

2 kN/ m A

3m

SOLUTION FR =

L

L 4 5 4

wdx =

1

0

L

Ans.

L 4 5 4

x dF =

0

1m

x + (16 + x2)2 dx

FR = 14.9 kN 4

x

B

(x)

1

x + (16x + x2 )2 dx

0

= 33.74 kN # m x =

33.74 = 2.27 m 14.9

Ans.


4–163.

Determine the resultant couple moment of the two couples that act on the assembly. Member OB lies in the x-z plane. z

y A 400 N

O

500 mm

150 N

SOLUTION

x 45

For the 400-N forces: 600 mm

1 2

MC1 = rAB * 400i

3

i

= 0.6 cos 45°

°

j

k

- 0.5

- 0.6 sin 45°

0

0

400

= - 169.7 j + 200k

3

B 400 mm

400 N

C 150 N

For the 150-N forces:

1 2

MC2 = rOB * 150 j

3

i

= 0.6 cos 45° 0

j

k

0 150

- 0.6 sin 45° 0

= 63.6i + 63.6k

3

M CR = M C1 + M C2 M CR =

#

63.6i - 170 j + 264k N m

Ans.


*4–164.

The horizontal 30-N force acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis?

z

200 mm B

A

45 45

30 N

10 mm

50 mm O

SOLUTION x

Position Vector And Force Vectors: rBA = { - 0.01i + 0.2 j} m rOA = [( - 0.01 - 0)i + (0.2 - 0) j + (0.05 - 0)k} m

= { - 0.01i + 0.2 j + 0.05k} m F = 30(sin 45°i - cos 45° j) N

= [21.213 i - 21.213 j] N Moment of Force F About z Axis: The unit vector along the z axis is k. Applying

Eq. 4–11, we have

#

Mz = k (rBA * F)

3

0 = - 0.01 21.213

0 0.2 - 21.213

1 0 0

3

= 0 - 0 + 1[( - 0.01)( - 21.213) - 21.213(0.2)] = - 4.03 N # m

Ans.

Or

#

Mz = k (rOA * F)

3

0 = - 0.01 21.213

0 0.2 - 21.213

1 0.05 0

3

= 0 - 0 + 1[( - 0.01)( - 21.213) - 21.213(0.2)] = - 4.03 N # m

Ans.

The negative sign indicates that Mz, is directed along the negative z axis.


y

4–165.

The horizontal 30-N force acts on the handle of the wrench. Determine the moment of this force about point O. Specify the coordinate direction angles a, b , g of the moment axis.

z

30 N

200 mm

B



A

45 45



10 mm

50 mm

O

SOLUTION x

Position Vector And Force Vectors: rOA

F

=

{( - 0.01

-

0)i

=

{ - 0.01i

+

0.2 j

=

30(sin 45°i

-

=

{21.213i

21.213 j} N

-

(0.2

+ +

0) j

-

+

(0.05

-

0)k} m

0.05k} m

cos 45° j) N

Moment of Force F About Point O: Applying Eq. 4–7, we have MO

=

rOA

F

=

3

=

{1.061i

=

{1.06i

*

i - 0.01 21.213

j 0.2 - 21.213 +

+

1.061 j

1.06 j

-

k 0.05 0

3

4.031k} N # m

-

4.03k} N # m

Ans.

The magnitude of MO is MO =

2 1.0612

+

1.0612

+

( - 4.031)2

=

4.301 N # m

The coordinate direction angles for MO are a = cos - 1

a 1.061 b 4.301

=

75.7°

Ans.

b = cos - 1

a 1.061 b 4.301

=

75.7°

Ans.

g = cos - 1

4.301 a 4.301 b -

=

160°

Ans.


y

4–166.

The forces and couple moments that are exerted on the toe and heel plates of a snow ski are Ft = - 50i + 80 j - 158k N, M t = - 6i + 4 j + 2k N m, and Fh = - 20i + 60 j - 250k N, M h = - 20i + 8 j + 3k N m, respectively. Replace this system by an equivalent force and couple moment acting at point P . Express the results in Cartesian vector form.

5 5

6

5

6

#

6

5

6

z

P

#

Fh Ft

O

Mh 800 mm

Mt

120 mm y x

SOLUTION FR = Ft + Fh = { - 70i + 140 j - 408k} MRP =

3

i

j

0.8 - 20

0 60

k

33

i

N j

+ 0.92 0 - 250 - 50

0 80

k

3

Ans.

+ ( - 6i + 4 j + 2k) + ( - 20i + 8 j + 3k) 0 - 158

MRP = (200 j + 48k) + (145.36 j + 73.6k) + ( - 6i + 4 j + 2k) + ( - 20i + 8 j + 3k)

#

MRP = { - 26i + 357.36 j + 126.6k} N m

#

MRP = { - 26i + 357 j + 127k} N m

Ans.


4–167.

Replace the force F having a magnitude of F = 50 lb and acting at point A by an equivalent force and couple moment at point C .

z

A

30 ft

C

F

SOLUTION FR FR

MRC

MA

=

=

50

B

O

(10i

+

15 j

2 (10)2

+

(15)2

{14.3i

*

+

21.4 j

=

rCB

F

=

=

{ - 1929i

+

=

{ - 1.93i

+

3

-

-

+

( - 30)2

10 ft

R

428.6 j

j 45 21.43 -

-

20 ft

x

42.9k} lb

i 10 14.29

0.429 j

30k)

Ans. k 0 - 42.86

15 ft

3

B

10 ft

428.6k} lb # ft 0.429k} kip # ft

Ans.


y

*4–168.

Determine the coordinate direction angles a, b , g of F,which is applied to the end A of the pipe assembly, so that the moment of F about O is zero.

20 lb

F

z

y

O

10 in.

6 in.

SOLUTION A

Require MO = 0. This happens when force F is directed along line OA either from point O to A or from point A to O. The unit vectors uOA and uAO are x uOA =

8 in.

6 in.

(6 - 0) i + (14 - 0) j + (10 - 0) k

2 (6 - 0)

2

+ (14 - 0)2 + (10 - 0)2

= 0.3293 i + 0.7683 j + 0.5488 k Thus, a

= cos - 1 0.3293 = 70.8°

Ans.

b

= cos - 1 0.7683 = 39.8°

Ans.

g

= cos - 1 0.5488 = 56.7°

Ans.

or uAO =

(0 - 6)i + (0 - 14) j + (0 - 10) k

2 (0 - 6)

2

+ (0 - 14)2 + (0 - 10)2

= - 0.3293 i - 0.7683 j - 0.5488 k Thus, a

= cos - 1 ( - 0.3293) = 109°

Ans.

b

= cos - 1 ( - 0.7683) = 140°

Ans.

g

= cos - 1 ( - 0.5488) = 123°

Ans.


4–169.

Determine the moment of the force F about point O. The force has coordinate direction angles of a = 60°, b = 120°, g = 45°. Express the result as a Cartesian vector.

20 lb

F

z

y

O

10 in.

6 in.

SOLUTION A

Position Vector And Force Vectors: 8 in.

rOA = {(6 - 0)i + (14 - 0) j + (10 - 0) k} in.

x

6 in.

= {6i + 14 j + 10k} in. F = 20(cos 60°i + cos 120° j + cos 45°k) lb

= (10.0i - 10.0 j + 14.142 k} lb Moment of Force F About Point O: Applying Eq. 4–7, we have MO = rOA * F

=

3

i

j

k

6 10.0

14 - 10.0

10 14.142

3

= 298i + 15.1 - 200k lb # in

Ans.


4–170.

Determine the moment of the force Fc about the door hinge at A. Express the result as a Cartesian vector.

z

C

1.5 m

2.5 m F C

250 N

a

SOLUTION Position Vector And Force Vector:

30

A

rAB = {[ - 0.5 - ( - 0.5)]i + [0 - ( - 1)] j + (0 - 0)k} m = {1 j} m

FC = 250

§

[ - 0.5 - ( - 2.5)]i + {0 - [ - (1 + 1.5 cos 30°)]} j + (0 - 1.5 sin 30°) k

B

[ -0.5 - (-2.5)]2 + {0 - [ - (1 + 1.5 cos 30°)]}2 + (0 - 1.5 sin 30°) 2

¥

a

B

1m

0.5 m

N x

y

= [159.33i + 183.15 j - 59.75 k]N Moment of Force Fc About Point A: Applying Eq. 4–7, we have MA = rAB * F

=

3

i

j

k

0 159.33

1 183.15

0 - 59.75

= - 59.7i - 159k N # m

3

Ans.


4–171.

Determine the magnitude of the moment of the force Fc about the hinged axis aa of the door.

z

C

1.5 m

2.5 m F C

250 N

a

SOLUTION Position Vector And Force Vectors:

30

A

rAB = {[ - 0.5 - ( - 0.5)]i + [0 - ( - 1)] j + (0 - 0)k} m = {1 j} m

FC = 250

§

{ - 0.5 - ( - 2.5)]i + {0 - [ - (1 + 1.5 cos 30°)]} j + (0 - 1.5 sin 30°) k

B

[ - 0.5 - ( - 2.5)]2 + {0 - [ - (1 + 1.5 cos 30°)]}2 + (0 - 1.5 sin 30°)2

¥

a

B

1m

0.5 m

N x

y

= [159.33 i + 183.15 j - 59.75k] N Moment of Force Fc About a - a Axis: The unit vector along the a – a axis is i.

Applying Eq. 4–11, we have

#

Ma - a = i (rAB * FC)

=

3

1 0 159.33

0 1 183.15

0 0 - 59.75

3

= 1[1( - 59.75) - (183.15)(0)] - 0 + 0 = - 59.7 N # m The negative sign indicates that Ma - a is directed toward the negative x axis.

#

Ma - a = 59.7 N m

Ans.


*4–172.

The boom has a length of 30 ft, a weight of 800 lb, and mass center at G. If the maximum moment that can be developed by the motor at A is M = 20 103 lb # ft, determine the maximum load W , having a mass center at G¿ , that can be lifted. Take u = 30°.

800 lb

1 2

14 ft ¿

G

W =

800(16 cos 30°)

319 lb

M

G

2 ft

SOLUTION =

A

u

W

20(103)

16 ft

+ W(30

cos 30°

+

2) Ans.


engineering mechanics 13e chap 04 solution

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