Engingeering Mechanics Dynamics in SI Units 14th Edition Hibbeler Solutions Manual

Engingeering Mechanics Dynamics in SI Units 14th Edition Hibbeler SOLUTIONS MANUAL Full download at: https://testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14thedition-hibbeler-solutions-manual/ 22–1. A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m > s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s.

SOLUTION $ mg - k(y + yst) = my

+ T ΣFy = may;

Hence

p =

=

k Bm

B

where kyst = mg

k $ y + y = 0 m 8(9.81) Where k = = 448.46 N > m 0.175

448.46 = 7.487 8 $ y + (7.487)2y = 0

6

$ y + 56.1y = 0

Ans.

The solution of the above differential equation is of the form: y = A sin pt + B cos pt

(1)

# v = y = Ap cos pt - Bp sin pt

(2)

At t = 0, y = 0.1 m and v = v0 = 1.50 m > s From Eq. (1) From Eq. (2)

0 .1 = A sin 0 + B cos 0 v0 = Ap cos 0 - 0

B = 0.1 m A =

v0 1.50 = = 0.2003 m p 7.487

Hence

y = 0.2003 sin 7.487t + 0.1 cos 7.487t

At t = 0.22 s,

y = 0.2003 sin [7.487(0.22)] + 0.1 cos [7.487(0.22)] = 0.192 m

Ans.

1193

Ans: y + 56.1 y = 0 y 0 t = 0.22 s = 0.192 m

1194

22–2. A spring has a stiffness of 800 N > m. If a 2-kg block is attached to the spring, pushed 50 mm above its equilibrium position, and released from rest, determine the equation that describes the block’s motion. Assume that positive displacement is downward.

SOLUTION p =

800 k = = 20 Am A 2

2-Kg

x = A sin pt + B cos pt x = - 0.05 m

when t = 0,

- 0.05 = 0 + B;

B = - 0.05

v = Ap cos pt - Bp sin pt v = 0 when t = 0, 0 = A(20) - 0;

A = 0

Thus, x = - 0.05 cos (20t)

Ans.

Ans: x = - 0.05 cos (20t)

1195

22–3. A spring is stretched 200 mm by a 15-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 0.75 m>s, determine the equation which describes the motion. What is the phase angle? Assume that positive displacement is downward.

SOLUTION k =

15(9.81) F = 735.75 N>m = 0.2 y

vn =

735.75 k = 7.00 = Am A 15

y = A sin vn t + B cos vn t y = 0.1 m when t = 0, 0.1 = 0 + B;

B = 0.1

v = A vn cos vn t - Bvn sin vn t v = 0.75 m>s when t = 0, 0.75 = A(7.00) A = 0.107 y = 0.107 sin (7.00t) + 0.100 cos (7.00t) f = tan - 1 a

Ans.

B 0.100 b = tan - 1 a b = 43.0° A 0.107

Ans.

Ans: y = 0.107 sin (7.00t) + 0.100 cos (7.00t) f = 43.0°

1196

*22–4 When a 2-kg block is suspended from a spring, the spring is stretched a distance of 40 mm. Determine the frequency and the period of vibration for a 0.5-kg block attached to the same spring.

SOLUTION k =

2(9.81) F = 490.5 N>m = 0.040 y

vn =

490.5 k = 31.321 = Am A 0.5

f =

31.321 vn = 4.985 = 4.98 Hz = 2p 2p

t =

1 1 = = 0.201 s f 4.985

Ans.

Ans.

Ans: f = 4.98 Hz t = 0.201 s

1197

22–5. When a 3-kg block is suspended from a spring, the spring is stretched a distance of 60 mm. Determine the natural frequency and the period of vibration for a 0.2-kg block attached to the same spring.

SOLUTION k =

vn =

3(9.81) F = = 490.5 N>m ¢x 0.060 490.5 k = 49.52 = 49.5 rad >s = m A A 0.2

f =

vn 49.52 = 7.88 Hz = 2p 2p

t =

1 1 = = 0.127 s f 7.88

Ans.

Ans.

Ans: vn = 49.5 rad > s t = 0.127 s

1198

Engingeering Mechanics Dynamics in SI Units 14th Edition Hibbeler SOLUTIONS MANUAL Full download at: https://testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14thedition-hibbeler-solutions-manual/ engineering mechanics: dynamics in si units pdf engineering mechanics statics 14th edition in si units pdf engineering mechanics: statics in si units (14e) pdf engineering mechanics dynamics hibbeler 14th edition pdf engineering mechanics dynamics 14th edition pdf free download hibbeler dynamics 14th edition solutions pdf engineering mechanics dynamics pdf engineering mechanics dynamics 14th edition solution manual pdf

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