Solutions Manual Engineering Mechanics: Dynamics 14th edition by Russell C. Hibbeler

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Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m>s2, where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s?

Solution a = 2t - 6 dv = a dt L0

v

dv =

L0

t

(2t - 6) dt

v = t 2 - 6t ds = v dt L0

s

ds =

s =

L0

t

(t2 - 6t) dt

t3 - 3t2 3

When t = 6 s, Ans.

v = 0 When t = 11 s,

Ans.

s = 80.7 m

Ans: s = 80.7 m

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1

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A golf ball is struck with a velocity of 80 ft>s as shown.

Determine the speed at which it strikes the ground at B and the time of flight from A to B.

vA 80 ft/s B A

45

10 d

Solution (vA)x = 80 cos 55° = 44.886 (vA)y = 80 sin 55° = 65.532 +2 1S

s = s0 + v0t

d cos 10° = 0 + 45.886t 1 1+ c 2 s = s0 + v0t + act 2 2

d sin 10° = 0 + 65.532 (t) +

1 ( - 32.2) ( t 2 ) 2

d = 166 ft Ans.

t = 3.568 = 3.57 s (vB)x = (vA)x = 45.886

1 + c2

v = v0 + act

(vB)y = 65.532 - 32.2(3.568) (vB)y = -49.357 vB = 2(45.886)2 + ( - 49.357)2

Ans.

vB = 67.4 ft>s

Ans: t = 3.57 s vB = 67.4 ft>s

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Solve Prob. $12–189 if the particle has an angular # acceleration u = 5 rad>s2 when u = 4 rad>s at u = p> 2 rad.

y

r

(8 u) ft

r

SOLUTION

u x

Time Derivatives: Here, p r = 8u = 8 a b = 4p ft 2 $ $ r = 8 u = 8(5) = 40 ft>s2

# # r = 8 u = 8(4) = 32.0 ft>s

Velocity: Applying Eq. 12–25, we have # vr = r = 32.0 ft>s # vu = r u = 4p(4) = 50.3 ft>s

Ans. Ans.

Acceleration: Applying Eq. 12–29, we have # $ ar = r - r u2 = 40 - 4p A 42 B = - 161 ft>s2 $ # # au = r u + 2r u = 4p(5) + 2(32.0)(4) = 319 ft>s2

Ans. Ans.

Ans: vr vu ar au

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= = = =

32.0 ft>s 50.3 ft>s - 161 ft>s2 319 ft>s2

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Determine the maximum constant speed at which the pilot can travel around the vertical curve having a radius of curvature r = 800 m, so that he experiences a maximum acceleration an = 8g = 78.5 m>s2. If he has a mass of 70 kg, determine the normal force he exerts on the seat of the airplane when the plane is traveling at this speed and is at its lowest point.

r 800 m

Solution an =

v2 v2 ; 78.5 = r 800 Ans.

v = 251 m>s + c ΣFn = man; N - 70(9.81) = 70(78.5)

Ans.

N = 6.18 kN

Ans: N = 6.18 kN

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The motor pulls on the cable at A with a force F = (e 2t) lb, where t is in seconds. If the 34-lb crate is originally at rest on the ground at t = 0, determine the crate’s velocity when t = 2 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate.

A

Solution F = e 2t = 34 t = 1.7632 s for crate to start moving

(+c)

mv1 + Σ 0#

L

Fdt = mv2

2

L1.7632

e 2tdt - 34(2 - 1.7632) =

34 v 32.2 2

2

1 2t e - 8.0519 = 1.0559 v2 2 L1.7632 Ans.

v2 = 2.13 m>s

Ans: v2 = 2.13 m>s

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499

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The blade divides the jet of water having a diameter of 3 in.

If one-fourth of the water flows downward while the other three-fourths flows upwards, and the total flow is Q = 0.5 ft3>s, determine the horizontal and vertical components of force exerted on the blade by the jet, gw = 62.4 lb>ft3.

3 in.

SOLUTION Equations of Steady Flow: Here, the flow rate Q = 0.5 ft2>s. Then, Q 0.5 dm 62.4 y = = = rw Q = (0.5) = 0.9689 slug>s. = 10.19 ft>s. Also, p 3 2 A dt 32.2 A B 4

12

Applying Eq. 15–25 we have ©Fx = ©

dm A youts - yins B ; - Fx = 0 - 0.9689 (10.19) dt

©Fy = ©

1 dm 3 A youty - yiny B ; Fy = (0.9689)(10.19) + (0.9689)(- 10.19) dt 4 4

Fx = 9.87 lb

Ans.

Ans.

Fy = 4.93 lb

Ans: Fx = 9.87 lb Fy = 4.93 lb

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599

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Determine the moment of inertia of the homogeneous triangular prism with respect to the y axis. Express the result in terms of the mass m of the prism. Hint: For integration, use thin plate elements parallel to the x–y plane and having a thickness dz.

z

–h (x – a) z = –– a

h

SOLUTION dV = bx dz = b(a)(1 -

z ) dz h

x

b

a

y

x dIy = dIy + (dm)[( )2 + z2] 2 =

x2 1 dm(x2) + dm( ) + dmz2 12 4

= dm(

x2 + z2) 3

= [b(a)(1 -

a2 z z )dz](r)[ (1 - )2 + z2] h 3 h

k

Iy = abr

a3 h - z 3 z ( ) + z2(1 - )]dz h h L0 3

= abr[ =

[

3 1 1 1 1 a2 (h4 - h4 + h4 - h4) + ( h4 - h4 )] 2 4 h 3 4 3h3

1 abhr(a2 + h2) 12

m = rV =

1 abhr 2

Thus, Iy =

m 2 (a + h2) 6

Ans.

Ans: Iy =

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799

m 2 ( a + h2 ) 6

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A 4-kg disk A is mounted on arm BC, which has a negligible mass. If a torque of M = 15e0.5t2 N # m, where t is in seconds, is applied to the arm at C, determine the angular velocity of BC in 2 s starting from rest. Solve the problem assuming that (a) the disk is set in a smooth bearing at B so that it rotates with curvilinear translation, (b) the disk is fixed to the shaft BC, and (c) the disk is given an initial freely spinning angular velocity of V D = 5 -80k6 rad>s prior to application of the torque.

250 mm A 60 mm

B M C

#

(5e0.5 t) N m

SOLUTION a) (Hz)1 + ©

L

Mz dt = (Hz)2

2

0+

L0

5e0.5 tdt = 4(vB)(0.25)

5 0.5 t 2 e 2 = vB 0.5 0 vB = 17.18 m>s Thus, 17.18 = 68.7 rad>s 0.25

vBC =

Ans.

b) (Hz)1 + © 2

0 +

L0

L

Mz dt = (Hz)2

1 5e0.5 tdt = 4(vB)(0.25) + c (4)(0.06)2 d vBC 2

Since vB = 0.25 vBC, then Ans.

vBC = 66.8 rad>s c) (Hz)1 + ©

L

Mz dt = (Hz)2

2

1 1 - c (4)(0.06)2 d(80) + 5e0.5 tdt = 4(vB)(0.25) - c (4)(0.06)2 d(80) 2 2 L0 Since vB = 0.25 vBC, Ans.

vBC = 68.7 rad>s

Ans: (a) vBC = 68.7 rad>s (b) vBC = 66.8 rad>s (c) vBC = 68.7 rad>s

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999

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The uniform rod of mass m is supported by a pin at A and a spring at B. If B is given a small sideward displacement and released, determine the natural period of vibration.

A

L

Solution 1 Equation of Motion. The mass moment of inertia of the rod about A is IA = mL2.

3 Referring to the FBD. of the rod, Fig. a, L 1 a + ΣMA = IAa ; - mg a sin u b - (kx cos u)(L) = a mL2 ba 2 3

B

k

However; x = L sin u. Then

- mgL 1 sin u - kL2 sin u cos u = mL2a 2 3 Using the trigonometry identity sin 2u = 2 sin u cos u, - mgL KL2 1 sin u sin 2u = mL2a 2 2 3

$ Here since u is small sin u ≃ u and sin 2u ≃ 2u. Also a = u . Then the above equation becomes $ mgL 1 mL2 u + a + kL2 bu = 0 3 2 $ 3mg + 6kL u + u = 0 2mL

Comparing to that of the Standard form, vn = t =

A

3mg + 6kL . Then 2mL

2p 2mL = 2p

vn A 3mg + 6kL

Ans.

Ans: t = 2p

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1199

2mL A 3mg + 6kL