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Spring Problems Part 2-Engineering Mechanics-Assignment Solution - Document's extract

Exercises, Dynamics

Post: July 16th, 2012
Extract
Engineering Mechanics - Dynamics

Chapter 15

e( −vB1y cos ( θ ) − vB1x sin ( θ ) ) = vB2n

⎛ vB1x ⎞ ⎜ ⎟ ⎜ vB1y ⎟ ⎜ vB2n ⎟ ⎜ ⎟ = Find ( vB1x , vB1y , vB2n , vB2t , t , R) ⎜ vB2t ⎟ ⎜t⎟ ⎜ ⎟ ⎝R⎠ ⎛ vB2n ⎞ ⎛ 1.70 ⎞ ft ⎜ ⎟=⎜ ⎟ vB2t ⎠ ⎝ 6.36 ⎠ s ⎝
*Problem 15-72

⎛ vB1x ⎞ ⎛ 3.00 ⎞ ft ⎜ ⎟=⎜ ⎟ ⎝ vB1y ⎠ ⎝ −6.00 ⎠ s
t = 0.19 s R = 0.79 ft

⎛ vB2n ⎞ ft ⎜ ⎟ = 6.59 s ⎝ vB2t ⎠

The drop hammer H has a weight WH and falls from rest h onto a forged anvil plate P that has a weight WP. The plate is mounted on a set of springs that have a combined stiffness kT . Determine (a) the velocity of P and H just after collision and (b) the maximum compression in the springs caused by the impact. The coefficient of restitution between the hammer and the plate is e. Neglect friction along the vertical guideposts A and B. Given: WH = 900 lb WP = 500 lb h = 3 ft Solution: kT = 500 g = 32.2 e = 0.6 lb ft ft s
2

δ st =
Guesses

WP kT

vH1 =

2g h

vH2 = 1 Given

ft s

vP2 = 1

ft s

δ = 2 ft

⎛ WH ⎞ ⎜ ⎟ vH1 = ⎝g⎠

⎛ WH ⎞ ⎛ WP ⎞ ⎜ ⎟ vH2 + ⎜ ⎟ vP2 ⎝g⎠ ⎝g⎠

e vH1 = vP2 − vH2

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Engineering Mechanics - Dynamics

Chapter 15

1 1 2 2 1 ⎛ WP ⎞ 2 kTδ st + ⎜ ⎟ vP2 = kTδ − WP( δ − δ st) 2 2⎝ g ⎠ 2

⎛ vH2 ⎞ ⎜ ⎟ ⎜ vP2 ⎟ = Find ( vH2 , vP2 , δ ) ⎜ ⎟ ⎝δ⎠

⎛ vH2 ⎞ ⎛ 5.96 ⎞ ft ⎜ ⎟=⎜ ⎟ ⎝ vP2 ⎠ ⎝ 14.30 ⎠ s

δ = 3.52 ft

Problem 15-73 It was observed that a tennis ball when served horizontally a distance h above the ground strikes the smooth ground at B a distance d away. Determine the initial velocity vA of the ball and the velocity vB (and θ) of the ball just after it strikes the court at B. The coefficient of restitution is e. Given: h = 7.5 ft d = 20 ft e = 0.7 g = 32.2 Solution: Guesses vA = 1 ft s ft s vB2 = 1 ft s t =1s ft s
2

vBy1 = 1

θ = 10 deg

Given

h=

12 gt 2

d = vA t vBy1 =
g t

e vBy1 = vB2 sin ( θ ) vA = vB2 cos ( θ )

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Engineering Mechanics - Dynamics

Chapter 15

⎛ vA ⎞ ⎜ ⎟ t⎟ ⎜ ⎜ vBy1 ⎟ = Find ( v , t , v , v , θ ) A By1 B2 ⎜ ⎟ ⎜ vB2 ⎟ ⎜ ⎟ ⎝θ⎠

vA = 29.30

ft s

vB2 = 33.10

ft s

θ = 27.70 deg

Problem 15-74 The tennis ball is struck with a horizontal velocity vA, strikes the smooth ground at B, and bounces upward at θ = θ1. Determine the initial velocity vA, the final velocity vB, and the coefficient of restitution between the ball and the ground. Given: h = 7.5 ft d = 20 ft

θ 1 = 30 deg
g = 32.2 Solution: Guesses ft s
2

θ = θ1
vA = 1 h= ft s t =1s d = vA t vBy1 = 1 ft s vB2 = 1 ft s e = 0.5

Given

12 gt 2

vBy1 = g t vA = vB2 cos ( θ )

e vBy1 = vB2 sin ( θ )

⎛ vA ⎞ ⎜ ⎟ ⎜t⎟ ⎜ vBy1 ⎟ = Find ( v , t , v , v , e) A By1 B2 ⎜ ⎟ ⎜ vB2 ⎟ ⎜e⎟ ⎝ ⎠
Problem 15-75

vA = 29.30

ft s

vB2 = 33.84

ft s

e = 0.77

The ping-pong ball has mass M. If it is struck with the velocity shown, determine how high h it rises above the end of the smooth table after the rebound. The coefficient of restitution is e.
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Engineering Mechanics - Dynamics

Chapter 15

Given: M = 2 gm e = 0.8 a = 2.25 m b = 0.75 m g = 9.81 m s
2

θ = 30 deg
m v = 18 s Solution:

Guesses

v1x = 1 t1 = 1 s

m s

v1y = 1 t2 = 2 s

m s

v2x = 1

m s

v2y = 1

m s

h =1m v1y = g t1 + v sin ( θ ) b = v2x t2 h = v2y t2 −

Given

v1x = v cos ( θ ) v2x = v1x

a = v cos ( θ ) t1 e v1y = v2y

⎛ g⎞t 2 ⎜ ⎟2 ⎝ 2⎠

⎛ v1x ⎞ ⎜⎟ ⎜ v1y ⎟ ⎜ v2x ⎟ ⎜⎟ ⎜ v2y ⎟ = Find ( v1x , v1y , v2x , v2y , t1 , t2 , h) ⎜t ⎟ ⎜ 1⎟ ⎜ t2 ⎟ ⎜⎟ ⎝h⎠

⎛ v1x ⎞ ⎛ 15.59 ⎞ ⎜⎟⎜ ⎟ ⎜ v1y ⎟ = ⎜ 10.42 ⎟ m ⎜ v2x ⎟ ⎜ 15.59 ⎟ s ⎜⎟⎜ ⎟ ⎝ v2y ⎠ ⎝ 8.33 ⎠

⎛ t1 ⎞ ⎛ 0.14 ⎞ ⎜ ⎟=⎜ ⎟s ⎝ t2 ⎠ ⎝ 0.05 ⎠
h = 390 mm

*Problem 15-76 The box B of weight WB is dropped from rest a distance d from the top of
the plate P of weight WP, which is supported by the spring having a stiffness k. Determine the maximum compression imparted to the spring. Neglect the mass of the spring.

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Engineering Mechanics - Dynamics

Chapter 15

Given:

WB = 5 lb lb k = 30 ft

WP = 10 lb d = 5 ft

g = 32.2 e = 0.6

ft s
2

Solution:

δ st =
Guesses

WP k

vB1 = ft s

2g d vP2 = 1 ft s

vB2 = 1

δ = 2 ft

Given

⎛ WB ⎞ ⎛ WB ⎞ ⎛ WP ⎞ ⎜ ⎟ vB1 = ⎜ ⎟ vB2 + ⎜ ⎟ vP2 ⎝g⎠ ⎝g⎠ ⎝g⎠

e vB1 = vP2 − vB2

1 12 2 1 ⎛ WP ⎞ 2 kδ st + ⎜ ⎟ vP2 = kδ − WP( δ − δ st) 2 2⎝ g ⎠ 2

⎛ vB2 ⎞ ⎜ ⎟ ⎜ vP2 ⎟ = Find ( vB2 , vP2 , δ ) ⎜ ⎟ ⎝δ⎠

⎛ vB2 ⎞ ⎛ −1.20 ⎞ ft ⎜ ⎟=⎜ ⎟ ⎝ vP2 ⎠ ⎝ 9.57 ⎠ s

δ = 1.31 ft

Problem 15-77 A pitching machine throws the ball of weight M towards the wall with an initial velocity vA as shown. Determine (a) the velocity at which it strikes the wall at B , (b) the velocity at which it rebounds from the wall and (c) the distance d from the wall to where it strikes the ground at C. Given: M = 0.5 kg vA = 10 m s a =3m b = 1.5 m e = 0.5

θ = 30 deg
g = 9.81 m s
2

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Engineering Mechanics - Dynamics

Chapter 15

Solution:

Guesses m s m s vBx2 = 1 vBy2 = 1 d =1m t2 = 1 s m s m s

vBx1 = 1 vBy1 = 1 h =1m t1 = 1 s Given

vA cos ( θ ) t1 = a vBy2 = vBy1 d = vBx2 t2 vA cos ( θ ) = vBx1

b + vA sin ( θ ) t1 −

1 2

g t1 = h

2

vA sin ( θ ) − g t1 = vBy1 h + vBy2 t2 − e vBx1 = vBx2 1 2 g t2 = 0
2

⎛ vBx1 ⎞ ⎜ ⎟ ⎜ vBy1 ⎟ ⎜ vBx2 ⎟ ⎜ ⎟ ⎜ vBy2 ⎟ = Find ( v , v , v , v , h , t , t , d) Bx1 By1 Bx2 By2 12 ⎜h⎟ ⎜ ⎟ ⎜ t1 ⎟ ⎜ t2 ⎟ ⎟ ⎜ ⎝d⎠

⎛ vBx1 ⎞ m ⎜ ⎟ = 8.81 s ⎝ vBy1 ⎠ ⎛ vBx2 ⎞ m ⎜ ⎟ = 4.62 s ⎝ vBy2 ⎠
d = 3.96 m

Problem 15-78 The box of weight Wb slides on the surface for which the coefficient of friction is μk. The box has velocity v when it is a distance d from the plate. If it strikes the pla
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