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**Unformatted text preview: **Engineering Mechanics - Dynamics
Chapter 15
Problem 15-31
The log has mass M and rests on the ground for which the coefficients of static and kinetic
friction are μs and μk respectively. The winch delivers a horizontal towing force T to its cable at A
which varies as shown in the graph. Determine the speed of the log when t = t2. Originally the
tension in the cable is zero. Hint: First determine the force needed to begin moving the log.
Given:
M = 500 kg
t1 = 3 s
μ s = 0.5
T1 = 1800 N
μ k = 0.4
g = 9.81
m
2
s
t2 = 5 s
Solution:
To begin motion we need
⎛ t0 2 ⎞
⎟ = μ Mg
2T 1 ⎜
s
⎜t 2⎟
⎝1 ⎠
t0 =
μs M g
2T 1
t1
t0 = 2.48 s
Impulse - Momentum
t1
⌠
⎮
0+⎮
⎮
⌡t
2
⎛t⎞
2T1 ⎜ ⎟ dt + 2T1 ( t2 − t1 ) − μ k M g( t2 − t0 ) = M v2
⎝ t1 ⎠
0
⎡⌠
1 ⎢⎮
v2 =
⎢⎮
M⎮
⎢⌡
t
t1
⎣
⎤
⎥
⎛t⎞
2T1 ⎜ ⎟ dt + 2T1 ( t2 − t1 ) − μ k M g( t2 − t0 )⎥
⎝ t1 ⎠
2
⎥
⎦
0
v2 = 7.65
m
s
*Problem 15-32
A railroad car having mass m1 is coasting with speed v1 on a horizontal track. At the same time
another car having mass m2 is coasting with speed v2 in the opposite direction. If the cars meet
and couple together, determine the speed of both cars just after the coupling. Find the difference
between the total kinetic energy before and after coupling has occurred, and explain qualitatively
what happened to this energy.
3
Units used: Mg = 10 kg
Given:
m1 = 15 Mg
3
kJ = 10 J
m2 = 12 Mg
321
...

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