Download ICW05D3-2 Group Problem: Car on Barge Solution

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
ICW05D3-2 Group Problem: Car on Barge Solution
A workman is unloading the last car (mass mc ) from a barge (mass mb ) tied to the pier by
a short cable. He drives with a constant acceleration from a standing start and attains a
speed v f in the time t f it takes to reach the end of the barge.
a) Does this put tension in the cable or does it push the barge against the dock?
b) Depending on your answer to a) determine an expression for either the tension in the
cable T (t) or the magnitude of the horizontal force of the pier on the barge F(t) during
the time t = 0 to t = t f . Express your answer in terms of some or all of the quantities mc ,
mb , v f and t f .
Solution:
a) Consider the car and the barge as the system. The momentum of the system is
increasing toward the right. Thus there must be an external force on the system pointing
right. That implies tension in the cable.
b) We begin by noting that the car moves with constant acceleration, so ac = v f / t f and
by integration
vc (t) = (v f / t f )t
using the fact that the car starts from rest, vc (0) = 0 .
There are several possible ways to approach this problem.
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First Approach: F = dp / dt applied to system consisting of the car and the barge.
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We shall apply F = dp / dt to this system that is under the influence of an external force,
the tension in the rope:
T (t) =
d
(m v + mc vc )
dt b b
Thus
T (t) = mc
dvc
= mc v f / t f .
dt
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Second Approach: We choose the car as our system and apply F = mca c to the car.
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Since the car is accelerating, it is experiencing a force by the barge on the car Fb,c = mca c .
Because the car and barge from an interaction pair , By Newton's Third Law, the force on
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the barge due to the car is Fc,b = !Fb,c = !mca c = !mc ac î . But the barge, is stationary, so
the horizontal forces must sum to zero. The barge experiences two forces: the force from
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the car Fc,b and the tension force of the rope on the barge Tr ,b (t) = T (t) î . Thus
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Tr ,b (t) + Fc,b = 0 implies that after taking the x -component !mc ac + T (t) = 0 . So the
tension in the rope is
T (t) = mc ac = mc v f / t f
in agreement with our result above.
Third Approach: Motion of the Center of Mass:
We can relate the acceleration of the center of mass to the external force, the tension,
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Tr ,b (t) = T (t) î . Let L be the length of the boat. Choose an origin at the left side of the
boat. Treat the barge as a point-like object located a distance L / 2 from the origin. Then
the x -component of the center of mass is
X cm (t) =
mc xc + mb xb mc xc + (L / 2)xb
.
=
mc + mb
mc + mb
Differentiating twice gives the x -component of the acceleration of the center of mass:
Acm (t) =
(mc v f / t f )
mc ac + mb ab
mc ac
.
=
=
mc + mb
mc + mb
mc + mb
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The tension in the rope is then found by using Fext = msystem A cm , where msystem = mc + mb
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and Fext = Tr ,b (t) = T (t) î . Thus
T (t) = (mc + mb ) Acm (t) = (mc + mb )
(mc v f / t f )
mc + mb
= mc v f / t f .
Remark: The x -component of momentum is not constant due to the presence of an
external force. Therefore mc vc (t) ! "mbvb (t) . It is incorrect to draw the following
conclusions that vb (t) = !mc vc (t) / mb hence ab (t) = !mc ac (t) / mb = !(mc / mb )(v f / t f ) ,
and finally T (t) = mb ab (t) = !mc v f / t f .Two things to note. First, the barge is stationary
so ab (t) = 0 . Second, this incorrect result for the tension differs only by a minus sign
from the correct result but it is a very important minus sign because the direction of the
tension force on the barge must be in the positive x -direction.