Solutions to Homework #4
Chapter 6
32. Picture the Problem: The forces exerted on the left and
right blocks are depicted at right.
Strategy: Because the pulley is ideal, the tension in the
string is equal to the weight of the hanging block.
This
can be verified by Newton’s Second Law in the vertical
direction for the hanging block:
.
Write Newton’s Second Law along the direction parallel
to the incline for the 6.7 kg block and substitute
into the resulting equation to find m.
Solution: 1. Write Newton’s
Second Law along the direction
parallel to the incline:
2. Substitute
into the
resulting equation to find m.
Insight: A larger m is required if the angle θ is increased.
If it is increased all the way to θ =90°, the
large mass will be hanging straight down and the mass m required to maintain equilibrium would be
6.7 kg.
40. Picture the Problem: Refer to the figure at right:
Strategy:
Write Newton’s Second Law for each block and
add the equations to eliminate the unknown tension
T
.
Solve the resulting equation for the acceleration
a
, and use
the acceleration to find the tension. Let
x
be positive in the
direction of each mass’s motion,
m
1
be the mass on the
table, and
m
2
be the hanging mass.
Solution: 1. (a)
The tension in the string is less than the
weight of the hanging mass.
If it were equal to the weight,
the hanging mass would not accelerate.
2. (b) Write Newton’s
Second Law
for each block and add
the equations:
3. Solve the resulting
equation for a:
4. (c) Use the first
equation to find T:
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View Full DocumentInsight: Note that the blocks move as if they were a single block of mass 6.30 kg under the influence of
a force equal to
The tension in the string would be zero if m
2
fell freely, 27.5 N if m
2
(and the entire system) were at rest.
46. Picture the Problem: The forces acting on the car are
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 Spring '08
 Ashkenkai
 Physics, Force, Work

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