Halliday ♦ Resnick ♦ Walker
FUNDAMENTALS OF PHYSICS
SIXTH EDITION
Selected Solutions
Chapter 5
5.9
5.29
5.43
5.53
9. In all three cases the scale is not accelerating, which means that the two cords exert forces of equal
magnitude on it. The scale reads the magnitude of either of these forces. In each case the tension force
of the cord attached to the salami must be the same in magnitude as the weight of the salami because
the salami is not accelerating. Thus the scale reading is mg, where m is the mass of the salami. Its
value is (11.0 kg)(9.8m/s2) = 108N.
29. The solutions to parts (a) and (b) have been combined here. The free-body diagram is shown below,
with the tension of the string
T, the force of gravity m
g, and the force of the air
F. Our coordinate
system is shown. The x component of the net force is T sin θ − F and the y component is T cos θ − mg,
where θ = 3 7◦.
Since the sphere is motionless the
net force on it is zero. We answer
the questions in the reverse order.
Solving T cos θ − mg = 0 for the
tension, we obtain T = mg/ cos θ =
(3.0 × 10−4)(9.8)/ cos 37◦ = 3.7 ×
10−3 N. Solving T sin θ−F = 0 for
the force of the air: F = T sin θ =
(3.7×10−3) sin 37◦ = 2.2×10−3 N.
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T
m
g

F
θ
•
+x
+y
43. The free-body diagram for each block is shown below. T is the tension in the cord and θ = 30◦ is the
angle of the incline. For block 1, we take the +x direction to be upthe incline and the +y direction
to be in the direction of the normal force N that the plane exerts on the block. For block 2, we take
the +y direction to be down. In this way, the accelerations of the two blocks can be represented by the
same symbol a, without ambiguity. Applying Newton’s second law to the x and y axes for block 1 and
to the y axis of block 2, we obtain
T − m1g sin θ = m1a
N − m1g cos θ = 0
m2g − T = m2a
respectively. The first and third of these equations provide a simultaneous set for obtaining values of a
and T. The second equation is not needed in this problem, since the normal force is neither asked for
nor is it needed as part of some further computation (such as can occur in formulas for friction).
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T
(+x)
m1g
N
θ
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T
m2g
(+y)
(a) We add the first and third equations above: m2g −m1g sin θ = m1a +m2a. Consequently, we find
a =
(m2 − m1 sin θ)g
m1 + m2
=
(2.30 kg) − 3.70 sin 30.0◦) (9.8)
3.70 + 2.30
= 0.735 m/s2
.
(b) The result for a is positive, indicating that the acceleration of block 1 is indeed up the incline and
that the acceleration of block 2 is vertically down.
(c) The tension in the cord is
T = m1a + m1g sin θ = (3.70)(0.735) + (3.70)(9.8) sin 30◦ = 20.8 N .
53. The forces on the balloon are the force of gravity mg (down) and the force of the air Fa (up). We take
the +y to be up, and use a to mean the magnitude of the acceleration (which is not its usual use in
this chapter). When the mass is M (before the ballast is thrown out) the acceleration is downward
and Newton’s second law is Fa − Mg = −Ma. After the ballast is thrown out, the mass is M − m
(where m is the mass of the ballast) and the acceleration is upward. Newton’s second law leads to
Fa − (M − m)g = (M − m)a. The earlier equation gives Fa = M(g − a), and this plugs into the new
equation to give
M(g − a) − (M − m)g = (M − m)a =⇒ m =
2Ma
g + a
.