P4.1 For the same force F, acting on different masses F = m1a1 and F = m2a2.
Setting these expressions for F equal to one another gives:
m 1 a2 1
(a)
m 2 a1 3
(b) F m1 m 2 a 4m1 a m1 3.00 m s 2
a 0.750 m/ s 2
P4.3 We use the particle under constant acceleration and particle under a net
force models. We first calculate the acceleration of the puck:
8.00ˆ
ˆj m / s 3.00ˆ m /s
t 8.00 s
0.625ˆ m/s 2 1.25ˆj m/s 2
In F ma , the only horizontal force is the thrust F of the rocket:
(a) F (4.00 kg) 0.625ˆ m/s 2 1.25ˆj m/s 2 2.50ˆ 5.00ˆj N
(b) Its magnitude is |F| (2.50 N) 2 (5.00 N) 2 5.59 N
P4.7 We use the particle under a net force model.
(a) F F1 F2 (20.0ˆi 15.0ˆj) N
Newton’s second law gives, with m = 5.00 kg,
4.00ˆ 3.00ˆj ?m/s 2
F
m
or a 5.00 m s 2 at 36.9
77
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78 The Laws of Motion
(b) F2 x 15.0cos 60.0 7.50 N
F2 y 15.0sin 60.0 13.0 N ANS. FIG. P4.7
F2 7.50ˆi 13.0ˆj N
F F1 F2 20.0ˆ 7.50ˆ 13.0ˆj N
(27.5ˆi 13.0ˆj) N
5.50ˆ
F ˆj m/s 2 6.08 m/s 2 at 25.3?
m
P4.9 Imagine a quick trip by jet, on which you do not visit the rest room and
your perspiration is just canceled out by a glass of tomato juice. By
subtraction, Fg p mg p and Fg mgC give C
Fg m g p gC
For a person whose mass is 90.0 kg, the change in weight is
Fg 90.0 kg 9.809 5 9.780 8 2.58 N
A precise balance scale, as in a doctor’s office, reads the same in
different locations because it compares you with the standard masses
on its beams. A typical bathroom scale is not precise enough to reveal
this difference.
P4.14 The two forces acting on the baseball are the force of the pitcher and the
force of gravity, Fg weight of ball mg .
The baseball starts from rest, and v release v. The time to accelerate the
baseball to its final speed is t:
v v vi v 0 ˆ v ˆ
a i i
t t t t
The distance is given by x vt: x t
v vt
(a)
2 2
(b) The force on the baseball is the force of the pitcher plus the force of
Fg v ˆ
gravity, or Fp Fg ˆj i.
gt
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Chapter 4 79
Solving for the force exerted by the pitcher gives:
Fg v ˆ
Fp i Fg ˆj
gt
P4.15 (a) 5.0 kg 9.8 m/s2 = 49 N up, to counterbalance the Earth’s force on
the block.
(b) (5.0 – 2.0)kg 9.8 m/s2 = 29 N up, the forces on the block are now
the Earth pulling down with 5.0 kg 9.8 m/s2 = 49 N and the rope
pulling up with 2.0 kg 9.8 m/s2 = 20 N. The forces from the floor
and rope together balance the weight.
(c) 0, the block now accelerates up away from the floor.
P4.19 (a) Assume the car and mass accelerate
horizontally. We consider the forces
on the suspended object.
Fy may : T cos mg 0
Fx max : T sin ma
ANS. FIG. P4.19
mg
Substitute T :
cos
mg sin
mg tan ma
cos
a g tan
(b) a 9.80 m s 2 tan 23.0 a 4.16 m s 2
P4.26 Refer to the free-body diagram. Choose the x axis pointing down the
slope so that the string makes the angle with the vertical. The
acceleration is obtained from vf = vi + at:
a (v f vi )/ t (30.0 m/s 2 0) / 6.00 s
a 5.00 m/s 2
Because the string stays perpendicular to the ceiling, we know that the
toy moves with the same acceleration as the van, 5.00 m/s2 parallel to
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80 The Laws of Motion
the hill. We take the x axis in this direction, so
ax 5.00 m/s 2 and a y 0
The only forces on the toy are the string tension in the y direction and
the planet’s gravitational force, as shown in the force diagram. The size
of the latter is mg = (0.100 kg)(9.80 m/s2) = 0.980 N
(a) Using Fx max gives (0.980 N) sin = (0.100 kg)(5.00 m/s2)
Then sin = 0.510 and = 30.7°
(b) Using Fy ma y gives +T − (0.980 N) cos = 0
T = (0.980 N) cos 30.7° = 0.843 N
ANS. FIG. P4.26
P4.29 Choose a coordinate system with î East and ĵ
North. The acceleration is
a 10.0 cos 30.0 ˆ 10.0sin 30.0 ˆj m/s 2
(8.66ˆ 5.00ˆj) m/s 2 ANS. FIG. P4.29
From Newton’s second law,
F ma (1.00 kg)(8.66ˆi m/s 2 5.00ˆj m/s 2 )
(8.66ˆi 5.00ˆj) N
and F F1 F2
so the force we want is
F1 F F2 (8.66ˆi 5.00ˆj 5.00ˆj) N
8.66ˆi N 8.66 N east
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Chapter 4 81
P4.32 We assume the vertical bar is in
compression, pushing up on the pin
with force A, and the tilted bar is in
tension, exerting force B on the pin
at –50°.
Fx 0:
2 500 N cos 30 Bcos 50 0 ANS. FIG. P4.32
B 3.37 103 N
Fy 0:
2 500 N sin 30 A 3.37 103 N sin 50 0
A 3.83 103 N
Positive answers confirm that
B is in tension and A is in compression.
P4.42 Choose the
+x-direction to
be horizontal
and forward
with the +y
ANS. FIG. P4.42
vertical and
upward. The
common acceleration of the car and trailer then has components of
ax 2.15 m s 2 and ay 0 .
(a) The net force on the car is horizontal and given by
Fx car F T mcar ax 1 000 kg 2.15 m s 2
2.15 103 N forward
(b) The net force on the trailer is also horizontal and given by
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82 The Laws of Motion
Fx trailer T mtrailer ax 300 kg 2.15 m s 2
645 N forw ard
(c) Consider the free-body diagrams of the car and trailer. The only
horizontal force acting on the trailer is T 645 N forward, and
this is exerted on the trailer by the car. Newton’s third law then
states that the force the trailer exerts on the car is
645 N tow ard the rear .
(d) The road exerts two forces on the car. These are F and n c shown in
the free-body diagram of the car. From part (a),
F T 2.15 103 N 2.80 103 N. Also,
Fy car nc Fgc mcar ay 0 , so nc Fgc mcar g 9.80 103 N.
The resultant force exerted on the car by the road is then
R car F2 n c2 2.80 103 N 2 9.80 103 N 2
1.02 10 4 N
at tan 1 c tan 1 3.51 74.1 above the horizontal and
n
F
forward. Newton’s third law then states that the resultant force
exerted on the road by the car is
1.02 104 N at 74.1 below the horizontal and rearward .
below the horizontal and rearward
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