CHAPTER 4
FORCES
AND
NEWTON'S LAWS OF MOTION
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9.
REASONING Let due east be chosen as the positive direction. Then, when both forces
point due east, Newton's second law gives
∑ F = FA + FB = ma1
where a1 = 0.50 m/s . When FA points due east and FB points due west, Newton's second
law gives
∑ F = FA – FB = ma 2
2
where a2 = 0.40 m/s 2 . Dividing the first equation by the second, we have
FA + FB a1
=
FA – FB a2
This expression can be solved for the ratio of the forces.
SOLUTION Solving for the ratio of the forces, we obtain
FA + FB (FA / FB ) +1 a1
=
=
FA – FB (FA / FB ) – 1 a2
FA
a F
 F a
a
+1 = 1  A – 1 = A 1 – 1
FB
a2  FB  FB a2 a2
Thus,
FA –1− (a1 / a2 ) –1 – (0.50 m/s 2 )/(0.40 m/s2 )
=
=
= 9.0
FB
1 – (a1 /a2 )
1 – (0.50 m/s 2 )/(0.40 m/s2 )
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17. REASONING We first determine the acceleration of the boat. Then, using Newton's
second law, we can find the net force ∑ F that acts on the boat. Since two of the three
forces are known, we can solve for the unknown force FW once the net force ∑ F is known.
SOLUTION Let the direction due east be the positive x direction and the direction due
north be the positive y direction. The x and y components of the initial velocity of the boat
are then
v 0 x = ( 2.00 m/s) cos 15.0° = 1.93 m/s
v 0 y = (2.00 m/s) sin 15.0° = 0.518 m/s
2
FORCES AND NEWTON'S LAWS OF MOTION
Thirty seconds later, the x and y velocity components of the boat are
v x = (4.00 m/s) cos 35.0° = 3.28 m/s
v y = ( 4.00 m/s) sin 35.0° = 2.29 m/s
Therefore, according to Equations 3.3a and 3.3b, the x and y components of the acceleration
of the boat are
v −v
3.28 m/s – 1.93 m/s
a x = x 0x =
= 4.50 × 10–2 m/s2
t
30.0 s
ay =
vy − v0y 2.29 m/s – 0.518 m/s
=
= 5.91 × 10–2 m/s2
t
30.0 s
Thus, the x and y components of the net force that act on the boat are
∑ Fx = max = (325 kg) (4.50 × 10 –2 m/s2 ) = 14.6 N
∑ Fy = may = (325 kg) (5.91 × 10–2 m/s 2 ) =19.2 N
The following table gives the x and y components of the net force ∑ F and the two known
forces that act on the boat. The fourth row of that table gives the components of the
unknown force FW .
Force
x-Component
y-Component
∑F
F1
F2
14.6 N
(31.0 N) cos 15.0° = 29.9 N
–(23.0 N ) cos 15.0° = –22.2 N
19.2 N
(31.0 N) sin 15.0° = 8.02 N
–(23.0 N) sin 15.0° = –5.95 N
FW = ∑ F − F1 − F2
14.6 N – 29.9 N + 22.2 N = 6.9 N
19.2 N – 8.02 N + 5.95 N = 17.1 N
The magnitude of FW is given by the Pythagorean theorem as
FW = (6.9 N) 2 + (17.1N) 2 = 18.4 N
Chapter 4
3
The angle θ that FW makes with the x axis is
 17.1 N 
θ = tan −1 
= 68°
6.9 N 
17.1 N
θ
Therefore, the direction of FW is 68°, north of east .
6.9 N
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33. REASONING AND SOLUTION There are two forces that act on the balloon; they are, the
combined weight of the balloon and its load, Mg, and the upward buoyant force FB . If we
take upward as the positive direction, then, initially when the balloon is motionless,
Newton's second law gives FB − Mg = 0 . If an amount of mass m is dropped overboard so
that the balloon has an upward acceleration, Newton's second law for this situation is
FB − (M − m)g = (M − m)a
But FB = mg , so that
Mg – ( M − m ) g = mg = ( M – m ) a
Solving for the mass m that should be dropped overboard, we obtain
m=
Ma
(310 kg )(0.15 m/s2 )
=
= 4.7 kg
g + a 9.80 m/s 2 + 0.15 m/s2
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73. REASONING The speed of the skateboarder at
the bottom of the ramp can be found by solving
Equation 2.9 (v 2 = v02 + 2ax where x is the distance
that the skater moves down the ramp) for v. The
figure at the right shows the free-body diagram for
the skateboarder. The net force ∑ F , which
accelerates the skateboarder down the ramp, is the
component of the weight that is parallel to the
incline: ∑ F = mg sinθ .
F
N
mg sinθ
mg cosθ
Therefore, we know from Newton's second law that the acceleration of the skateboarder
down the ramp is
a=
∑ F mg sinθ
=
= g sinθ .
m
m
SOLUTION Thus, the speed of the skateboarder at the bottom of the ramp is
v = v 20 + 2ax = v02 + 2gx sinθ = (2.6 m/s)2 + 2(9.80 m/s2 )(6.0 m) sin18° = 6.6 m/s
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