Download HW #6

5.
REASONING AND SOLUTION An object will not necessarily accelerate when two or
more forces are applied to the object simultaneously. The applied forces may cancel so the
net force is zero; in such a case, the object will not accelerate. The resultant of all the forces
that act on the object must be nonzero in order for the object to accelerate.
6.
REASONING AND SOLUTION
Since the father and the daughter are standing on ice skates, there is virtually no friction
between their bodies and the ground. We can assume, therefore, that the only horizontal
force that acts on the daughter is due to the father, and similarly, the only horizontal force
that acts on the father is due to the daughter.
a. According to Newton's third law, when they push off against each other, the force exerted
on the father by the daughter must be equal in magnitude and opposite in direction to the
force exerted on the daughter by the father. In other words, both the father and the daughter
experience pushing forces of equal magnitude.
b. According to Newton's second law, ¦ F = ma . Therefore, a = ¦ F / m . The magnitude of
the net force on the father is the same as the magnitude of the net force on the daughter, so
we can conclude that, since the daughter has the smaller mass, she will acquire the larger
acceleration.
10. REASONING AND SOLUTION The mass of an object is a quantitative measure of its
inertia. The mass of an object is an intrinsic property of the object and is independent of the
location of the object. The weight of an object is the gravitational force exerted on the
object by the earth. The gravitational force depends on the distance between the object and
the center of the earth. Therefore, when an object is moved from sea level to the top of a
mountain, its weight will change, while the mass of the object remains constant.
11. REASONING AND SOLUTION The weight of the ball always acts downward. The force
of air resistance will always act in the direction that is opposite to the direction of motion of
the ball. The net force on the ball is the resultant of the weight and the force of air
resistance.
a. As the ball moves upward, the force of air resistance acts downward. Since air resistance
and the weight of the ball act in the same direction in this case, the net force on the ball will
be greater in magnitude than the weight of the ball.
b. As the ball falls downward, the force of air resistance is upward. Since air resistance and
the weight of the ball act in opposite directions, the net force that acts on the ball will be
smaller in magnitude than the weight of the ball.
Note that in both cases, the net force points downward since the object accelerates
downward in both cases.
16. REASONING AND SOLUTION The apparent weight will differ from the true weight only
in an accelerating elevator. When the scale in an elevator reads the true weight, the only
conclusion that can be made is that the elevator has zero acceleration. Therefore, one cannot
conclude whether the elevator is moving with any constant velocity upward, any constant
velocity downward, or whether the elevator is at rest, since each of these conditions involves
zero acceleration.
10. REASONING AND SOLUTION
FE = F cos θ = (720 N) cos 38° = 570 N
FN = F sin θ = (720 N) sin 38° = 440 N
11. REASONING Newton’s second law gives the acceleration as a = (ȈF)/m. Since we seek
only the horizontal acceleration, it is the x component of this equation that we will use;
ax = (ȈFx)/m. For completeness, however, the free-body diagram will include the vertical
forces also (the normal force FN and the weight W).
SOLUTION The free body diagram is shown
at the right, where
+y
F1 = 59.0 N
F2 = 33.0 N
ș = 70.0°
FN
When F1 is replaced by its x and y
components, we obtain the free body
diagram in the second drawing.
F2
F1
W
Choosing right to be the positive direction, we
have
ΣF
F cos θ − F2
ax = x = 1
m
m
ax =
( 59.0 N ) cos 70.0° − ( 33.0 N ) =
7.00 kg
−1.83 m/s 2
The minus sign indicates that the horizontal
acceleration points to the left .
+x
ș
+y
FN
F2
F1cos ș
F1sin ș
W
+x
1
§
·
16. REASONING For both the tug and the asteroid, Equation 2.8 ¨ x = v 0 t + at 2 ¸ applies
©
¹
2
with v0 = 0 m/s, since both are initially at rest. In applying this equation, we must be careful
and use the proper acceleration for each object. Newton’s second law indicates that the
acceleration is given by a = ΣF/m. In this expression, we note that the magnitudes of the net
forces acting on the tug and the asteroid are the same, according to Newton’s action-reaction
law. The masses of the tug and the asteroid are different, however. Thus, the distance
traveled for either object is given by, where we use for ΣF only the magnitude of the pulling
force
1
1 § ΣF · 2
¸t
x = v 0 t + at 2 = ¨
2
2© m ¹
SOLUTION Let L be the initial distance between the tug and the asteroid. When the two
objects meet, the distances that each has traveled must add up to equal L. Therefore,
1
1
2
2
L = x T + x A = aT t 2 + a A t 2
§ 1
1 · 2
1 § ΣF · 2
1 § ΣF · 2
1
L= ¨
+
¸t + ¨
¸ t = ΣF ¨
¸t
2 © mT ¹
2 © mA ¹
2
© mT m A ¹
Solving for the time t gives
t=
2L
§ 1
1 ·
ΣF ¨
+
¸
© mT mA ¹
=
2 ( 450 m )
§
·
1
1
(490 N )¨
+
¸
© 3500 kg 6200 kg ¹
= 64 s
18. REASONING Newton’s law of universal gravitation indicates that the gravitational force
that each uniform sphere exerts on the other has a magnitude that is inversely proportional to
the square of the distance between the centers of the spheres. Therefore, the maximum
gravitational force between two uniform spheres occurs when the centers of the spheres are
as close together as possible, and this occurs when the surfaces of the spheres are touching.
Then, the distance between the centers of the spheres is the sum of the two radii.
SOLUTION When the bowling ball and the billiard ball are touching, the distance between
their centers is r = rBowling + rBilliard. Using this expression in Newton’s law of universal
gravitation gives
F=
GmBowling mBilliard
r2
=
GmBowling mBilliard
(r
Bowling
=
( 6.67 ×10
−11
+ rBilliard
)
2
N ⋅ m 2 / kg 2 ) ( 7.2 kg )( 0.38 kg )
( 0.11 m + 0.028 m )
2
= 9.6 × 10−9 N
20. REASONING AND SOLUTION The forces that act on the rock are shown
at the right. Newton's second law (with the direction of motion as positive)
is
R
ΣF = mg – R = ma
Solving for the acceleration a gives
a=
mg – R
=
m
(
)
( 45 kg ) 9.80 m/s 2 – ( 250 N )
45 kg
mg
= 4.2 m/s 2
28. REASONING AND SOLUTION The figure at the right
shows the three spheres with sphere 3 being the sphere
of unknown mass. Sphere 3 feels a force F31 due to the
presence of sphere 1, and a force F32 due to the
presence of sphere 2. The net force on sphere 3 is the
resultant of F31 and F32.
Note that since the spheres form an equilateral triangle,
each interior angle is 60°. Therefore, both F31 and F32
make a 30° angle with the vertical line as shown.
3
F
F
31
32
1.20 m
30°
2
1
Furthermore, F31 and F32 have the same magnitude
given by
F=
GMm 3
r2
where M is the mass of either sphere 1 or 2 and m3 is the mass of sphere 3. The components
of the two forces are shown in the following drawings:
F
31
F cos θ
30.0°
F cos θ
F sin θ
F
32
30.0°
F sin θ
Clearly, the horizontal components of the two forces add to zero. Thus, the net force on
sphere 3 is the resultant of the vertical components of F31 and F32:
F3 = 2F cos θ = 2
GMm 3
cos θ.
r2
The acceleration of sphere 3 is given by Newton's second law:
a3 =
F3
(6.67 × 10 −11 N ⋅ m 2 /kg 2 ) (2.80 kg )
GM
= 2 2 cosθ = 2
cos 30.0°
m3
(1.20 m )2
r
= 2.25 × 10 –10 m/s 2
32. REASONING AND SOLUTION
acceleration, it follows that
Since both motions are characterized by constant
1
yJ
yE
=
2
1
2
a J t 2J
a E t 2E
where the subscripts designate those quantities that pertain to Jupiter and Earth. Since both
objects fall the same distance, the above ratio is equal to unity. Solving for the ratio of the
times yields
tJ
tE
=
aE
aJ
=
GM E / R E2
GM J / R J2
=
RJ
ME
RE
MJ
= (11.2 )
1
= 0.628
318