Download chapter 4 forces and newton`s laws of motion

CHAPTER 4 FORCES AND
NEWTON'S LAWS OF MOTION
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION When the car comes to a sudden halt, the upper part of the
body continues forward (as predicted by Newton's first law) if the force exerted by the lower back
muscles is not great enough to give the upper body the same deceleration as the car. The lower
portion of the body is held in place by the force of friction exerted by the car seat and the floor.
When the car rapidly accelerates, the upper part of the body tries to remain at a constant
velocity (again as predicted by Newton's first law). If the force provided by the lower back
muscles is not great enough to give the upper body the same acceleration as the car, the upper
body appears to be pressed backward against the seat as the car moves forward.
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2.
REASONING AND SOLUTION When the birdfeeder is hanging freely and no one is pulling
on the dangling (lower) cord, there is a tension in the cord between the birdfeeder and the tree limb
(the upper cord), because the upper cord supports the weight of the birdfeeder. When the lower
cord is pulled down with a slow continuous pull, the tension in both cords increases slowly. Since
the upper cord has a larger tension to begin with, it always has the greater tension as the lower
cord is pulled. Thus, the upper cord snaps first.
On the other hand, when the child gives the lower cord a sudden, downward pull, the tension in
the lower cord increases suddenly. However, the tension in the upper cord does not increase as
suddenly. The reason is that the birdfeeder has a large mass, so it accelerates very slowly. Thus,
the upper cord is stretched slowly and, consequently, the tension in the upper cord rises slowly.
Since the tension rises much faster in the lower cord, it is the first to snap.
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3.
REASONING AND SOLUTION If the net external force acting on an object is zero, it is
possible for the object to be traveling with a nonzero velocity. According to Newton’s second law,
ΣF = ma, if the net external force ΣF is zero, the acceleration a is also zero. If the acceleration is
zero, the velocity must be constant, both in magnitude and in direction. Thus, an object can move
with a constant nonzero velocity when the net external force is zero.
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4.
REASONING AND SOLUTION According to Newton's second law, a net force is required
to give an object a non-zero acceleration.
a. If an object is moving with a constant acceleration of 9.80 m/s2, we can conclude that there is a
net force on the object.
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b. If an object moves with a constant velocity of 9.80 m/s, its acceleration is zero; therefore, we
can conclude that the net force acting on the object is zero.
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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.
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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.
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7.
REASONING AND SOLUTION
a. The force of the gymnast on the trampoline causes the elastic surface of the trampoline to
deform.
b. The reaction force exerted by the trampoline on the gymnast causes the gymnast to decelerate
and come to a momentary stop.
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8.
REASONING AND SOLUTION The two oppositely directed pushing forces predicted by
Newton's third law act on different bodies; one force acts on the crate, while the other acts on the
person who is pushing on the crate. Since the two forces act on different objects, they cannot
cancel each other. Whether or not the crate moves depends on the net force that acts on the crate.
If the crate does not move under the action of a single pushing force, the only reasonable
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FORCES AND NEWTON'S LAWS OF MOTION
conclusion is that there must be another force acting on the crate that cancels the pushing force.
The other force is the force of static friction.
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9.
REASONING AND SOLUTION The magnitude of the gravitational force between any two of
the particles is given by Newton's law of universal gravitation: F = Gm1 m2 / r 2 where m1 and m2
are the masses of the particles and r is the distance between them. Since the particles have equal
masses, we can arrange the particles so that each one experiences a net gravitational force that has
the same magnitude if we arrange the particles so that the distance between any two of the particles
is the same. Therefore, the particles should be placed at the corners of an equilateral triangle with
all three sides of equal length.
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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.
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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.
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12. REASONING AND SOLUTION From Equation 4.5, W = mg , we know that the weight of
an object is directly proportional to its mass. The proportionality constant on a given planet is g,
the magnitude of the acceleration due to gravity on that planet.
If object A weighs twice as much as object B at the same spot on the earth, then the mass of
object A is twice as much as the mass of object B. The mass of any object is independent of the
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object's location in the universe. Since the weight of an object is directly proportional to its mass,
object A will weigh twice as much as object B at the same spot on any planet. While the values of
the weights will differ from planet to planet, as the value of g varies from planet to planet, the ratio
of the weights (which equals the ratio of their masses) will be the same on any planet.
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13. REASONING AND SOLUTION
Equations 4.4 and 4.5 may be combined to give the
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acceleration due to gravity at a distance r from the center of the earth: g = GM E / r . Since this
expression depends on r, the acceleration of a freely falling object does depend on its location.
The acceleration due to gravity will be greater at Death Valley, California (where r is smaller) than
it is on the top of Mt. Everest (where r is greater). Since r is measured from the center of the
earth, however, these values will be very close.
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14. REASONING AND SOLUTION Assuming that the accelerating mechanism remains attached
to the rocket, the acceleration will be greater when the rocket is fired horizontally.
The
accelerating mechanism provides an acceleration that points in the initial direction of motion of the
rocket. The net acceleration is the resultant of the accelerating mechanism and the acceleration due
to gravity. When the rocket is fired horizontally, these accelerations will be at right angles to each
other. When the rocket is fired straight up, these accelerations will be in opposite directions. The
magnitude of the resultant will be greater when these two accelerations are at right angles rather
than when they are in opposition. Therefore, the acceleration will be greater when the rocket is
fired horizontally.
If we assume that the accelerating mechanism is not attached to the rocket, then once the rocket
is fired, the only force on the rocket is that due to gravity, and the rocket has the acceleration due
to gravity regardless of its orientation. In this case, the acceleration of the rocket is the same
regardless of whether it is fired straight up or fired horizontally.
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15. REASONING AND SOLUTION If the elevator were at rest, or moving with a constant
velocity, the scale would read the true weight of mg = 98 N. When the elevator is accelerating, the
scale reading will differ from 98 N and will display the apparent weight, FN, which is given by
Equation 4.6: FN = mg + ma where a, the acceleration of the elevator, is positive when the
elevator accelerates upward and negative when the elevator accelerates downward.
a. When the apparent weight is FN = 75 N, the apparent weight is less than the true weight (mg =
98 N) so a must be negative. The elevator is accelerating downward.
b. When the apparent weight is FN = 120 N, the apparent weight is greater than the true weight
(mg = 98 N) so a must be positive. The elevator is accelerating upward.
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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.
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17. REASONING AND SOLUTION If the elevator were at rest, or moving with a constant
velocity, the scale will read the true weight mg. When the elevator is accelerating, the scale
reading will differ from the true weight and will register the apparent weight, FN, which is given by
Equation 4.6: FN = mg + ma where a, the acceleration of the elevator, is positive when the
elevator accelerates upward and negative when the elevator accelerates downward.
Since the scale registers 600 N when the elevator is moving with constant velocity, we know that
the true weight is 600 N.
a. The elevator is moving upward. When the elevator slows down, its acceleration vector points
downward. The term ma will be negative; therefore, the scale reading will be less than 600 N.
b. When the elevator is stopped, the scale will register the true weight; therefore, the scale reads
600 N.
c. The elevator is moving downward and speeding up. The acceleration vector points downward,
and the term ma is negative. Therefore, the scale registers a value that is less than 600 N.
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18. REASONING AND SOLUTION Since the sled moves with constant velocity, the force of
kinetic friction is present. The magnitude of this force is given by µk FN , where µk is the
coefficient of kinetic friction and FN is the magnitude of the normal force that acts on the sled.
Furthermore, the horizontal component of the applied force must be equal in magnitude to the force
of kinetic friction, since there is no acceleration.
When the person pulls on the sled, the vertical component of the pulling force tends to decrease
the magnitude of the normal force relative to that when the sled is not being pulled or pushed. On
the other hand, when the person pushes on the sled, the vertical component of the pushing force
tends to increase the normal force relative to that when the sled is not being pulled or pushed.
Therefore, when the sled is pulled, the magnitude of the force of kinetic friction, and therefore the
magnitude of the applied force, is less than when the sled is pushed.
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19. REASONING AND SOLUTION We know that µs = 2.0 µk for a crate in contact with a
cement floor. The maximum force of static friction is f sMAX = µs FN while the force of kinetic
friction is f k = µk FN . As long as the crate is on the cement floor, we can conclude that the
magnitude of the maximum static frictional force acting on the crate will always be twice the
magnitude of the kinetic frictional force on the moving crate, once the crate has begun moving.
However, the force of static friction may not have its maximum value. Thus, the magnitude of the
static frictional force is not always twice the magnitude of the kinetic frictional force.
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20. REASONING AND SOLUTION A box rests on the floor of a stationary elevator. Because
of static friction, a force is required to start the box sliding across the floor of the elevator. The
magnitude of this force is given by f sMAX = µs FN , where FN is the magnitude of the normal force
exerted on the box by the floor of the elevator. When the elevator is stationary, the magnitude of
the normal force exerted on the box is mg where m is the mass of the crate and g is the magnitude
of the acceleration due to gravity. When the crate accelerates upward, the floor of the elevator will
push against the box to accelerate it upward; therefore, when the crate accelerates upward, the
magnitude of the normal force will be greater than mg. When the elevator accelerates downward,
the normal force between the box and the floor of the elevator will be less than when the elevator is
stationary; therefore, the magnitude of the normal force will be less than mg. Thus, the magnitude
of the force required to start the box sliding across the floor of the elevator are ranked as follows in
ascending order: (c) elevator accelerating downward; (a) elevator stationary; (b) elevator
accelerating upward.
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21. REASONING AND SOLUTION When the rope is tied to a tree and pulled by the ten people,
the tension in the rope is twice as great as it was when it was used in a tug-of-war with five people
on each team. Therefore, the rope is more likely to break when it is tied to the tree and pulled by
ten people.
Note that when the rope is tied to the tree and pulled by ten people, the situation is equivalent to a
tug-of-war with ten people on each team.
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22. REASONING AND SOLUTION An object is in equilibrium when its acceleration is zero.
When a stone is thrown from the top of a cliff, its acceleration is the acceleration due to gravity;
therefore, the stone is not in equilibrium.
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23. REASONING AND SOLUTION An object is in equilibrium when its acceleration is zero.
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FORCES AND NEWTON'S LAWS OF MOTION
a. If a single nonzero force acts on an object, the object will accelerate according to Newton's
second law. The object is not in equilibrium.
b. If two forces that point in mutually perpendicular directions act on an object, the object will
experience a net force. By Newton's second law, the object will, therefore, have a nonzero
acceleration. The object is not in equilibrium.
c. If two forces that point in directions that are not perpendicular act on the object, the object may
or may not be in equilibrium, depending on how the forces are oriented. In general, the resultant of
two such forces is nonzero, the object will accelerate, and it is not in equilibrium. In the special
case where the two forces point in opposite directions and have the same magnitude, the net force
is zero, the object has zero acceleration, and the object is, therefore, in equilibrium.
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24. REASONING AND SOLUTION A circus performer hangs from a stationary rope. Since
there is no acceleration, the tension in the rope must be equal in magnitude to the weight of the
performer. She then begins to climb upward by pulling herself up, hand-over-hand. Whether the
tension in the rope is greater than or equal to the tension when she hangs stationary depends on
whether or not she accelerates as she moves upward. When she moves upward at constant
velocity, the tension in the rope will be the same. When she accelerates upward, the rope must
support the net upward force in addition to her weight; therefore, in this case, the tension in the
rope will be greater than when she hangs stationary.
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25. REASONING AND SOLUTION If a sky diver with an open parachute approaches the
ground with a constant velocity, the acceleration of the sky diver is zero, and the sky diver is,
therefore, in equilibrium. The two forces responsible for the equilibrium are the weight of the sky
diver and the force exerted on the sky diver by the parachute. These forces must be equal in
magnitude and opposite in direction so that their resultant is zero.
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26. REASONING AND SOLUTION There are three forces that act on the ring as shown in the
figure below. The weight of the block, which acts downward, and two forces of tension that act
along the rope away from the ring. Since the ring is at rest, the net force on the ring is zero. The
weight of the block is balanced by the vertical components of the tension in the rope. Clearly, the
rope can never be made horizontal, for then there would be no vertical components of the tension
forces to balance the weight of the block.
Chapter 4 Conceptual Questions
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T
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27. REASONING AND SOLUTION A freight train is accelerating on a level track. The tension in
the coupling between the engine and the first car depends on the total mass being pulled by the
engine. Therefore, other things being equal, if some of the cargo in the last car were transferred to
any one of the other cars, the tension in the coupling between the engine and the first car would
remain the same. This is because the transfer does not change the total mass being pulled by the
engine.
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