everyday forces

1. ©Photo
Researchers,
Inc.
Everyday Forces Main Ideas
Explain the difference between
mass and weight.
Find the direction and
magnitude of normal forces.
Describe air resistance as a
form of friction.
Use coefficients of friction to
calculate frictional force.
Weight on the Moon On the moon, astronauts
weigh much less than they do on Earth.
FIGURE 4.1
weight a measure of the gravitational
force exerted on an object; its value can
change with the location of the object in
the universe
normal force a force that acts on a
surface in a direction perpendicular to
the surface
Key Terms
weight static friction coefficient of friction
normal force kinetic friction
Weight
How do you know that a bowling ball weighs more than a tennis ball? If you
imagine holding one ball in each hand, you can imagine the downward
forces acting on your hands. Because the bowling ball has more mass than
the tennis ball does, gravitational force pulls more strongly on the bowling
ball. Thus, the bowling ball pushes your hand down with more force than
the tennis ball does.
The gravitational force exerted on the ball by Earth, Fg, is a vector
quantity, directed toward the center of Earth. The magnitude of this
force, Fg, is a scalar quantity called weight. The weight of an object can
be calculated using the equation Fg = mag, where ag is the magnitude
of the acceleration due to gravity, or free-fall acceleration. On the surface
of Earth, ag = g, and Fg = mg. In this book, g = 9.81 m/s2 unless other-
wise specified.
Weight, unlike mass, is not an inherent property of an
object. Because it is equal to the magnitude of the force due
to gravity, weight depends on location. For example, if the
astronaut in Figure 4.1 weighs 800 N (180 lb) on Earth, he
would weigh only about 130 N (30 lb) on the moon. The
value of ag on the surface of a planet depends on the planet’s
mass and radius. On the moon, ag is about 1.6 m/s2—much
smaller than 9.81 m/s2.
Even on Earth, an object’s weight may vary with location.
Objects weigh less at higher altitudes than they do at sea
level because the value of ag decreases as distance from the
surface of Earth increases. The value of ag also varies slightly
with changes in latitude.
The Normal Force
Imagine a television set at rest on a table. We know that the gravitational
force is acting on the television. How can we use Newton’s laws to explain
why the television does not continue to fall toward the center of Earth?
An analysis of the forces acting on the television will reveal the forces
that are in equilibrium. First, we know that the gravitational force of
Earth, Fg, is acting downward. Because the television is in equilibrium,
we know that another force, equal in magnitude to Fg but in the opposite
direction, must be acting on it. This force is the force exerted on the
television by the table. This force is called the normal force, Fn.
Forces and the Laws of Motion 133
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Differentiated Instruction
Preview Vocabulary
Visual Vocabulary Students may have
difficulty understanding the difference
between the weight of an object and its
mass. Sketch a diagram showing an
astronaut with a weight on Earth of
180 lb. Then sketch the same astronaut
on different planets with different
gravitational forces. Make the astronaut
the same size to emphasize that mass is
the same everywhere. A person who
weighs 180 lb on Earth would have these
weights on different bodies in the solar
system: Mercury: 68 lb, Venus: 163 lb,
moon: 30 lb, Jupiter: 426 lb.
FIGURE 4.1 Point out that it is easier to
lift a massive object on the moon than
on Earth because the object weighs less
on the moon, even though its mass
remains the same. Also, an object’s
inertia is the same regardless of the
magnitude of free-fall acceleration.
Ask Will a dart shot from a dart gun go
farther horizontally on Earth or on the
moon? Disregard air resistance.
Answer: The dart will travel farther
on the moon. Because the dart is
accelerated downward more slowly
on the moon than on Earth, it is in
motion for a longer time on the
moon. The horizontal velocity will
be the same in each case.
Teaching Tip
For practical purposes, the gravitational
field near the surface of Earth is
constant. For example, a person who
weighs 180.0 lb at sea level would weigh
179.5 lb at an altitude of 9.0 km above
sea level. This difference in weight is
only 0.3%.
Plan and Prepare
Teach
TEACH FROM VISUALS
English Learners
English learners might have difficulty differen-
tiating between homophones. For example,
this page includes terms that have homo-
phones with similar sounds but very different
meanings. For example:
weigh and way
night and knight
When discussing concepts as a class, check
occasionally to be sure that students know
which meaning of a term is used when
homophones are discussed. While students
may eventually figure it out themselves, they
may also miss valuable information in the
meantime.
Forces and the Laws of Motion 133
SECTION 4

2. F Fs F Fk
PHYSICS
Spec. Number PH 99 PE C04-004-001-A
Boston Graphics, Inc.
617.523.1333
(a)
Fn
Fg
Fn
Fg
The word normal is used because the direction of the contact force is
perpendicular to the table surface and one meaning of the word normal
is “perpendicular.” Figure 4.2 shows the forces acting on the television.
The normal force is always perpendicular to the contact surface but is
not always opposite in direction to the force due to gravity. Figure 4.3
shows a free-body diagram of a refrigerator on a loading ramp. The
normal force is perpendicular to the ramp, not directly opposite the force
due to gravity. In the absence of other forces, the normal force, Fn , is
equal and opposite to the component of Fg that is perpendicular to the
contact surface. The magnitude of the normal force can be calculated as
Fn = mg cos θ. The angle θ is the angle between the normal force and a
vertical line and is also the angle between the contact surface and a
horizontal line.
The Force of Friction
Consider a jug of juice at rest (in equilibrium) on a table, as in Figure 4.4(a).
We know from Newton’s first law that the net force acting on the jug is
zero. Newton’s second law tells us that any additional unbalanced force
applied to the jug will cause the jug to accelerate and to remain in motion
unless acted on by another force. But experience tells us that the jug will
not move at all if we apply a very small horizontal force. Even when we
apply a force large enough to move the jug, the jug will stop moving
almost as soon as we remove this applied force.
Friction opposes the applied force.
When the jug is at rest, the only forces acting on it are the force due to
gravity and the normal force exerted by the table. These forces are equal
and opposite, so the jug is in equilibrium. When you push the jug with a
small horizontal force F, as shown in Figure 4.4(b), the table exerts an equal
force in the opposite direction. As a result, the jug remains in equilibrium
and therefore also remains at rest. The resistive force that keeps the jug
from moving is called the force of static friction, abbreviated as Fs .
Normal Force
In this example, the
normal force, Fn, is
equal and opposite
to the force due to
gravity, Fg.
FIGURE 4.2
Normal Force When an
Object Is on a Ramp
The normal force is not
always opposite the force
due to gravity, as shown
by this example of
a refrigerator on
a loading ramp.
FIGURE 4.3
(b) When a small force is applied, the jug
remains in equilibrium because the
static-friction force is equal but
opposite to the applied force.
(a) Because this jug of juice is in
equilibrium, any unbalanced
horizontal force applied to it will
cause the jug to accelerate.
(c) The jug begins to accelerate as soon
as the applied force exceeds the
opposing static-friction force.
static friction the force that resists
the initiation of sliding motion between
two surfaces that are in contact and
at rest
Overcoming the Force of Friction
FIGURE 4.4
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Differentiated Instruction
FIGURE 4.2 Tell students that the
television is in equilibrium, so the normal
force from the table must be equal in
magnitude and opposite in direction
to the gravitational force exerted on
the television.
Instruct students to draw free-body
diagrams for the television, table, and
Earth and to identify the third-law pairs.
Ask Do the forces in Figure 4.2
constitute an action-reaction pair
(Newton’s third law)?
Answer: No, both forces act on the
television and therefore cannot be an
action-reaction pair.
Teaching Tip
Be sure students understand why the
two angles labeled θ in Figure 4.3 are
equal. Recognizing equal angles in
free-body diagrams is an important
problem-solving skill for this chapter.
Teach continued
TEACH FROM VISUALS
Below Level
Hands-on exploration will help students to
better understand the difference between
static friction and kinetic friction. Encourage
them to explore friction by gently pushing
objects of varying mass across their desks.
Which objects required greater force to
overcome static friction? Students may say
larger or heavier objects. Remind them that
these objects have greater mass.
Encourage students to discuss examples of
friction in everyday life. How is static friction
involved in walking? Static friction holds your
foot in place while you move your other foot.
Compare the friction of walking on an icy
surface compared with a dry one. Less static
friction on ice means that the friction is easier
to overcome, which makes you slip and fall.
When pushing something large, why is it easier
to keep it moving than to start it moving?
Static friction is greater than kinetic friction.
134 Chapter 4

3. CO4-004-007-A
As long as the jug does not move, the force of static friction is always
equal to and opposite in direction to the component of the applied force
that is parallel to the surface ( Fs = -Fapplied ). As the applied force
increases, the force of static friction also increases; if the applied force
decreases, the force of static friction also decreases. When the applied
force is as great as it can be without causing the jug to move, the force of
static friction reaches its maximum value, Fs,max.
Kinetic friction is less than static friction.
When the applied force on the jug exceeds Fs,max, the jug begins to move
with an acceleration to the left, as shown in Figure 4.4(c). A frictional force
is still acting on the jug as the jug moves, but that force is actually less
than Fs,max. The retarding frictional force on an object in motion is called
the force of kinetic friction (Fk). The magnitude of the net force acting on
the object is equal to the difference between the applied force and the
force of kinetic friction (Fapplied - Fk).
At the microscopic level, frictional forces arise from complex interactions
between contacting surfaces. Most surfaces, even those that seem very
smooth to the touch, are actually quite rough at the microscopic level, as
illustrated in Figure 4.5. Notice that the surfaces are in contact at only a few
points. When two surfaces are stationary with respect to each other, the
surfaces stick together somewhat at the contact points. This adhesion is
caused by electrostatic forces between molecules of the two surfaces.
The force of friction is proportional to the normal force.
It is easier to push a chair across the floor at a constant speed than to
push a heavy desk across the floor at the same speed. Experimental
observations show that the magnitude of the force of friction is approxi-
mately proportional to the magnitude of the normal force that a surface
exerts on an object. Because the desk is heavier than the chair, the desk
also experiences a greater normal force and therefore greater friction.
Friction can be calculated approximately.
Keep in mind that the force of friction is really a macroscopic effect
caused by a complex combination of forces at a microscopic level.
However, we can approximately calculate the force of friction with certain
assumptions. The relationship between normal force and the force of
friction is one factor that affects friction. For instance, it is easier to slide a
light textbook across a desk than it is to slide a heavier textbook. The
relationship between the normal force and the force of friction provides a
good approximation for the friction between dry, flat surfaces that are at
rest or sliding past one another.
Microscopic View of Surfaces
in Contact On the microscopic level,
even very smooth surfaces make contact
at only a few points.
FIGURE 4.5
kinetic friction the force that opposes
the movement of two surfaces that are
in contact and are sliding over
each other
Tips and Tricks
In free-body diagrams, the force of friction is always parallel to the surface of contact.
The force of kinetic friction is always opposite the direction of motion. To determine
the direction of the force of static friction, use the principle of equilibrium. For an
object in equilibrium, the frictional force must point in the direction that results in a
net force of zero.
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Static vs. Kinetic Friction
Purpose Show that kinetic friction is
less than static friction.
Materials rectangular block, hook,
spring scale
Procedure Use the spring scale to
measure the force required to start the
rectangular block moving. Then, use the
spring scale to measure the frictional
force for constant velocity. Perform
several trials. Have students record all
data and find the average for each. Point
out that the normal force and the
surfaces remain the same, so the only
difference in the two average values is
due to motion.
Friction of Different Surfaces
Purpose Show students that the force
of friction depends on the surface.
Materials large cube with different
materials (such as glass, carpeting, and
sandpaper) covering each of four sides,
with two sides left uncovered; hook;
spring scale
Procedure Attach the hook to one of
the two uncovered sides of the block.
Pull the block across the table with the
spring scale. Repeat the demonstration
with a new surface of the cube exposed
to the table. Repeat the demonstration
for the two remaining covered sides.
Have students summarize the results
and reach a conclusion concerning the
nature of the surfaces in contact and
the frictional force.
Demonstration
Demonstration
Forces and the Laws of Motion 135

4. ©Mark
Gallup/Pictor/Image
State
The force of friction also depends on the composition and qualities of
the surfaces in contact. For example, it is easier to push a desk across a
tile floor than across a floor covered with carpet. Although the normal
force on the desk is the same in both cases, the force of friction between
the desk and the carpet is higher than the force of friction between the
desk and the tile. The quantity that expresses the dependence of frictional
forces on the particular surfaces in contact is called the coefficient of
friction. The coefficient of friction between a waxed snowboard and the
snow will affect the acceleration of the snowboarder shown in Figure 4.6.
The coefficient of friction is represented by the symbol μ, the lowercase
Greek letter mu.
The coefficient of friction is a ratio of forces.
The coefficient of friction is defined as the ratio of the force of friction to
the normal force between two surfaces. The coefficient of kinetic friction is
the ratio of the force of kinetic friction to the normal force.
μk =
Fk
_
Fn
The coefficient of static friction is the ratio of the maximum value of the
force of static friction to the normal force.
μs =
Fs,max
_
Fn
If the value of μ and the normal force on the object are known, then
the magnitude of the force of friction can be calculated directly.
Ff = μFn
Figure 4.7 shows some experimental values of μs and μk for different
materials. Because kinetic friction is less than or equal to the maximum
static friction, the coefficient of kinetic friction is always less than or equal
to the coefficient of static friction.
Minimizing Friction
Snowboarders wax their boards to
minimize the coefficient of friction
between the boards and the snow.
FIGURE 4.6
coefficient of friction the ratio of the
magnitude of the force of friction
between two objects in contact to the
magnitude of the normal force with
which the objects press against
each other
FIGURE 4.7
COEFFICIENTS OF FRICTION (APPROXIMATE VALUES)
μs μk μs μk
steel on steel 0.74 0.57 waxed wood on wet snow 0.14 0.1
aluminum on steel 0.61 0.47 waxed wood on dry snow – 0.04
rubber on dry concrete 1.0 0.8 metal on metal (lubricated) 0.15 0.06
rubber on wet concrete – 0.5 ice on ice 0.1 0.03
wood on wood 0.4 0.2 Teflon on Teflon 0.04 0.04
glass on glass 0.9 0.4 synovial joints in humans 0.01 0.003
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Problem Solving
Friction and Surface Area
Purpose Show the relation between
surface area and frictional forces.
Materials rectangular block, hook,
spring scale
Procedure Attach the hook to the
block. Pull the block across the table
with the spring scale. Have students
note the force required to pull the block
at a constant velocity. Repeat the
demonstration for another surface area
in contact with the table and have
students note the force. Ask students to
summarize the results and to reach a
conclusion concerning the areas in
contact and frictional forces.
FIGURE 4.6 Remind students that
frictional force depends on the coeffi-
cient of friction and the normal force.
Ask What changes in environment
might cause a change in the frictional
force experienced by the snowboarder
on the way down the hill?
Answer: Answers will vary but could
include the following: surface conditions
(such as wet or dry snow, ice, and dirt),
whether the snowboarder is moving or
not moving, and the angle of the hill.
Teach continued
Demonstration
TEACH FROM VISUALS
Reality Check
Encourage students to rely on their common
sense and experience with real objects when
checking answers. For example, if they are
asked to calculate and compare the coeffi-
cient of static friction of the same object on a
rough surface and a smooth surface, and their
answer gives a greater value for the smooth
surface, then they should recognize that there
is an error in the calculation. They know from
experience that friction is less on smoother
surfaces compared with rougher surfaces, and
they should apply this real-life experience to
the mathematical model represented by the
equation. If they are calculating the coefficient
of kinetic friction and the coefficient of static
friction for an object being dragged across a
paved surface, they should expect the
coefficient of kinetic friction to be the lesser
value, because they have experienced that it
takes less force to keep an object moving than
to start it moving.
136 Chapter 4

5. Coefficients of Friction
Sample Problem D A 24 kg crate initially at rest on a
horizontal floor requires a 75 N horizontal force to set it in
motion. Find the coefficient of static friction between the crate
and the floor.
ANALYZE Given: Fs,max = Fapplied = 75 N
m = 24 kg
Unknown: μs = ?
SOLVE Use the equation for the coefficient of
static friction.
μs =
Fs,max
_
Fn
=
Fs,max
_
mg
μs = 75 N
__
24 kg × 9.81m/s2
μs = 0.32
1. Once the crate in Sample Problem D is in motion, a horizontal force of 53 N keeps
the crate moving with a constant velocity. Find μk, the coefficient of kinetic
friction, between the crate and the floor.
2. A 25 kg chair initially at rest on a horizontal floor requires a 165 N horizontal force
to set it in motion. Once the chair is in motion, a 127 N horizontal force keeps it
moving at a constant velocity.
a. Find the coefficient of static friction between the chair and the floor.
b. Find the coefficient of kinetic friction between the chair and the floor.
3. A museum curator moves artifacts into place on various different display surfaces.
Use the values in Figure 4.7 to find Fs,max and Fk for the following situations:
a. moving a 145 kg aluminum sculpture across a horizontal steel platform
b. pulling a 15 kg steel sword across a horizontal steel shield
c. pushing a 250 kg wood bed on a horizontal wood floor
d. sliding a 0.55 kg glass amulet on a horizontal glass display case
μs =
μs =
μ =
PREMIUM CONTENT
Interactive Demo
HMDScience.com
Tips and Tricks
Because the crate is on
a horizontal surface, the
magnitude of the normal
force (Fn) equals the
crate’s weight (mg).
Forces and the Laws of Motion 137
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Classroom Practice
Coefficients of Friction
A refrigerator is placed on a ramp. The
refrigerator begins to slide when the
ramp is raised to an angle of 34°. What is
the coefficient of static friction?
Answer: 0.67
PROBLEM guide D
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = Sample Problem Set II (online)
Solving for:
µ SE
Sample, 1–2; Ch. Rvw.
35, 36*, 37*, 49
PW 4–7, 10*
PB 8–10
Ff
SE
3
PW
Sample, 1–3, 7, 10*
PB 5–7
Fn
, m PW 8–9
PB Sample, 1–4
*Challenging Problem
Answers
Practice D
1. 0.23
2. a. 0.67
b. 0.52
3. a. 8.7 × 102
N, 6.7 × 102
N
b. 1.1 × 102
N, 84 N
c. 1 × 103
N, 5 × 102
N
d. 5 N, 2 N
Forces and the Laws of Motion 137

6. ©Houghton
Mifflin
Harcourt
Overcoming Friction
Sample Problem E A student attaches a rope to a 20.0 kg box
of books. He pulls with a force of 90.0 N at an angle of 30.0° with
the horizontal. The coefficient of kinetic friction between the box
and the sidewalk is 0.500. Find the acceleration of the box.
ANALYZE Given: m = 20.0 kg μk = 0.500
Fapplied = 90.0 N at θ = 30.0°
Unknown: a = ?
Diagram:
applied
F
g
F
k
F
n
F
PLAN Choose a convenient coordinate system, and find the x and y
components of all forces.
The diagram at left shows the most convenient coordinate
system, because the only force to resolve into components
is Fapplied.
Fapplied,y = (90.0 N)(sin 30.0°) = 45.0 N (upward)
Fapplied,x = (90.0 N)(cos 30.0°) = 77.9 N (to the right)
Choose an equation or situation:
A. Find the normal force, Fn, by applying the condition of
equilibrium in the vertical direction: ΣFy = 0.
B. Calculate the force of kinetic friction on the box:
Fk = μkFn.
C. Apply Newton’s second law along the horizontal
direction to find the acceleration of the box:
ΣFx = max.
SOLVE Substitute the values into the equations and solve:
A. To apply the condition of equilibrium in the vertical
direction, you need to account for all of the forces in the
y direction: Fg , Fn , and Fapplied,y. You know Fapplied,y and
can use the box’s mass to find Fg.
Fapplied,y = 45.0 N
Fg = (20.0 kg)(9.81 m/s2) = 196 N
30°
Fg
Fapplied
Fn
Fk
Continued
Interactive Demo
HMDScience.com
PREMIUM CONTENT
Choose a convenient coordinate system, and find the and
Chapter 4
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Problem Solving
Classroom Practice
Overcoming Friction
Two students are sliding a 225-kg sofa at
constant speed across a wood floor.
One student pulls with a force of 225 N
at an angle of 13° above the horizontal.
The other student pushes with a force
of 250 N at an angle of 23° below the
horizontal. What is the coefficient of
kinetic friction between the sofa and
the floor?
Answer: 0.22
How could the students make moving
the sofa easier?
Answer: They could change the angles,
put the sofa on rollers, or wax the floors.
Teach continued
Alternative Approaches
When evaluating the measures of Fapplied, x
and
Fapplied, y
, remind students that they may use the
Pythagorean theorem. This is the equation
that compares the three sides of a right
triangle. The theorem can be written as:
a2
+ b2
= c2
When solving for Fapplied, x
and Fapplied, y
, use the
equation as so:
(Fapplied
)2
= (Fapplied, x
)2
+ (Fapplied, y
)2
= (77.9)2
+ (45)2
Fapplied
= 90 N
138 Chapter 4

7. Overcoming Friction (continued)
Next, apply the equilibrium condition, ΣFy = 0, and solve for Fn.
ΣFy = Fn + Fapplied,y - Fg = 0
Fn + 45.0 N - 196 N = 0
Fn = -45.0 N + 196 N = 151 N
B. Use the normal force to find the force of kinetic friction.
Fk = μk Fn = (0.500)(151 N) = 75.5 N
C. Use Newton’s second law to determine the horizontal
acceleration.
ΣFx = Fapplied, x - Fk = max
ax =
Fapplied, x - Fk
__
m = 77.9 N - 75.5 N
__
20.0 kg
= 2.4 N
_
20.0 kg
=
2.4 kg• m/s2
__
20.0 kg
a = 0.12 m/s2 to the right
CHECK
YOUR
WORK
The normal force is not equal in magnitude to the weight because the y
component of the student’s pull on the rope helps support the box.
1. A student pulls on a rope attached to a box of books and moves the box down the
hall. The student pulls with a force of 185 N at an angle of 25.0° above the
horizontal. The box has a mass of 35.0 kg, and μk between the box and the floor is
0.27. Find the acceleration of the box.
2. The student in item 1 moves the box up a ramp inclined at 12° with the horizontal.
If the box starts from rest at the bottom of the ramp and is pulled at an angle of
25.0° with respect to the incline and with the same 185 N force, what is the
acceleration up the ramp? Assume that μk = 0.27.
3. A 75 kg box slides down a 25.0° ramp with an acceleration of 3.60 m/s2.
a. Find μk between the box and the ramp.
b. What acceleration would a 175 kg box have on this ramp?
4. A box of books weighing 325 N moves at a constant velocity across the floor when
the box is pushed with a force of 425 N exerted downward at an angle of 35.2°
below the horizontal. Find μk between the box and the floor.
Tips and Tricks
Remember to pay attention to the direction
of forces. Here, Fg is subtracted from Fn and
Fapplied,y because Fg is directed downward.
Tips and Tricks
Fk is directed toward the left, opposite
the direction of Fapplied,x. As a result,
when you find the sum of the forces in
the x direction, you need to subtract
Fk from Fapplied, x.
Forces and the Laws of Motion 139
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PROBLEM guide E
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = Sample Problem Set II (online)
Solving for:
Ff
, a SE Sample, 1–3; Ch. Rvw.
38*, 39*, 47a–b*, 48c
PW 5–7
PB 4–7
Fn
, m SE Ch. Rvw. 21, 29, 41, 50,
52*
PW Sample, 1–3
PB 8–10
µ SE 3, 4; Ch. Rvw. 36–37,
48b*
PW 4
PB Sample, 1–3
*Challenging Problem
Answers
Practice E
1. 2.7 m/s2
in the positive x direction
2. 0.77 m/s2
up the ramp
3. a. 0.061
b. 3.61 m/s2
down the ramp (Note
that m cancels in the solution,
and a is the same in both cases;
the slight difference is due to
rounding.)
4. 0.609
Forces and the Laws of Motion 139

8. ©The
Goodyear
Tire
Rubber
Company
Air resistance is a form of friction.
Another type of friction, the retarding force produced by air resistance, is
important in the analysis of motion. Whenever an object moves through a
fluid medium, such as air or water, the fluid provides a resistance to the
object’s motion.
For example, the force of air resistance, FR, on a moving car acts in
the direction opposite the direction of the car’s motion. At low speeds,
the magnitude of FR is roughly proportional to the car’s speed. At higher
speeds, FR is roughly proportional to the square of the car’s speed.
When the magnitude of FR equals the magnitude of the force moving the
car forward, the net force is zero and the car moves at a constant speed.
A similar situation occurs when an object falls through air. As a
free-falling body accelerates, its velocity increases. As the velocity
increases, the resistance of the air to the object’s motion also constantly
increases. When the upward force of air resistance balances the
downward gravitational force, the net force on the object is zero and the
object continues to move downward with a constant maximum speed,
called the terminal speed.
A
ccelerating a car seems simple to the driver. It is
just a matter of pressing on a pedal or turning a
wheel. But what are the forces involved?
A car moves because as its wheels turn, they push
back against the road. It is actually the reaction force of
the road pushing on the car that causes the car to
accelerate. Without the friction between the tires and
the road, the wheels would not be able to exert this force
and the car would not experience a reaction force.
Thus, acceleration requires this friction. Water and snow
provide less friction and therefore reduce the amount of
control the driver has over the direction and speed of
the car.
As a car moves slowly over an area of water on the
road, the water is squeezed out from under the tires.
If the car moves too quickly, there is not enough time for
the weight of the car to squeeze the water out from
under the tires. The water trapped between the tires and
the road will lift the tires and car off the road, a
phenomenon called hydroplaning. When this situation
Driving and Friction
occurs, there is very little friction between the tires and
the water, and the car becomes difficult to control.
To prevent hydroplaning, rain tires, such as the ones
shown above, keep water from accumulating between
the tire and the road. Deep channels down the center of
the tire provide a place for the water to accumulate, and
curved grooves in the tread channel the water outward.
Because snow moves even less easily than water,
snow tires have several deep grooves in their tread,
enabling the tire to cut through the snow and make
contact with the pavement. These deep grooves push
against the snow and, like the paddle blades of a
riverboat, use the snow’s inertia to provide resistance.
Chapter 4
140
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Differentiated Instruction
Why It Matters
Driving and Friction
The coefficient of friction between the
ground and the tires of a car is less when
rain or snow is on the ground. Snow and
rain tires are excellent examples of ways
that tires are adapted to regain some of
the necessary frictional forces.
Point out to students that the
friction between a tire and pavement is
more complex than the simple sliding
friction between dry surfaces, which
they have been studying. The force of
friction on a car tire is not necessarily
simply proportional to the normal force.
The fact that there is not a simple
proportion between the frictional and
normal forces is due in part to the fact
that the tires are rolling, so they peel
vertically away from the surface rather
than continuously slide across it. Also,
when the road is covered with water or
snow, other factors such as viscosity
come into play.
Teach continued
Pre-AP
Objects moving in outer space do not
experience air resistance. Thus, Earth continu-
ally orbits the sun without slowing down.
(Earth’s speed is actually decreasing because of
frequent collisions with small masses such as
meteoroids, but this effect is minor.) Earth’s
velocity does, however, change slightly as it
orbits the sun. Have students do research to
find out what factor affects Earth’s velocity
in its orbit.
Earth moves slightly faster in its orbit when
it is closer to the sun in January than when it is
farthest from the sun in July. This difference is
a corollary to Kepler’s second law of planetary
motion, the equal-area law, which describes
the movement of planets in elliptical orbits.
140 Chapter 4

9. There are four fundamental forces.
At the microscopic level, friction results from interactions between the
protons and electrons in atoms and molecules. Magnetic force also
results from atomic phenomena. These forces are classified as
electromagnetic forces. The electromagnetic force is one of four
fundamental forces in nature. The other three fundamental forces are
gravitational force, the strong nuclear force, and the weak nuclear
force. All four fundamental forces are field forces.
The strong and weak nuclear forces have very small ranges, so their
effects are not directly observable. The electromagnetic and gravitational
forces act over long ranges. Thus, any force you can observe at the
macroscopic level is either due to gravitational or electromagnetic forces.
The strong nuclear force is the strongest of all four fundamental
forces. Gravity is the weakest. Although the force due to gravity holds the
planets, stars, and galaxies together, its effect on subatomic particles is
negligible. This explains why electric and magnetic effects can easily
overcome gravity. For example, a bar magnet has the ability to lift another
magnet off a desk.
Reviewing Main Ideas
1. Draw a free-body diagram for each of the following objects:
a. a projectile accelerating downward in the presence of air resistance
b. a crate being pushed across a flat surface at a constant speed
2. A bag of sugar has a mass of 2.26 kg.
a. What is its weight in newtons on the moon, where the acceleration due
to gravity is one-sixth that on Earth?
b. What is its weight on Jupiter, where the acceleration due to gravity is
2.64 times that on Earth?
3. A 2.0 kg block on an incline at a 60.0° angle is held in equilibrium by a
horizontal force.
a. Determine the magnitude of this horizontal force. (Disregard friction.)
b. Determine the magnitude of the normal force on the block.
4. A 55 kg ice skater is at rest on a flat skating rink. A 198 N horizontal force
is needed to set the skater in motion. However, after the skater is in mo-
tion, a horizontal force of 175 N keeps the skater moving at a constant
velocity. Find the coefficients of static and kinetic friction between the
skates and the ice.
Critical Thinking
5. The force of air resistance acting on a certain falling object is roughly pro-
portional to the square of the object’s velocity and is directed upward. If the
object falls fast enough, will the force of air resistance eventually exceed the
weight of the object and cause the object to move upward? Explain.
Forces and the Laws of Motion 141
SECTION 4 FORMATIVE ASSESSMENT
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Answers to Section Assessment
1. a. An arrow labeled Fg
should point down,
and an arrow labeled Fair
should point
opposite the direction of motion. The
arrow Fg
should be longer than the
arrow Fair
.
b. Fg
points down, Fn
points up, Fapplied
is
horizontal, and Ffriction
points in the
opposite direction. The two vertical
arrows are equal in length, as are the
two horizontal arrows.
2. a. 3.70 N
b. 58.5 N
3. a. 34 N
b. 39 N
4. 0.37, 0.32
5. no; Once at equilibrium, the velocity will
not increase, so the force of air resistance
will not increase.
Assess Use the Formative Assessment
on this page to evaluate student
mastery of the section.
Reteach For students who need
additional instruction, download the
Section Study Guide.
Response to Intervention To reassess
students’ mastery, use the Section Quiz,
available to print or to take directly
online at HMDScience.com.
Assess and Reteach
Forces and the Laws of Motion 141