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Transcript
```4
Key Concepts
After completing this chapter you will be
able to
 describe the relationship among mass,
gravitational field strength, and the force of
gravity
 analyze and solve problems involving the
force of gravity and free fall
 analyze and solve problems involving
friction and coefficients of friction
 conduct an inquiry that applies Newton’s
laws
 plan and conduct an inquiry to analyze the
effects of forces acting on an object
 analyze a technology that applies
Newton’s laws
 evaluate the impact on society and the
environment of technologies that use the
principles of forces
Applications of Forces
How Do We Use Forces to Understand and Design
Technology?
The Indy race car involves a tremendous amount of technology
related to forces. Technology such as the powerful engine and the
large wide tires is involved with increasing force. Other parts of the
car that involve technology are the streamlined shape and the oil in
the engine, which both help to decrease frictional forces.
One of the most prominent features of Indy cars are the wings at
the front and back of the cars. These components act like inverted
airplane wings, creating downward instead of upward forces on the
cars. What effect, if any, does a feature like this have on the
performance of the car?
Another interesting design feature is the wide tires that have no
treads. These tires are quite different from the all-season tires used
on most vehicles in Ontario. At the start of the race, drivers drive
slowly around the track together while swerving back and forth. How
does this improve performance?
A lot of technology is involved in making more powerful and
efficient engines for these cars. Ideally, a driver wants a car that
can move fast and accelerate quickly yet is easy to control and
uses as little fuel as possible. Some of this can be accomplished by
making the car, including the engine, lighter. The unusual external
design of the car reduces air friction.
New innovations might one day involve coating engine parts with
near-frictionless carbon. This type of coating is very easy to apply to
almost any surface, and it almost eliminates friction while being very
durable. However, it is very expensive and might only be used in
race cars in the near future.
In this chapter you will explore the force of gravity and frictional
forces, as well as how these forces affect motion, in more detail.
You will also learn how gravity, friction, and Newton’s laws can be
used to design and understand different types of technology. Some
of this technology, such as that used in cars, may be familiar.
However, some of the technology, such as near-frictionless carbon
and mechanical arms, may be new to you.
Chapter 3 Introduction 1
STARTING POINTS
Answer these questions using your current knowledge. You will have an opportunity to revisit these
questions later, applying concepts and skills from the chapter.
1. (a) Explain in your own words why the wings on the Indy car in the photograph increase the
downward force on the car.
(b) What effect, if any, does this have on the normal force acting on the car? Draw a free-body
diagram (FBD) to help explain.
(c) What effect, if any, does this have on the force of friction acting on the tires? Is this effect
2. What effect does the surface area of the tire in contact with the road have on the magnitude of
friction acting on the tire?
3. (a) Why are race car tires wider than tires on passenger cars?
(b) Why do many car races stop when it rains?
(c) What advantages do you think tires on passenger cars have over race car tires?
4. (a) What are some other possible applications of near-frictionless carbon?
(b) Explain how near-frictionless carbon can help to make a car engine more efficient and last
longer.
[Formatter: this is the chapter opener photo, and should fill the page up to the
Mini Investigation (below)]
[CATCH: C04-P001-OP11SB (size CO): photo of an Indy car ]
Mini Investigation
Friction from Shoes
Skills: Performing, Observing, Analyzing, Evaluating, Communicating
You can measure the force of friction acting on a stationary object by
pulling on it with a spring scale or a force meter. If you measure the
largest force required to move the object, you will determine the
maximum amount of friction the surface exerts on the object just before
it starts to move. In this activity, you will measure the maximum force
of friction that must be overcome to start moving several different types
of shoes on the floor. Before performing this activity, make sure that
you know how to zero the spring scale or calibrate the force meter.
Equipment and Materials: balance; spring scale or force meter; four
different types of shoes (dress shoe, running shoe, loafers, highheeled shoes, etc.)
1. Examine each shoe and arrange them, from largest to smallest
maximum force of friction, based on what you guess the force of
friction might be.
2. Make a table with these headings: Type of shoe, Mass of shoe, Fg
on shoe, Maximum force of friction, and Maximum force of
friction/Fg.
3. Measure and record the mass of each shoe in your table.
4. Calculate the force of gravity on each shoe using the equation Fg =
mg. Record the force of gravity in the table.
5. Measure the maximum force of friction for the first type of shoe. To
do this, place the shoe on the floor and pull horizontally with the
spring scale or the force meter, as shown in Figure 1. You can
hook the scale or meter to the heel of the shoe or to the laces. The
maximum force of friction is equal to the largest force reading
before the shoe starts moving. Record the reading in the table.
[CATCH: C04-P002-OP11SB (size C): set up photo of a spring scale
hooked onto the front part of a shoe being pulled by a person’s
hand ]
[caption] Figure 1
6. Repeat step 5 for the other shoes and record your results in the
table.
7. Some shoes are heavier than others and so will be harder to start
moving. The only way to compare the shoes in a fair way is to
divide each maximum force of friction by the force of gravity acting
A. To get a fair comparison between the shoes, you divided the
maximum force of friction by the force of gravity on each shoe. How
does this make the comparison of the results fair? [T/I]
B. Some people might argue that running shoes are designed to
increase the force of friction more than other kinds of shoes. [T/I]
[C]
(a) Do you agree with this statement? Explain your reasoning.
(b) Do the results of this investigation support this statement? Explain
Chapter 3 Introduction 2
[START SECTION 4.1: 6 pages]
4.1
[CATCH: C04-P003-OP11SB (size D):
photo of a skydiver falling at terminal
velocity]
Gravitational Force Near
Earth
Some people try skydiving just for the thrill. Others skydive as part
of their military training. Still others like the view and the feeling of
of gravity and how it can be used to study the physics of activities
such as skydiving.
Figure 1 A skydiver is affected by the
force of gravity and air friction.
The only forces acting on a skydiver jumping out of a plane are
gravity and air friction. Since the force of gravity is so much greater
in magnitude than the air friction at the beginning of the dive, the
skydiver accelerates downward, initially at 9.8 m/s2 [down]. As the
[CATCH: C04-P004A-OP11SB Size E.
Multiflash photo of falling ball]
[C04-P004B-OP11SB Size E, multi-flash
photograph of a large, light object that is
falling at terminal velocity]
skydiver continues to fall, the acceleration decreases until the
skydiver reaches a constant or terminal speed (Figure 1). At this
terminal speed, the skydiver is no longer accelerating. In this
section, you will investigate the forces that act on a falling object
and why that object eventually reaches a constant terminal speed.
Figure 2 (a) Multi-flash photograph of a
ball in free fall starting from rest (b)
Multi-flash photograph of a light object
with large surface area in free fall.
Which object is accelerating? How can
you tell?
Gravity and Free Fall
Figure 2(a) shows a small, heavy ball falling from rest. The
photograph was taken using a flash that goes on and off at equal
time intervals. The image of the ball was then captured at different
positions during its fall. Notice that the distance between the starting
free fall the motion of a falling object where
the only force acting on the object is gravity
position and the position in the second image is small, while the
images get farther and farther apart at later times. This uneven
spacing means that the ball is accelerating while it is falling.
Now look at the falling object in Figure 2(b). The photograph
shows a light object with a large cross-sectional area (the area you
see if you look directly up at the falling object). Notice that the
distance between successive images in this photograph is constant.
This means that the object is not accelerating, but is falling with
constant velocity. So the net force acting on the object is zero. In
this situation, another force must be acting on the falling object other
terminal speed the maximum constant
speed of a falling object
than gravity. The force that balances gravity is air friction—friction
caused by the air. In many instances in this chapter, we will study
objects that are falling and we will ignore air friction. When an
object is falling under the influence of the gravitational force only,
the object is said to be in free fall.
The force of air friction acting on an object depends on many
Section Title 3
factors. One factor is the cross-sectional area of the object. Larger
cross-sectional areas experience more air friction than smaller crosssectional areas. To test this, hold a sheet of paper horizontally in
one hand and an identical sheet of paper crumpled up into a ball in
the other hand. Then drop both objects from the same height. You
will notice that the crumpled paper falls much more quickly than the
sheet of paper. Both objects have the same mass, and the force of
gravity acting on both objects has the same magnitude. However,
the air friction acting on each object is quite different. The horizontal
sheet of paper has a larger cross-sectional area, so it experiences
more air friction. Another factor that affects the force of air friction is
the speed of the object. Faster-moving objects experience more air
friction. Air friction acts opposite to the direction of motion of the
object if there is no wind.
jumps out of the plane and starts falling. The instant the skydiver
leaves the plane, her initial velocity in the vertical direction is zero
and she experiences very little air friction (Figure 3(a)). As the
skydiver continues to fall, her speed increases and she is now
accelerating. As the speed of the skydiver increases, the magnitude
of air friction acting on her also increases (Figure 3(b)). During the
time that the air friction is increasing, the net force on the skydiver
is decreasing. This means that the acceleration of the skydiver is
force field a region of space surrounding an
object that can exert a force on other objects
that are placed within that region and are
able to interact with that force
decreasing. Eventually the magnitude of air friction becomes great
enough that it equals the magnitude of the force of gravity (Figure
3(c)). At this moment, the skydiver is moving at constant speed.
That speed is called the terminal speed.
[CATCH: C04-F002-OP11SB (size D):
Earth’s gravitational field using vector
arrows]
To slow down, the skydiver must increase the force of air friction
acting on her body. To do this, the skydiver must increase her
cross-sectional area moving through the air. So she opens her
Figure 4 The gravitational force field
surrounding Earth attracts all other objects
placed within this field. The magnitude of
Earth’s gravitational field decreases as an
object moves farther and farther away from
Earth’s surface.
parachute. As soon as she opens the parachute, the upward force
of air friction is much greater in magnitude than the downward force
of gravity (Figure 3(d)). So the skydiver begins to slow down
because the net force is directed upward. Since the speed of the
skydiver is decreasing, the force of air friction acting on the skydiver
gravitational field strength the force per
unit mass acting on an object when placed in
a gravitational field
also decreases (Figure 3(e)). Eventually the force of air friction
decreases to the point where its magnitude again equals the force
of gravity (Figure 3(f)). Now the skydiver is moving at a much
Section Title 4
slower terminal speed than before and she can land safely on the
ground.
[CATCH FORMATTER: Set the following 6 pieces of art side by side
at full page measure. Each is size C1. Label (a) through (f)]
[CATCH: C04-F001A-OP11SB]
[CATCH: C04-F001B-OP11SB]
[CATCH: C04-F001C-OP11SB]
[CATCH: C04-F001D-OP11SB]
[CATCH: C04-F001E-OP11SB]
[CATCH: C04-F001F-OP11SB]
Figure 3 FBDs of a skydiver in free fall under various conditions. (a) The magnitude of air friction is
small at the beginning because the speed is small. (b) As the speed increases, so does the
magnitude of air friction. (c) Eventually the skydiver reaches a terminal speed. (d) The parachute
opens and the magnitude of air friction increases, and the skydiver slows down. (e) As the skydiver
continues to slow down, the magnitude of air friction decreases. (f) The terminal speed is lower with
the parachute open.
Gravitational Field Strength
[CATCH: C04-P005-OP11SB (size D):
photo showing the variation of Earth’s
gravitational field strength at different
locations]
How can Earth pull on objects and make them fall toward its
centre? Gravity is an action-at-a-distance force that pulls on objects
without making any contact with them. This occurs because Earth is
surrounded by a gravitational force field. A force field is a region of
space surrounding an object that can exert a force on other objects;
Figure 5 The magnitude of the
gravitational field strength varies
depending on the location on Earth’s
surface.
the field exerts a force only on objects placed within that region that
are able to interact with that force. To represent the force field
around Earth, we draw lines of force that point toward Earth’s centre
(Figure 4). No matter how large or small the mass of an object is,
the object will always be attracted to Earth when it interacts with
[margin icon for Investigation 4.1.1]
Investigation 4.1.1: Acceleration due
to Gravity and Terminal Velocity (p.
xxx)
In this investigation, you will explore
some of the factors that affect the
acceleration due to gravity at your
location. You will also measure its value
this force field.
To determine the magnitude of Earth’s gravitational force field at a
particular location near its surface, physicists use a quantity called
gravitational field strength. The gravitational field strength is the
force, per kilogram of mass, acting on an object within a
gravitational field. The gravitational field strength is a vector quantity
because it has a direction. The gravitational field strength due to
Earth always points toward Earth’s centre. Gravitational field strength
has units of newtons per kilogram. At Earth’s surface, the
gravitational field strength is 9.8 N/kg [down]. Notice that this has
the same magnitude as the acceleration due to gravity at Earth’s
surface. In other words, the gravitational field strength at a location
has the same magnitude as the acceleration due to gravity at that
location. To determine the gravitational field strength at a particular
Section Title 5
location, you can measure the force of gravity acting on an object
[CATCH: C04-P006-OP11SB (size D):
photo of astronauts Julie Payette and Robert
Thirsk aboard ISS. Photo should clearly
show astronauts floating in
microgravity/free fall. If photo of Julie
Payette and Robert Thirsk is not available,
please try to get a photo showing other
using a newton spring scale or force meter and divide by the mass
of the object. Then you can use the equation F g  mg to solve for g :
g
Fg
m
Another way to find the gravitational field strength in a region is
to drop an object and measure the acceleration due to gravity. This
method only gives accurate results if the air friction is small.
Payette and Robert Thirsk on the ISS.
These astronauts appear to be floating
or weightless when they are actually in
free fall with the space station.
Careful measurements of Earth’s gravitational field strength show
that its magnitude decreases as an object is moved farther away
from Earth’s surface. The distance between the object and Earth’s
surface must be huge before the difference becomes measurable.
However, if the distance is large enough, Earth’s gravitational field
strength becomes very weak. For example, if the distance between
GO TO NELSON SCIENCE
an object and Earth’s surface (at sea level) is equal to one Earth
radius, the gravitational field strength is only 2.45 N/kg [down]. Of
course, objects are not normally this far away from Earth’s surface.
Even at the top of Mount Everest at an altitude of 8848 m above
sea level, the magnitude of the gravitational field strength is 9.7647
N/kg.
Since Earth is not a perfect sphere, the magnitude of the
gravitational field strength at Earth’s surface varies according to the
geographic location of the object (Figure 5). Earth bulges out slightly
at the equator due to the rotation of the planet. At the poles, an
object at sea level is 21 km closer to Earth’s centre than if it were
at sea level at the equator. This means that the magnitude of the
gravitational field strength is slightly greater at the poles than at the
equator. For example, at the North Pole the magnitude of the
gravitational field strength is 9.8322 N/kg, whereas at the equator it
is 9.7805 N/kg. The magnitude of the gravitational field strength
gradually increases with latitude as you travel from the equator
toward either pole. On average, the gravitational field strength on
Earth’s surface is 9.8 N/kg [down], to two significant digits.
The Difference between Mass and Weight
The terms “mass” and “weight” are used interchangeably in everyday
language, but, in fact, these two words have different meanings.
Mass is the quantity of matter in an object. The only way to change
Section Title 6
the mass of an object is to add or remove matter. The mass of an
object does not change due to location or changes in gravitational
field strength. The units of mass are kilograms, and mass is
measured using a balance.
Weight is a measure of the force of gravity, Fg , acting on an
object. Since weight and the force of gravity are the same thing, the
weight of an object depends on location and the magnitude of
Earth’s gravitational field at that location. Weight is a vector, and its
magnitude is measured in newtons with a spring scale or a force
sensor. When using a force sensor or a spring scale to measure
weight, the object must either be at rest or moving with a constant
velocity while being supported by the scale or sensor.
On the moon or a planet other than Earth, your weight is different
but your mass is the same. For example, the magnitude of the
gravitational field strength on the surface of the moon is
approximately one sixth that on Earth’s surface. This means that the
force of gravity is weaker on the moon and the magnitude of your
weight is less. However, your body contains the same amount of
matter at either location, so your mass is unchanged.
Astronauts aboard the International Space Station (ISS) appear to
float within the station while they are performing tasks. This is often
referred to as “weightlessness” or “microgravity.” These terms are
misleading because the force of gravity still acts on the astronauts
because they are being pulled down toward Earth’s centre. In fact,
gravity is the force that keeps the station and the astronauts in orbit.
In reality, the astronauts and the space station are in free fall
(Figure 6). [CATCH CAREER GLOBE]
You can experience a sensation of microgravity by travelling in an
elevator. When the elevator is moving at a constant velocity, either
up or down, everything appears normal. However, if the elevator
accelerates upward, you feel heavier. If the elevator accelerates
downward, you feel lighter.
The following Tutorial will help to clarify how the vertical
acceleration of an object such as an elevator affects how light or
heavy you feel.
Tutorial 1: The Normal Force Is Not Always Equal in Magnitude to the Force of
Gravity
The following Sample Problems demonstrate the relationship between the force of gravity and the
normal force.
Section Title 7
Sample Problem 1: Cart on an Incline
A cart rolls down an incline. Assume that friction is negligible. Draw an FBD for the cart. In which
direction are the normal force and the force of gravity on the cart?
Solution
First draw the FBD of the cart (Figure 7).
[CATCH: C04-F003-OP11SB (size C1): draw the FBD of a cart on an incline]
[caption] Figure 7
The force of gravity on the cart is down since Earth attracts all masses toward its centre. However,
the normal force is perpendicular to the surface. Since the surface of the ramp is tilted, the normal
force is not directly up. This example clearly shows that the normal force and the force of gravity are
not always in opposite directions.
Sample Problem 2: Person Accelerating Up in an Elevator
A 50 kg person is standing on a bathroom scale inside an elevator. The scale is calibrated in
newtons. What is the reading on the scale when the elevator is accelerating up at 2.2 m/s 2?
Solution
Step 1. Determine the force of gravity acting on the person.
Fg  mg
 (50 kg)(9.8 N/kg [down])
 490 N [down]
Step 2. Now draw the FBD of the person (Figure 8). Choose up as positive and down as negative.
[CATCH: C04-F004-OP11SB (size C1): draw the FBD of a person on a bathroom scale in an
elevator that is accelerating up]
[caption] Figure 8
Step 3. Determine the normal force acting on the person. Since the elevator is accelerating up, the
person is also accelerating up. The person’s acceleration is equal to the elevator’s acceleration.
FN  Fg  ma
FN  (490 N)  ma
FN  490 N  (50 kg)(2.2 m/s 2 )
FN  600 N
So the reading on the scale is equal to the normal force. The reading on the scale is 600 N.
Sample Problem 3: Pushing on a Person Standing on a Bathroom Scale
A 60.0 kg person is standing on a bathroom scale calibrated in newtons. A friend pushes down on
the person with a force of 72.0 N. What is the reading on the scale?
Solution
Step 1. Determine the force of gravity acting on the person.
Fg  mg
 (60 kg)(9.8 N/kg [down])
 588 N [down]
Step 2. Now draw the FBD of the person (Figure 9). Choose up as positive and down as negative.
[CATCH: C04-F005-OP11SB (size C1): draw the FBD of a person on a bathroom scale with a friend
pushing down on the person. The friend is not standing on the scale.]
Section Title 8
[caption] Figure 9
Step 3. Determine the normal force acting on the person. Because the person is at rest and not
accelerating, the net force must be zero.
FN  (588 N)  (72.0 N)  0
FN  660 N
The reading on the scale is equal to the normal force. The reading on the scale is 660 N.
The last two examples show that the normal force is not always equal in magnitude to the force of
gravity.
Practice
1. A 12 kg box sits on top of a 38 kg box. [T/I][C]
(a) Draw an FBD for each box.
(b) Find the normal force acting on the 12 kg box. [ans: 120 N]
(c) Find the normal force acting on the 38 kg box due to the floor. [ans: 490 N]
2. A child has a mass of 36 kg and is sitting on a seat on an amusement park ride. The ride makes
the seat move up and down. Find the normal force acting on the child when the child is
(a) moving up at a constant velocity of 12 m/s [ans: 350 N]
(b) moving down at a constant velocity of 14 m/s [ans: 350 N]
(c) accelerating down at a constant acceleration of 1.8 m/s2 [ans: 420 N] [T/I][C]
4.1 SUMMARY
• Air friction increases with the cross-sectional area of the object
and the speed of the object. Air friction acts opposite to the velocity
of the object.
• Force fields cause action-at-a-distance forces.
• The gravitational field strength is the force per unit mass acting on
an object placed in a gravitational field. The gravitational field at
Earth’s surface is g = 9.8 N/kg [down]. This value decreases with
distance from the surface of Earth. It is greater at the poles and
less at the equator because the Earth is not a perfect sphere.
[FORMATTER: don’t break section 4.1 questions over the page
break: keep them all on the same page]
4.1 QUESTIONS
1. Use Newton’s second law and the force of gravity to explain why all objects fall with the same
acceleration in the absence of air friction. [K/U]
2. Explain why a person with an open parachute has a lower terminal speed than a person with a
closed parachute. [K/U]
3. Why do light objects with large cross-sectional areas fall more slowly in air than dense objects
with small cross-sectional areas? [K/U]
4. In an action movie, a military plane releases a heavy box while in flight. The box is attached to a
parachute that opens as soon as it leaves the plane. Part of the way down to the ground the
parachute malfunctions and the box breaks free. Describe the forces acting on the box while it is
falling and use them to describe the velocity and acceleration of the box. Use FBDs to help you
5. An astronaut with mass 74 kg goes up to the ISS on a mission. During her stay, the gravitational
field strength on the station is 8.6 N/kg. [T/I]
(a) What is the mass of the astronaut on the station?
(b) What is the difference between the astronaut’s weight on Earth’s surface and her weight on the
station?
(c) Why does the weight of the astronaut change but not her mass when moving from the surface of
Earth to the station?
(d) Why does the astronaut appear weightless on the station?
6. Copy and complete Table 1 by calculating the weight of an object of mass 20.000 kg at different
latitudes on Earth. Use the results to answer the following questions. [T/I]
[FORMATTER: set table at C to fit within one column]
Table 1
Latitude (o)
Weight of mass (N)
Distance from
g (N/kg [down])
Earth’s centre (km)
Section Title 9
0 (equator)
9.7805
6378
30
9.7934
6373
60
9.8192
6362
90 (North Pole)
9.8322
6357
(a) What is the difference in the weight of the object from the equator to the North Pole?
(b) Why does the weight change at different latitudes?
(c) Explain why the gravitational field strength increases with latitude.
7. A cargo box on a rocket has a mass of 32.00 kg. The rocket will travel from Earth to the moon.
[K/U][T/I]
(a) What will happen to the mass of the cargo box during the mission? Explain your reasoning.
(b) Determine the weight of the box at the surface of Earth.
(c) The weight of the box on the moon is 52.06 N. Find the gravitational field strength on the moon.
8. Summarize the differences between mass and weight by copying and completing Table 2. [K/U]
[FORMATTER: set table at C with to fit within one column]
Table 2
Quantity
Definition
Symbol
SI unit
Method of
Variation
measuring
with
location
mass
weight
9. Copy Table 3 into your notebook and complete it for a 57 kg instrument on each planet. [T/I]
[FORMATTER: set table at C with to fit within one column]
Table 3
Planet
Weight (N)
g (N/kg)
Mercury
188
Venus
462
Jupiter
26
10. A 24 kg mass sits on top of a scale calibrated in newtons. Find the reading on the scale if
(a) the mass is at rest and no one is pushing on it
(b) the mass is at rest and someone is pushing down on it with a force of 52 N
(c) the mass is at rest and someone is pulling up on it with a force of 74 N [T/I]
11. To test the forces acting on a person during an amusement park ride, a physics student sits on
a bathroom scale calibrated in kilograms while on the ride. Before the ride starts moving, the
reading on the scale is 58 kg. Calculate the reading on the scale when the person is
(a) moving down at a constant velocity of 8.0 m/s
(b) accelerating at 2.7 m/s2 [up]
(c) accelerating at 3.8 m/s2 [down] [T/I]
12. Two heavy boxes are at rest as shown in Figure 10. [T/I]
[CATCH FORMATTER: please insert C04-F006-OP11SB here, before parts a, b, c]
[CATCH: C04-F006-OP11SB (size B1): two boxes as shown in art ms]
[caption] Figure 10
(a) Calculate the normal force on box 1 and the normal force on box 2 from the floor.
(b) Calculate the normal force on box 1 and the normal force on box 2 from the floor if a person
pulls up on box 1 with a force of 55 N.
(c) Calculate the normal force on box 1 and the normal force on box 2 from the floor if a person pulls
up on box 2 with a force of 55 N.
(d) Compare and contrast your answers to (b) and (c) and explain why one normal force changed
while the other did not.
13. A 72 kg person jumps up off a bathroom scale. Find the acceleration of the person when the
14. An electrician holds a 3.2 kg chandelier up against a ceiling with a force of 53 N. What is the
normal force acting on the chandelier from the ceiling? [T/I]
15. Newton used a diagram similar to Figure 11 to explain how a cannonball could be put into orbit
around Earth. Explain how this diagram shows that the cannonball is actually in constant free fall.
Which of the trajectories shown is the fastest? Explain your reasoning. [C] [A]
[CATCH: C04-F007-OP11SB (size C):cannonball in orbit ]
Section Title 10
[caption] Figure 11
[END PAGE 6 of 6]
Section Title 11
[START Section 4.2: 5 pages]
4.2
Friction
Sometimes friction is a problem and we try to minimize its effects.
By adding wax to skis (Figure 1(a)), a skier can reduce friction
between the skis and the snow. The type of wax used depends on
the snow temperature. In the flexible joints of the human body
(Figure 1(b)), synovial fluid helps reduce friction between the moving
bones. When there is too much friction, these joints can become
very painful.
[FORMATTER: Please place C04-P007-OP11SB and C04-F008-OP11SB side-by-side]
[CATCH: C04-P007-OP11SB (size B1): photo labelled (a) showing a person placing wax on a ski ]
[CATCH: C04-F008-OP11SB (size B1): art labelled as (b) showing a typical flexible joint in human
body]
Figure 1 (a) Waxing skis helps to reduce friction. Different types of wax are used for different snow
temperatures. (b) A typical flexible joint in the human body. Like oil in a car engine or wax on a ski,
synovial fluid helps reduce friction between the layers of cartilage lining a flexible joint.
Other times, friction is necessary. When a car is starting to move
[CATCH: C04-P008-OP11SB (size D):
set up photo of a force sensor being
used to pull an object horizontally.]
or a person is walking or running, friction is needed. Without friction,
cars could not slow down or turn corners. In this section you will
learn about the factors that affect the force of friction acting on
objects and ways to control friction.
Figure 2 A force sensor is used to pull a
mass horizontally. Static friction keeps
the mass at rest.
static friction ( F S ) the force exerted by a
surface on a stationary object that prevents it
from moving
The Difference between Static and Kinetic Friction
To help clarify the difference between static and kinetic friction,
consider the following experiment. We use a force sensor to pull an
object horizontally on a horizontal surface. At first the sensor exerts
no force on the object, but we gradually pull with more force (Figure
2). At first the object does not move due to static friction. Static
kinetic friction ( F K ) the force exerted on
an object by a surface opposite to the
velocity of the object relative to the surface
friction ( FS ) is the force exerted on an object by a surface that
prevents a stationary object from starting to move. In this case, the
object remains at rest because the static friction is equal in
[CATCH: C04-F009-OP11SB (size D):
draw a graph showing the force of
friction acting on an object]
magnitude and opposite in direction to the applied force, keeping the
object at rest.
Eventually the applied force from the force sensor becomes large
Figure 3 A graph of the magnitude of
enough to start moving the object. This means a maximum amount
Section Title 12
friction versus the magnitude of the
applied force. Once the object starts to
move, the friction drops suddenly
because the coefficient of kinetic friction
is usually less than the coefficient of
static friction.
of static friction ( FSmax ) must be overcome to cause a stationary
object to move.
Once the object starts moving, kinetic friction replaces static
friction. Kinetic friction ( FK ) is the force exerted on an object by a
surface opposite to the direction of motion of the object. As the
[margin icon for Investigation 4.2.1]
Investigation 4.2.1: Factors That
Affect Friction (p. xxx)
In this investigation, you will explore
some of the factors that affect the
maximum force of static friction and the
force of kinetic friction acting on an
object. You will design your own
procedure and take an average of
several measurements before forming a
applied force from the force sensor continues to increase, the object
begins to accelerate. If the applied force decreases and the object
starts moving at a constant velocity, then the applied force must be
equal in magnitude to the kinetic friction. Figure 3 shows the graph
of the force of friction versus the applied force during this
experiment. Notice that during the time that static friction acts on the
object, the magnitude of the applied force equals the magnitude of
coefficient of friction (μ) the ratio of Ff
to the normal force
static friction. So, during that time the graph is a straight line
starting from the origin with a slope of 1.
Different types of kinetic friction apply to different situations. If an
object is scraping or sliding across a surface, then we call it sliding
[CATCH: C04-F010-OP11SB (size D):
draw an FBD of an object being pulled
by a horizontal force]
friction. If the object is round and it rolls across a surface, then it is
called rolling friction. Fluid friction (or drag) is involved when a boat
goes through water or a plane moves through the air. In this
chapter we will deal with sliding friction for most of the problems.
Figure 4 An FBD of an object pulled by
a horizontal force. The magnitudes of
the force of friction and the normal force
are used to find the coefficient of friction.
However, we will use the generic term “kinetic friction” under most
circumstances.
Many factors affect the force of friction acting on an object.
Coefficients of Friction
The friction acting on an object is a complicated topic. The
magnitude of friction acting on an object may depend on the mass
of the object, the type of material the object is made of, and the
type of surface the object is in contact with. When dealing with air
friction, the speed of the object and the shape of the object also
have an effect. In this section we will deal only with frictional forces
acting on an object in contact with horizontal surfaces. The only
forces applied to the object will be horizontal.
Many experiments have been done to measure the magnitude of
the force of kinetic friction and the maximum force of static friction.
The amounts of kinetic friction and static friction depend on the
types of materials in the two surfaces that are in contact. However,
coefficient of static friction (μS ) the
ratio of FS to the normal force
for a particular pair of surfaces, the ratio of the frictional force
max
Section Title 13
coefficient of kinetic friction (μK) the
ratio of FK to the normal force
(kinetic or static) to the normal force is a constant. This has led to
the definition of a quantity called the coefficient of friction. The
coefficient of friction is the ratio of the magnitude of the force of
friction, Ff, acting on an object to the magnitude of the normal force,
[margin icon for Investigation 4.2.2]
Investigation 4.2.2: Coefficients of
Friction (p. xxx)
In this investigation, you will explore
how the coefficient of static friction
compares to the coefficient of kinetic
friction. You will also determine some of
the factors that affect these values.
FN, acting on the object (Figure 4). The Greek letter μ is used to
represent the coefficient of friction. We define the coefficient of
friction mathematically with the following equation:
[CATCH FORMATTER: set the following equation in blue screened box]

Ff
FN
where Ff is the magnitude of the force of friction acting on an object
in newtons, FN is the magnitude of the normal force acting on the
object in newtons, and μ is the coefficient of friction. The coefficient
of friction is just a number, with no direction or units.
To calculate the coefficient of static friction for an object on a
surface, we need to determine the maximum force of static friction in
that particular situation. For almost all situations, the force required
to start an object moving is greater than the kinetic friction acting on
the object when it is moving. This means that FSmax is usually slightly
greater than FK. Since there are two types of friction (static and
kinetic), there are two different coefficients of friction. One is the
coefficient of static friction, which represents the ratio of FSmax to the
normal force. The coefficient of static friction is represented by the
symbol μS. The other is the coefficient of kinetic friction, which
represents the ratio of the kinetic friction to the normal force. The
coefficient of kinetic friction is represented by the symbol μK. The
maximum force of static friction is usually greater than the kinetic
friction, so the coefficient of static friction is usually greater than the
coefficient of kinetic friction. The corresponding equations are
[CATCH FORMATTER: set the following equations in blue screened box]
S 
FSmax
FN
and K 
FK
FN
The coefficient of friction between an object and a surface
depends only on the type of materials. These coefficients of friction
can only be determined experimentally. The results of such
experiments are often inconsistent, even when performed carefully.
Results can be affected by the condition of the surface, including the
cleanliness of the surface, whether the surface is wet or dry, and
Section Title 14
the roughness of the surface. This means one scientist may obtain a
different coefficient of friction than another scientist even when
neither one has made any mistakes. So, in many cases, a range of
values is given for coefficients of friction. Table 1 gives several
coefficients of kinetic and static friction for some common materials.
Table 1 Approximate coefficients of kinetic and static friction
Material
μS
rubber on concrete (dry)
rubber on concrete (wet)
rubber on asphalt (dry)
rubber on asphalt (wet)
steel on steel (dry)
0.78
steel on steel (greasy)
0.05–0.11
leather on oak
0.61
ice on ice
0.1
steel on ice
0.1
rubber on ice
wood on dry snow
0.22
wood on wet snow
0.14
Teflon on Teflon
0.04
near-frictionless carbon
synovial joints in humans
0.01
μK
0.6–0.85
0.45–0.75
0.5–0.80
0.25–0.75
0.42
0.029–0.12
0.52
0.03
0.01
0.005
0.18
0.10
0.04
0.02–0.06
0.003
In the following Tutorial, we will apply the equations for the
coefficient of friction to determine the force of friction acting on a
wooden block.
Tutorial 1: Determining the Forces of Friction
The following Sample Problem will help to clarify how the normal force and the force of friction are
related to the coefficients of friction.
Sample Problem 1: Analyzing the Frictional Forces on a Wooden Block
A block of wood of mass 3.0 kg sits on a horizontal wooden floor. The largest horizontal force that
can be applied to the block before it will start moving is 14.7 N. Once the block starts moving, it only
takes 8.8 N to keep it moving at a constant velocity.
(a) Calculate the coefficient of static friction for the block and the floor.
(b) Determine the force of friction acting on the block if a horizontal force of 6.8 N [E] acts on the
block.
(c) Calculate the maximum magnitude of static friction acting on the block if a 2.1 kg object is placed
on top of it.
(d) Determine the coefficient of kinetic friction.
Solution
(a) Given: m = 3.0 kg; Fa = 14.7 N
Required: μS
Analysis: Since the block is not moving, the net force on the block is zero. This means that
FSmax
on the block must also be 14.7 N acting in the opposite direction to keep the block at rest. To
calculate the coefficient of static friction, we can use the equation
S 
FSmax
FN
. Also, the force of
gravity must be equal to the normal force.
FS
Solution: S  max
FN

FSmax
mg
14.7 N
2
(3.00 kg)(9.8 m/s )
 0.50
Notice that 1 kg∙m/s2 = 1 N, so the units cancel.
Statement: The coefficient of static friction is 0.50.

Section Title 15
(b) The horizontal applied force (6.8 N [E]) is less than the maximum force of static friction (14.7 N),
so the block will remain at rest. This means the net force on the block is still zero and the applied
force must be cancelled by the static friction. The static friction on the block must be 6.8 N [W].
(c) Given: mT = 3.0 kg + 2.1 kg = 5.1 kg
Required: FS
max
Analysis: Since the same materials are still in contact, μS = 0.50. To calculate the maximum force
of static friction, we can use the equation
S 
FSmax
FN
. Since the total mass is now 5.1 kg, we can
calculate the normal force using the equation FN = mTg.
FS
Solution: S  max
FN
FSmax  S FN
FSmax  S mT g
FSmax  (0.50)(5.1 kg)(9.8 m/s )
2
FSmax  25 N
Statement: The maximum magnitude of static friction acting on the block is 25 N.
(d) Given: m = 3.0 kg; Fa = 8.8 N
Required: μK
Analysis: The block is moving with a constant velocity, so the net force on the block is zero. This
means the kinetic friction, FK , on the block must also be 8.8 N acting in the opposite direction. To
find the coefficient of kinetic friction we can use the equation
Solution:  K 
FK
FN

FK
mg
K 
FK
.
FN
8.8 N
2
(3.00 kg)(9.8 m/s )
 0.30
Statement: The coefficient of kinetic friction is 0.30.
Practice
1. Determine the coefficient of friction for each situation. [T/I]
(a) It takes a horizontal force of 62 N to get a 22 kg box to just start moving across the floor. [ans:
0.29]
(b) It only takes 58 N of horizontal force to move the same box at a constant velocity. [ans: 0.27]
2. A 75 kg hockey player glides across the ice on his skates with steel blades. What is the force of
friction acting on the skater? Use Table 1 to help you complete this question. [T/I] [ans: 7.4 N]
3. A 1300 kg car skids across an asphalt road. Use Table 1 to calculate the force of friction acting
(a) dry [ans: 6400 N to 10 000 N]
(b) wet [ans: 3200 N to 9600 N]
(c) covered with ice [T/I] [ans: 64 N]

4.2 SUMMARY
• Static friction ( FS ) is the force exerted on an object by a surface
that prevents a stationary object from starting to move.
• The maximum force of static friction ( FSmax ) is the amount of force
that must be overcome to start a stationary object moving.
• Kinetic friction ( FK ) is the force of friction acting on an object
moving across a surface. The force of kinetic friction acts in the
opposite direction to the velocity of the object. The horizontal applied
force required to keep an object moving at a constant velocity
across a horizontal surface is equal in magnitude and opposite in
Section Title 16
direction to the kinetic friction.
• The coefficient of friction is the ratio of the force of static or
kinetic friction to the normal force. The equations for the coefficient
of static friction and the coefficient of kinetic friction are  S 
FSmax
FN
F
and K  K .
FN
• Coefficients of friction are determined experimentally and depend
only on the types of materials in contact. The force of static or
kinetic friction exerted by a surface on an object depends only on
the coefficient of friction and the magnitude of the normal force.
4.2 QUESTIONS
1. For each situation, determine if friction is helpful, makes the action more difficult, or both. Explain
(a) turning a door knob
(b) pushing a heavy box across a rough surface
(c) gliding across smooth ice to demonstrate uniform motion
(d) tying a knot
2. A typical bicycle braking system involves a lever on the handle bars that you pull and a brake pad
near the rim of the wheel, as shown in Figure 5. Describe how the braking system works using the
concepts of normal force and friction. [K/U] [C]
[CATCH FORMATTER: please wrap text on one side of the image if necessary to save space, or
crop image as needed]
[CATCH: C04-P009-OP11SB (size C): photo of a typical bicycle braking system consisting of two
brake sides on either side of the wheel]
[caption] Figure 5
3. Some people think that if a surface is polished to the point where it is very smooth, it can be
made virtually frictionless. Figure 6 shows a metal surface that appears smooth to the unaided eye.
Describe what this photograph reveals about friction. [K/U] [C]
[CATCH FORMATTER: please crop this photo if necessary to reduce vertical height]
[CATCH: C04-P010-OP11SB (size C): polished metal surface SEM.]
[caption] Figure 6
4. A 1.4 kg block on a horizontal surface is pulled by a horizontal applied force during a physics
experiment. It takes 5.5 N to start it moving and 4.1 N to keep it moving at a constant velocity. [T/I]
(a) Calculate the coefficients of friction.
(b) Which of the following changes to the experiment will affect the coefficients of friction? Explain.
(i) turning the block onto another side (ii) changing the surface (iii) putting a mass on top of the
block
(c) What effect will each situation have on the static and kinetic friction acting on each object?
Explain.
(i) putting a mass on the block (ii) pulling up on the block (iii) putting slippery grease on the surface
5. Examine the coefficients of friction in Table 1 to answer the following. [A]
(a) Roads in Canada are typically made out of asphalt or concrete. Is one material significantly safer
than the other? Explain your reasoning.
(b) Explain why drivers should reduce their speed on wet roads.
(c) Why do we salt roads in the winter, especially when there is freezing rain?
6. You are dragging a large 110 kg trunk across the basement floor with a horizontal force of 380 N
at a constant velocity. [T/I]
(a) Find the coefficient of kinetic friction.
Section Title 17
(b) A friend feels sorry for you and decides to help by pulling straight up on the trunk with a force of
150 N. Will this help? Calculate the force required to pull the trunk at a constant velocity to help you
decide.
(c) Instead of pulling up on the trunk, your 55 kg friend just sits on it. What force is required keep the
trunk moving at a constant velocity?
7. A large 26 kg desk is at rest on the floor. The coefficient of static friction is 0.25. One person pulls
on the desk with 52 N [E] and another pulls with 110 N [W]. Will the desk move? Explain your
reasoning. [T/I]
8. A large 12 000 kg bin is sitting in a parking lot. A truck pushes the bin with a force of 32 000 N
[forward]. The coefficient of static friction for the bin is 0.50 and the coefficient of kinetic friction is
0.40. A tractor pulls on the bin and it starts to move. Find the minimum force exerted by the tractor
to
(a) start the bin moving
(b) keep the bin moving at a constant velocity [T/I]
9. A gradually increasing horizontal force is applied to a mass initially at rest on a horizontal surface.
Draw a graph of the force of friction versus the applied force under the following circumstances. [T/I]
[C]
(a) the coefficient of static friction is slightly greater than the coefficient of kinetic friction
(b) the coefficient of static friction is equal to the coefficient of kinetic friction
[FORMATTER: Omit questions 10-12 if there is insufficient space]
10. A doorstop keeps a door open when the doorstop is wedged underneath the door. Use the
concepts from this section to explain how doorstops work. [C] [A]
11. Describe how to determine each of the following experimentally. [T/I] [C]
(a) the coefficient of static friction
(b) the coefficient of kinetic friction
12. Explain why the manufacturer of a running shoe might be more concerned about having a high
coefficient of friction than the manufacturer of a dress shoe. [C] [A]
[END page 5 of 5]
Section Title 18
[START Section 4.3: 7 pages]
4.3
[CATCH: C04-P011-OP11SB (size D):
photo showing the sole of a running
shoe]
Solving Friction Problems
Sometimes friction is desirable and we want to increase the
coefficient of friction to help keep objects at rest. For example, a
running shoe is typically designed to have a large coefficient of
friction between the sole and the floor or ground (Figure 1). The
design of the sole involves many different factors, such as choosing
materials with high coefficients of friction and shaping the sole to
Figure 1 The sole of a running shoe is
designed to increase friction to help
runners accelerate quickly.
[CATCH: C04-P012-OP11SB (size D):
photo of a hockey stick shooting a
puck.]
increase friction under, for example, wet conditions. Even the
appearance is important because people want their shoes to look
good.
However, sometimes we want as little friction as possible. A
hockey player shooting a puck toward the net does not want the
puck to slow down as it moves across the ice (Figure 2). In
transportation technologies, you want to minimize the force of friction
acting on accelerating railway cars and trailers pulled by trucks.
Figure 2 In some cases, such as when
a puck moves across the ice, it is better
to have less friction.
In this section you will explore how the force of friction affects the
motion of an object, whether the object is at rest or in motion. You
will also learn how to solve problems involving friction.
Static Friction Problems
Sometimes when you push on an object, it does not move. The
reason is that static friction is acting on the object. If two or more
objects are attached, you must overcome the combined maximum
force of static friction to cause the objects to move. For example,
when a train pulls several railway cars forward, it must exert enough
force to overcome the combined maximum force of static friction
acting on all the railway cars. If the applied force acting on the
railway cars is less than this combined force of static friction, the
cars will not move. In the following Tutorial, we will examine the
effect of static friction when it is acting on more than one object at
a time.
Tutorial 1: Static Friction Acting on Several Objects
In the following Sample Problem, we will use the coefficient of friction to determine the forces acting
on sleds being pulled by an adult.
Sample Problem 1: Friction Acting on Two Sleds
Two sleds are tied together with a rope (Figure 3). The coefficient of static friction between each
sled and the snow is 0.22. A small child is sitting on sled 1 (total mass of 27 kg) and a larger child
sits on sled 2 (total mass of 38 kg). An adult pulls on the sleds.
(a) What is the greatest horizontal force that the adult can exert on sled 1 without moving either
sled?
(b) Calculate the magnitude of the tension in the rope between sleds 1 and 2 when the adult exerts
Section Title 19
this greatest horizontal force.
[CATCH: C04-F011-OP11SB (size C): draw two children sitting on two separate sleds attached by a
rope. Draw a rope on the first sled being pulled by an adult.]
[caption] Figure 3
Solution
(a) The two sleds do not move when the adult pulls on sled 1. This means that the net force acting
on the sleds is zero and the applied force must be cancelled by the total maximum force of static
friction acting on the two sleds. To calculate the static friction, we combine the two masses and treat
the sleds as one single object.
Given: mT = 27 kg + 38 kg = 65 kg; μS = 0.22
Required: FS
max
Analysis:
S 
FSmax
FN
Solution: FSmax  S FN
 S mg
 (0.22)(65 kg)(9.8 m/s )
2
 140 N
Statement: The greatest horizontal force the adult can exert on sled 1 without moving either sled is
140 N [forward].
(b) Draw the FBD for sled 2 (Figure 4). Keep in mind that sled 2 does not move, which means the
net force is zero. In this case the tension and the static friction acting on sled 2 will cancel.
[CATCH: C04-F012-OP11SB (size C1): draw the FBD of sled 2.]
[caption] Figure 4
Given: m = 38 kg; μS = 0.22
Required: FT
Analysis: Since FT = FS , calculate FS
using  S 
max
max
FSmax
FN
.
Solution: FSmax  S FN
 S mg
 (0.22)(38 kg)(9.8 m/s )
2
 82 N
Statement: The magnitude of the tension in the rope is 82 N.
Practice
1. Two large trunks sit side by side on the floor. The larger trunk (52 kg) is to the left of the smaller
trunk (34 kg). A person pushes on the larger trunk horizontally towards the right. The coefficient of
static friction between the trunks and the floor is 0.35. [T/I]
(a) Determine the magnitude of the maximum force the person can exert without moving either
trunk. [ans: 290 N]
(b) Calculate the force the larger trunk exerts on the smaller trunk. [ans: 120 N]
(c) Would either answer change if the person pushed in the opposite direction on the smaller trunk?
2. A 4.0 kg block of wood sits on a long board (Figure 5). A string is tied to the wood, running over a
pulley and down to a mass. The greatest mass that can be hung from the string without moving the
block of wood is 1.8 kg. Find the coefficient of static friction between the block of wood and the
board. [T/I] [ans: 0.45]
[CATCH: C04-F013-OP11SB (size D): draw a block on a board attached with a string to another
block hanging from a pulley]
[CATCH: C04-P013-OP11SB (size D):
Section Title 20
close-up photo of a runner’s feet in
position on starting blocks]
[caption] Figure 5
Figure 6 When a track and field athlete
starts a race using starting blocks, the
athlete can achieve a greater forward
acceleration by pushing backwards on
the blocks.
Static Friction and Motion
It is important to understand that static friction acts on an object
when the object is at rest on a surface. This does not mean that
static friction cannot be used to make an object move. For example,
when you take a step forward, the bottom of your shoe is at rest
with respect to the ground. The foot you are standing on is not
actually moving forward with the rest of your body. The action force
is you pushing backwards on the ground. According to Newton’s
third law, the ground pushes back on your foot with a force of equal
magnitude but opposite in direction (the reaction force). This reaction
force is the force of static friction, and it is this force that actually
pushes you forward. Keep in mind that the force of static friction is
usually greater in magnitude than the force of kinetic friction. This
means that in a race from a standing start without starting blocks
(Figure 6), you ideally want to push backwards with both feet,
making sure your shoes do not slip. In the following Tutorial, we will
examine the effect of static friction that causes an object to move.
Tutorial 2: Static Friction Can Cause Motion
The following Sample Problem will help to clarify how static friction can cause motion.
Sample Problem 1: Friction Acting on a Person When Starting a Race
The coefficient of static friction between a person’s shoe and the ground is 0.72. Determine the
maximum magnitude of acceleration of the 62 kg person, if he starts running on a horizontal surface
from rest.
Solution
Given: μS = 0.70; m = 62 kg
Required: a
Analysis: Draw the FBD of the person (Figure 7). In this case, the maximum force of static friction
is the net force since the normal force and gravity cancel. First we need to calculate FS .
max
Fnet  FSmax ; FSmax  SFN
[CATCH: C04-F014-OP11SB (size C1): FBD of a male person running]
[caption] Figure 7
Solution: FSmax  S FN
 S mg
 (0.70)(62 kg)(9.8 m/s )
2
 425.3 N
Now calculate the magnitude of the acceleration:
[CATCH: C04-F015A-OP11SB and
C04-F015B-OP11SB (each size E):
draw two FBDs labelled (a) and (b). Part
Section Title 21
(a) shows Fa and FK acting in the
same direction. Part (b) shows Fa and
FK acting in opposite directions.]
Figure 8 (a) Both the applied force and
friction act in the same direction,
producing a large net force. (b) The
applied force and kinetic friction are in
opposite directions, producing a smaller
net force.
Fnet  FSmax
ma  425.3 N
(62 kg)a  425.3 N
a  6.9 m/s
Statement: The maximum magnitude of the acceleration is 6.9 m/s2.
Practice
1. Two people start running from rest. The first person has a mass of 59 kg and is wearing dress
shoes with a coefficient of static friction of 0.52. The other person is wearing running shoes with a
coefficient of static friction of 0.66. [T/I]
(a) Calculate the maximum possible initial acceleration of the person wearing dress shoes. [ans: 5.1
m/s2]
(b) Explain why we do not really need the mass of either person when finding the initial maximum
possible acceleration.
(c) Determine the ratio of the two accelerations and compare it to the ratio of the two coefficients of
friction. [ans: 0.79; they are equal]
2. A skater with mass 58 kg is holding one end of a rope and standing at rest on ice. Assume no
friction. Another person with mass 78 kg is standing just off the ice on level ground and is holding
the other end of the rope. The person standing on the ground pulls on the rope to accelerate the
skater forward. The coefficient of static friction between the ground and the off-ice person is 0.65.
Calculate the maximum possible acceleration of the skater. [T/I] [ans: XXXX]
2
Kinetic Friction Problems
When an object is sliding across a surface, it experiences kinetic
friction in a direction opposite to the direction of motion. One
common misconception is that kinetic friction always reduces the net
force acting on an object. For example, in Sample Problem 1 in
Tutorial 3, a person pushes against a sliding box to stop it from
sliding any farther. In this situation, both the applied force and the
force of kinetic friction act in the same direction (Figure 8(a)). So
the magnitude of the net force is greater than if the person applies
a force in the opposite direction (Figure 8(b)). Since the magnitude
of the net force is greater, the acceleration of the object is also
greater and the object slows down faster. In the following Tutorial,
we will examine how kinetic friction causes a moving object to come
to rest.
Tutorial 3: Kinetic Friction and Motion
The following Sample Problems will help to clarify the effect of kinetic friction on motion and the
magnitude of the acceleration.
Sample Problem 1: Stopping a Sliding Box
A heavy 250 kg box slides down a ramp and then onto a level floor toward a storage room. The
coefficient of kinetic friction is 0.20. A person sees the box moving at 1.0 m/s [right] and pushes on it
with a horizontal force of 140 N [right].
(a) How far does the box move before coming to rest?
(b) How will the results change if the box is moving left and the person still pushes with the same
force?
Solution
(a) Given: μK = 0.20; m = 250 kg; F a  140 N [R]
Required: Δd
Analysis: Draw the FBD of the box (Figure 9). The kinetic friction must be opposite to the velocity.
We will first find the kinetic friction. We can then use the FBD to find the acceleration. Once we have
the acceleration, we can use one of the kinematics equations to calculate the distance.
v 22  v12  2ad ; FK  KFN; Fnet  Fa  FK
[CATCH: C04-F016-OP11SB (size C1): draw the FBD of a box sliding across a floor]
Section Title 22
[caption] Figure 9
Solution: FK   K FN
  K mg
 (0.20)(250 kg)(9.8 m/s 2 )
 490 N
Now calculate the acceleration:
Fnet  Fa  FK
ma  140 N  490 N
(250 kg)a  350 N
a  1.4 m/s
Next calculate the distance travelled. Since we do not know the time, we use the kinematics
2
equation
v 22  v12  2ad .
v2  v1  2ad
2
2
v2  v1  2ad
2
2
v2  v1
2a
2
d 
2
2
0  (1.0 m/s )
2
2(1.4 m/s )
d  0.36 m
Statement: The box moves 0.36 m before coming to rest.
(b) In this situation, friction is directed toward the right. When added to the applied force, it produces
a larger net force toward the right. The resulting acceleration is also larger and the box travels a
shorter distance before coming to rest.
2
2
d 
Sample Problem 2: Friction and Moving Sleds
Two sleds tied together are pulled east across an icy surface with an applied force of 150 N [E]
(Figure 10). The mass of sled 1 is 18.0 kg and the mass of sled 2 is 12.0 kg. The coefficient of
kinetic friction for each sled is 0.20.
(a) Calculate the acceleration of the sleds.
(b) Determine the tension in the rope between the sleds.
[CATCH: C04-F017-OP11SB (size C): draw two sleds attached by a rope being pulled by a person]
[caption] Figure 10
Solution
(a) Both sleds move together with the same acceleration so we can treat them as one large object.
The total mass of the two sleds is mT = 18.0 kg + 12.0 kg = 30.0 kg. There is no need to worry about
tension at this point because it is an internal force and does not contribute to the acceleration of the
total mass.
Draw the FBD of the two sleds (Figure 11).
[CATCH: C04-F018-OP11SB (size C1): draw the FBD of the two sleds]
[caption] Figure 11
Given: μK = 0.20; m = 30.0 kg; F a  150 N [E]
Required: a
Analysis: Fnet  ma ; Fnet  Fa  FK
Section Title 23
Solution: First we need to determine the force of kinetic friction.
FK  K FN
 K mg
 (0.20)(30.0 kg)(9.8 m/s 2 )
 58.8 N
Now calculate the acceleration:
Fnet  Fa  FK
ma  150 N  58.8 N
(30.0 kg)a  91.2 N
a  3.04 m/s
Statement: The acceleration of both sleds is 3.0 m/s2 [E].
(b) Using the FBD for sled 2,
2
[CATCH: C04-F019-OP11SB (size C1): draw the FBD of sled 2]
Fnet  FT  FK 2
m2 a  FT   K m2 g
(12.0 kg)(3.04 m/s 2 )  FT  (0.20)(12.0 kg)(9.8 m/s 2 )
FT  60 N
Statement: The tension in the rope is 60 N.
Practice
1. A 0.170 kg hockey puck is initially moving at 21.2 m/s [W] along the ice. The coefficient of kinetic
friction for the puck and the ice is 0.005. [T/I]
(a) What is the speed of the puck when it reaches the other end of the rink 58.5 m away? [ans: 21.3
m/s]
(b) After playing for a while the ice becomes rougher and the coefficient of kinetic friction increases
to 0.47. What will happen if you repeat the same shot down the rink?
2. An electric motor is used to pull a large box of mass 125 kg across a floor in a factory using a
long cable. The tension in the cable is 350 N and the box accelerates at 1.2 m/s 2 [forward] for 5.0 s.
The cable breaks and the box slows down and stops. [T/I] [C]
(a) Calculate the coefficient of kinetic friction. [ans: 0.29]
(b) Determine the total displacement of the box. [ans: 21 m]
3. A snowmobile is used to pull two sleds across the ice. The mass of the snowmobile and the rider
is 320 kg. The mass of the first sled behind the snowmobile is 120 kg and the mass of the second
sled is 140 kg. The ground exerts a force of 1500 N [forward] on the snowmobile. The coefficient of
kinetic friction for the sleds on ice is 0.15. Assume that no other frictional forces act on the
snowmobile. Calculate the acceleration of the snowmobile and sleds. [T/I] [C] [ans: XXXX]
4. A string is tied to a 3.2 kg object on a table and a 1.5 kg hanging object (Figure 12). The 1.5 kg
object is falling down and the 3.2 kg object is moving right. The coefficient of kinetic friction between
the 3.2 kg object and the table is 0.30. [T/I]
[CATCH: C04-F020-OP11SB (size D): draw two objects connected by a rope over a pulley. One
object is on a table while the other is hanging.]
[caption] Figure 12
(a) Calculate the acceleration of each object. [ans: XXXX]
(b) Determine the tension in the string. [ans: XXXX]
(c) How far will the objects move in 1.2 s if the initial velocity of the 3.2 kg object is 1.3 m/s [right]?
[ans: XXXX]
4.3 SUMMARY
[FORMATTER: set icon
beside 4.3 Summary]
• Static friction exists between an object and a surface when the
object is not sliding on the surface.
• Static friction can be used to move objects.
Section Title 24
[margin icon for Investigation 4.3.1]
Investigation 4.3.1: Predicting Motion
with Friction (p. xxx)
In this investigation, you will calculate
the acceleration of a system of objects
using Newton’s laws. Then you will
measure the actual acceleration by
doing the experiment. Afterward, you
will compare your calculated value with
the measured value.
• Kinetic friction always acts in a direction that is opposite to the
motion of the object.
• The kinematics equations from Unit 1 can be used to solve some
problems involving friction and other forces.
4.3 QUESTIONS
1. During fundraising week at school, students decide to hold a competition of strength. Contestants
pay \$1 to try to move a heavy object. If they can move the object, they win \$10. The object is a
large box of books of mass 250 kg, with a coefficient of static friction of 0.55. One student has a
mass of 64 kg and a coefficient of friction of static friction of 0.72 between his shoes and the floor.
[T/I]
(a) What is the maximum force of static friction for the student?
(b) What is the maximum force of static friction for the box?
(c) Is the competition fair? Explain your reasoning.
2. In an action movie, an actor is lying on an ice shelf and holding onto a rope. The rope runs over a
cliff and down to another actor who is hanging on in midair. The actor on the ice shelf has a mass of
55 kg and the actor hanging in midair has a mass of 78 kg. Neither actor can grab onto anything to
help stop their motion, yet in the movie neither one is moving. [T/I] [A]
(a) Calculate the minimum coefficient of static friction.
(b) Is your answer to part (a) reasonable considering that the surface is ice? Explain.
(c) What could the director do to make the scene more realistic? Explain your reasoning.
3. In a physics experiment on static friction, two objects made of identical material are tied together
with string. The first object has a mass of 5.0 kg and the second object has a mass of 3.0 kg.
Students use a force meter to pull the objects forward across a horizontal surface. The students
measure the maximum force of static friction as 31.4 N to move both objects. [T/I]
(a) What is the coefficient of static friction?
(b) What is the tension in the string?
(c) Another student pushes down on the 3.0 kg mass with a force of 15.0 N. What are the maximum
force of static friction and the tension in the string now?
(d) Will any answer to question (c) change if the second student pushes down on the 5.0 kg mass
4. A student puts a 0.80 kg book against a vertical wall and pushes horizontally on the book toward
the wall with a force of 26 N (Figure 13). The book does not move. [T/I]
(a) Calculate the minimum coefficient of static friction.
(b) Describe two ways the student could make the book accelerate down without changing the
applied force.
[CATCH: C04-F021-OP11SB (size C1): draw a hand pushing a book against the wall ]
[caption] Figure 13
5. A string is tied to a 4.4 kg block and a 120 g hanging bucket (Figure 14). Students add 20 g
washers one at a time to the bucket. The students are unaware that the coefficient of static friction
is 0.42. [T/I]
[CATCH: C04-F022-OP11SB (size D): draw a block attached to a bucket by a string. The bucket is
hanging over a pulley.]
[caption] Figure 14
(a) What is the maximum force of static friction for the block?
(b) How many washers can the students add to the bucket without moving the block?
(c) Will this investigation yield an accurate result if they use it to find the coefficient of static friction?
(d) The coefficient of kinetic friction is 0.34. Find the acceleration of the block when the final washer
is placed in the bucket and the masses start to move.
6. In a tug-of-war contest on a firm, horizontal sandy beach, Team A has six players with an
average mass of 65 kg and team B has five players with an average mass of 84 kg. Team B, pulling
with a force of 3.2 kN, dislodges team A and then decreases its force to 2.9 kN to pull Team A
Section Title 25
across the sand at a constant velocity. Determine Team A’s coefficient of
(a) static friction
(b) kinetic friction [T/I]
7. Two students push a 260 kg grand piano across the classroom floor. Kathy pushes with 280 N
[forward] while Matt pushes with 340 N [forward]. The piano accelerates at 0.30 m/s2 [forward]. [T/I]
(a) What is the coefficient of kinetic friction?
(b) How long will it take the piano to stop moving after pushing it for 6.2 s from rest?
8. A 65 kg sprinter accelerates from rest into a strong wind that exerts a frictional force of 62 N. The
ground applies a constant forward force of 250 N on the sprinter’s feet. [T/I]
(a) Calculate
(i) the sprinter’s acceleration
(ii) the distance travelled in the first 2.0 s
(iii) the minimum coefficient of friction between the sprinter’s shoes and the track
(d) Is the friction static or kinetic? Explain.
9. A homeowner accelerates a 15.0 kg lawnmower uniformly from rest to 1.2 m/s in 2.0 s. The
coefficient of kinetic friction is 0.25. Calculate the horizontal applied force acting on the lawnmower.
[T/I]
10. A 75 kg baseball player is running at 2.8 m/s [forward] when he slides into home plate for a
distance of 3.8 m before coming to rest. Calculate the coefficient of kinetic friction. [T/I]
11. For a school project, a student builds a toy airplane with a mass of 2.4 kg. She pulls the plane
forward at a constant velocity of 1.5 m/s, while the plane experiences an upward lift force of 11 N.
The coefficient of friction between the plane and the ground is 0.40. [T/I]
(a) Determine the normal force acting on the plane.
(b) Calculate the magnitude of the force the student applies to the plane.
12. A mover slides a heavy 120 kg box across the floor at a constant velocity of 0.8 m/s [E]. The
mover pushes on the box with a horizontal force of 353 N [E]. [T/I]
(a) Calculate the coefficient of kinetic friction.
(b) As a joke, another mover decides to “help” and pushes horizontally on the box with a force of
180 N [W]. How far will the box move before coming to rest?
(c) What will happen to the motion of the box if the second mover pulls directly up on the box with a
force of 180 N? Calculate the new acceleration.
13. In a science project, students use a 3.0 kg electric motor to lift a 1.2 kg object hanging from a
string and move a 1.6 kg object sitting on a surface (Figure 15). The coefficient of static friction
between the motor and the surface is 0.50. The coefficient of kinetic friction between the 1.6 kg
object and the surface is 0.30. Assume that the motor starts the objects moving. What is the
maximum acceleration of the objects? [T/I]
[CATCH: C04-F023-OP11SB (size D): draw an electric motor attached to a block on a table by a
string. This object is also attached by a string to another object hanging over a pulley.]
[caption] Figure 15
[END Page 7 of 7]
Section Title 26
[START Section 4.4: 6 pages]
4.4
[CATCH: C04-P014-OP11SB (size D):
photo showing a crash test with dummy,
air bag and seat belt]
Forces Applied to
Automotive Technology
Throughout this unit we have addressed automotive safety features
such as seat belts and headrests. In this section you will learn how
forces apply to other safety features in cars. Some of these features
involve increasing friction with different types of road surfaces. For
example, you will learn why different tires are used for specific
Figure 1 A typical crash test. How many
vehicle safety features are shown in this
picture?
applications and road conditions, and how anti-lock brakes and
electronic stability controls help drivers maintain control of vehicles
on slippery surfaces. Another car safety feature is crumple zones
that become crushed when a car collides with another object (Figure
1). These different technologies have advantages, but they also have
limitations. By understanding the physics involved in these safety
[CATCH: C04-P015-OP11SB (size D):
photo of a race car focusing on the tires]
features, you will become a safer driver and a more informed car
consumer.
The Physics of Car Tires
In Investigation 4.2.1, you learned that for many pairs of materials
Figure 2 Race cars have wide tires to
increase the surface area in contact with
the road. This greater contact area
helps increase the magnitude of the
force of friction acting on the tires.
the force of static friction or kinetic friction acting on an object is not
affected by the amount of surface area in contact with a surface.
Instead, the coefficients of friction depend on the types of materials
in contact. The magnitude of static friction or kinetic friction only
depends on the coefficient of friction and the magnitude of the
[CATCH: C04-F024-OP11SB (size D):
draw a close up of a car tire showing the
normal force. However, rubber is an exception. For example, the
magnitude of static friction acting on a rubber tire on a road surface
depends on the surface area of the tire that is in contact with the
road. That is one reason that race cars have very wide tires with no
treads (Figure 2). These types of tires are designed to maximize the
Figure 3 The tread on tires is designed
to work under various road conditions.
Beneath the rubber tread are various
layers that provide added safety and
strength.
surface area in contact with the road, which increases the magnitude
of static friction acting on the tires, and also dissipates heat more
quickly. This, in turn, helps race cars move along curved paths
without slipping. If it starts raining during a race, the event is
sometimes postponed because these types of tires have no treads.
Unlike race car tires, the tires on passenger vehicles need to
provide good traction (force of friction) on road surfaces under all
types of weather conditions, not just on sunny days. The tread area
on a tire contains small and large grooves (Figure 3). Grooves
Section Title 27
provide pathways for water, such as rain and snow, to pass beneath
the tire as it rolls over a wet road. This helps the tire maintain
contact with the road. As a result, the magnitude of the force of
static friction acting on the tires is usually large enough for safe
driving.
When driving at a low safe speed on a wet road, the water level
in front of a passenger car tire is low (Figure 4(a)). The water has
time to move through the grooves in the tire tread and be squeezed
out behind the tire. In this situation, friction from the road still acts
on the tire. If the driver starts to speed up, some of the water in
front of the tire will not have enough time to pass through the
grooves toward the back of the tire (Figure 4(b)). This causes the
water level in front of the tire to increase. This excess water causes
the tire to start to lose contact with the road. If the driver’s speed
continues to increase, the water level in front of the tire will increase
to the point where the tire no longer directly touches the road
surface (Figure 4(c)). Now the tire experiences no friction from the
road and very little friction from the water. The two materials in
contact are now rubber and water, not rubber and the wet road.
This situation is called hydroplaning or aquaplaning and is very
dangerous. It results in the driver losing control of the vehicle and
being unable to slow down due to the very low friction acting on the
tires. The photographs in Figure 4(a) to (c) show an actual tire
losing contact with the road as the speed increases. In each case
the force of friction acting on the tire decreases. The only way to
avoid hydroplaning is to slow down when the roads are wet.
[CATCH: C04-F026-OP11SB (size D):
diagram showing the parts of a disc
brake.]
[CATCH FORMATTER: SET FIGURE SIDE-BY-SIDE ACROSS PAGE AND LABEL EACH PART
AS (a), (b), AND (c) RESPECTIVELY AS SHOWN WITH APPROPRIATE PHOTO OF TIRE
TREADS INTERACTING WITH WATER ON A ROAD SURFACE. Each piece is SIZE E]
[CATCH: C04-F025A-OP11SB to C04-F025C-OP11SB]
[C04-P017A-OP11SB to C04-P017C-OP11SB) (each size E): diagram or photo of water passing
through the grooves of a car tire travelling at different speeds ]
Figure 5 The piston pushes the brake
pads against the rotor, creating a force
of friction. The friction slows down the
wheel.
Figure 4 (a) When a car travels at low speeds on a wet road, water levels are low in front of and
high behind the tire (the normal stage). (b) If the speed of the car increases, water levels increase in
front of the tire (the transition stage). (c) If the speed of the car continues to increase, water levels in
front of the tire increase to the point where the tire no longer makes direct contact with the road
Section Title 28
surface (the hydroplane stage). This situation is very dangerous because the magnitude of the force
of friction acting on the tires is very low.
The tread depth on passenger car tires is usually 7 mm to 8 mm
for a new tire. As the tires wear, the tread depth decreases and
less water is able to move through the grooves. Driving with worn
tires increases the chance of hydroplaning. When the tread depth
gets too low, the tires should be changed.
The Physics of Car Brakes
mechanic or automotive engineer,
GO TO NELSON SCIENCE
There is no perfectly safe car crash, but engineers continually work
at protecting car occupants as much as possible. Even though
vehicles are becoming safer every year, many people across the
world still die in car crashes.
The best way to survive an accident is to avoid one. Nothing will
ever replace sensible driving at the proper speed. However, sooner
or later every driver gets into a situation where an accident might
occur. One safety feature found in many cars is anti-lock brakes. A
disc brake on a car is very similar to the brakes on a bicycle.
However, in disc brakes, a piston squeezes the brake pads against
a rotor (Figure 5). Braking harder increases the magnitude of the
[CATCH: C04-F027A-OP11SB and
C04-F027B-OP11SB (each size D):
diagram showing what understeering
and oversteering look like in terms of a
car’s motion along a curved road. ]
normal force, which in turn increases the force of friction acting on
the rotor. The rotor is attached to the wheel, and as the rotor slows
down so does the wheel.
One problem with disc brakes on cars is that the brake can stop
the wheel from turning a lot faster than the friction from the road
can stop the car from moving. This results in the wheels becoming
locked and sliding across the road. So the car skids and the driver
Figure 6 Both understeering and
oversteering are dangerous and can
cause a car to go off the road or end up
hitting another vehicle.
is unable to steer the car. The anti-lock braking system (ABS) on
cars helps solve this problem.
An ABS uses a computer to monitor the readings of speed
sensors on the wheels of the car. If a wheel experiences a sudden
large decrease in speed, the computer quickly reduces the force on
the brake pads until the wheel moves at an acceptable speed again.
The computer can change the force on the brake pads very rapidly,
which allows the car to slow down as fast as possible without the
tires skidding. This helps to slow down the car rapidly while allowing
the driver to maintain control and steer the car.
One problem with ABS is that drivers feel a chattering action in
Section Title 29
[CATCH: C04-P018-OP11SB (size D):
photo of a crash test dummy after a
crash test]
the brake pedal when slowing down to avoid an accident. Drivers
often think this means something is wrong and lift their foot off the
brake pedal or, even worse, start pumping the brakes. However, this
chattering action in the brake pedal is normal for ABS and drivers
should continue to apply firm pressure to the brake pedal. Under no
Figure 7 Crash test dummies have
many sensors inside them to help
provide engineers with important
circumstances should the driver pump the brakes in a car with ABS.
Pumping the brakes will just make the car travel farther before it
stops. There is some evidence that the ABS technology does not
actually improve the safety of car crashes because drivers do not
use it properly. In some cases drivers who maintain control due to
ABS actually steer off the road to avoid an accident, which often
results in more serious injuries. [CATCH CAREER GLOBE]
[CATCH: C04-F028-OP11SB (size D):
diagram showing the gear mechanism
and piston in a modern seat belt ]
Electronic Stability Controls on Cars
Another safety system on cars very similar to ABS is traction
control. Traction control is actually the reverse of ABS. ABS comes
into play when a car is slowing down and the tires start sliding on
the road. Traction control is used when the car is speeding up,
especially when starting from rest, and the tires start sliding. If the
Figure 8 When a computer detects a
crash, it ignites the gas in the chamber,
which pushes a piston up. A rack gear
attached to the top of the piston turns
another gear, which pulls in the seat
belt.
driver presses down hard on the accelerator, one or more of the
wheels may start turning faster than the car is moving. The force of
friction on the fast-turning wheels decreases and the driver can lose
control of the vehicle and have an accident. Traction control involves
sensors that detect the sliding tires. This information is sent to a
computer, which decreases the amount of fuel to the engine. The
computer may even use ABS to slow the wheels down. The main
thing is that traction control helps the driver maintain control of the
vehicle while accelerating.
Electronic stability control (ESC) uses both traction control and
ABS to help increase the safety of a vehicle. The sensor for ESC,
called a yaw, is usually located in the centre of the car. Its purpose
is to detect if the car is tilting too much one way or another when a
car is making a sharp turn. The two basic scenarios where ESC
comes into play are when the car is experiencing understeering or
oversteering (Figure 6). Understeering occurs when the force of
friction acting on the front wheels is not enough to prevent the car
from travelling in a straight line while the driver is trying to turn.
Oversteering is the opposite. When oversteering occurs, the car
Section Title 30
turns more than the driver intended and the back wheels start to
slide sideways, spinning the car around.
To help prevent both of these situations, ESC can activate one or
more brakes using ABS or adjust the speed of the car using traction
control. The end result is that the driver has a better chance of
controlling the motion of the car with ESC than without it. However,
ESC does not actually drive the car for you. Only responsible driving
can keep the driver and passengers safe. A driver who is driving
too fast or faster than road conditions allow can easily create a
dangerous situation, whether the car has ESC or not. Research has
shown that ESC has had a significant impact in reducing singlevehicle accidents, especially with larger vehicles like SUVs.
Crash Testing
Crash tests are all about making sure cars are safe and helping to
make them even safer. Yet car crashes are still one of the leading
causes of death and injury in North America. One of the most
important tools used during a crash test is a crash test dummy. A
crash test dummy is designed to accurately simulate what will
happen to a person during an accident (Figure 7). To collect data
during a car crash, the dummy has three different kinds of sensors.
The dummy is typically covered with accelerometers that measure
the acceleration of different parts, like the head and torso. A motion
sensor in the chest measures how much it gets compressed during
the crash. This information can be used to estimate the severity of
injuries to the torso. Load sensors all over the dummy are used to
measure other forces acting on it. This extra information helps
engineers determine the nature and severity of injuries.
Seat Belts
Possibly the most important way to protect a person during an
accident is the seat belt. You have already studied how the seat
belt is involved in keeping a person in a car seat. However, early
seat belt designs actually caused injuries due to the tremendous
force exerted by the belt on the person. Today two design features
for seat belts are used to help prevent these types of injuries.
The first seat belt design feature is called a pretensioner. The
pretensioner actually pulls in on the belt when a computer detects a
crash (Figure 8). During a sudden stop or a mild crash, the seat
Section Title 31
belt is reeled in, causing the person to be pulled into the optimal
crash position in the seat. This reduces the forces acting on the
person and helps keep the person in the car seat. However, if the
accident is very violent, the forces exerted by the seat belt on the
person may cause serious injury, even if the person is sitting in the
optimal position. A safety feature to help in this situation is a load
limiter. Load limiters release some of the belt if a tremendous force
starts to pull on it. One simple way to do this is to sew a fold into
the belt material. When a person in an accident pushes forward on
the belt material with a large force, the stitching breaks and a
greater length of seat belt material becomes available. This
decreases the force acting on the person. Another way of making a
load limiter is to use a torsion bar. This bar keeps the belt in place
by restricting the turning of the spool that is used to wind up the
belt material. If a large force acts on the belt, the torsion bar can
twist slightly, releasing some of the material and reducing the force
acting on the person.
Car Body Design
The most noticeable safety feature during a crash test is the
crumple zone of the car. At one time, cars were built with rigid
frames that were inflexible even during severe accidents. With these
cars, drivers and passengers experienced huge accelerations as the
car stopped almost immediately during the collision. Crumple zones
on cars are designed to crush during an accident. This crushing
action increases the time it takes to stop the car during a collision.
By increasing the time interval during the crash, the acceleration of
the people within the car and the forces acting on those people are
significantly reduced. Lower forces mean that it is more likely that
the people in the car will survive the accident uninjured.
Another car safety feature involves some parts of the car body
being made of plastic. The plastic body parts serve two main
purposes during a collision. First, they will not interfere with the
crumple zones since plastic is very flexible. Second, the lightweight
plastic keeps the mass of the car low enough to make the car
easier to stop to avoid a collision.
Air Bags
Another car safety feature is air bags. These devices are deployed
Section Title 32
during a car accident to help prevent the people inside a car from
hitting a hard surface like the dashboard or steering wheel (Figure
9). A thin nylon air bag is folded in the steering wheel or
dashboard. A sensor that detects a collision causes sodium azide to
react with potassium nitrate to produce nitrogen gas. The reaction
happens so fast that the air bag is pushed out at over 300 km/h.
During a car crash, a driver or passenger continues to move forward
even though the car comes to a sudden stop. With a deployed air
bag, the person collides with the air bag instead of the steering
wheel or dashboard. As the person collides with the air bag, it
compresses. This increases the time interval of the collision, which
further reduces the magnitude of the force acting on the person.
Even when an air bag is deployed, a person can still become
injured. If a person is sitting too close to the air bag, the rapidly
inflating air bag can cause serious injury to the head and arms. It is
for this reason that children should never sit on the passenger seat.
Instead, children are more protected from injury when seat belted on
the back seat. It is important to keep in mind that a seat belt should
be worn at all times, whether or not the vehicle has air bags.
[CATCH: C04-F029-OP11SB (size B): diagram showing an air bag before and after deployment]
Figure 9 An air bag before and after a collision
4.4 SUMMARY
• Unlike with other materials, the force of friction acting on rubber
tires depends on the surface area in contact with a road surface.
For most other substances, the force of friction depends only on the
coefficient of friction and the magnitude of the normal force.
• Hydroplaning occurs when a vehicle tire loses direct contact with
the road surface. This situation occurs when a driver is moving too
fast on a very wet road.
• The anti-lock braking system (ABS) uses a computer to adjust the
braking on each car wheel to prevent the wheels from locking up.
This helps a driver maintain control of a vehicle when slowing down
quickly.
Section Title 33
• Electronic stability control (ESC) uses both ABS and traction
control to help prevent oversteering or understeering.
• The crumple zones of a car are designed to crush during an
accident. The crushing action increases the time it takes the car to
come to a complete stop during the collision. This longer time
interval, in turn, decreases the magnitude of both the acceleration
and the forces acting on the people inside the car.
[FORMATTER: the following questions cannot break over a page:
keep them all on the same page]
4.4 QUESTIONS
1. Tire X has treads 8 mm deep. Tire Y has the same design but the treads are 2 mm deep. [A]
(a) Which tire is more likely to experience hydroplaning when driving on a wet road? Explain your
reasoning.
(b) What happens to the force of friction acting on the tire as the surface area decreases during the
transition stage of hydroplaning? Explain your reasoning.
2. Figure 10 shows the wear pattern for improperly inflated tires. Describe what effect, if any, these
wear patterns might have on a car travelling on a wet road. [C] [A]
[CATCH: C04-F030-OP11SB (size C): diagram showing tire wear on tires with different levels of tire
pressure.]
[caption] Figure 10
3. Tires are rated according to their traction, temperature, and tread wear. Research this rating
system and briefly describe the meaning of each rating characteristic. [T/I] [C]
4. Describe what happens to the water levels near a tire as it travels on a wet road at everincreasing speeds. Explain how these water levels affect the force of friction acting on the tires.
[K/U]
5. Compare the disc brake on a car to the brakes on a bicycle. How are the two braking systems
different? [A]
6. Some stationary exercise bikes have a control that adjusts the amount of friction on the wheel.
[T/I] [C] [A]
(a) What is the purpose of this device?
(b) Describe a possible design for this device. Include a diagram in your description.
(c) Describe how you would perform an investigation to determine the relationship between the
setting on the control and the maximum force of static friction for the wheel.
7. Explain how disc brakes work in terms of forces and Newton’s laws. [K/U]
8. Discuss the validity of the following statement: “Car brakes slow the wheels down but friction from
the road slows the car down.” Explain your reasoning and give examples to support your
arguments. [C] [A]
9. Explain how ABS helps to achieve each result listed below, even though it periodically reduces
the friction on the wheel. [K/U]
(a) reduce the stopping distance
(b) help the driver maintain control of the car
10. (a) Compare and contrast ABS and traction control.
(b) Explain how and when ESC uses ABS and traction control. [T/I]
11. Examine the crash test in Figure 11. [T/I]
(a) Describe at least two safety features visible in this photograph.
(b) Describe the type of information that an engineer would gather during a crash test.
(c) What evidence is there in the photograph that the crumple zone was very effective in protecting
the people in the car?
[CATCH: C04-P019-OP11SB (size B): photo of a car after a crash test]
Section Title 34
[caption] Figure 11
12. Explain how crumple zones help to reduce the force on people inside a car during an accident.
[K/U]
13. Describe two new safety innovations involving seat belts. [K/U]
14. (a) What is the purpose of an air bag?
(b) Explain how an air bag works.
(c) Explain how air bags help protect people in cars during a collision. [K/U]
15. From 2008 to 2010, Toyota made several recalls on different models of cars. Research the
reasons for the recalls and how Toyota fixed these problems. [T/I]
16. Research the function of a side curtain air bag. Explain how it works, when it is used, and how
successful it is in preventing injuries. [T/I]
17. In 2009, the Ontario provincial government made it illegal to use a hand-held cellphone while
driving a vehicle. [C][A]
(a) List three reasons why you think the government passed this legislation.
(b) What effect, if any, do you think this legislation will have on traffic safety?
[END page 6 of 6]
Section Title 35
[START Section 4.5: 5 pages]
4.5
[CATCH: C04-P020-OP11SB (size D):
stroboscopic photo showing a golf swing
]
Forces Applied to Sports
and Research
Newton’s laws of motion and the principles of gravity and friction are
involved in many activities and technologies all around us. Some of
these applications involve reducing forces, such as Teflon coatings
Figure 1 A stroboscopic photograph of
a typical golf swing. What forces are
involved in making the golf ball move as
far as possible?
on frying pans or applying wax to skis or snowboards. Others
involve increasing forces, as in the design of golf clubs (Figure 1) or
running shoes. In this section you will learn about many new
applications for forces in sports and research. To understand these
applications, you will use everything you have learned in this unit.
Forces Applied to Sports
In many different sports, there are times when you want to increase
forces and other times when you want to decrease forces. In this
section we will examine a few examples of these applications of
forces.
Golf Club Design
In many sports, a player has to move a ball or puck as quickly as
possible by striking it. In volleyball, basketball, and football, players
use their arms to do this. In soccer, players use their legs. In
hockey, cricket, baseball, and golf, players use another object such
[CATCH: C04-F031-OP11SB (size D):
diagram showing the parts of a golf club
]
as a stick, bat, or club. Here we will examine the physics of a
typical golf club.
The typical golf swing for a long-distance shot involves making
Figure 2 The parts of a golf club
the head of the club hit the ball as fast as possible without twisting
the head of the club. The faster the club head is moving when it
strikes the ball, the farther the ball will travel. However, if you swing
the club harder, it does not always work. If the centre of the head
[CATCH: C04-P021-OP11SB (size D):
photo of volleyball player spiking the
ball.]
(called the sweet spot) does not make contact with the ball, the
head may twist and the ball will travel in the wrong direction. A
better-designed club can typically increase distance and control, but
it cannot make an inexperienced golfer into a good one.
The different parts of a golf club are shown in Figure 2. The grip
Figure 3 The physics involved in a
volleyball spike is similar to the physics
involved in a golf swing.
is made of either leather or rubber. Its purpose is to increase the
force of static friction acting on the player’s hands, increasing
control. The shaft connects the grip to the head and is made of
Section Title 36
steel or a carbon-fibre resin composite. The carbon fibre is lighter
but more expensive. The length of the shaft varies with the type of
club. Clubs that are designed to move the ball over long distances
have a longer shaft and are called woods. Clubs that are meant for
shorter distances have shorter shafts and are called irons. The
[CATCH: C04-F032-OP11SB (size D):
diagram showing the slushy water layer
between a hockey skate and an ice
surface. The water should cover the
entire surface under the blade and
should look like a very thin layer. Make
the bottom of the skate blade less
curved.]
shafts vary in stiffness as well. A less flexible shaft is meant for
players who can move the club quickly. More flexible shafts are
meant for players who tend to whip the head of the club around to
gain speed. The head of the club is the part that makes direct
contact with the ball. Heavier heads tend to twist less when they hit
the ball off centre, but they are harder to swing and hit the ball
more slowly.
Figure 4 A layer of slushy water that
forms between the bottom of a skate
blade and the ice reduces the force of
friction. The bottom of the blade is
curved to help a skater dig into the ice to
accelerate.
In other sports, a player must also use good technique to hit a
ball or puck as fast as possible, as in the case of a good hit in
baseball, a hard shot in hockey, or a hard spike in volleyball (Figure
3). In the end, a better hockey stick helps, but nothing replaces skill
and practice.
Footwear Design
In many sports, the footwear worn by an athlete is just as important
as any other piece of equipment used in the game. Sometimes the
[CATCH: C04-F033-OP11SB (size D):
diagram showing the main parts of a
running shoe]
footwear is designed to decrease the force of friction, as with
skates. At other times, the frictional forces must be increased to
help athletes stop and start quickly, as with running shoes.
The coefficient of kinetic friction of a skate blade on ice may be
as low as 0.005, when the ice is perfectly smooth. Physicists
thought that this very low coefficient of friction was due to a very
Figure 5 The parts of a running shoe
thin layer of water forming between the blade and the frozen ice.
They thought that the pressure of the skate pushing down on the
ice caused the ice to melt, making it more slippery. Physicists now
know that this is only a factor when the ice is at exactly 0 °C.
When you push down on ice, you can decrease its melting point
without changing its temperature. The decreased melting point results
in the ice melting. However, a 90 kg skater only decreases the
footwear designer,
GO TO NELSON SCIENCE
melting point by about 0.1 °C. Ice that is colder than 0 °C will not
become more slippery due to this effect.
Physicists have actually found that a thin liquid layer of slushy
water exists naturally on the surface of ice rinks. It is this slushy
layer that reduces the coefficient of kinetic friction. The layer is
Section Title 37
usually very thin (10–8 m) and it gets thinner as the ice is cooled. At
–250 °C, the slushy layer does not exist at all, and the coefficient of
kinetic friction of a steel skate blade on ice is 0.6, which is
comparable to most other coefficients of friction.
Modern skate blades are slightly curved at the bottom (Figure 4).
This allows the skater to dig the edge of the blade into the ice and
use the normal force to push forward when accelerating.
Running shoes are designed to increase the force of static friction
exerted by the ground on the athlete. This allows the athlete to
accelerate at greater rates. This design is extremely important in
sports where quick starts and stops are required. It also helps an
athlete have greater manoeuvrability. Running shoes are also
designed to protect ankles and knees from injuries during strenuous
activities and to be as light as possible to reduce weight and help
mechanical engineer,
GO TO NELSON SCIENCE
increase speed.
The typical design of a running shoe is shown in Figure 5. The
bottom tread of the shoe (called the outsole) is made from a
durable carbon-rubber compound to help increase the coefficient of
static friction. The midsole is made from a foam substance to help
reduce the magnitude of forces acting on the ankles and knees as a
person is running. The inner part of the midsole is often made of a
harder material to provide stability, while the outer part is made of a
softer material to help cushion the impact. It is not unusual for the
magnitude of the normal force to triple during various sports
activities. The cushion helps absorb some of this impact. Some
midsoles include air or gels to help further absorb the impacts when
running.
The upper part of the shoe is designed for comfort and to
decrease movement of the foot within the shoe. The heel counter is
a hard cup-shaped device used to promote stability and the
movement of the heel. The forces exerted by an athlete on the
ground while running can be very large in magnitude, but from
Newton’s third law, a force of equal magnitude is exerted by the
ground on the feet of the athlete in the opposite direction. The best
running shoes allow an athlete to quickly accelerate, while at the
same time absorbing some of these forces to protect the athlete
from injury. [CATCH CAREER LINK ICON]
Section Title 38
[CATCH: C04-P023-OP11SB (size D):
photo of Oscar Pistorius during a race ]
Bearing Design
One way to reduce friction and increase efficiency of devices such
as generators, motors, and fans is to use bearings. Bearings are
Figure 8 Oscar Pistorius has been
referred to as the “blade runner” from
South Africa.
devices that allow surfaces to slide or roll across each other while
reducing the force of friction. Many different types of bearings can
be used under different circumstances. A plain bearing involves
sliding two surfaces across each other while they are lubricated with
oil or graphite. A rolling element bearing uses balls or rollers to
reduce friction (Figure 6). The balls or rollers are usually lubricated
with oil. Both of these types of bearings have been used for years
and can have very low coefficients of friction.
Some newer types of bearings can actually reduce friction to
[CATCH: C04-P024-OP11SB (size D):
photo of a person with a mechanical
arm ]
negligible levels in some cases. For example, fluid bearings use a
film of fluid, such as air or oil, to separate two surfaces (Figure 7).
The fluid film reduces the force of friction drastically in a similar way
that a thin layer of water separates a skate blade on ice. Fluid
bearings require a seal to keep the fluid in place and a pump to
replace the fluid when it leaks. These bearings can be made
cheaply, and they create less noise when operating than rolling
Figure 9 The Proto 1 mechanical arm
element bearings. However, they can fail suddenly without warning if
the seal breaks. These types of bearings also use energy to keep
the lubricant in place. A typical use for fluid bearings is in the hard
prosthesis technician or biomechanical
engineer,
GO TO NELSON SCIENCE
drive of a computer. [CATCH CAREER LINK ICON]
[FORMATTER: set the following two images side by side]
[CATCH: C04-P022-OP11SB (size B1): photo showing rolling element bearings.]
[CATCH: C04-F034-OP11SB (size D): cross-sectional diagram showing a fluid bearing]
[CATCH: C04-P025-OP11SB (size D):
photo of a researcher examining a
coating of near-frictionless carbon.]
Figure 6 A typical rolling element bearing
Figure 7 A typical fluid bearing
Magnetic bearings use magnetic fields, instead of fluids, to keep
two surfaces separated. The two surfaces do not make direct
Figure 10 A researcher examining a
coating of near-frictionless carbon
materials scientist,
GO TO NELSON SCIENCE
contact with each other, so this technology is often called magnetic
levitation. Magnetic bearings often require a back-up bearing system
in case they fail. Electricity is required to operate the electromagnets
that keep the surfaces separated. So energy is required to keep the
bearings working. However, the bearings need no maintenance and
Section Title 39
have no known upper limit on speed. Typical uses of magnetic
bearings are in motors, turbines, generators, Maglev trains, and
household meters.
Forces Applied to New Innovations
One area of interest in current physics involves the design of
artificial limbs. Another area of current research is in developing
materials that have very low coefficients of friction.
Prosthesis Design
A prosthesis is an artificial device that replaces a missing body part.
This application of forces involves replacing parts of the body that
have been damaged by accident or disease or are the result of birth
defects. These artificial body parts could be dentures, hip
replacements, heart valves, or even artificial hearts. In this section
we will examine artificial limbs.
Artificial limbs have seen many recent improvements, involving
new materials that are lighter, more durable, and more flexible than
previous materials. Some claim that these new materials and
designs actually give an amputee a physical advantage over other
athletes. For example, a newly designed artificial leg is lighter and
stronger than a human leg. This might allow an athlete to run faster
and longer than they normally could. For example, Oscar Pistorius of
South Africa (Figure 8) was not eligible to compete in the 2008
Summer Olympic games because it was believed that his artificial
leg gave him an unfair advantage. The ruling was eventually
overturned; however, Oscar did not end up qualifying. He did very
well, though, in the Paralympics.
The design of artificial limbs has progressed even further than just
making the materials lighter and stronger. Scientists and engineers
are working on making them look and act more like real limbs. One
technique involves targeted muscle reinnervation (TMR). This method
uses a robotic arm or leg that is connected to different muscles in
the body. When the patient thinks about moving the artificial limb,
the muscles contract and the sensors detect the contraction, making
the limb move. The U.S. Pentagon’s research division, Defense
Advanced Research Projects Agency (DARPA), is working on
connecting the artificial limb directly to the human nervous system.
The Proto 1 is the DARPA prototype that uses TMR to control an
Section Title 40
artificial mechanical arm (Figure 9). These artificial limbs are
designed for people who have lost a limb in combat. The Proto 1
can complete tasks previously thought impossible by a mechanical
arm. The arm can perform complex tasks such as taking a credit
card out of a wallet or stacking cups using sensory feedback from
the arm rather than vision. One of the main advantages of this type
of arm over other similar devices is the ability to closely monitor the
force of the grip of the hand and to easily make adjustments to the
amount of force.
New designs are expected to have greater degrees of freedom
and behave even more like a human arm. The next design will have
80 different sensors that will provide feedback about touch, position,
and even temperature. [CATCH CAREER LINK ICON]
Near-Frictionless Carbon
One of the major problems in mechanical devices is wear and loss
of efficiency due to friction. At one time, Teflon seemed to be the
answer to this problem. Teflon has a low coefficient of friction and
can be used as a coating on non-stick frying pans and pots to help
reduce the amount of oil required when cooking. However, near-
frictionless carbon is now the frontrunner in the race to produce a
solid substance with the lowest coefficient of friction.
Near-frictionless carbon has a coefficient of friction less than
0.001, whereas Teflon has a coefficient of 0.04 when tested under
the same conditions (Figure 10). The coating of near-frictionless
carbon is very hard and resists wear from sliding and rolling. It can
be applied to many different types of surfaces and reduces the need
to replace and repair moving parts in many different types of
machines. High-tech applications are being researched in the space
program and in aircraft design, for example. However, any type of
bearing could make use of this new carbon coating. This technology
may eventually become cheap enough to be used in automobiles
and compressors such as air conditioners. [CATCH CAREER LINK
ICON]
4.5 SUMMARY
• The study of forces can be applied to improve sports technology.
• Bearings are used to decrease the force of friction and improve
efficiency.
Section Title 41
• New research into innovations with forces involves fields of study
such as prosthesis design and materials such as near-frictionless
carbon.
4.5 QUESTIONS
1. The physics involved in striking a golf ball and making it move as fast as possible is similar in
many ways to the physics involved in striking a puck or ball in many other sports. [K/U] [A]
(a) Describe how a spike in volleyball is similar to hitting a golf ball.
(b) How is a slap shot in hockey similar to striking a golf ball?
2. A baseball bat has a sweet spot when it strikes a baseball. [K/U] [T/I] [A]
(a) What does the term “sweet spot” imply about the motion of the ball after the swing?
(b) Research what happens to the motion of the bat and the batter if the ball hits the bat at positions
other than the sweet spot.
3. Examine the typical golf swing in Figure 1 on page XXXX. What evidence do you have from this
photograph that the golfer is using good technique when striking the golf ball? [K/U]
4. At one time physicists thought that the blade of a skate pushing down on the ice created a thin
layer of slushy water that decreased the force of friction acting on the blade. [K/U]
(a) Explain why physicists now know this is not true.
(b) Why do skates actually slide so easily over ice?
(c) Why will blades not slide easily over extremely cold ice?
5. The Therma Blade (Figure 11) uses a small battery to heat the skate blade when the player is on
the ice. This type of skate blade experiences significantly less friction than a normal hockey blade.
[T/I] [C] [A]
(a) Explain why the Therma Blade experiences less friction than a normal hockey blade on ice.
(b) A battery must be inserted into the hollow plastic above the top of the blade. Why might this be
(c) Research the Therma Blade. Describe how it works and its current level of use. Discuss the
[CATCH: C04-P026-OP11SB (size C): photo of a Therma Blade]
[caption] Figure 11
6. (a) How are forces involved in the production of electricity?
(b) How can improved bearing technology help with electricity production? [K/U]
7. The ancient Egyptians used rolling logs to move large blocks of stone when building the
pyramids. How is the physics of a rolling element bearing similar to this ancient method of reducing
the force of friction? [A]
8. (a) Describe a typical fluid bearing, and explain how it reduces the force of friction acting on two
surfaces.
(b) Compare the physics of a fluid bearing to that of the blade of a skater on ice. [K/U]
9. (a) Describe how a magnetic bearing works.
(b) What disadvantages do magnetic bearings have over rolling element bearings? [K/U]
10. (a) Describe some of the widely used improvements to artificial limbs.
(b) In what ways are these artificial limbs better than human limbs? [K/U] [A]
11. Some people with artificial limbs want to compete with athletes in the Olympics and in other
sports events. [C] [A]
(a) Give three reasons why this should be allowed and even encouraged.
(b) As technology improves, artificial limbs may provide a significant advantage to people using
them. Should professional athletes using artificial limbs be allowed to compete with those who do
not use these limbs? Explain your reasoning.
12. (a) Describe the Proto 1 and explain how it works.
(b) What improvements are being planned for artificial limbs? [K/U]
13. There has been some debate recently that some people might wish to replace properly working
parts of the body with new improved prostheses. This may become more prevalent as the
technology begins to improve and prostheses become better at performing tasks than the actual
human body. Experiments involving the nervous system are already underway and artificial hearts
have been around for years. [T/I] [A]
(a) Research some of the new advances in prostheses.
(b) How do you feel about this possible trend in the use of this technology? Should governments get
involved with this issue? What kind of legislation should be passed, if any? Explain your reasoning.
Section Title 42
14. A common use of prostheses today is the artificial hip. Figure 12 shows a typical design used
during a hip replacement. What characteristics do you think the materials used in a hip replacement
must have to be able to function properly? [A]
[CATCH: C04-P027-OP11SB (size C): an artificial hip joint ]
[caption] Figure 12
15. Describe the properties of near-frictionless carbon and discuss how it might be used in the
future. [K/U]
[END page 5 of 5]
Section Title 43
[START Explore an Issue Feature: 2 pages]
4.6
Defining the Issue
Researching
Identifying Alternatives
Analyzing the Issue
Defending a Decision
Communicating
Evaluating
[CATCH: C04-P028-OP11SB (size D):
photo of a car in a snowy ditch after
Explore an Issue in Forces
Mandatory Snow Tires in the Winter?
It is a typical winter morning in Ontario, with many people getting
into their cars and leaving for work. Fresh snow has fallen overnight,
and the roads are covered in a thick blanket of snow The
snowploughs have not yet cleared the roads. So people are forced
to make their way through the snow-covered highways and streets.
A common sight is a car off the side of the highway stuck in the
snow (Figure 1).
Many accidents like this could be prevented by drivers changing
their driving habits during bad weather conditions. Just slowing down
and driving more carefully can help. Some drivers think that
Figure 1 A car that has lost control on a
changing their tires to snow tires during the winter months can also
drastically reduce accidents.
In Ontario, all-season tires and snow tires are the two main types
[CATCH: C04-P030A-OP11SB and
C04-P030B-OP11SB (each size D):
photo of a snow tire (labelled (a) ) and
an all -season tire (labelled (b) ). ]
of tires used on vehicles (Figure 2). Some drivers use their all-
season tires year round. Others switch between all-season tires and
snow tires. Snow tires are used when the temperature gets colder
and there is a chance of snow or ice covering the roads. These
tires are marked at stores with a special symbol (Figure 3). This, of
course, means that car owners must buy two sets of tires for each
Figure 2 (a) Typical treads on a snow
tire (b) Typical treads on an all-season
tire
car. In addition, if you do not want to pay the cost of changing the
tires on your rims every year, it also means that you have to buy
an extra set of rims. The extra set of rims allows for easier
installation every time you change the tires.
[CATCH: C04-P029-OP11SB (size D):
photo of logo used on snow tires]
Many people in Ontario think that all-season tires are suitable for
all weather conditions and can be used year round. They also think
that the added expense of another set of tires and rims is not worth
the money, and that many people cannot afford it. Some people
even go as far as to claim that snow tires are just a marketing tool
Figure 3 This logo is placed on any new
snow tire that has met the required
performance standards for snowy
conditions.
or scare tactic used to increase tire sales. Others believe that snow
tires are essential for winter driving. They think that these tires make
a drastic difference in winter driving and the extra cost is
outweighed by the improvement in safety. Others claim that snow
tires should be mandatory for all vehicles in Ontario during the
winter months.
Section Title 44
As of 2008, from November 15 to April 15, all drivers in Quebec
must put snow tires on their vehicles or risk being fined. The fines
can be very expensive, but the police often have trouble spotting
vehicles without snow tires. Jean-Marie de Koninck, the head of a
snow tire law was passed about 10 % of drivers did not use snow
tires but were involved in 38 % of the vehicle accidents during
winter.
The Issue
The Ontario government is considering passing a law making snow
tires mandatory for all cars and trucks on the road. This means all
company and personal vehicles must use snow tires during the
winter months.
snow tires,
GO TO NELSON SCIENCE
Role
You have been assigned to an Ontario task force to research the
tires in Ontario during the winter. Part of your task is to investigate
the differences in the two types of tires; the difference in
performance in cold, snowy, and icy conditions; and whether or not
they are useful in a typical Ontario winter. You will be required to
produce a report on the subject that includes what scientific
evidence you have gathered, your reasoning, and your
recommendations.
Audience
You will present your findings to the government and citizens of
Ontario.
Goal
To decide whether a law should be passed in Ontario making snow
tires mandatory
Research
performance of all-season tires and snow tires during the winter.
Consider opinions, as well as scientific evidence, on this issue.
Research these questions:
Section Title 45
• What is the range of temperature where each type of tire is most
effective?
• How different are the treads on these two types of tires?
• How do the tires compare in stopping distance when a vehicle
stops on ice or snow?
• Are all-wheel-drive vehicles another option instead of using snow
• Are all-season tires a viable option that can be used safely all
year round?
• What are some disadvantages of using each type of tire?
• How much does the average snow tire cost? Should cost of the
tires be a factor in the decision? [CATCH WEBLINK ICON]
Identify Solutions
Is a new law required in Ontario to make snow tires mandatory
during the winter months? Can a compromise be reached? If so,
what is the compromise and how will it be implemented? Can
drivers just replace two all-season tires with two winter tires?
Make a Decision
Communicate
Complete a presentation of your findings that can be shared with the
that most of your target audience are not engineers or scientists but
consumers and government officials. However, these people will
make the final decision on whether or not to proceed with your
recommendation.
Plan for Action
Now that you have an opinion about using snow tires in Ontario, what will you do? One suggestion
is to have a talk with your family to help them decide if they should purchase snow tires. Write one
or two paragraphs outlining what you would say to your family members to make them more
[END Explore an Issue]
[END page 2 of 2]
Section Title 46
[START INVESTIGATIONS: 5 pp]
CONTROLLED EXPERIMENT
4.1.1
Acceleration Due to
Gravity and Terminal
Speed
In this investigation, you will drop different objects and
Questioning
Hypothesizing
Predicting
Planning
Controlling Variables
Performing
Observing
Analyzing
Evaluating
Communicating
measure the acceleration of each object. You will then study
objects that are affected by air friction to a greater degree
and investigate the concept of terminal speed.
Testable Questions
[CATCH: C04-P031-OP11SB (size C2): set-up photo. Student’s hand
should be holding the tape above the timer.]
(a) What is the acceleration due to gravity at
(b) What factors, if any, affect the acceleration
due to gravity?
Hypothesis/Prediction
[caption] Figure 1
Equipment and Materials
After reading through the Experimental Design
• ticker tape timer or motion sensor
Testable Question given in part (b) above. Your
• clamp
and Procedure, write a hypothesis for the
• retort stand
hypothesis should include predictions and
• two objects (one 50 g and the other 100 g)
• coffee filter
Variables
Procedure
Identify the controlled, manipulated, and
Part A: Acceleration Due to Gravity
responding variables in this experiment.
Experimental Design
There are two possible methods of performing
this investigation. One uses a motion sensor
and the other uses a ticker tape timer. Your
teacher will let you know before you begin
which procedure you will use. You will attach a
motion sensor to a retort stand and drop two
different objects below the sensor. Another
method involves using a similar set-up with a
1. Set up the equipment as shown in Figure 1.
If you are using ticker tape, make sure that the
tape will not catch on anything as the object
falls. [CATCH CAUTION HAND ICON]
[CATCH SAFETY ICON]
Take care to not drop the objects near your feet.
2. Drop the 50 g object and obtain the data
necessary to find its acceleration. Repeat with
the 100 g object.
ticker tape timer (Figure 1).
Section Title 47
3. Drop all three objects (50 g, 100 g, and
object during free fall? Why does this happen?
which one(s) land first. Record your
(h) What effect would each situation below
coffee filter) at the same time, and observe
observations. If you are using a motion sensor,
graph the falling motion of the coffee filter.
Part B: Simulating Free Fall
simulate an object in free fall, use it to examine
[T/I]
have on the terminal speed of an object? [T/I]
• increase the mass but keep the crosssectional area the same
• increase the cross-sectional area but keep
the mass the same
the velocity of a falling object with a large
mass. If possible, vary the mass of the object
and observe the effect of mass on the terminal
speed. Also, change the cross-sectional area of
the object without changing its mass, and
determine the effect of cross-sectional area on
terminal speed.
Analyze and Evaluate
(a) Answer the Testable Questions. [T/I] [C]
(b) In terms of the variables in this experiment,
what type of relationship was being tested? [T/I]
(c) Calculate the acceleration of the 50 g and
100 g objects from the data gathered in the
experiment. What effect does the mass of an
object have on the acceleration due to gravity?
[T/I]
(d) Compare the motion of the coffee filter to
that of the other two masses. Explain why the
motion is different. [T/I]
(e) List some possible sources of error. What
could be done to reduce these sources of error
if you were to repeat the experiment? [T/I]
(f) Comment on the accuracy of your hypothesis.
[C]
Apply and Extend
(g) Describe the velocity-time graph for an
object in free fall for an extended period of
time. What happens to the acceleration of the
Section Title 48
CONTROLLED EXPERIMENT
4.2.1
Factors That Affect
Friction
In this investigation you will design and perform several
controlled experiments where you will measure the maximum
force of static friction and the kinetic friction acting on an
Questioning
Hypothesizing
Predicting
Planning
Controlling Variables
Performing
Observing
Analyzing
Evaluating
Communicating
object. To measure the forces you can use a spring scale
or a force meter. Keep in mind that the coefficients of
friction are less exact than most other measurements in
physics. To get meaningful results, complete each
measurement several times and use an average value,
rather than a single measurement.
Testable Question
Experimental Design
How do the following factors affect the
You will design your own procedure for this
object?
magnitude of frictional forces acting on an
• the size of the contact area of the object with
• the smoothness of the contact area and of
different in magnitude than kinetic friction? If so,
• the types of materials in the contact area and
• How does the mass of an object being pulled
magnitude of the force of friction acting on an
investigation. Several factors affect the
• the mass of the object
a surface
• Is the maximum force of static friction
the surface
which one has the greater magnitude?
in the surface
affect the magnitude of kinetic friction acting on
Hypothesis/Prediction
magnitude of static friction?
the object? Does it affect the maximum
After reading through the experiment, write a
• How does the contact area of the object
Your hypothesis should include predictions and
object?
hypothesis to answer the Testable Question.
affect the magnitude of friction acting on the
• How do the types of materials in contact with
Variables
on an object?
Identify the controlled, manipulated, and
responding variables in this experiment.
each other affect the magnitude of friction acting
In your group, decide how you will test the
preceding questions. Your teacher will tell you if
you are to test all the questions or only a few.
Write out the steps in your procedure and
Section Title 49
prepare any data tables that you will need. List
(f) List some possible sources of error. What
HAND ICON]
if you were to repeat the experiment? [T/I]
any safety precautions. [CATCH CAUTION
[CATCH SAFETY ICON]
Take care not to drop the objects near your feet.
Equipment and Materials
• spring scale or force sensor
• balance
• blocks of wood or other materials
• various objects of different mass
• string
Copy this list and add to it any other materials
needed.
Procedure
When your teacher has approved your
procedure, carry out the experiment.
Analyze and Evaluate
(a) Answer the Testable Question. [T/I] [C]
(b) In terms of the variables in this experiment,
what type of relationship was being tested? [T/I]
(c) How does the magnitude of the maximum
force of static friction compare to the magnitude
of kinetic friction on an object? [T/I]
(d) Plot a graph of the magnitude of the force
could be done to reduce these sources of error
(g) Comment on the accuracy of your
hypothesis. [C]
Apply and Extend
(h) Why should you use the greatest force
required to start an object moving to measure
the maximum force of static friction acting on
the object? [T/I]
(i) To measure the kinetic friction, you could
pull an object at a constant velocity. Use
Newton’s laws to explain why the applied force
measured by the spring scale or force meter is
equal in magnitude to the force of kinetic
friction. [T/I]
(j) When determining the magnitude of kinetic
friction acting on an object, why is it better to
use the average applied force required to pull
the object at a constant velocity rather than a
maximum applied force? [T/I] [C]
(k) In this experiment, why was it important to
repeat your measurements several times for
each factor and use average values rather than
a single measurement? [T/I]
of kinetic friction versus the mass of an object.
What is the relationship between these two
variables? Does a similar relationship exist
between the mass of an object and the
maximum magnitude of static friction? Explain
(e) Did the maximum magnitude of static friction
and the magnitude of kinetic friction change
significantly when you changed the contact
area? [T/I]
Section Title 50
CONTROLLED EXPERIMENT
4.2.2
Coefficients of
Friction
Coefficients of friction are determined by experiments rather
than by calculations. In this investigation, you will measure
the maximum magnitude of static friction and the magnitude
Questioning
Hypothesizing
Predicting
Planning
Controlling Variables
Performing
Observing
Analyzing
Evaluating
Communicating
of kinetic friction for a wood block being pulled on several
different surfaces. You will then apply the equations for
static and kinetic friction to determine the coefficients of
friction in each situation. The applied force acting on the
block should always be parallel to the surface.
Testable Question
How does the type of material in a surface
affect the coefficient of static friction and the
coefficient of kinetic friction for a block of wood,
and how do these coefficients compare with
each other?
Hypothesis/Prediction
After reading through the experiment, write a
hypothesis to answer the Testable Question.
Your hypothesis should include predictions and
[caption] Figure 1
To measure the maximum magnitude of static
friction acting on the block, slowly increase the
applied force on the spring scale or force
meter. The greatest reading before the object
starts to move is the maximum magnitude of
static friction. To measure the magnitude of
kinetic friction, use the average reading required
to pull the block slowly at a constant velocity
across the surface.
Equipment and Materials
Variables
Identify the controlled, manipulated, and
responding variables in this experiment.
Experimental Design
You will use a spring scale or a force meter to
pull a wood block horizontally along various
horizontal surfaces of your choice (Figure 1).
[CATCH: C04-P032-OP11SB (size C2): set-up photo of a block of
wood on a surface being pulled by a person’s hand pulling on a spring
scale]
• balance
• spring scale or force sensor
• block of wood
• various flat surfaces of different materials
such as wood, metal, or plastic ramps; a
desktop; a floor; sheets of plastic
• string
Procedure
1. Make a table with these headings: surface,
FSmax, and FK.
Section Title 51
2. Measure the mass of the block of wood.
3. Using the first surface, measure the
maximum magnitude of static friction three
separate times by slowly increasing the applied
horizontal force as you pull the spring scale or
force meter. Record the results in your table.
Apply and Extend
(g) Would it make sense if a student claimed
that their data showed that the coefficient of
kinetic friction was greater than the coefficient of
static friction? Explain your reasoning. [T/I]
4. Using the same flat surface, pull the block at
a slow constant velocity across the surface with
the spring scale or force meter. Use the
average reading for the magnitude of kinetic
friction. Repeat the measurement two more
times. Record the results in your table.
5. Repeat steps 3 and 4 for two other surfaces.
Try to use one surface you think might exert a
low force of friction on the block and at least
one that might exert a high force of friction.
Analyze and Evaluate
(a) Answer the Testable Question. [T/I] [C]
(b) In terms of the variables in this experiment,
what type of relationship was being tested? [T/I]
(c) For each surface, calculate the average of
FSmax and FK acting on the block of wood. Using
these values, calculate the coefficient of static
friction and the coefficient of kinetic friction. How
do these coefficients compare for a particular
surface material? [T/I]
(d) Did the coefficients of friction change when
the block was placed on a different surface?
[T/I]
(e) List some possible sources of error. What
could be done to reduce these sources of error
if you were to repeat the experiment? [T/I]
(f) Comment on the accuracy of your
hypothesis. [C]
Section Title 52
CONTROLLED EXPERIMENT
4.3.1
Predicting Motion with
Friction
In this investigation, you will predict the acceleration of a
system of objects given the total mass of the system and
the coefficient of kinetic friction for the system. You may
Questioning
Hypothesizing
Predicting
Planning
Controlling Variables
Performing
Observing
Analyzing
Evaluating
Communicating
use a surface and a block of wood from a previous
investigation where you already know the coefficient of
kinetic friction, or you can determine the coefficient of kinetic
friction for another surface and block of wood.
Testable Question
How does friction affect the acceleration of the
system shown in Figure 1?
Hypothesis/Prediction
After reading through the experiment, write a
hypothesis to answer the Testable Question.
Your hypothesis should include predictions and
Variables
Identify the controlled, manipulated, and
responding variables in this experiment.
[caption] Figure 1
Equipment and Materials
• ticker tape timer or motion sensor
• balance
• pulley
• spring scale or force meter (optional)
• block of wood
• flat surface such as a wood, metal, or plastic
ramp or a desktop
• 100 g object
• string
Experimental Design
You will use a ticker tape timer, a motion
Procedure
of the wood block shown in Figure 1. Before
Let the 100 g object pull on the block to make
calculate the acceleration using Newton’s laws.
Release the block to see if the system
the acceleration that you measure from the
objects from the string until you see that the
sensor, or a similar device to record the motion
1. Set up the equipment as shown in Figure 1.
you perform the experiment, you will first
sure that the system is starting from rest.
Then you will compare the calculated value to
accelerates. If it does not, hang additional
experiment.
system accelerates when you release the block.
[CATCH: C04-F035-OP11SB (size C2): diagram of a block of wood
on a surface attached to another object hanging over a pulley using a
piece of string. Draw a motion sensor behind the block of wood.]
[CATCH CAUTION HAND ICON]
[CATCH SAFETY ICON]
Take care not to drop the objects near your feet.
Section Title 53
2. Measure the mass of the block of wood.
Record the total mass that is hanging from the
string. Remember that the sum of all the
[END INVESTIGATIONS page 5 of 5]
masses will give you the mass of the system.
3. Obtain the coefficient of kinetic friction from a
previous lab or determine the coefficient of
kinetic friction using a spring scale or force
meter.
4. Calculate the acceleration of the system
using the known information and Newton’s laws.
Assume that the system is starting from rest.
This will be your calculated acceleration. If you
5. Perform the experiment and obtain the data
necessary to measure the acceleration.
Analyze and Evaluate
(a) Answer the Testable Question. [T/I] [C]
(b) Determine the acceleration of the system
from the data gathered in the experiment. [T/I]
(c) Calculate the percentage error for the actual
and calculated accelerations. [T/I]
(d) List some possible sources of error. What
could be done to reduce these sources of error
if you were to repeat the experiment? [T/I]
(e) Comment on the accuracy of your
hypothesis. [C]
Apply and Extend
(f) What effect would each of these situations
have on the acceleration of the system? [T/I]
• placing another object on top of the wood
block
• increasing the mass that is hanging from the
string
Explain your reasoning. If time permits, test
each situation and comment on your results.
Section Title 54
Chapter 4
Summary
Summary Questions
1. Use the Key Concepts in the margin on page XXX to create a
study guide for this chapter. For each point, create three or four
sub-points that provide further information, relevant examples,
explanatory diagrams, or general equations.
2. Look back at the Starting Points questions on page XXX. Answer
these questions using what you have learned in this chapter.
Vocabulary
free fall
terminal velocity
force field
gravitational field strength
static friction
kinetic friction
coefficient of friction
coefficient of static friction (μS)
coefficient of kinetic friction (μK)
Career Pathways
Grade 11 Physics can lead to a wide range of careers. Some
require a college diploma or a B.Sc. degree. Others require
specialized or post-graduate degrees. This graphic organizer
shows a few pathways to careers mentioned in this chapter.
1. Select an interesting career that relates to Applications of
Forces. Research the educational pathways you would need to
follow to pursue this career. What is involved in the required
educational programs?
2. What is involved in becoming a prosthetics fitter or designer?
Are there different pathways you could take to this career?
Research at least two programs and summarize your findings in a
brief report.
[INSERT PLACEHOLDER for Self-Quiz: 1 page]
Chapter 4 Self-Quiz
[QUESTIONS TO COME]
[INSERT PLACEHOLDER for Chapter Review: 6 pages]
Section Title 55
Chapter 4 Review
[QUESTIONS TO COME]
Section Title 56
[START UNIT 2 TASK: 2 pages]
Egg Crash Test
In this unit, you have studied forces and how
they are related to technology. Some of the
technological applications have dealt with
vehicles and vehicle safety. One of the most
important tests performed on a vehicle is the
crash test. In each crash test, a dummy is
placed inside the vehicle to help engineers
determine any vehicle safety issues that may
occur during a crash.
Equipment and Materials
Many different materials can be used for this
task. You can probably find most of the
materials at home. Following are some
suggestions:
• a dynamics cart or similar toy car
• ruler
• metre stick
• scissors
• egg or other fragile object
• construction paper
• cardboard
• balsa wood
• tape
In this task, you will perform a similar test,
• glue
crash test dummy and a dynamics cart instead
Procedure
except that you will use an egg instead of a
of a vehicle (Figure 1). The egg must be
PART A: DESIGN
unbroken after the crash. You will be required
your teacher or a similar toy vehicle. The rules
moving from rest and have a reliable way to
• The egg must be attached to the front of the
Higher marks will be awarded for greater
or barrier during the collision.
impact, and an undamaged egg.
other device to help lessen the forces acting on
[CATCH: C04-P033-OP11SB (size C): set-up photo of two students
up the eggshell by covering it or chemically
attached to the front of the cart and yet remain
1. You may use a dynamics cart provided by
to design a device that can start the cart
for the cart design are as follows:
determine the velocity of the cart upon impact.
cart, and it must make full contact with the wall
velocities upon impact, greater total mass on
• You may build a bumper-like structure or any
working on building a bumper on a dynamics cart.]
the egg during impact, but you cannot toughen
treating it.
• The egg should be easy to insert and remove
from the bumper.
• Consider your skills and safety while
2. Design a method of launching the cart so
that it will hit the wall with a predictable
Figure 1
velocity. To do this, you may design a device
Section Title 57
that is simple or more elaborate, but it should
by the damage factor. An egg with absolutely
3. Design your procedure. You must have a
minor cracking a score of 2, an egg with major
provide consistent velocities with all trials.
procedure that you can use to determine the
velocity of the cart just before striking the
no damage gets a score of 3, an egg with
cracking but still intact 1, and all others get 0.
barrier.
Analyze and Evaluate
force acting on the cart during the collision (if
5. Make a flow diagram showing the steps you
encountered during the design and construction
6. Draw a diagram of your bumper with labels
(c) Explain how you solved design and
be inserted.
could have done differently to improve your
launch the dynamics cart.
(d) Create a poster that displays the safety
you must first present the design ideas to your
works. Include the drawings you created in Part
4. Develop a procedure for analyzing the net
(a) Discuss the role of friction on the motion of
you are using a force sensor or motion sensor).
(b) Describe any difficulties that you
used to build the bumper.
of the cart. [T/I] [C]
showing how it will work and how the egg will
construction problems, and discuss what you
7. Draw another diagram showing how you will
crash test score. [T/I] [C]
8. Before attempting to build any of your ideas,
features of your bumper, explaining why it
teacher. [CATCH CAUTION HAND ICON]
[CATCH CAUTION]
• Use only tools that you are comfortable with, or get help from
someone more experienced.
• Check with your teacher before using any tool in the lab.
• Take care to not drop the cart near your feet.
procedure. [T/I] [C]
Apply and Extend
(e) Examine the flow diagrams of the design
9. Keep a log of the ideas you tested and any
procedures used by other groups. Compare your
diagrams or brief explanations of the steps
(f) Explain how you could have improved the
scores that you calculated. You can include
procedure to those used by other groups.
used.
PART B: PERFORMING THE CRASH TEST
10. Perform three trials of your crash test
bumper design. The first trial should involve a
lower-velocity collision, and the last two should
be higher-velocity collisions. Record the velocity
of the cart. Your score will be calculated using
the trial that has the least significant damage to
the egg.
11. Determine your score by multiplying the
velocity of the cart by the mass of the cart and
Section Title 58
[CATCH ASSESSMENT CHECKLIST]
ASSESSMENT CHECKLIST
criteria:
Knowledge/Understanding
• Demonstrate knowledge of forces.
• Demonstrate knowledge of the safety features of vehicles.
Thinking/Investigation
• Design a cart according to given specifications.
• Design a procedure for launching a cart and record relevant
information.
• Evaluate the cart design and procedure and modify it as needed to
improve performance.
• Analyze cart performance and evaluate based on performance of
other designs.
Communication
• Prepare a log of designs, test results, and modifications.
• Communicate design procedure in the form of drawings and a flow
chart.
• Create a poster demonstrating the design, safety features,
development, and function of the cart.
• Demonstrate understanding of design and construction of the cart.
Application
• Use equipment and materials effectively to design, test, build, and
modify the cart.
• Demonstrate an understanding of how forces can be applied to
design.