Hp04win 1221714291335420-8

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Forces and the Laws of Motion
Chapter 4
Table of Contents
Section 1 Changes in Motion
Section 2 Newton's First Law
Section 3 Newton's Second and Third Laws
Section 4 Everyday Forces

Chapter 4
Objectives
• Describe how force affects the motion of an object.
• Interpret and construct free body diagrams.

Chapter 4
Force

Chapter 4
Force
• A force is an action exerted on an object which may
change the object’s state of rest or motion.
• Forces can cause accelerations.
• The SI unit of force is the newton, N.
• Forces can act through contact or at a distance.

Chapter 4
Comparing Contact and Field Forces

Chapter 4
Force Diagrams
• The effect of a force depends on both magnitude
and direction.Thus, force is a vector quantity.
• Diagrams that show force vectors as arrows are
called force diagrams.
• Force diagrams that show only the forces acting on a
single object are called free-body diagrams.

Chapter 4
Force Diagrams, continued
In a force diagram, vector
arrows represent all the
forces acting in a
situation.
A free-body diagram
shows only the forces
acting on the object of
interest—in this case, the
car.
Force Diagram Free-Body Diagram

Chapter 4
Drawing a Free-Body Diagram

Section 2 Newton’s First Law
Chapter 4
Objectives
• Explain the relationship between the motion of an
object and the net external force acting on the object.
• Determine the net external force on an object.
• Calculate the force required to bring an object into
equilibrium.

Chapter 4
Newton’s First Law
• An object at rest remains at rest, and an object in
motion continues in motion with constant velocity
(that is, constant speed in a straight line) unless the
object experiences a net external force.
• In other words, when the net external force on an
object is zero, the object’s acceleration (or the
change in the object’s velocity) is zero.

Chapter 4
Net Force
• Newton's first law refers to the net force on an
object.The net force is the vector sum of all forces
acting on an object.
• The net force on an object can be found by using the
methods for finding resultant vectors.
Although several forces are
acting on this car, the vector sum
of the forces is zero. Thus, the
net force is zero, and the car
moves at a constant velocity.

Chapter 4
Sample Problem
Determining Net Force
Derek leaves his physics book on top of a drafting
table that is inclined at a 35° angle. The free-body
diagram below shows the forces acting on the book.
Find the net force acting on the book.

Chapter 4
Sample Problem, continued
1. Define the problem, and identify the variables.
Given:
Fgravity-on-book = Fg = 22 N
Ffriction = Ff = 11 N
Ftable-on-book = Ft = 18 N
Unknown:
Fnet = ?

Chapter 4
2. Select a coordinate system, and apply it to the
free-body diagram.
Tip: To simplify the problem, always
choose the coordinate system in
which as many forces as possible lie
on the x- and y-axes.
Choose the x-axis parallel to and the y-axis perpendicular to
the incline of the table, as shown in (a). This coordinate
system is the most convenient because only one force
needs to be resolved into x and y components.

Chapter 4
3. Find the x and y components of all vectors.
Add both components to the free-body diagram, as shown in (c).
cos θ =
Fg,x
Fg
Fg,x = Fg cos θ
Fg,x = (22 N)(cos 55°)
Fg,x = 13 N
sin θ =
Fg,y
Fg
Fg,y = Fg sin θ
Fg,x = (22 N)(sin 55°)
Fg,x = 18 N
Draw a sketch, as shown in (b), to help find the
components of the vector Fg. The angle θ is equal to
180°– 90° – 35° = 55°.

Chapter 4
For the y direction:
ΣFy = Ft – Fg,y
ΣFy = 18 N – 18 N
ΣFy = 0 N
4. Find the net force in both the x and y directions.
Diagram (d) shows another free-body
diagram of the book, now with forces
acting only along the x- and y-axes.
For the x direction:
ΣFx = Fg,x – Ff
ΣFx = 13 N – 11 N
ΣFx = 2 N

Chapter 4
5. Find the net force.
Add the net forces in the x and y directions together as
vectors to find the total net force. In this case, Fnet = 2 N in the
+x direction, as shown in (e). Thus, the book accelerates
down the incline.

Chapter 4
Inertia
• Inertia is the tendency of an object to resist being
moved or, if the object is moving, to resist a change
in speed or direction.
• Newton’s first law is often referred to as the law of
inertia because it states that in the absence of a net
force, a body will preserve its state of motion.
• Mass is a measure of inertia.

Chapter 4
Mass and Inertia

Chapter 4
Inertia and the Operation of a Seat Belt
• While inertia causes
passengers in a car to
continue moving forward as
the car slows down, inertia
also causes seat belts to lock
into place.
• The illustration shows how
one type of shoulder harness
operates.
• When the car suddenly slows
down, inertia causes the large
mass under the seat to
continue moving, which
activates the lock on the
safety belt.

Chapter 4
Equilibrium
• Equilibrium is the state in which the net force on an
object is zero.
• Objects that are either at rest or moving with
constant velocity are said to be in equilibrium.
• Newton’s first law describes objects in equilibrium.
Tip: To determine whether a body is in equilibrium, find the net
force. If the net force is zero, the body is in equilibrium. If there
is a net force, a second force equal and opposite to this net
force will put the body in equilibrium.

Section 3 Newton’s Second and
Third LawsChapter 4
Objectives
• Describe an object’s acceleration in terms of its
mass and the net force acting on it.
• Predict the direction and magnitude of the
acceleration caused by a known net force.
• Identify action-reaction pairs.

Chapter 4
Newton’s Second Law
The acceleration of an object is directly
proportional to the net force acting on the
object and inversely proportional to the
object’s mass.
ΣF = ma
net force = mass × acceleration
Third Laws
ΣF represents the vector sum of all external forces
acting on the object, or the net force.

Chapter 4
Newton’s Second Law
Third Laws

Chapter 4
Newton’s Third Law
• If two objects interact, the magnitude of the force
exerted on object 1 by object 2 is equal to the
magnitude of the force simultaneously exerted on
object 2 by object 1, and these two forces are
opposite in direction.
• In other words, for every action, there is an
equal and opposite reaction.
• Because the forces coexist, either force can be
called the action or the reaction.
Third Laws

Chapter 4
Action and Reaction Forces
• Action-reaction pairs do not imply that the net
force on either object is zero.
• The action-reaction forces are equal and opposite,
but either object may still have a net force on it.
Third Laws
Consider driving a nail into wood with
a hammer. The force that the nail
exerts on the hammer is equal and
opposite to the force that the hammer
exerts on the nail. But there is a net
force acting on the nail, which drives
the nail into the wood.

Chapter 4
Newton’s Third Law
Third Laws

Chapter 4
Objectives
• Explain the difference between mass and weight.
• Find the direction and magnitude of normal forces.
• Describe air resistance as a form of friction.
• Use coefficients of friction to calculate frictional force.

Chapter 4
Weight
• The gravitational force (Fg) exerted on an object
by Earth is a vector quantity, directed toward the
center of Earth.
• The magnitude of this force (Fg) is a scalar
quantity called weight.
• Weight changes with the location of an object in
the universe.

Chapter 4
Weight, continued
• Calculating weight at any location:
Fg = mag
ag = free-fall acceleration at that location
• Calculating weight on Earth's surface:
ag = g = 9.81 m/s2
Fg = mg = m(9.81 m/s2
)

Chapter 4
Comparing Mass and Weight

Chapter 4
Normal Force
• The normal force acts on a surface in a direction
perpendicular to the surface.
• The normal force is not always opposite in
direction to the force due to gravity.
– In the absence of other forces, the
normal force is equal and opposite
to the component of gravitational
force that is perpendicular to the
contact surface.
– In this example, Fn = mg cos θ.

Chapter 4
Normal Force

Chapter 4
Friction
• Static friction is a force that resists the initiation
of sliding motion between two surfaces that are in
contact and at rest.
• Kinetic friction is the force that opposes the
movement of two surfaces that are in contact and
are sliding over each other.
• Kinetic friction is always less than the maximum
static friction.

Chapter 4
Friction

Chapter 4
Friction Forces in Free-Body Diagrams
• In free-body diagrams, the force of friction is always
parallel to the surface of contact.
• The force of kinetic friction is always opposite the
direction of motion.
• To determine the direction of the force of static
friction, use the principle of equilibrium. For an
object in equilibrium, the frictional force must point
in the direction that results in a net force of zero.

Chapter 4
The Coefficient of Friction
• The quantity that expresses the dependence of
frictional forces on the particular surfaces in
contact is called the coefficient of friction, µ.
• Coefficient of kinetic friction:
µk =
Fk
Fn
µs =
Fs,max
Fn
• Coefficient of static friction:

Chapter 4 Section 4 Everyday Forces

Chapter 4
Sample Problem
Overcoming Friction
A student attaches a rope to a 20.0 kg box of
books.He pulls with a force of 90.0 N at an angle of
30.0° with the horizontal. The coefficient of kinetic
friction between the box and the sidewalk is 0.500.
Find the acceleration of the box.

Chapter 4
1. Define
Given:
m = 20.0 kg
µk = 0.500
Fapplied = 90.0 N at θ = 30.0°
Unknown:
a= ?
Diagram:

Chapter 4
The diagram on the right shows the
most convenient coordinate system,
because the only force to resolve
into components is Fapplied.
2. Plan
Choose a convenient coordinate system, and
find the x and y components of all forces.
Fapplied,y = (90.0 N)(sin 30.0º) = 45.0 N (upward)
Fapplied,x = (90.0 N)(cos 30.0º) = 77.9 N (to the right)

Chapter 4
Choose an equation or situation:
A. Find the normal force, Fn, by applying the condition of
equilibrium in the vertical direction:
ΣFy = 0
B. Calculate the force of kinetic friction on the box:
Fk = µkFn
C. Apply Newton’s second law along the horizontal direction to
find the acceleration of the box:
ΣFx = max

Chapter 4
3. Calculate
A. To apply the condition of equilibrium in the vertical direction,
you need to account for all of the forces in the y direction:
Fg, Fn, and Fapplied,y. You know Fapplied,y and can use the box’s mass to
find Fg.
Fapplied,y = 45.0 N
Fg = (20.0 kg)(9.81 m/s2
) = 196 N
Next, apply the equilibrium condition,
ΣFy = 0, and solve for Fn.
ΣFy = Fn + Fapplied,y – Fg = 0
Fn + 45.0 N – 196 N = 0
Fn = –45.0 N + 196 N = 151 N
Tip: Remember to
pay attention to the
direction of forces.
In this step, Fg is
subtracted from Fn
and Fapplied,y because
Fg is directed
downward.

Chapter 4
B. Use the normal force to find the force of kinetic friction.
Fk = µkFn = (0.500)(151 N) = 75.5 N
C. Use Newton’s second law to determine the horizontal
acceleration.
ax =
Fapplied,x − Fk
m
=
77.9 N − 75.5 N
20.0 kg
=
2.4 N
20.0 kg
=
2.4 kg • m/s2
20.0 kg
a = 0.12 m/s2
to the right
ΣFx = Fapplied − Fk = max

Chapter 4
4. Evaluate
The box accelerates in the direction of the net
force, in accordance with Newton’s second law.
The normal force is not equal in magnitude to the
weight because the y component of the student’s
pull on the rope helps support the box.

Chapter 4
Air Resistance
• Air resistance is a form of friction. Whenever an
object moves through a fluid medium, such as air or
water, the fluid provides a resistance to the object’s
motion.
• For a falling object, when the upward force of air
resistance balances the downward gravitational
force, the net force on the object is zero. The object
continues to move downward with a constant
maximum speed, called the terminal speed.

Chapter 4
Fundamental Forces
• There are four fundamental forces:
– Electromagnetic force
– Gravitational force
– Strong nuclear force
– Weak nuclear force
• The four fundamental forces are all field forces.

Multiple Choice
Chapter 4
a =
F
m1
a =
F
m2
a =
F
m1 + m2
a =
F
(m1)(m2 )
Use the passage below to answer questions 1–2.
Two blocks of masses m1 and m2 are placed in contact with each
other on a smooth, horizontal surface. Block m1 is on the left of
block m2. A constant horizontal force F to the right is applied to m1.
1. What is the acceleration of the two blocks?
A. C.
B. D.

Multiple Choice, continued
Two blocks of masses m1 and m2 are placed in contact with each
other on a smooth, horizontal surface. Block m1 is on the left of
block m2. A constant horizontal force F to the right is applied to m1.
2. What is the horizontal force acting on m2?
F. m1a
G. m2a
H. (m1 + m2)a
J. m1m2a
Chapter 4

Chapter 4
3. A crate is pulled to the right with a force of 82.0 N, to the left
with a force of 115 N, upward with a force of 565 N, and
downward with a force of 236 N. Find the magnitude and
direction of the net force on the crate.
A. 3.30 N at 96° counterclockwise from the positive x-axis
B. 3.30 N at 6° counterclockwise from the positive x-axis
C. 3.30 x 102
at 96° counterclockwise from the positive x-axis
D. 3.30 x 102
at 6° counterclockwise from the positive x-axis

Chapter 4
4. A ball with a mass of m is thrown into the air, as shown in the
figure below. What is the force exerted on Earth by the ball?
A. mballg directed down
B. mballg directed up
C. mearthg directed down
D. mearthg directed up

Chapter 4
5. A freight train has a mass of 1.5 x 107
kg. If the locomotive
can exert a constant pull of 7.5 x 105
N, how long would it take
to increase the speed of the train from rest to 85 km/h?
(Disregard friction.)
A. 4.7 x 102
s
B. 4.7s
C. 5.0 x 10-2
s
D. 5.0 x 104
s

A truck driver slams on the brakes and skids
to a stop through a displacement ∆x.
Chapter 4
6. If the truck’s mass doubles, find the truck’s skidding distance in terms of
∆x. (Hint: Increasing the mass increases the normal force.)
A. ∆x/4
B. ∆x
C. 2∆x
D. 4∆x

A truck driver slams on the brakes and skids
to a stop through a displacement ∆x.
Chapter 4
7. If the truck’s initial velocity were halved, what would be the
truck’s skidding distance?
A. ∆x/4
B. ∆x
C. 2∆x
D. 4∆x

Chapter 4
8. What is the relationship between the forces at point A?
F. Fs=Fapplied
G. Fk=Fapplied
H. Fs<Fapplied
I. Fk>Fapplied
Use the graph at right to
answer questions 8–9. The
graph shows the relationship
between the applied force
and the force of friction.

Chapter 4
9. What is the relationship between the forces at point B?
A. Fs, max=Fk
B. Fk> Fs, max
C. Fk>Fapplied
D. Fk<Fapplied
Use the graph at right to
answer questions 8–9. The
graph shows the relationship
between the applied force
and the force of friction.

Short Response
Base your answers to questions 10–12 on the
information below.
A 3.00 kg ball is dropped from rest from the
roof of a building 176.4 m high.While the ball
is falling, a horizontal wind exerts a constant
force of 12.0 N on the ball.
10.How long does the ball take to hit the ground?
Chapter 4

Short Response
information below.
10.How long does the ball take to hit the ground?
Answer: 6.00 s
Chapter 4

Short Response, continued
Chapter 4
information below.
11. How far from the building does the ball hit the ground?

information below.
11. How far from the building does the ball hit the ground?
Answer: 72.0 m
Chapter 4

information below.
12. When the ball hits the ground, what is its speed?
Chapter 4

information below.
12. When the ball hits the ground, what is its speed?
Answer: 63.6 m/s
Chapter 4

passage.
A crate rests on the horizontal bed of a pickup truck.
For each situation described below, indicate the
motion of the crate relative to the ground, the motion of
the crate relative to the truck, and whether the crate
will hit the front wall of the truck bed, the back wall, or
neither. Disregard friction.
13. Starting at rest, the truck accelerates to the right.
Chapter 4

Chapter 4
passage.
For each situation described below, indicate the motion of the
crate relative to the ground, the motion of the crate relative to
the truck, and whether the crate will hit the front wall of the truck
bed, the back wall, or neither. Disregard friction.
13. Starting at rest, the truck accelerates to the right.
Answer: at rest, moves to the left, hits back wall

Chapter 4
passage.
For each situation described below, indicate the motion of the crate relative to the
ground, the motion of the crate relative to the truck, and whether the crate will hit the front
wall of the truck bed, the back wall, or neither. Disregard friction.
14. The crate is at rest relative to the truck while the
truck moves with a constant velocity to the right.

passage.
14. The crate is at rest relative to the truck while the
truck moves with a constant velocity to the right.
Answer: moves to the right, at rest, neither
Chapter 4

passage.
15. The truck in item 14 slows down.
Chapter 4

passage.
15. The truck in item 14 slows down.
Answer: moves to the right, moves to the right,
hits front wall
Chapter 4

16. A student pulls a rope attached to a 10.0 kg wooden
sled and moves the sled across dry snow. The student
pulls with a force of 15.0 N at an angle of 45.0º.
If µk between the sled and the snow is 0.040, what
is the sled’s acceleration? Show your work.
Chapter 4
Extended Response

16. A student pulls a rope attached to a 10.0 kg wooden
sled and moves the sled across dry snow. The student
pulls with a force of 15.0 N at an angle of 45.0º.
If µk between the sled and the snow is 0.040, what
is the sled’s acceleration? Show your work.
Answer: 0.71 m/s2
Chapter 4
Extended Response

Chapter 4
Extended Response, continued
17. You can keep a 3 kg book from dropping by pushing
it horizontally against a wall. Draw force diagrams,
and identify all the forces involved. How do they
combine to result in a zero net force? Will the force
you must supply to hold the book up be different for
different types of walls? Design a series of
experiments to test your answer. Identify exactly
which measurements will be necessary and what
equipment you will need.

17. You can keep a 3 kg book from dropping by pushing
it horizontally against a wall. Draw force diagrams,
and identify all the forces involved. How do they
combine to result in a zero net force? Will the force
you must supply to hold the book up be different for
different types of walls? Design a series of
experiments to test your answer. Identify exactly
which measurements will be necessary and what
equipment you will need.
Answer: Plans should involve measuring forces such
as weight, applied force, normal force, and friction.
Chapter 4
Extended Response, continued

Chapter 4
Force Diagrams

Chapter 4
Inertia and the Operation of a Seat Belt

Hp04win 1221714291335420-8

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