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Forces and the Laws of MotionChapter 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



Section 1 Changes in MotionChapter 4

Objectives

• Describe how force affects the motion of an object.

• Interpret and construct free body diagrams.



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

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



Section 2 Newton’s First LawChapter 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.

Section 2 Newton’s First Law



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 Fg,x

Fg

Fg,x Fg cos

Fg,x (22 N)(cos 55)

Fg,x 13 N

sin Fg,y

Fg

Fg,y Fg sin

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

is equal to 180– 90 – 35 = 55.



Chapter 4



For the y direction:

SFy = Ft – Fg,y

SFy = 18 N – 18 N

SFy = 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:

SFx = Fg,x – Ff

SFx = 13 N – 11 N

SFx = 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

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 withconstant 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.

SF = ma

net force = mass acceleration


Third Laws

SF represents the vector sum of all external forces

acting on the object, or the net force.



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.



Section 4 Everyday ForcesChapter 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

Normal Force


• The normal force acts on a surface in a directionperpendicular 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 .



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 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, m.

• Coefficient of kinetic friction:

mk Fk

Fn

ms 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

mk = 0.500

Fapplied = 90.0 N at = 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:

SFy = 0

B. Calculate the force of kinetic friction on the box:

Fk = mkFn

C. Apply Newton’s second law along the horizontal direction to

find the acceleration of the box:

SFx = max



Chapter 4



3. CalculateA. 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,

SFy = 0, and solve for Fn.

SFy = 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 = mkFn = (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

SFx Fapplied Fk max



Chapter 4



4. EvaluateThe 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

Standardized Test PrepChapter 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


a F

m1

a F

m2

a F

m1 m2

a F

(m1)(m2 )



1. What is the acceleration of the two blocks?

A. C.

B. D.



Multiple Choice, continued



2. What is the horizontal force acting on m2?

F. m1a

G. m2a

H. (m1 + m2)a

J. m1m2a







2. What is the horizontal force acting on m2?

F. m1a

G. m2a

H. (m1 + m2)a

J. m1m2a






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





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





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





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





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 102s

B. 4.7s

C. 5.0 x 10-2s

D. 5.0 x 104s





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 102s

B. 4.7s

C. 5.0 x 10-2s

D. 5.0 x 104s





A truck driver slams on the brakes and skids

to a stop through a displacement Dx.


6. If the truck’s mass doubles, find the truck’s skidding distance in terms of Dx. (Hint: Increasing the mass increases the normal force.)

A. Dx/4

B. Dx

C. 2Dx

D. 4Dx








6. If the truck’s mass doubles, find the truck’s skidding distance in terms of Dx. (Hint: Increasing the mass increases the normal force.)

A. Dx/4

B. Dx

C. 2Dx

D. 4Dx








7. If the truck’s initial velocity were halved, what would be the truck’s skidding distance?

A. Dx/4

B. Dx

C. 2Dx

D. 4Dx








7. If the truck’s initial velocity were halved, what would be the truck’s skidding distance?

A. Dx/4

B. Dx

C. 2Dx

D. 4Dx





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.





8. What is the relationship between the forces at point A?

F. Fs=Fapplied

G. Fk=Fapplied

H. Fs<Fapplied

I. Fk>Fapplied






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






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




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?




Short Response


information below.





10.How long does the ball take to hit the ground?

Answer: 6.00 s




Short Response, continued



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






information below.





12. When the ball hits the ground, what is its speed?






information below.





12. When the ball hits the ground, what is its speed?

Answer: 63.6 m/s






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.









13. Starting at rest, the truck accelerates to the right.

Answer: at rest, moves to the left, hits back wall







14. The crate is at rest relative to the truck while the

truck moves with a constant velocity to the right.





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







15. The truck in item 14 slows down.







15. The truck in item 14 slows down.

Answer: moves to the right, moves to the right,

hits front wall





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 mk between the sled and the snow is 0.040, what

is the sled’s acceleration? Show your work.


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 mk between the sled and the snow is 0.040, what

is the sled’s acceleration? Show your work.

Answer: 0.71 m/s2


Extended Response




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.


Extended Response, continued



Chapter 4

Force Diagrams




Chapter 4

Inertia and the Operation of a Seat Belt


Documents

Chapter 4 Forces and the Laws of Motion Table of Contentsgilsonscience.weebly.com/uploads/2/1/1/4/21140528/chapter_4_powe… · Fg,x 18 N Draw a sketch, as ... Copyright © by Holt,