Chapter 4

Forces and Newton’s Laws of Motion

4.1 The Concepts of Force and Mass

A force is a push or a pull.Arrows are used to represent forces. The length of the arrowis proportional to the magnitude of the force.

15 N

5 N

4.3 Newton’s Second Law of Motion

SI Unit for Force

This unit of force is called a newton (N).

4.1 The Concepts of Force and Mass

Mass is a measure of the amount of “stuff” contained in an object.

SI Unit of Mass: kilogram (kg)

4.2 Newton’s First Law of Motion

An object continues in a state of restor in a state of motion at a constant Speed unless changed by a net force

The net force is the SUM of allof the forces acting on an object.

Newton’s First Law (Law of Inertia)

4.2 Newton’s First Law of Motion

The net force on an object is the sum of all forces acting on that object.

Individual Forces Net Force

10 N4 N 6 N

4.2 Newton’s First Law of Motion

Individual Forces Net Force

3 N

4 N

5 N36

4.3 Newton’s Second Law of Motion

F

Mathematically, the net force is written as

where the Greek letter sigma denotes the vector sum.

4.3 Newton’s Second Law of Motion

Newton’s Second Law

When a net force acts on an objectof mass m, the acceleration that results is directly proportional to the net force and hasa magnitude that is inversely proportional tothe mass. The direction of the acceleration isthe same as the direction of the net force.

mF

a

aF

m

4.4 The Vector Nature of Newton’s Second Law

xx maFyy maF

The direction of force and acceleration vectorscan be taken into account by using x and ycomponents.

aFm

is equivalent to

4.4 The Vector Nature of Newton’s Second Law

4.4 The Vector Nature of Newton’s Second Law

Force x component y component

+17 N

+(15 N) cos67

0 N

+(15 N) sin67

+23 N +14 N

The net force on the raft can be calculatedin the following way:

P

A

4.4 The Vector Nature of Newton’s Second Law

2sm 018.0kg 1300

N 23

m

Fa

x

x

2sm 011.0kg 1300

N 14

m

Fa

y

y

4.5 Newton’s Third Law of Motion

Newton’s Third Law of Motion

Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body.

4.5 Newton’s Third Law of Motion

Suppose that the magnitude of the force is 36 N. If the massof the spacecraft is 11,000 kg and the mass of the astronautis 92 kg, what are the accelerations?

4.5 Newton’s Third Law of Motion

.astronaut On the

. spacecraft On the

PF

PF

2sm0033.0kg 11,000

N 36

ss m

Pa

2sm39.0kg 92

N 36

AA m

Pa

4.7 The Gravitational Force

Newton’s Law of Universal Gravitation

Every particle in the universe exerts an attractive force on everyother particle.

4.7 The Gravitational Force

For two particles that have masses m1 and m2 and are separated by a distance r, the force has a magnitude given by

221

r

mmGF

2211 kgmN10673.6 G

4.7 The Gravitational Force

N 104.1

m 1.2

kg 25kg 12kgmN1067.6

8

22211

221

r

mmGF

4.7 The Gravitational Force

4.7 The Gravitational Force

Definition of Weight

The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downwards, toward the center of the earth.

SI Unit of Weight: newton (N)

4.8 The Normal Force

Definition of the Normal Force

The normal force is the force that a surfaceexerts on an object with which it is in contact – It is ALWAYS perpendicular to the surface.

4.8 The Normal Force

N 4

0N 15N 11

N

N

F

F

A hand pushes down on a block with a force of 11 N. The block weighs 15N. What is the magnitude of the normal force?

A rope is now used to lift the block with a force of 11 N. What is the magnitude of the normal force?

N 26

0N 15N 11

N

N

F

F

4.8 The Normal Force

Apparent Weight

The apparent weight of an object is the reading of the scale.

It is equal to the normal force the man exerts on the scale.

4.8 The Normal Force

mamgFF Ny

mamgFN

apparent weight

trueweight

4.9 Static and Kinetic Frictional Forces

When an object is in contact with a surface there is a forceacting on that object. The component of this force that is parallel to the surface is called the frictional force.

4.9 Static and Kinetic Frictional Forces

When the two surfaces are not sliding across one anotherthe friction is called static friction.

Nss Ff

4.9 Static and Kinetic Frictional Forces

Note that the magnitude of the frictional force doesnot depend on the contact area of the surfaces.

What does it depend on???

4.9 Static and Kinetic Frictional Forces

Static friction opposes the impending relative motion betweentwo objects.

Kinetic friction opposes the relative sliding motion motions thatactually does occur.

Nkk Ff

10 k is called the coefficient of kinetic friction.

4.9 Static and Kinetic Frictional Forces

Page 104

4.9 Static and Kinetic Frictional Forces

The sled comes to a halt because the kinetic frictional forceopposes its motion and causes the sled to slow down.

4.9 Static and Kinetic Frictional Forces

Suppose the coefficient of kinetic friction is 0.05 and the total mass is 40kg. What is the kinetic frictional force?

N 20sm80.9kg4005.0 2

mgFf kNkk

4.3 Newton’s Second Law of Motion

A free-body-diagram is a diagram that represents the object and the forces that act on it.

4.3 Free Body Diagrams

The net force in this case is:

275 N + 395 N – 560 N = +110 N

and is directed along the + x axis of the coordinate system.

4.6 Types of Forces: An Overview

Examples of different forces:

Friction forceTension forceNormal force

Weight (Force due to gravity)Gravitational Force

Magnetic ForceElectric Force

4.10 The Tension Force

Cables and ropes transmit forces through tension.

This diagram shows four forces acting upon an object. There aren’t always four forces, For example, there could be one or more forces.

Problem 1

A book is at rest on a table top. Diagram the forces acting on the book.

Problem 1

In this diagram, there are normal and gravitational forces on the book.

Problem 2

An egg is free-falling from a nest in a tree. Neglect air resistance. Draw a free-body diagram showing the forces involved.

Gravity is the only force acting on the egg as it falls.

Problem 3

A rightward force is applied to a book in order to move it across a desk. Consider frictional forces. Neglect air resistance. Construct a free-body diagram. Let’s see what this one looks like.

Note the applied force arrow pointing to the right. Notice how friction force points in the opposite direction. Finally, there is still gravity and normal forces involved.

Problem 4

• A block is at rest on an incline and stays in place because of static friction. Draw a free-body diagram

• A skier is standing motionless on a horizontal tow rope, which is about to pull her forward. The skier’s mass is 59kg and the coefficient of static friction between the skis and the snow is 0.14. Draw a free body diagram. What is the magnitude of the maximum force that the tow rope can apply to the skier without causing her to move.

Objects in Equilibrium• Objects that are either at rest or moving with

constant velocity are said to be in equilibrium• This means that the net force acting on the

object is zero• Equivalent to the set of component equations

given by 0 F

0 xx maF 0 yy maF

4.11 Equilibrium Application of Newton’s Laws of Motion

Reasoning Strategy

• Draw a free-body diagram.

• Choose a set of x, y axes for each object and resolve all forcesin the free-body diagram into components that point along theseaxes.

4.11 Equilibrium Application of Newton’s Laws of Motion

What is the Tension in Rope 1 and 2?

The engine has a weight of 3150 N

4.11 Equilibrium Application of Newton’s Laws of Motion

N 3150W

Force x component y component

1T

2T

W

0.10sin1T

0.80sin2T

0

0.10cos1T

0.80cos2T

W

4.11 Equilibrium Application of Newton’s Laws of Motion

00.80sin0.10sin 21 TTFx

00.80cos0.10cos 21 WTTFy

The first equation gives 21 0.10sin

0.80sinTT

Substitution into the second gives

00.80cos0.10cos0.10sin

0.80sin22

WTT

4.11 Equilibrium Application of Newton’s Laws of Motion

0.80cos0.10cos0.10sin0.80sin

2

WT

N 5822 T N 1030.3 31 T

4.12 Nonequilibrium Application of Newton’s Laws of Motion

xx maF

yy maF

When an object is accelerating, it is not in equilibrium.

4.12 Nonequilibrium Application of Newton’s Laws of Motion

The acceleration is along the x axis so 0ya

The mass of the supertanker is 1.5 x 108 kg. There is a drive force with magnitude 75000 N and an opposing force of 40000 N. The tanker moves with an acceleration of 2.0 x 10-3 m/s2. Solve for the Tension in each of the ropes.

4.12 Nonequilibrium Application of Newton’s Laws of Motion

Force x component y component

1T

2T

D

R

0.30cos1T

0.30cos2T

0

0

D

R

0.30sin1T

0.30sin2T

4.12 Nonequilibrium Application of Newton’s Laws of Motion

00.30sin0.30sin 21 TTFy

21 TT

x

x

ma

RDTTF

0.30cos0.30cos 21

4.12 Nonequilibrium Application of Newton’s Laws of Motion

TTT 21

N 1053.10.30cos2

5

DRmaT x