Chapter 3

Newton's laws of motion

Forces

3 .1 Force,weight,and gravitational mass

Kinematics describes motion, but does not explain why motion occurs. In order to explain the cause of motion,

we must learn about Statics and Dynamics … and FORCES.What is a Force?

•There are two types of forces that we can see•Contact Forces•At a distance forces

•the force causes a change in velocity, or an acceleration…

•a push or pull that one body exerts on another•Vector

Fkick

Fgrav

mass

Pushing forceFriction force

Contact ForceForces in which the two interacting objects are physically in contact with each other.EXAMPLESHitting - Pulling with a rope - Lifting weights - Pushing a couch

Frictional Force - Tensional Force - Normal Force- Air Resistance ForceApplied Force- Spring ForceAt A Distance Force (field force)•Forces in which the two interacting objects are not in physical contact with each other, but are able to exert a push or pull despite the physical separation.Example:Gravitational Force - Electrical Force - Magnetic Force- Nuclear Forces

Measuring Forces

• Forces are measured in newtons (kg . m/s2).

• Forces are measured using a spring scale.

If the forces on an object are equal and opposite, they are said to be balanced, and the object experiences no change in motion. If they are not equal and opposite, then the forces are unbalanced and the motion of the object changes.

What is meant by unbalanced force?

• Are forces that results in an object’s motion being changed. +velocity changes (object accelerates)

Ffriction

W

Fpull

Fnet

NN

Balanced Force• A force that produces no

change in an object’s motion because it is balanced by an equal, opposite force.

• no change in velocity

FALSE! A net force does not cause motion. A net FALSE! A net force does not cause motion. A net force causes a force causes a changechange in motion, or acceleration. in motion, or acceleration.

ExampleOn a horizontal, frictionless surface, the blocks above are being acted upon by two opposing horizontal forces, as shown. What is the magnitude of the net force acting on the 3kg block?

A. zero B. 2NC. 1.5 ND. 1NE. More information is needed.

Example :Example :

a=3.02 m / s

01

2

9.3620

15tan

/02.35

13.15

smam

F

0

o25

Weight

The weight of an object is the gravitational force that the planet exerts on the object. The weight always acts downward, toward the center of the planet.

SI Unit of Weight: Newton (N)

gmW Weight = Mass x Gravity

– m: mass of the body (units: kg)– g: gravitational acceleration

(9.8m/s2, • As the mass of a body increases, its’

weight increases proportionally

• Weight– the force of gravity on an object

MASSalways the same

(kg)

WEIGHTdepends on gravity

(N)

W = mgW: weight (N)m: mass (kg)g: acceleration due

to gravity (m/s2)

Mass …

Note that mass is involved in the force of gravity! This is a separate property from that of inertia, so we give this property the name gravitational mass.

• is a measure of how much matter there is in something -- usually measured in kilograms in science

• causes an object to have weight in a gravitational field• A measure of the resistance of an object to changes in its

motion due to a force (describes how difficult it is to get an object moving)

• Scalar

Mass = Inertia – an object’s resistance to motion

Would it be more difficult to pull an elephant or a mouse?

• what is the weight of a 2 kg mass?

• W = Fg = mxg = 2 kg x 9.8 m/s2

= 19.6 N

• What is the mass of a 1000 N person?• W = Fg = mxg m = Fg/g = 1000 N / 9.8 m/s2

=102 kg

• A girl weighs 745 N. What is his mass

m = F ÷ g

m = (745 N) ÷ (9.8 m/s2)

m = 76.0 kg

3.3 Newton's first law Every object continues in a state of rest , or of uniform motion in a straight line , unless it is compelled to change that state by forces acting upon it.

An equivalent statement of the first law is that :

An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force.

This, at first, does not seem obvious. Most things on earth tend to slow down and stop. However, when we consider the situation, we see that there are lots of forces tending to slow the objects down such as friction and air resistance.

– “Law of Inertia”

• Inertia– tendency of an object to resist any change in its motion– increases as mass increases

Inertia• Inertia is a term used to measure the ability of an object

to resist a change in its state of motion. • An object with a lot of inertia takes a lot of force to start

or stop; an object with a small amount of inertia requires a small amount of force to start or stop.

• The word “inertia” comes from the Latin word inertus, which can be translated to mean “lazy.”

A railway engine pulls a wagon of mass 10 000 kg along a straight track at a steady speed. The pull force in the couplings between the engine and wagon is 1000 N.b.A. What is the force opposing the motion of the wagon? c.B .If the pull force is increased to 1200 N and the resistance to movement of the wagon remains constant, what would beThe acceleration of the wagon?

Solution a) When the speed is steady, by Newton’s first law, the resultant force must be zero. The pull on the wagon must equal the resistance to motion. So the force resisting motion is 1000 N. b) The resultant force on the wagon is 1200 – 1000 = 200 N By Equation

example

3.5Newton’s Third Law of Motion

• Identifying Newton’s third law pairs– Each force has the same magnitude– Each force acts along the same line but in opposite directions– Each force acts at the same time– Each force acts on a different object– Each force is of the same type

– When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first.

FF BAAB

3rd LawAccording to Newton, whenever objects A and B interact with each other, they exert forces upon each other. When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body.

All forces come in action-reaction pairsEx: feet push backward on floor, the floor pushes forward

on feet

• Problem: How can a horse How can a horse

pull a cart if the cart pull a cart if the cart is pulling back on is pulling back on the horse with an equal but opposite force? the horse with an equal but opposite force?

NO!!!

Aren’t these “balanced forces” resulting in no Aren’t these “balanced forces” resulting in no acceleration?acceleration?

forces are equal and opposite but act on different objects

– they are not “balanced forces”– the movement of the horse

depends on the forces acting on the horse

Explanation:Explanation:

a) Find the acceleration of a 20 kg crate along a horizontal floor when it is pushed with a resultant force of 10 N parallel to the floor. b) How far will the crate move in 5s (starting from rest)?

Solution

b) Distance travelled

Δ X

Δ X

Δ X

example

A 1kg stone fall freely from rest from a bridge. a -What is the force causing it to accelerate?b -What is its speed 4s later?c -How far has it fallen in this time?

Solution

The force causing it to fall is its weight. As it is falling with acceleration due to gravity

b)

c)

example

• Action-Reaction Pairs

The hammer exerts a force on the nail to the right.

The nail exerts an equal but opposite force on the hammer to the left.

• Single isolated force cannot exist• For every action there is an equal and

opposite reaction

• Action-Reaction Pairs The rocket exerts a downward force

on the exhaust gases. The gases exert an equal but opposite

upward force on the rocket.FG

FR

Flying gracefully through the air, birds depend on Newton’s third law of motion. As the birds push down on the air with their wings, the air pushes their wings up and gives them lift.

Other examples of Newton’s Third Law

• The baseball forces the bat to the left (an action); the bat forces the ball to the right (the reaction).

Define the OBJECT (free body)• Newton’s Law uses the forces acting

ON object• N and Fg act on object

• N’ and Fg’ act on other objects

Normal Force Is Not Always Equal to the Weight

3.6 Newton’s Second Law• Newton’s Second Law of Motion• The force FF needed to produce an acceleration aa is

– The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

F = ma

am

F

a

Fm

F: force (N)m: mass (kg)a:acceleration

(m/s2)

1 N = 1 kg ·m/s2

Newton’s Third Law

• Action-Reaction Pairs Both objects accelerate. The amount of acceleration depends on the

mass of the object.

a Fm

Small mass more acceleration Large mass less acceleration

Newton’s Second Law

F = maF: force (N)m: mass (kg)a:acceleration

(m/s2)

1 N = 1 kg ·m/s2

a

m

F

a

Fm

More about F = ma

If you double the mass, you double the force. If you double the acceleration,

you double the force.

What if you double the mass and the acceleration?

(2m)(2a) = 4F

Doubling the mass and the acceleration quadruples the force.

So . . . what if you decrease the mass by half? How much force would the object have now?

Equilibrium• The condition of zero acceleration is called

equilibrium. • In equilibrium, all forces cancel out leaving

zero net force.• Objects that are standing still are in

equilibrium because their acceleration is zero.

• Objects that are moving at constant speed and direction are also in equilibrium.

• A static problem usually means there is no motion.

Example :

a2

example

8 How much force is needed to accelerate a 1,300 kg car at a rate of 1.5 m/s2?

A- 867 Nb 1,950 Nc 8,493 Nd 16,562 N

F = m a or

= 1300Kg x 1.5m/s2

F = 1950 N

•ΣF = m a a =0 ΣF = 0

Balancing Forces (Statics)• There are often situations where a number of forces are acting on something, and the object has no motion – it is STATIC or in EQUILIBRIUM or at rest.• This means the NET FORCE on the object is zero, or in other words the forces balance each other out.

Objects in equilibrium (no net external force) also move at constant velocity

In a two-dimensional problem, we can separate this equation into its two component, in the x and y directions:Σ Fx = 0Σ Fy = 0

unbalanced Forces (Dynamics)•There are other situations where all the forces acting on something do not cancel each other out completely.•This means the NET FORCE on the object is not zero, the object will change its motion and accelerate proportional to the object’s mass.

•ΣF = m a

• The acceleration of an object is equal to the force you apply divided by the mass of the object.

• If you apply more force to an object, it accelerates at a higher rate.

• If an object has more mass it accelerates at a lower rate because mass has inertia.

What force would be required to accelerate a 40 kg mass by 4 m/s2?

F = ma

F = (40 kg)(4 m/s2)

F = 160 N

A 4.0 kg shot-put is thrown with 30 N of force. What is its acceleration?

a = F ÷ m

a = (30 N) ÷ (4.0 kg)

a = 7.5 m/s2

Newton’s Second Law of Motion

Note that this is a vector equation, and should really be worked in component form:

Fx = max

Fy = may .

We can now see that Newton’s First Law of Motion is really just a special case of his Second Law of Motion.

ExampleA railway engine pulls a wagon of mass 10 000 kg along a straight track at a steady speed. The pull force in the couplings between the engine and wagon is 1000 N. a.) What is the force opposing the motion of the wagon? b.) If the pull force is increased to 1200 N and the resistance to movement of the wagon remains constant, what would be the acceleration of the wagon?

Solution a) When the speed is steady, by Newton’s first law, the resultant force must be zero. The pull on the wagon must equal the resistance to motion. So the force resisting motion is 1000 N. b) The resultant force on the wagon is 1200 – 1000 = 200 N By Equation

Example

Examplea) Find the acceleration of a 20 kg crate along a horizontal floor when it is pushed with a resultant force of 10 N parallel to the floor. b) How far will the crate move in 5s (starting from rest)?

Solution

Distance travelled

mx

x

mmFa

amF

vs

25.655.05.0

ta (1/2) t

/5.020/10/

2

20

2

A 1kg stone fall freely from rest from a bridge. a -What is the force causing it to accelerate?b -What is its speed 4s later?c -How far has it fallen in this time?

Solution

The force causing it to fall is its weight. As it is falling with acceleration due to gravity

mx

x

atvx

smv

v

atvv

smF

maF

4.78

)4(8.92

10

2

1

/2.39

48.90

/8.91

2

20

0

Example :

3 . 7 the significance of Newton's laws of motion

3 .7 some examples of Newton's lawsMake a diagram (conceptualize)Categorize: no acceleration (at rest )

accelerating object:

Isolate each object and draw a free body diagram for each object. Draw in all forces that act on the object.Establish a convenient coordinate system.Write Newton’s law for each body and each coordinate component. set of equations.Finalize by checking answers.

0F

maF 0F

Free body diagram

A lift with its load has a mass of 2000 kg. It is supported by a steel cable. Find the tension in the cable when it: a -is at restb - accelerates upwards uniformly at 1m/s2

C - move upwards at a steady speed of 1 m/sd - moves downwards at a steady speed of 1 m/se - accelerates downwards with uniform acceleration of 1 m/s2

Example :

a) When at rest we can use Newton’s first law which says that the resultant force on the lift is zero. Force acting down is the lifts weight, the force acting up is the tension in the cable. These two must be equal and opposite to give a resultant force of zero.

b) As the lift is accelerating upwards so T must exceed the weight mg. So the resultant acceleration force by Newton second law, F = ma, so

c) As in (a), by Newton’s first law, the resultant force on the lift must be zero, so

NT

mamgT

mamgT

amF

21600196002000

NmgT

mgT

F

avelocitytcons

19600

0

0

0 tan

d) As in (c) the tension in the cable will still equal mg since the change in direction of motion does not alter the fact that there is no acceleration.

e ) If the lift accelerates downwards, then mg must exceed the tension T. So the resultant accelerating force is

NmgT

mgT

F

avelocitytcons

19600

0

0

0 tan

NT

mamgT

mamgT

amF

17600200019600

Two hanging objects connected by a light string passing over a frictionless pulleyWhat are the acceleration of the objects and the tension on the string?

Example :

m1 a1 = T - m1 g- m2 a2 = T – m2 g

- m2 a = T – m2 g m⇒ 2 a = -T + m2 g

a = g (m2 – m1) / (m1 + m2) = ( 7-4 ) x9.8 / ( 7+4 ) = 3x 9.8 / 11 =29.4 /11 a = 2 . 6 5 m/s2

m1 a = T - m1 gT= m1a+ m1gT = m1 ( a+ g)T = 4( 2.65 + 9.8 ) N

( m1 = 4 kg m2 = 7 kg )

A body of mass 5kg lies on a smooth horizontal table. It is connected by a light in extensible string, which passes over a smooth pulley at the edge of the table, to another body of mass 3kg which is hanging freely. The system is released from rest. Find the tension in the string and the acceleration.

Example:

N 18.4 3.685s /m 3.68 29.4/883x9.8/ a

3x9.8 3a 5a

3a- 3g - 5a

) 2 ( and ) 1 ( solving

) 2 ( 33

) (1 5

2

1

T

agT

mamgT

aT

amT

amF

Example:

N 5.313.65s / 3.6

14/8.99

995

5a T

99

2

2

11

1

T

ma

a

aga

amF

agT

amgmT

amF

Example:

3-12 FrictionFriction results from relative motion between objects.Frictional forces are forces that resist or oppose motion.

Sliding Friction:• When two solid surfaces slide over each other.• The amount of friction between two surfaces depends on two

factors.1. The kinds of surfaces.2. The force pressing the surfaces together (weight).• Static friction is the frictional force that prevents two surfaces from

sliding past each other.

Friction force (N)Ff = NNormal force (N)

Coefficient of friction

S

S

Friction forceTo first order, we consider two effects on friction1. Normal force of contact between the surfaces2. Types of surfaces in contact (texture of surfaces

The following empirical laws hold true about friction:

- Friction force, f, is proportional to normal force, n.

- s and k: coefficients of static and kinetic friction, respectively

- Direction of frictional force is opposite to direction of relative motion

- Values of s and k depend on nature of surface.

- s and k don’t depend on the area of contact.

- s and k don’t depend on speed.

- s, max is usually a bit larger than k.

- Range from about 0.003 (k for synovial joints in humans) to 1 (s for rubber on concrete). See table 5.2 in book.

nf ss nf kk

Table of friction coefficients

ExampleA 10 N force pushes down on a box that weighs 100 N. As the box is pushed horizontally, the coefficient of sliding friction is 0.25.Determine the force of friction resisting the motion.

Fs = μ s x N

(10+100)x0.25 =27.5

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

kg20sm80.9kg4005.0 2

mgFf kNkk

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

Calculate acceleration

• Three people are pulling on a wagon applying forces of 100 N,150 N, and 200 N.

• The wagon has a mass of 25 kilograms.• Determine the acceleration and the direction the wagon

moves. (2m/s2 ) to the left

Calculate force

• An airplane needs to accelerate at 5 m/sec2 to reach take-off speed before reaching the end of the runway.

• The mass of the airplane is 5,000 kilograms.

• How much force is needed from the engine? (F=25000N)

A 1200 kg car moving at 100 km/h coasts to a stop in 25 seconds. What is the value of the static coefficient of friction?

m = 1200 kgvo = 100km/h = 27.8 m/s

t = 25 s

µ=F/Nwhere F = ma & N=mg

= ma/mg =(m(|v – vo|)/t)/mg

=(27.8m/s/25 s )/ 9.8 m/s2) µ = 0.113

vo

f

From v = vo + at,

a = (v-vo)/t

example

April 20, 2023 75Norah Ali Al-moneef

A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N.

a) Calculate the Force of Frictionb) Calculate the Force Normal

mg

FNFa

Ff

NFF fa 50

NNmg 98)8.9)(10(

April 20, 2023 76Norah Ali Al-moneef

Suppose the same box is now pulled at an angle of 30 degrees above the horizontal.

a) Calculate the Force of Friction

b) Calculate the Force Normal

mg

FN Fa

Ff30

NFF

NFF

axf

aax

3.43

3.4330cos50cos

Fax

Fay

NN

FmgN

mgFN

mgN

ay

ay

73

30sin50)8.9)(10(

!

April 20, 2023 77Norah Ali Al-moneef

What is the weight of an object that is being pulled at a constant velocity by a force of 25 N if the coefficient of sliding

friction between the object and the surface is 0.75 ?

∑Fy = 0N - W =0 N = W

∑Fx = ma ( a = 0 constant velocity ) ∑Fx = 0

Fk= F = 25 NFk= μk NN = Fk / μk

W = N = 25 / 0.75 = 33.3 NW = 33.3 N

Example

April 20, 2023 78Norah Ali Al-moneef

exampleSuppose 35 kg crate was not moving at a constant speed, but rather

accelerating at 0.70 m/s2. Calculate the applied force. The coefficient of kinetic friction is still 0.30.

mg

N

Fa

Ff

April 20, 2023

Example 3- 29

A block is at rest on an inclined plane the coefficient of static fraction is µ .what is the maximum possible angle of inclination θ ( max ) of the surface for which the

block will remain at rest

s

F = w sin θ

s

s

N = w cos θ

tanθ

April 20, 2023 80Norah Ali Al-moneef

Summary of Chapter 3• Newton’s first law: If the net force on an object is zero, it will

remain either at rest or moving in a straight line at constant speed.

• Newton’s second law:

• Newton’s third law:

• Weight is the gravitational force on an object W= m g.

• Free-body diagrams are essential for problem-solving. Do one object at a time, make sure you have all the forces, pick a coordinate system and find the force components, and apply Newton’s second law along each axis.

•Kinetic friction: Fk = μkN

• Static friction: Fs ≤ μsN .

• Fk < Fs μk < μs