Lecture 3

Newton’s laws

Forces

Friction

table

object

Fweight object

Freaction of the table

Fweight object = Freaction table

Faction

Freaction

NEWTON’S THIRD LAW

“For every action there is an equal and

opposite reaction”

Magnitude of force

If one object exerts a force on a

second object, the second object exerts a

force back on the first that is equal in

magnitude and opposite in direction.

System 2 (trolley with patient): External force

provided by the attendant.

System 1 (trolley, patient, and attendant):

External force provided by the floor: reaction to

the force the attendant exerts on the floor.

define the system and

determine the external forces

Define system and external force

Application: Newton’s 2nd Law

Fa

1/ If the attendant exerts a force of 100 N on the

floor, what is the magnitude of acceleration of

the trolley and of the attendant?

Mass of patient = 50 kg

Mass of trolley = 20 kg

Mass of attendant=85 kg

System 1: F=ma

100 N = (Mass System 1)a

Application: system & external force

Mass System 1 = 155 kg

Mass System 2 = 70 kg

a = 100N/155kg = 0.645 ms-2

Fa

3/ Why is the force exerted by the attendant on

the floor bigger than the force he exerts on the

trolley?

2/ What force does the attendant then exert on

the trolley?

System 2: Fattn = (Mass System 2)*a

= (70kg)(0.645ms-2) = 45.2 N

Because he also accelerates.

Application: system & external force

Mass System 2

= 70 kg

F=ma = 85kg.0.645ms-2=54.8N

Mass of attendant = 85 kg

Fa

“co-linear”

F1

F2

F=F1+F2

In general we use vectors or graphical method

to determine the resultant force

“perpendicular”

F1

F2 F

q

Adding Forces

2 2

1 2F F F

Direction

Magnitude

F is the Resultant force

Length of arrow represents magnitude of force

1 2

1

tanF

Fq

2

1

tanF

Fq

*

Often need to split a force into two

perpendicular components.

Projection

perpendicular

to the plane

Projection

parallel to

the plane

F1 = mgCosa

F2 = mgSina

W=mg

a

mgCosa

F1

F2

Projection of a force

w=mg

Cosa = F1/mg

Sina = F2/mg a

a

Example

A gardener pushes a lawnmower with a force

of 500N directed along its handle which

makes an angle of 60º with the ground in order

for it to move at a constant velocity. Determine

(a). The component of this force which is directed

horizontally H, in order to overcome the frictional

forces and maintain a constant velocity, and,

(b). the component of the force which is

directed vertically downwards, V .

Use method of components

600

500N

H

V

H = 500 cos600

= 250N

V =500Sin600

= 433N

TENSION (T) is any force

carried by a flexible string,

rope etc. acts all along the

string

WEIGHT (w=mg) due to gravity

w=mg

NORMAL FORCE (N) normal

to a plane

w = mg

N=w=mg

FRICTION (f ) parallel to a plane

F f

w

T= w

Classes of Forces

Origin:

Plane deforms under

load. The elastic

force attempting to

restore the plane to its original shape is the

“normal force”.

The normal force is the reaction by the plane to

the perpendicular (or normal) component of

force applied. N=mg

W=mg

W=mg a

Normal Force

a mgCosa

EXAMPLE

A constant horizontal force F = 4N to the right,

is applied to a box of mass 2kg resting on a

level, frictionless surface.

What is the acceleration of the box?

Mark the forces acting on the box in the

diagram. There are three forces acting

on this box

1. The force F pulling the box

2. The weight (w) of the box

3. The normal reaction (N) (The third law says

that if the box pushes down on the table, the

table pushes up on the box)

Let the x axis (right) and y axis (up) be the

positive directions.

We assume the positive x axis is the

direction of the force.

Two children pull in opposite directions on a

toy wagon of mass 8.0kg. One exerts a

force of 30N, the other a force of 45N. Both

pull horizontally and friction is negligible.

What is the acceleration of the wagon?

Exercise:

w=mg

N=mg

F1=30N F2=45N System: the toy

wagon;

x positive to the

right

Fext, net =F2 - F1 = 45N - 30N = 15N (right)

a = Fext, net/m = (15/8)ms-2=1.875ms-2

to the right

Biting Force

Force between bottom and top teeth

Bite-force estimation for Tyrannosaurus rex

Nature 382, 706 - 708 (22 August 1996);

6,410 N to 13,400 N

Muscles in the human jaw

can provide a typical biting

force of ≈ 600N

muscle

pivot

force

Biting Force

Average maximum human biting force ≈ 750 N

Biting force varies depending on region of mouth

Molar region 400-850 N

Premolar region 220-440 N

Cuspid region 130-330 N

Incisor region 90-110 N

Energy of a bite is absorbed by

Food

Teeth

Periodontal ligament

bone

Tooth design is such that it can absorb large

static and impact energies

force

Friction is a force that always acts to oppose

the motion of one object sliding on another.

Schematic showing the nature

of forces on a very small (atomic)

scale. Surface roughness results

in frictional forces

Applications of Newton’s Laws

Friction is a contact force

resulting from surface roughness

Surfaces slide over each other,

rough bits catch each other

and impede the motion.

Surfaces not perfectly smooth.

Friction and Normal reaction force are two

types of contact forces

Applications of Newton’s Laws

The normal reaction force, N, is described

by Newton’s third law.

The third law says that for every force there

is an equal but opposite force.

If one object is sliding on another,

the frictional force, F, always acts in the

direction opposed to the motion,

the normal force always acts in a direction

perpendicular to the surface.

A

N

W

F

Applications of Newton’s Laws

Imagine an object lying on a frictional surface

and a very small horizontal force, A, acts on it.

The force diagram is drawn as follows.

The forces acting on the object are its

weight w, the normal force N, the frictional

force F and the applied force A.

F = µN

where the constant of proportionality is known

as the Coefficient of friction µ.

Applications of Newton’s Laws

To a good approximation the frictional force (F)

of an object on a surface is proportional to the

normal force of the surface on the object

The most familiar thing about friction is that the

heavier an object the greater the friction. It

is more difficult to slide a full box than an empty

one.

Wf=mfg We=meg

N=mfg N=meg

Large force required to

move heavy object

Small force required to

move light object

F N

It is proportional to the normal force:

k is the coefficient of kinetic friction

(between moving substances)

Const. speed: (A=F)

Acceleration: (A>F)

Deceleration: (A<F)

F= kN

w=mg

N=mg

F= kN A

Kinetic Friction

Kinetic friction is the force that opposes

motion due to physical contact between

substances, and is parallel to the contact

surfaces.

It behaves like the normal force:

F=A up to a maximum of sN

s is the coefficient of static friction

(between immobile substances)

A F = A

Generally k < s

If A< sN: static friction prevents motion

If A> sN: static friction isn’t strong enough

motion kinetic friction.

Static Friction

Static friction is the force that prevents motion

due to physical contact between substances, and

which is parallel to the contact surface.

Static Kinetic

A

F = kN

Fs,max

F

Applications of Newton’s Laws

Friction– Graphical representation

Typical values for coefficient of static friction s

of some substances

Rubber on concrete s = 1

Synovial joints s ≈0.016

Values of coefficients of friction s and k

depend on the nature of the surfaces

Typically 0<k<s<1

= sN

EXAMPLE

You press a book flat against a vertical wall.

In what direction is the frictional force exerted

by the wall on the book?

(a)downward

(b) upward

(c) into the wall

(d) out of the wall

EXAMPLE

A crate is sitting in the centre of a flatbed truck.

The truck accelerates eastwards, and the crate

moves with it (not sliding on the bed of the

truck). In what direction is the frictional force

exerted by the bed of the truck on the crate?

(a)to the west

(b) to the east

(c) there is no frictional force

because the crate does

not slid

H N

w

F

Abrasion or wear due to surface roughness

Friction

Dental restorations and appliances should

have smooth surfaces

Abrasive resistance of material depends on

Hardness

Surface roughness

Excessive wear of tooth enamel by opposing

ceramic crown caused by

High biting force

Rough ceramic surface

Example

Remedy

Adjust occlusion-

•broader contact areas---reduces stresses

Polish ceramic to reduce abrading surface

Dental considerations Abrasion

Lubrication in the mouth is provided by saliva

Friction

Saliva greatly reduces the frictional force and

helps prevent excessive wear of teeth

Reduced friction→ reduces wear

ensures good contact between

opposite teeth is not impaired and surfaces

used for chewing are maintained.

Saliva, also protects the tongue and other soft

tissues inside the mouth

Saliva reduces friction force

by up to a factor of 20.

Lubricating properties of saliva are maintained

over the range of jaw biting forces (0→600N)

muscle

pivot

force

A crate slides at constant velocity down a plane

inclined at an angle a to the horizontal. If the

coefficient of kinetic friction between the surfaces

of the crate and the plane is 0.6,

determine the angle a.

W=mg a

a mgCosa

k kF N mgCos a

kF N mgSin a

kmgCos mgSin a a k

mgSinTan

mgCos

a a

a

1 1 0tan tan 0.6 31ka

EXAMPLE

Constant velocity

EXAMPLE

You need to move an object across a

horizontal surface by pushing it. Its mass is

50kg and the coefficients of friction are µs=0.5

and µk=0.4.

How much force A do you have apply

horizontally to a) get it moving?

b) To keep it moving?

The object will start to move when the applied

force A becomes very slightly greater than the

maximum frictional force Fmax=µsN. For our

purposes we will say this happens when A=Fmax

When the object is stationary the friction that

impedes motion is determined by µs. The

forces on the box are outlined below

w=mg

N = mg

A Fmax = sN