FORCES

Intuitively, a force is like

What is a force?

a push or a pullwhich produces or tends to produce motion

Forces experienced in Daily Life

• Weight• Normal reaction• Friction• Viscous force• Tension• Upthrust• Lift• Electrical force• Magnetic force

Weight

W

• weight is not the same as mass; it is a force• it is the gravitational force exerted by the Earth• it passes through the centre of gravity of the body

Normal reaction

N

• two bodies in contact with each other• perpendicular to the surface of contact

Friction

• friction is exerted two surfaces slide across one another• direction is along the surface of contact

Cause of friction

F

• hollows and humps all over the surface• actual contact area only a fraction 1/10000 of total area• extreme high pressure at contact points causes welding

of surfaces• forces are needed to overcome these adhesive forces

when trying to slide over the surface

movement

• It is harder to move a stationary object than to move the object while it is moving

• Static friction is the friction exerted by the ground in order to prevent the object from moving

• Kinetic friction is the friction exerted by the ground to oppose the motion of the object while it is moving

Static and kinetic friction

• Static friction is not constant; it varies in magnitude

Limiting friction

1 N1 N

• If P is 1 N, F will also be 1 N to prevent object moving

• Suppose a force P is applied trying to move the object

• If P increased to 2 N, F also increased to 2 N• But there is limit to how much F can increase to• Maximum possible static friction is called limiting friction• P must exceed limiting friction in order to move object

2 N2 NP

F

In Fig 1.1, an object was moving to the right on a rough surface. In Fig 1.2, an object rests in equilibrium on a rough slope. In both cases, draw the friction force acting on each object.

Example 1

friction

friction

Fig 1.1 Fig 1.2

Viscous force

• When body moves in fluid, it experiences resistance

• such resistance is known as viscous force

• examples: air resistance and water resistance• viscous force depends on the speed of the body• the greater the speed, the greater the viscous force

Terminal velocity

release

W gathering speed

W

F

v

gathering more speed

W

F

V

finally reaches constant terminal velocity

W

F

VT

Tension

• Tension is exerted by a stretched rope, string or spring.

• When a body is attached to a string, the tension in the taut string would tend to pull the body.

Hooke’s Law

F

F (l - lo)

=> F e

=> F = ke

where k is force constant

(elastic constant, spring constant or stiffness F constant)

F

xe

Strain energy in a Deformed wire

In general, work done by a force F in extending a wire from x1 to x2 is the area under the force-extension graph.

=>Work done in extension or strain energy stored in wire, W = ½ Fe = ½ ke2

Assume that Hooke’s Law is obeyed. =>For a force-extension graph, it will be a straight line.

Example 2 A vertical wire suspended from one end is

stretched by attaching weight of 20 N to the lower end. If the extension is 1 x 10-3 m, what is

(a) the force constant;(b) the energy stored in the wire;(c) the gravitational potential energy loss by the weight in dropping a distance of

1 x 10-3 m?

Assuming Hooke’s law is obeyed,(a) F = ke

k = F/e = 20/(1x10-3) Nm-1

= 2 x 104 Nm-1

(b) energy stored in wire, W = ½ Fe = ½ (20)(1x10 -3)

= 1 x10-2 J

43

10210

20x

e

F== −

(c)Gravitational potential energy lost by weight

= mgh = 2 x 10-2 J

By conservation of energy, P.E. lost = Energy stored in wire

+ heat dissipated when weight at end of wire comes to rest after

vibrating.

Upthrust

• Upthrust is an upward push on a body when it is immersed in a fluid (gas or liquid)

• Upthrust is exerted by the fluid

• Upthrust is due to pressure difference of fluid at the top and bottom of immersed portion of the body

upthrust

Consider an object partially immersed in a fluid of density . The area of the top surface of the object is A and the immersed depth is h.

Example 3

h

h

(a) What is the pressure difference across the immersed portion of the object?

(b) Hence write down the expression for the upthrust acting on the object.

(c) What is the volume of fluid displaced by the object?

(d) Hence write down the expression for the weight of fluid displaced.

(e) Comment on your answers to (b) and (d).

h g

h g A

h A

h A g

They are the same.

Example 3 shows that

Upthrust = weight of fluid displaced

This is actually the Achimedes’ Principle

Archimedes’ Principle states that

the upthrust on a body in a fluid

is equal and opposite to the

weight of the fluid displaced by the body.

Archimedes’ Principle states that

the upthrust on a body in a fluid

is equal and opposite to the

weight of the fluid displaced by the body.

Lift

The answer is the upward lift force exerted on their wings when in motion

What helps birds and aeroplanes maintain its flight?

Electric force

like charges repel

Electric force is exerted between two electric charges

+ -+ +

unlike charges attract

Magnetic force

like poles repel

Magnetic force is exerted between two magnetic materials or between electric currents

unlike poles attract

N N N S

Forces experienced in Daily Life

• Weight• Normal reaction• Friction• Viscous force• Tension• Upthrust• Lift• Electrical force• Magnetic force

Different forces

weight

normal reaction

weight

upthrust

weight

normal reaction

Different forces

weight

lift

weight

tension

Different forces

forwardforce

air resistance

frictionweight

normal reaction

normal reaction

speed

How did this ‘forward force’ come about?

air resistance

thrust

Different forces

weight

lift

How did this ‘thrust’ come about?

Who exerts on who

A force is always exerted by some body on some other body.

Friction exerted by ground on tires

Gases expelled by rocket

What makes a car move? What makes a rocket fly?

Test Yourself. Identify the forces

weight

weight

weight

normal reaction

FGM

Fundamental types of force

When scientists examined all the forces,

they found that many of them are similar in nature.

Scientists have identified 4 fundamental types of force:

• gravitational force

• electromagnetic force

• nuclear force

• weak force

All forces in our daily life can be classified into one of the fundamental types. In the following table, identify the nature of each force:

Force Nature

Weight (W) of an object Gravitational

Attraction between two oppositely -charged bodies Electromagnetic

Attraction by a magnet on a piece of iron Electromagnetic

Tension (T) in a string pulling an object Electromagnetic

Pushing a person wi th your hands Electromagnetic

Normal reaction (N) by the table on an object resting on it Electromagnetic

Friction (F) experienced by an object moving on a rough surface Electromagnetic

gravitational

electromagnetic

electromagnetic

electromagnetic

electromagnetic

electromagnetic

electromagnetic

2 Addition of VectorsParallelogram Rule

A

B

RA

B

Triangle Rule

A

B

A

B

R

Finding resultant force

The magnitude of resultant force can be found by

• drawing vector diagram to scale

• calculation (pythagoras theorem, cosine rule, etc)

• resolution

Two forces are given below:

Example 4

Find the magnitude of the resultant force.

5 N4 N

30º

70º

Scale used is 1 cm : 1 N

Method 1 Drawing vector diagram to scale

From the vector diagram,magnitude of resultant R is 5.8 N

5 N(5 cm)

4 N(4 cm)

R(5.8 cm)

What is missing in the answer?

• Using Cosine rule: x2 = 52 + 42-2 (4) (5) cos800

=>x =5.84 N• Using Sine rule:

5 N

304 N

7080

x

€

α

€

sin800

5.836=

sinα

4=> α = 42.40

Method 2: By calculation

Method 3 By resolving vectors

Magnitude of resultant R is given by

R2 = (5.70)2 + (1.26)2 R = 5.8 N

Rx = 5 cos 30° + 4 cos 70° = 5.70 N

R

30º

70º5 N

4 N5 cos 30°

4 cos 70°

5 sin 30°4 sin 70°

Ry = 5 sin 30° - 4 sin 70° = -1.26 N

Rx 5.70 N

Ry

1.26 N

Two horizontal forces act at a point to produce a resultant force of magnitude 40 N in the eastward direction. Given that one of the forces is in the northward direction and has a magnitude of 30 N, find the magnitude and direction of the second force.

Example 5

Magnitude of second force F = 302 + 402 = 50 NAngle = tan-1 (40/30) = 53°,

direction of F is 53° east of south (or bearing 127°)

30 N

40 N

FN

E

Centre of Gravity

• The centre of gravity of a body is the single point at which the entire weight of the body can be considered to act.

Centre of gravity and Free body diagram

Free body diagram (Important)

• is a diagram showing all the forces acting on a particular object

• is an important tool for solving problems

An object A of weight w rests on top of another object B of weight W placed on the ground, as shown.

Example 6

Draw separate free body diagrams showing forces acting on

(a) A only

(b) B only, and

(c) A and B together.

N1 = normal reaction exerted by B on A

N2 = normal reaction exerted by A on B

N3 = normal reaction exerted by ground on B

Answer

N1

w

N2

WN3

W+wN3

N1 is numerically equal to N2 (action / reaction pair)

Common forces in free body diagramsForce Applicable when Direction of force Example

weightW Object has a mass

verticallydownwards

through centre ofgravity

tensionT

Object is attachedto a string

along the stringpulling towardsthe centre of the

string

normalreaction

N

Object is in contactwith a surface

normal to andaway from the

contact surface

frictionF

Object tries tomove across arough surface

along the roughsurface

N

F

Total force R exerted by surface on moving

object consists of two components

- normal reaction N

- frictional force F

R is also known as the contact force

Force exerted by surface (only)

Motion

R

An object of weight W, resting on a rough surface, is connected to a suspended object of weight w by a string over a smooth pulley. Draw and label the forces acting on each object.

A Non lecture Note Example

W

Normalreaction

tension

friction

tension

w

Turning effect of a forceConsider a water wheel which is free to rotate about its centre.

Water flowing to the right exerts force on lower blades.

This force causes the wheel to rotate about its centre.

We say that the force has a turning effect.

Turning effect of a force is also known as its moment.

Amount of moment depends on force and distance away.

Moment of a force

The moment of a force about an axis is defined as the product of the force and the perpendicular distance between the axis and the line of action of the forces.

The moment of a force is also known as the torque.

Al

F

Moments about A= F l (clockwise)

Al

F

300

Moment of a forceExample: Not in lecture notes

Moments about A= F l sin 30 0 (clockwise)= 1/2 F l

Al

F

300

Mtd 2

lsin 30

0

F

Al

300

Mtd 1

l sin 30 0

F sin 300

F cos 300

Find the moments of the following forces about point A.

Example 7

Moment of 30 N about A =

Moment of 40 N about A =

Moment of 20 N about A =

4 m

5 m

3 m

30 N 40 N

20 N40

60

A

30 × 4 = 120 N m (anticlockwise)

40 × 3 sin 60= 104 N m (clockwise)

= 77 N m (clockwise)

20 × 5 cos 40

Torque of a couple

Couple = pair of equal and opposite forces whose lines of action do not coincide

dx

A

F

F

Taking moment about any arbitrary point, say A,

total anticlockwise moment = F × (d+x) - F × x

= F d

The torque of a couple is equal to the product of one of its forces and the perpendicular distance between the lines of action of the two forces.

Example 8Calculate the torque acting on the rod 2.0 m long in Figs 9.1 and 9.2.

Fig. 9.1:

Torque = F d = 10 × 2.0 = 20 N m

30º2.0 m

10 N

10 N

Fig. 9.110 N

10 N

Fig. 9.2

Fig. 9.2:

Perpendicular distance between 10 N forces = 2.0 cos 30º

Torque = 10 × 2.0 cos 30º = 17 N m

2.0 cos 30º

5 System in equilibrium

A system is in equilibrium when there isno resultant force and no resultant torque.

First condition: Resultant force is zero• forces would form a closed triangle or polygon• sum of components resolved in any direction is zero• system is said to be in translational equilibrium• is either at rest or moving with constant velocity• has constant linear momentum

Second condition: Resultant torque is zero

• total clockwise moment = total anticlockwise moment• if there are only 3 forces,

they would intersect at a common point• system is said to be in rotational equilibrium• is at rest or rotating with constant angular velocity• has constant angular momentum

Example 9A horizontal force F is exerted on the pendulum of weight W, causing the pendulum to be suspended at an angle to the vertical, as shown. Find F in terms of W and .

From the vector diagram,

tan = F / W F = W tan

F

F

W TT

W

Example 10A body of weight 200 N is suspended by two cords, A and B, as shown in the diagram. Find the tension in each cord.

From the vector diagram,

tan 60º = W / TA TA = W / tan 60º = 200 / tan 60º = 115 N

sin 60º = W / TB TB = W / sin 60º = 200 / sin 60º = 231 N

TA

TA

WTBTB

W

60º

60º cord B

cord A

Example 11A uniform rod is supported with the fulcrum exactly at the centre of the rod. Two masses were placed on the rod and the system is in equilibrium. Find m.

Taking moments about the fulcrum,

clockwise moments = anticlockwise moments

m g × 0.30 = 2.0 × g × 0.45

m = 3.0 kg

m2.0 kg

m g2.0 × g

0.45 m 0.30 mN

W

Example 12A uniform rod XY of weight 20 N is freely hinged to a wall at X. It is held horizontal by a string attached at Y at an angle of 20º to the rod, as shown.

Find

(a) the tension in the string,

(b) the magnitude of the force exerted by the hinge.

string

X Y20º

Example 12

(a) Let be the length of the rod and

T be the tension in the string

Taking moments about X,

anticlockwise moments = clockwise moments

T sin 20º × = 20 × ( / 2)

T = 29 N

20 N

T 20 N

T29 N

R

R

X Y20º 70º

(continued)

string

(b) Let R be the force exerted by the hinge

From the vector diagram, using cosine rule,

R2 = 202 + 292 - 2(20)(29) cos 70º

R = 29 N

Example 13A heavy uniform beam of length is supported by two vertical cords as shown.

weight

(7/10) (3/10)

cord Bcord A

TA TB

Find the ratiotension in cord B

tension in cord ATaking moments about the centre of gravity,

clockwise moments = anticlockwise moments

TA × (2/10) = TB × (5/10)

ratio TA / TB = 5 / 2

The End