2.9 Equilibrium

8/8/2019 2.9 Equilibrium

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2.9 FORCES IN EQUILIBRIUM

Learning Outcome

Describe the situationswhere forces are in

equilibrium

State what a resultant

force is

Add two forces to

determine the

resultant force.



Newton¶s Law of Motion

First Law:

In a sate of equilibrium , a stationaryobject will continue to remain stationary

and an object in motion will continue to

move at a uniform velocity




Second Law:

The rate of change of momentum of amoving body is proportional to and in

the same direction as the net force

acting on it.




Third Law:

To every action there is an equal but

opposite reaction.



EQUILIBRIUM

R

Object resting on the

table is in equilibrium.

The forces that act on

them are balanced.

The forces cancel

each other so that the

resultant force is zeroWeight = Normal reaction

W = R

Stationary object



Force + Weight = Normal

reaction

R = 26 N

F + W = R

F = R ± W

= 26 ± 15

= 11 N

Equilibrium with pushing force

F



Equilibrium with tension in a rope

W = R + T

W = T

R

T=11N

W = 15N

R = W ± T

= 15-11

= 4N



Pushing

force, F

Frictional

force, f

Weight, W

Normal

reaction , R

F = f

Object Moving with

constant velocity is inequilibrium.

The forces that act on

them are balanced.

The forces cancel

each other so that the

resultant force is zero

EQUILIBRIUM

R = W

Objects in motion with

constant velocity





Box resting on an

inclined plane

Weight of box

Normal

reactionFriction

The box is in

equilibrium.

How do

these forces

cancel each

other ?

EQUILIBRIUM



ADDITION OF TWO FORCESParallelogram method

1. The two forces are drawn from

the same point.

2. Complete a parallelogram using r uler,

protractor and compasses.

3. Resultant force is represented by thediagonal from the common point

80o

1 cm : 20 N

80o



ADDITION OF TWO FORCESTriangle method

1. The two forces are drawn after one

another

2. Complete a triangle.

3. Resultant force is represented by

the third line jointing the startingpoint to the ending point.

80o

1 cm : 20 N

80o



Refer information given in workedexample (pg 55)

Using a scale of 1 cm to represent 10 kN, draw a

parallelogram of forces .

Determine(a) the magnitude of the resultant force on

the oil rig.

(b) the angle the resultant force made with the100 kN force

Repeat using the triangle method.

Activity



Home work

Masterypractice 2.9 pg 57



2.9 FORCES IN EQUILIBRIUMLearning Outcome

Resolve a force into the

effective componentforces.

Solve problems involving

forces in equilibrium.



RESOLUTION OF FORCES

To break up or resolve a single force into two

components. (Usually perpendicular)

One force

F

U

Two effective components

F1

F2F1

F= sin U

F2

F= cos U

F1 = F sin U F2 = F cos U

F

U



RESOLVE WEIGH

T ON INCLINED

P

LA

NE

Weight of box50o

W1

W2

W

50o

Component of

weight along plane ,

W1

W= sin 50o

W1 = W sin 50o

Component of weightperpendicular to plane ,

W2

W= cos 50o

W2 = W cos 50o



ANALYSING FORCES IN EQUILIBRIUM

The addition

and resolutionof forces can

help us analyse

forces in

equilibrium.Weight, W

Normal

reaction,R Friction,F

U

The box is at rest



Weight, W

Normal


U

U

W1

W2

The box is at rest

Resultant force=0

F = W1

= W sin U

R = W2

= W cos U

Resolve forces along & perpendicular

to plane



By drawing a triangle of forces

The box is at rest

Resultant force=0

Alternative

Weight, W

Normal


U

UW

R

F

U



Worked Example 1pg 56

Fx

120

120 N

40o

Fy

Fx

= cos 40o

Fx = 120 cos 40o

= 91.9 N

Fy

120 = sin 40o

Fx = 120 sin 40o

= 77.1 N

Fx pushes the roller forward.

Fy helps to press the roller

onto the ground.



Worked Example 2

A lorry is moving with a

constant velocity up a

slope. The frictional force

between lorry and slopeis 500 N . what is the

thrust exerted by the

engine?

Solution

20000 N

R

F

T

Free body diagram

Component of weight

down the slope = W sin U

The lorry is in equilibrium, so

T = F + W sin U

= 500 + 20000 sin 10o

= 500 + 3473 N

= 3973 N



Worked Example 3

The ring is at rest. The

weight of the engine is

3150 N. Draw a vector

diagram to scale todetermine T1 and T2

Scale : 1 cm : 500 N

6.3 cm

80o

10o

A

B

C

T1= AC

T2= BC



2.9 FORCES IN EQUILIBRIUMLearning Outcome

Resolve a force into the

effective componentforces.

Solve problems involving

forces in equilibrium.



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2.9 Equilibrium