Teach A Teach A Level MathsLevel Maths

BodiesBodies

in in EquilibriumEquilibrium

"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

© Christine Crisp

Volume 4: Mechanics 1Volume 4: Mechanics 1Bodies in EquilibriumBodies in Equilibrium

Since the method of resolving forces can be applied to any of these problems, we’ll use it in the following examples.

using a triangle of forces,

We have used 2 methods for solving problems with 3 forces in equilibrium:

However, if we have more than 3 forces, or the forces are not in equilibrium, we cannot use a triangle of forces, so we then resolve the forces.

finding the components of the forces by resolving.

40

P

e.g.1. The diagram shows a particle of weight of2 newtons that is tied to a light inextensible string attached to a wall.

The particle is held in equilibrium, as shown, by a horizontal force of magnitude P newtons.

2

Find the tension in the string and the value of P.Solution:The first step is to show the forces acting on the particle.

T

P

2

T

Equilibrium.

Find T and P.

4040 Tcos 40 20

Tcos 40 2

Resolving:

We should never miss out this stage as doing so leads to errors in later problems.

T 2cos 40

2·61 ( 3 s.f. )

P0

P Tsin 40

P 2·61sin 40 P 1·68 newtons ( 3 s.f. )

Tsin 40

The tension is 2·61 newtons.

Decide with your partner what the component of T is, without drawing a separate

diagram.across the angle means cos

35

Find the magnitudes of the tension in the string and the contact force between the particle and plane.

e.g.2. A particle of weight 10 newtons rests on a smooth plane inclined at 35 to the horizontal.

The particle is supported, in equilibrium, by a light inextensible string parallel to the slope.

35

Weight 10 newtons.

Solution:

10

TR EquilibriumSmooth plane

The plane is smooth so there is no friction.

Agree with your partner which directions you would use to resolve the forces.

Ans: There are 2 reasons for preferring parallel and perpendicular to the plane:• We will only need to resolve 1 force, the

weight.• T will appear in one equation and R in the other, so we won’t have to solve simultaneous equations.

Find T and R.

and using the right angle between the slope and the perpendicular,

3555

35 10

TR

35

To find the components of the weight, we need an angle.

90 55 35

Using this right angled triangle . . .

the 3rd angle is 90 35 55

3555

35 10

TR

35

We can just use 35 ( the angle of the slope ) without needing to subtract.

35 10

TR

10 cos 35R R 10 cos 35

T

8·19 newtons ( 3 s.f. )

Resolving:

1035

10sin 35

10cos 35

0

Solution:

Find T and R.

5·74 newtons ( 3 s.f. )

T 10sin 3510sin 35 0

35

Tip: Lots of problems you will meet in M1 involve objects on slopes so it is well worth remembering the component of the weight down the slope:

W

Component of W down the

slope :W sin

P

14 25

10

R

Solution:

Can you see what the horizontal component of the pushing force is, without drawing a separate diagram ?

Ans: Pcos 25

Constant velocity equilibrium

e.g.3 A box of weight 10 newtons is being pushed at a constant speed in a straight line across a horizontal surface by a force of magnitude P newtons at 25 to the surface. There is a constant resisting force of magnitude 14 newtons. Find P and the magnitude of the normal reaction.

P

14 25

e.g.3 A box of weight 10 newtons is being pushed at a constant speed in a straight line across a horizontal surface by a force of magnitude P newtons at 25 to the surface. There is a constant resisting force of magnitude 14 newtons. Find P and the magnitude of the normal reaction.Solutio

n:Resolving:

Pcos 25 0

14

P

14cos 25P15·4 ( 3

s.f. )R

R 16·5 newtons ( 3

s.f. )

Psin 25 10 010 15·4sin 25

10

R

A body in equilibrium is either at rest or moving with a constant velocity. To solve equilibrium problems we resolve the forces and form equation(s) using X 0 and/or Y 0. For bodies on a slope we usually resolve parallel and perpendicular to the slope.

The component of the weight, W, down a slope is W sin where is the angle of the slope.

SUMMARY

EXERCISE1. A particle of weight 2 newtons rests on a

smooth plane inclined at 42 to the horizontal. It is supported by a force of magnitude P newtons acting parallel to the slope.

42 2

PFind the value of P and the magnitude of the normal reaction.

42 2

PR

Solution:

2cos 42R R 2cos 42

P

1·49 newtons ( 3 s.f. )

P 2sin 42 1·34 ( 3 s.f. )

Resolving: 2sin 42 0

0

EXERCISE

2. A particle of weight W is held in equilibrium by two inextensible strings AC and BC at 60 and 30 to the horizontal as shown in the diagram.If the tension in BC is 1 newton, find the value of W and the tension in AC.

60 30

A

C

B

EXERCISE

60 30

AB

C

Solution:

W

T1

Method 1: Resolving horizontally and vertically

1 cos 30 0

T cos 60

01 sin 30 T sin 60W

cos 30T cos 60

T

cos 30cos 60

T 1·73 ( 3 s.f. )

1 sin 30 1·73 sin

60

W 2

W

The weight is 2 newtons and the tension in AC is 1·73 newtons.

EXERCISE

60 30

AB

C

W

T1

Solution:

10

W cos 60

0T

W sin

60 T2 sin

60

Wcos

601

W

W 2

W 1·73 ( 3 s.f. )

The weight is 2 newtons and the tension in AC is 1·73 newtons.

EXERCISE

Method 2: Resolving parallel and perpendicular to CB

60 1cos 60

The following page contains the summary in a form suitable for photocopying.

Summary

BODIES IN EQUILIBRIUM

TEACH A LEVEL MATHS – MECHANICS 1

A body in equilibrium is either at rest or moving with a constant velocity.

To solve equilibrium problems we resolve the forces and form equation(s) using X 0 and/or Y 0.

For bodies on a slope we usually resolve parallel and perpendicular to the slope.

The component of the weight, W, down a slope is W sin , where

is the angle of the slope.