Teach A Level
Maths
Bodies
in Equilibrium
Volume 4: Mechanics 1
Bodies in Equilibrium
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We have used 2 methods for solving problems with 3
forces in equilibrium:
 using a triangle of forces,
 finding the components of the forces by
resolving.
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.
Since the method of resolving forces can be applied
to any of these problems, we’ll use it in the following
examples.
e.g.1. The diagram shows a particle of weight of
2 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.
Find the tension in the string
and the value of P.
40
T
P
2
Solution:
The first step is to show the
forces acting on the particle.
Equilibrium. Find T and P.
Resolving:
40
Tcos 40- 2 = 0
T
P
Decide
with
your
We
never
 should
Tcos
40 miss
=partner
2 out what
the
of T so
is,leads
without
this component
stage as doing
T = cos2 40
diagram.
todrawing
errors ainseparate
later problems.
= 2·61 ( 3 s.f. )
2
across the angle
The tension is 2·61 newtons.
means cos
P - Tsin 40 = 0

P = Tsin 40
P = 2·61sin 40


P = 1·68 newtons ( 3 s.f. )
e.g.2. A particle of weight 10 newtons rests on a
smooth plane inclined at 35 to the horizontal.
35
The particle is supported, in equilibrium, by a light
inextensible string parallel to the slope.
Find the magnitudes of the tension in the string and
the contact force between the particle and plane.
R
T
Equilibrium
Smooth plane
Weight 10 newtons.
35
10
Find T and R.
Solution:
Agree
with your partner which directions you would
use to
thesmooth
forces.
Theresolve
plane is
so there is no friction.
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.
R
T
35
55
35
10
35
To find the components of the weight, we need an angle.
Using this right angled triangle . . . the 3rd angle is
90 - 35 = 55
and using the right angle between the slope and the
perpendicular,
90 - 55 = 35
R
T
35
55
35
10
35
We can just use 35 ( the angle of the
slope ) without needing to subtract.
Solution:
Find T and R.
R
T
35
35
35 10cos 35
10
10
10sin 35
Resolving:
T - 10sin 35 = 0
T = 10sin 35

= 5·74 newtons ( 3 s.f. )
R - 10 cos 35 = 0

R = 10 cos 35
= 8·19 newtons ( 3 s.f. )
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:
Component of W
down the slope :
W sin a
a
W
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.
Solution:
P
14
R
25
10
Constant velocity  equilibrium
Can you see what the
horizontal component of the
pushing force is, without
drawing a separate diagram ?
Ans: Pcos 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.
Solution:
R
P
Resolving:
14
Pcos 25 - 14 = 0
25

10
14
cos 25
P = 15·4 ( 3 s.f. )
P =
R - 10 - Psin 25 = 0
R = 10 + 15·4sin 25

= 16·5 newtons ( 3 s.f. )
SUMMARY
 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 a where a is the angle of the slope.
EXERCISE
1. 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.
P
42
2
Find the value of P
and the magnitude of
the normal reaction.
EXERCISE
Solution:
R
42
Resolving:
P
2
P - 2sin 42 = 0

P = 2sin 42
= 1·34 ( 3 s.f. )
R - 2cos 42 = 0

R = 2cos 42
= 1·49 newtons ( 3 s.f. )
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.
A
B
60
C
30
If the tension in BC is 1 newton, find the value of
W and the tension in AC.
EXERCISE
Solution:
A
B
T
1
60
30
C
W
Method 1: Resolving
horizontally and vertically
1 cos 30 - T cos 60 = 0
 -T cos 60 = -cos 30
T = cos 30

cos 60

T = 1·73
( 3 s.f. )
1 sin 30 + T sin 60 - W = 0
 1 sin 30 + 1·73 sin 60 = W

W =2
The weight is 2 newtons and the tension
in AC is 1·73 newtons.
EXERCISE
Solution:
A
B
T
1
60
60
30
C
W
Method 2: Resolving parallel
and perpendicular to CB
1 - W cos 60 = 0

1 = Wcos 60

1
=W
cos 60

W =2
T - W sin 60 = 0

T = 2sin60

W = 1·73 ( 3 s.f. )
The weight is 2 newtons and the tension
in AC is 1·73 newtons.
The following page contains the summary in a form
suitable for photocopying.
TEACH A LEVEL MATHS – MECHANICS 1
BODIES IN EQUILIBRIUM
Summary

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,
angle of the slope.
W,
down a slope is
W sin a ,
where
a
is the