Teach A Level
Maths
Moment of
a Force
Volume 4: Mechanics 1
Moment of a Force
We can balance a ruler on an outstretched finger if
the middle of the ruler is on our finger.
We can model the ruler as
a thin, uniform rod; its
mass is evenly distributed.
A small movement of the
ruler results in it starting
to tip ( and then it will
probably slide ).
It’s impossible to balance the ruler if we place our
finger away from the centre.
When we draw a force diagram, because the body is
uniform, we can show the weight acting at the centre.
Drawing the forces on
the ruler, this is
but this is not.
possible . . .
R
W
R
W
However the forces are the same in the 2 situations.
In the 2nd situation, the ruler is not in equilibrium
because there is a turning effect.
The turning effect of a force is called the moment
of the force.
The moment of the weight in this diagram has an
anti-clockwise moment about A.
R
A
W
By also supporting the ruler to the left of W, we
can again make it rest in equilibrium.
R2
B
R2
W
A
The normal reaction at B has a clockwise moment
about A.
R2
B
R2
W
A
The force at B is half the size of the weight, yet its
moment balances that of the weight.
Distance as well as magnitude effects the size of the
moment of a force.
The moment of a force, F, about a point A is defined
as
Fd
where d is the perpendicular distance from A to the
line of action of F.
e.g.
x
P
d
x
W
x
A
An anti-clockwise moment is taken as positive, so
the
of W about
Themoment
line of action
of W A is . . . +Wd
The moment of W about P is . . .
-Wx.
The moment of a force, F, about a point A is defined
as
Fd
where d is the perpendicular distance from A to the
line of action of F.
If supports are placed at
A and P, there will be
reaction forces at these
points.
R
N
A
P
W
What
do you
notice of
about
the moment
ofPRso
about
The line
of action
R passes
through
the P.
Which
forcefrom
has P
the
moment
aboutisAzero.
?
distance
tosame
the line
of action
Therefore the moment of R about P is zero.
Similarly, the moment of N about A is zero.
Watch out for perpendicular distances.
We need this
length.
A
x
This isn’t perpendicular
to the line of action.
Watch out for perpendicular distances.
pull
fix
A
x
Tip: If you are not sure whether the moment of a force
is clockwise or anti-clockwise, imagine the perpendicular
line to be fixed at the point you are taking moments
about . . . and pull the other end of the perpendicular
in the direction of the force. In your mind you will
see it start to turn.
When we are solving problems involving a body such as
a ruler, plank or ladder, we model the body as a rod.
The rod is an example of a rigid body, one which
cannot sag or bend.
For a rigid body to rest in equilibrium,
• the resultant force is zero, and
• the sum of the moments about any point is zero.
e.g.1. A plank of mass 18 kg and length 4 m rests in
equilibrium on smooth supports at A and B.
A is at one end of the plank and B is 1 m from the
other end.
A
B
1
Modelling the plank as a uniform rod, find the
magnitudes of the reactions on the plank at A and
B, giving the answers in terms of g.
mass: 18 kg
length: 4 m
R
A
N
2
B
1
18g
The rod is uniform, so the weight acts at the centre.
Solution:
R + N - 18g = 0 - - - - - - (1)
Resolving:
N(3) - 18g(2) = 0 - - - - - - (2)
36g
From “anti-clockwise
(2),
N =
This means
3
moments about
A”. N = 12g newtons
Subs. in (1):
R + 12g - 18g = 0  R = 6g newtons
A :
e.g.2. A package of mass 6 kg is placed at B on a
uniform plank AB of length 2 m and mass 9 kg. The
plank rests on supports at A and P as shown.
A
P
B
Modelling the plank as a uniform rod and the
package as a particle at B, find how far P is from
B if the plank is on the point of tipping.
Package: mass 6 kg
Uniform plank AB:
length 2 m, mass 9 kg
A
R
N=0
1
P
9g
x
1
B
6g
How far is P from B if the plank is about to tip ?
Solution:
What
the wordingin“the
about
We
aredoes
not interested
the plank
forcesison
the to tip”
tell us about
forcethe
at reaction
A?
package,
so wethe
ignore
between the
plank and package and draw the weight as if it is
Ans:ofAs
the
plank tips, it turns about P and the
part
the
plank.
force at A disappears.
We can take the force at A as zero to get the
point when the plank is about to tip, since even the
smallest movement of P will cause it to tip.
Package: mass 6 kg
Uniform plank AB:
length 2 m, mass 9 kg
A
R
N=0
1
P
9g
1
x
B
6g
How far is P from B if the plank is about to tip ?
Solution:
We can take moments about A, B, P or even the
centre of the rod.
Choosing to take moments about P means we only
need one equation but it’s a bit more awkward
than using B, so I’ve chosen B.
Package: mass 6 kg
Uniform plank AB:
length 2 m, mass 9 kg
A
R
N=0
1
P
9g
x
1
B
6g
How far is P from B if the plank is about to tip ?
Solution:
9g(1) - Rx = 0
B :
- - - - - - (1)
R - 9g - 6g = 0 - - - - - - (2)
R = 15g

Resolving:
Subst. in (1):
9g - 15gx = 0
9

x = 15
P is 0·6 m from B.
 x = 0·6
A non-uniform rod does not have its mass spread
evenly along its length.
The point at which we can balance a non-uniform
rod is called the centre of mass. On a diagram we
show the weight acting through this point.
e.g.
centre of mass
B
A
mg
e.g.3. A non-uniform rod AB of weight 70 newtons
and length 5 m hangs horizontally in equilibrium held
by ropes at A and B.
2
A
x
G
B
The centre of mass is at a point G, 2 m from A.
Modelling the ropes as light inextensible strings,
find the tension in each rope.
weight: 70 newtons
length: 5 m
centre of mass at G
Solution:
A :
Resolving:
T1
3
2
x
A
70
G
T2
B
T2(5) - 70(2) = 0 - - - - - - (1)
T2 = 28

T1 + T2 - 70 = 0 - - - - - - (2)
T1 = 70 - T2

T1 = 42

The tension at A is 42 newtons and at B is 28 newtons.
SUMMARY
 The moment of a force, F, about a point A is
defined as
Fd




where d is the perpendicular distance from A to
the line of action of F.
For a rigid body to be in equilibrium,
• the resultant force is zero and
• the sum of the moments about any point is zero.
We usually solve problems by resolving and taking
moments about one point.
Moments can be taken about any convenient point.
The centre of mass is the point where we can
assume all the weight acts.
EXERCISE
1. A uniform rod AB of weight 20 newtons and length
6 m rests horizontally in equilibrium on supports at
A and P, where P is 4 m from A.
A
4
P
B
Find the magnitudes of the forces on the rod at
A and P.
EXERCISE
Rod: weight 20 newtons
length 6 m
Find R and N.
Solution:
R
N
4
A
3
P
20
N(4) - 20(3) = 0 - - - - - - (1)
4N = 60

N = 15

R + N - 20 = 0 - - - - - - (2)
Resolving:
R = 20 - 15

R =5

The force at A has magnitude 5 newtons and
at P magnitude 15 newtons.
A :
B
EXERCISE
2. A non-uniform rod AB of weight 20 newtons and
length 5 m rests horizontally in equilibrium on
supports at A and B.
A
x
5
B
G
The reaction on the rod at A is 12 newtons.
The centre of mass is at G. Find,
(a) the reaction at B, and
(b) the distance of G from A.
EXERCISE
non-uniform rod, AB
centre of mass: G
weight: 20 newtons
length: 5 m
reaction at A: 12 newtons
12
R
A
x
5
B
G
20
Solution:
(a) Find the reaction at B
Resolving:
12 + R - 20 = 0
R =8

The reaction at B is 8 newtons.
- - - - - - (1)
EXERCISE
non-uniform rod, AB
centre of mass: G
weight: 20 newtons
length: 5 m
reaction at A: 12 newtons
12
A
R
x
x
5
B
G
20
Solution:
(b) Find the distance of G from A.
R(5) - 20(x) = 0
Subs. R = 8  40 - 20x = 0
x =2

A :
G is 2 m from A.
R =8
- - - - - - (2)
The following page contains the summary in a form
suitable for photocopying.
TEACH A LEVEL MATHS – MECHANICS 1
MOMENT OF A FORCE
Summary

The moment of a force,
F,
about a point
A
is defined as
Fd
where
d
is the perpendicular distance from

A rigid body cannot sag or bend.

For a rigid body to be in equilibrium,
A
to the line of action of
•
the resultant force is zero and
•
the sum of the moments about any point is zero.
F.

We usually solve problems by resolving and taking moments about one point.

Moments can be taken about any convenient point.

The centre of mass is the point where we can assume all the weight acts.