Download moment of a force

M1 – MOMENTS
When you apply a force to a particle, there is only one point at which the force can act.
With larger objects, applying the force at different points can have different effects.
For example, if you close a door, it is easier if you push from the edge furthest from the hinges and
much harder if you push at the point close to the hinges. The strength of the turning effect
depends not only on how hard you push, but also on where and in which direction you push – in
other words it depends on the line of action of the force.
MOMENT OF A FORCE
A force, F acts at a distance d from a point A.
The turning effect of the force about the point has
a magnitude of |F| x d
The turning effect is called the moment of the force
or the torque.
As |F| is in newtons and d is in metres, the unit for the
moment of a force is the newton metre (Nm).
F
d
A
You need to consider the direction (the sense) of a moment:
F
CLOCKWISE
MOMENT
ANTICLOCKWISE
MOMENT
d
d
F
A
A
In these diagrams, the magnitude of the moment is the same, but they are in opposite senses.
An anticlockwise moment is generally regarded as positive and a clockwise one as negative.
In most situations you have two or more forces, each with its own turning effect. You can
combine these to give an overall moment.
Example 1
Forces of 10N, 15N and 18N act as shown in the rectangular
lamina ABCD, with AB = 6m and BC = 4m.
Find the total moment of the forces about A.
18N
D
C
6m
4m
A
10N
15N
B
LAMINA: a thin
plane object – in
other words it has
length and width
but we can neglect
its thickness
M1 – MOMENTS
Example 2
The diagram shows a triangular lamina ABC, with AB = AC = 4m. D is the mid-point of AC. Forces
of 6N, 8N, 12N and P act as shown. The total moment about A is 10Nm.
Find (a)
the force P
C
12N
(b)
the total moment about C.
P
2m
D
8N
2m
4m
A
EXERCISE 1
6N
B
M1 – MOMENTS
Forces at an angle
Suppose you have a rod AB of length a. You apply a force of magnitude F at B at an angle θ to the
rod. You want to find the moment of the force about A.
There are two possible methods:
METHOD 1:
The perpendicular distance from A to
The line of action F is AC.
F
a
θ
A
θ
From the triangle ABC, AC = a sinθ.
Therefore:
Moment of F about A = F x AC = Fasinθ
C
B
M1 – MOMENTS
METHOD 2:
Resolve F into two components.
Fcosθ in the direction AB
Fsinθ perpendicular to AB
F
Fsinθ
a
The first of these components is along a line
Through A, so its moment about A is zero.
The moment of F is produced entirely by the
Second component.
Hence,
Moment of F about A = Fsinθ x a = Fasinθ
A
B
θ
Fcosθ
Example 3
Find the total moment about the point A of the forces shown in the diagram.
12N
3.2m
A
1.7m
23°
55°
B
25N
20N
Example 4
The diagram shows a rectangular lamina ABCD.
A force of 20N is applied at C, as shown.
Find the moment of this force about A.
D
5.4m
C
3.2m
A
B
35°
M1 – MOMENTS
Example 5
The force F = (5i + 2j)N acts at the point Q with position vector (4i + 5j)m.
Find the moment of F about the point P with position vector (i + 3j)m.
x
2N
6
5
5N
Q
4
3
P
2
1
1
EXERCISE 2
2
3
4
5
6
y
M1 – MOMENTS
M1 – MOMENTS
Equilibrium of parallel forces
If forces act on a particle, the particle accelerates unless the forces are in equilibrium. They are in
equilibrium of their resultant is zero.
When forces act on a larger object, the forces may not be concurrent (in other words, acting
through the same point). In this case, you need more information to decide whether they are in
equilibrium.
The simplest example of non-concurrent forces is when they are parallel. Such forces may have a
resultant force, but they may also have a turning effect – a total moment – about some point.
This system of forces has a resultant force
Forces UP = 18N
Forces DOWN = 12N
Resultant force = 6N UP
10N
There is a turning effect about the point A.
Clockwise moments = 12 x 3 = 36Nm
Anti-clockwise moments = 8 x 4 = 32Nm
Turning effect = 4Nm clockwise
A
8N
3m
1m
12N
M1 – MOMENTS
In general, an object acted upon by a system of parallel forces may have both linear acceleration
and rotational acceleration. Both of these accelerations must be zero for the object to be in
equilibrium.
A system of parallel forces is in equilibrium if:


The resultant of the forces is zero AND
The total moment of the forces about any point is zero.
Both conditions must be met if forces are to be in equilibrium.
For this system
Forces UP = 12N
Forces DOWN = 12N
Resultant force = 0N
Turning effect about the point A.
Clockwise moments = 12 x 3 = 36Nm
Anti-clockwise moments = 9 x 4 = 36Nm
Turning effect = 0Nm
3N
A
9N
3m
1m
12N
Therefore this system of forces is in equilibrium.
A system may have a zero resultant, yet may not be in equilibrium.
For this system
Forces UP = 12N
Forces DOWN = 12N
Resultant force = 0N
Turning effect about the point A.
Clockwise moments = 12 x 3 = 36Nm
Anti-clockwise moments = 10 x 4 = 40Nm
Turning effect = 4Nm anti-clockwise
2N
A
10N
3m
1m
12N
Therefore this system of forces is NOT in equilibrium.
A system of forces like the last one which only has a turning effect is called a couple.
The moment about any point will be the same.
A uniform beam means that its weight acts through the centre of the beam, which is its centre of
gravity.
If a beam is non-uniform its centre of gravity is not at its centre. You will either be told where the
centre of gravity is or you will have to find it.
M1 – MOMENTS
Example 6
A uniform beam AB of mass 10kg and length 4m rests in a horizontal position on a single support
C, 1m from A. The other end of the beam is supported by a vertical string as shown.
4m
A
1m
B
C
Find the reaction at the support and the tension in the string.
Example 7
The diagram shows a light rod AB of length 3m. The point C divides AB in the ratio 2:1 and the
forces of 6N, 7N and 8N act at A, B and C as shown.
6N
PN
x
A
7N
C
D
B
8N
A fourth force, PN, is applied to a point D on the rod, where AD = xm, so that the system is then in
equilibrium. Find the values of P and of x.
M1 – MOMENTS
Example 8
A uniform beam AB, of mass 20kg and length 3m, rests horizontally on two supports at A and at C,
where AC = 2m. Find the reactions at both supports.
A
B
C
Example 9
A non-uniform beam AB, of length 4m and weight 500N rests horizontally on supports at A and B.
The reaction at B is 140N more than the reaction at A. Find the position of the centre of gravity of
the beam.
M1 – MOMENTS
Example 10
The diagram shows a uniform beam AB, of weight 360N and length 10m. It is supported
horizontally at A and at C, where AC = 6m. An object of weight 120N is attached to the beam
between B and C at a distance xm from C
x
A
B
C
120N
(a)
(b)
Find the reactions at A and C in terms of x.
Explain what happens as x increases.
M1 – MOMENTS
Tilting
RB
RA
A heavy rod of mass M placed across two
supports at A and B which are at the same
horizontal level will be in equilibrium
provided its centre of mass C is between A
and B as shown.
C
B
A
Mg
RB
RA
If an extra vertical force, F is applied to the
rod at some point other than A or B the rod
may or may not remain in equilibrium,
depending on the magnitude and point of
application of F.
C
B
A
Mg
F
As F increases, to remain in equilibrium the magnitude of the reaction at A must decrease to retain
the equilibrium. At the time when the value of RA is just zero, the rod is said to be on the point of
tilting about B.
Example 11
A uniform rod AB has length 6m and mass 4kg.
It is resting in equilibrium in a horizontal position on supports at points X and Y where AX = 2m and
AY = 4.5m.
A particle of mass M kg is placed at point C where AC = 5m.
Given that the rod is on the point of tilting about Y, calculate the value of M.
M1 – MOMENTS
EXERCISE 3
M1 – MOMENTS
M1 – MOMENTS
M1 – MOMENTS
M1 – MOMENTS
Answers:
Exercise 1
Exercise 2
Exercise 3