Physics Factsheet
www.curriculum-press.co.uk
Number 180
Moments
A force acting on a body will cause that body to accelerate in a
straight line. This is what we tend to confine ourselves to at GCSE
level.
The unit of the moment of a force is the newton-metre
(Nm). Generally the force will be given in newtons. However it
may be necessary to convert a mass (in kg) into a weight (in N)
using w = mg.
a
The distance involved may be given in cm (or mm, etc). It may
or may not be necessary to change this to metres. Quite often it
is possible to work in Ncm throughout the calculation.
F
However a force acting on a body can also cause a body to rotate
about an axis, causing a rotational acceleration.
Notice that a moment is a vector. It has a magnitude
and direction (clockwise or anticlockwise).
α
Example 1
O
F
Moment
= (2.5 × 9.81) × 30 × sin120 = 640Ncm
or 6.4Nm clockwise (to 2sf).
O
30 cm
Another factsheet will deal with rotational mechanics (or rotational
dynamics).
2.5 kg
1200
In this Factsheet we are going to confine ourselves to the moments
caused by a turning force or torque.
W
Exam Hint:- Sometimes it is useful to redraw the setup, showing
just forces and distances. There are two ways that moments
can be drawn with the perpendicular distances shown. Both
give the same results, so you can use either. There is no need
for confusion.
Turning force
The turning effect of a force about an axis depends on two factors:
(a) the magnitude of the force
(b) the perpendicular distance between the line of action of the
force and the axis.
Examples:
O
d
Small torque
(F small, d small)
F
O
F sinθ
F
F
Large torque
(F large, d large)
d
(a) Redrawn showing
the perpendicular
component of the
distance, d
O
(b) Redrawn
showing the
perpendicular
component of
the force, F
In both cases, the moment equals Fd sinθ. In general you can
just use the moment equation without any need for further
drawing.
F
Small torque
(F and d not perpendicular)
θ
d
F
θ
d
O
d sinθ
d
Example 2
O
Calculate the moment of the force about axis O:
Definition of moment of a force:
F Moment of force = F × d × sin θ
d
O
20
“The moment of a force equals the product
of the force and the perpendicular distance
from the line of action of the force to the axis.”
θ
0
75cm
Example:
Moment
= Fd sinθ = 15 × 0.75 × sin70
= 11 Nm anticlockwise.
Notice that the angle supplied
(200) is not angle θ.
Exam Hint:- Take care that the angle you use in your
calculation is the angle between the line of action of the force
and the line to the axis.
6N
2m
F = 15N
O
Moment = 2 × 6 = 12Nm
1
Physics Factsheet
180. Moments
Example 3
Typical Exam Question 1
A cyclist of mass 55kg uses his weight to push down on the pedal
at points A, B, and C.
F
A
B
C
20 cm
20 cm
450
The long bar shown has a fixed axis at X. A pair of forces is
applied as shown. Find the resultant moment about X.
210 cm
X
12N
F
200
12N
270 cm
F
Answer:
Moment = (12 × 270) – (12 × 210) = 720 Ncm or 7.2 Nm clockwise
20 cm
(a) Find the moment applied about the axle in each position.
(b) Explain why the greatest turning effect occurs at point B.
*Notice that this is exactly the same couple as exerted on the
steering wheel, and results in exactly the same turning force
about X (as predicted in the keypoint above). The fact that the
two forces in the couple work against each other does not affect
the resultant moment.
Answer:
(a) A: moment = (55×9.81)×0.20×sin45 = 76Nm clockwise
B: moment = (55×9.81)×0.20 = 110Nm clockwise
C: moment = (55×9.81)×0.20×sin70 = 100Nm clockwise
Moment is a vector. In calculations it is often convenient
to make clockwise moments positive and anticlockwise moments
negative, as in this example.
*Notice that θ=700 (not 200) at C.
(b) Greatest turning effect at B, as W and d unchanged, and sinθ
= 1 (perpendicular situation).
Principle of Moments
Turning forces are often applied in pairs, where the two forces are
equal and opposite. The two forces are antiparallel and often act on
opposite sides of the axis, so their effects add together.
F
Often a body is acted on by several forces, each of which can exert
a turning force about an axis. If the body is to have no rotational
acceleration, these turning effects must balance each other.
The statement of the Principle of Moments says:
“For a body to be in rotational equilibrium, the total clockwise moment
about any axis must equal the total anticlockwise moment about the
same axis.”
O
F
Both forces contribute to the clockwise moment about the axis in
this situation.
Example 4
F1 = 25 N
8.0 cm
A simple example is the use of our hands on the steering wheel of a
car when making a right turn.
O
left hand
pushes up
6.0 cm
F2
Find the value of F2 for equilibrium about axis O.
Answer:
Clockwise moments = anticlockwise moments
25 × 8.0 = F2 × 6.0
F2 = 33N
right hand
pulls down
We can put some numbers on these forces and distances. For this
steering wheel:
And in a real-life situation:
Example 5
12 N
30 cm
A
A uniform ladder of mass 20kg and length 8.0m leans against a
shiny wall (no friction) at an angle of 250 to the vertical. The
bottom of the ladder is fixed to the floor (it cannot slide). Find
the force of reaction of the wall against the ladder.
30 cm
B
C
12 N
Moment = (12 × 0.30) + (12 × 0.30) = 7.2 Nm clockwise.
R
Answer:
The weight of the uniform ladder is at its
centre.
If we forget about this being a steering wheel, and choose point A
as a fixed axis:
8.0
m
Moment about A = 12 × 0.60 = 7.2Nm clockwise again (the upward
force through A has no turning effect about A).
Clockwise moment about O
= (20 × 9.81) × (4.0 × cos65) = 332Nm.
650
Or if we choose C to be a fixed axis, we will find that the moment
about C has the same value once again. Check this yourself.
O
The total moment of a couple is always the same, no
matter what fixed axis you use.
2
W
Anticlockwise moment about O
= R × (8.0 × sin65) = 7.3R Nm
7.3R = 332, R = 45N
Physics Factsheet
180. Moments
Practice Questions
3. An accurate balance is used to calculate the mass of an object
by balancing moments. A small mass slides along a scale to
achieve equilibrium.
1. In which situation, A or B, does the force F exert a greater moment
about the axis?
A
B
10 cm
O
13 cm
65
1.50 mm
220 mm
0
2.50g
F
F
M
2. A screwdriver is used to pry off the lid of a paint tin.
The balance position is shown. Calculate the unknown mass in
kg. (Ignore the mass of the scale itself.)
16.0 cm
4. A van of mass 520 kg has its weight equally shared on each
wheel. It is stationary on a small uniform bridge as shown. (The
mass of the bridge is 200kg.)
20 N
(a) In which direction is the 20N force best applied, assuming
the screwdriver is horizontal?
(b) What acts as the axis of rotation?
(c) Calculate the moment exerted about this axis by the 20N
force?
(d) If the length of the tip of the screwdriver to the left of the
pivot is 0.5cm, what is the upward force on the edge of the
lid just before it lifts?
A
B
1.5 m
2.5 m
3.0 m
Find the upward force of the ground on the bridge at A and at B.
Notice:
(a) The forces on each end of the bridge are different.
(b) FA + FB = 2990 + 4080 = 7070 N
Weight of bridge + van = 7060 N
So the vertical forces on the bridge are in balance (as expected).
Take moments about point B (and ignore force acting through B).
7.0FB = (260 × g × 3.0) + (200 × g × 3.5) + (260 × g ×5.5) = 28500
FB = 4080 N
4. Take moments about point A (and ignore force acting through A).
7.0FB = (260 × g × 1.5) + (200 × g × 3.5) + (260 × g × 4.0) = 20,900
FB = 2990 N
3. Principle of Moments
1.50 × M × g = 2.5×10-3 × g ×220
M = 0.37 kg
(Notice that g cancels out of the calculation. We don’t need to use it.)
The moment is greater in B.
B: moment = 13F × sin 65 = 12 Ncm clockwise
1. A: moment = 10F Ncm clockwise
Answers
3
2. (a)
(b)
(c)
(d)
Vertically downward (perpendicular).
The point on the screwdriver touching the top of the wall of the tin.
moment = 20 × 16.0 = 320 Ncm clockwise
Upward force on lid = Downward force on tip of screwdriver
Use Principle of Moments.
20 × 16.0 = F × 0.5, F = 640 N
Acknowledgements:
This Physics Factsheet was researched and written by Paul Freeman
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU
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