Notes · MathematicsUK · A-Levels

Moments

Moments measure the turning effect of a force. The A-Level covers the moment of a force about a point, the sense of a moment, the principle of moments and the full equilibrium of a rigid body (zero resultant force and zero total moment), applied to uniform and non-uniform rods, reactions at supports, and tilting or limiting equilibrium.

4 sections~11 min reading time3 competenciesLevel Standard 2 · Advanced 2

T·202020 / 20

Exam profile

AO1 · Calculate moments and apply equilibrium conditions to rigid bodiesAO2 · Justify the conditions for equilibrium and the sense of momentsAO3 · Model rods, beams and ladders and analyse tilting

Operators:findshow thatdeterminecalculatemodel

basic level

AS-Level covers the moment of a force and the principle of moments for a body in equilibrium.

higher level

The full A-Level adds non-uniform rods, reactions at supports, and tilting or limiting equilibrium.

Depth

Text

Contents · 4 sections

Moments

The moment of a force#

StandardDfE GCE Mathematics — momentsAQA 7357 (Mechanics)

Moment of a force about a pivot

Fig. 1The downward force

F

at perpendicular distance

d

from the pivot has moment

Fd

about it.

The moment of a force about a point measures its turning effect about that point. For a force of magnitude

F

acting at a perpendicular distance

d

from the point, the moment is

M = Fd

, measured in newton metres (N m). The larger the force or the further from the pivot it acts, the greater the turning effect — which is why a longer spanner loosens a bolt more easily.

A moment has a sense: it turns either clockwise or anticlockwise about the point. When taking moments you choose one sense as positive (commonly anticlockwise) and give each moment its correct sign. Keeping the sense consistent is essential, because equilibrium will require the clockwise and anticlockwise moments to balance.

The distance in

M = Fd

is the perpendicular distance from the point to the line of action of the force, not the distance to the point of application. When a force acts at an angle, you use either the perpendicular distance from the pivot to the line of the force, or equivalently the component of the force perpendicular to the arm times the arm length:

M = Fd\sin\theta

. Getting this perpendicular distance right is the crux of angled-force moment problems.

Moments are additive: the total moment about a point is the sum of the moments of all the individual forces, each with its sense. A force whose line of action passes through the point has zero moment about it (its perpendicular distance is zero), a fact often exploited by taking moments about a point where an unknown force acts, so that unknown does not appear in the equation.

M = Fd \quad (\text{N m}), \qquad M = Fd\sin\theta \text{ for a force at angle } \theta

Moment of a force

Force times perpendicular distance to its line of action; a force through the point has zero moment.

Worked example

Moment of an angled force

A force of $20\ \text{N}$ acts at a point $0.8\ \text{m}$ from a pivot $O$ , directed at $30^\circ$ to the arm $O\!P$ . Find the moment about $O$ .

Perpendicular component
The component of the force perpendicular to the arm is $20\sin 30^\circ = 10\ \text{N}$ .
Moment
$M = 10 \times 0.8$ .
$M = 20\sin 30^\circ \times 0.8 = 8\ \text{N m}$
State
$M = 8\ \text{N m}$ .

Result: The moment about $O$ is $8\ \text{N m}$ .

Typical mistakes

Using the distance to the point of application rather than the perpendicular distance to the line of action.
Losing track of the sense (clockwise or anticlockwise) of a moment.

Active revision

A force of $12\ \text{N}$ acts at the end of a horizontal rod $0.5\ \text{m}$ from a pivot, at $40^\circ$ above the horizontal. Find the moment of the force about the pivot.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

The principle of moments and equilibrium#

StandardDfE GCE Mathematics — momentsAQA 7357 (Mechanics)

A balanced beam (seesaw)

Fig. 2The beam balances when

W_1 d_1 = W_2 d_2

: the two moments about the pivot are equal and opposite.

The principle of moments states that a body in equilibrium under a set of forces has, about any point, its total clockwise moment equal to its total anticlockwise moment. Equivalently, the sum of the moments about any point is zero. This is the rotational counterpart of the force-balance conditions and is what stops a balanced body from turning.

For a rigid body, full equilibrium requires two things together: the resultant force is zero (so it does not accelerate) and the total moment about any point is zero (so it does not rotate). These give up to three equations in two dimensions — resolve horizontally, resolve vertically, and take moments about a chosen point — which are enough to find the unknowns in a typical problem.

A powerful feature is that the moment equation may be taken about any point, and a clever choice simplifies the algebra. Taking moments about a point where an unknown force acts removes that unknown, because its moment there is zero. This lets you solve directly for one reaction without first finding the others.

The classic application is a seesaw or balanced beam: two weights balance about a pivot when

W_1 d_1 = W_2 d_2

, the clockwise moment of one equalling the anticlockwise moment of the other. Recognising this balance condition, and extending it to several forces, is the foundation of all the rod and beam problems that follow.

\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}} \quad (\text{about any point})

The principle of moments

In equilibrium the turning effects balance; equivalently the total moment about any point is zero.

Worked example

Balancing a beam

A light rod is pivoted at its centre. A weight of $40\ \text{N}$ hangs $0.6\ \text{m}$ to the left of the pivot. At what distance to the right must a $30\ \text{N}$ weight hang to balance it?

Anticlockwise moment
The left weight gives $40 \times 0.6 = 24\ \text{N m}$ .
Clockwise moment
The right weight gives $30 \times d$ .
Balance
$30d = 24 \Rightarrow d = \dfrac{24}{30} = 0.8\ \text{m}$ .
$30d = 24 \Rightarrow d = 0.8\ \text{m}$

Result: The $30\ \text{N}$ weight must hang $0.8\ \text{m}$ to the right of the pivot.

Typical mistakes

Balancing only the forces, forgetting the moment condition (or vice versa).
Mixing clockwise and anticlockwise senses inconsistently within one equation.

Active revision

A light beam $4\ \text{m}$ long rests on a pivot at its centre. A $30\ \text{N}$ weight hangs $1.5\ \text{m}$ from the pivot on one side. Where must a $45\ \text{N}$ weight hang on the other side to balance it?

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Rods, beams and reactions#

AdvancedDfE GCE Mathematics — momentsAQA 7357 (Mechanics)

A rod on two supports

Fig. 3The upward reactions

R_A

and

R_B

balance the weight

W

; moments about one support give the other reaction.

A rod or beam resting on supports is the central application of moments. A uniform rod has its mass evenly distributed, so its weight acts at its centre (the midpoint); this single downward weight, together with the upward reactions at the supports and any additional loads, forms the system of forces to be balanced. The first step is always a clear diagram showing every force and its position.

For a beam on two supports, there are two unknown reactions, found using the equilibrium conditions. Resolving vertically gives the sum of the reactions (they support the total weight); taking moments about one support gives the other reaction directly, because the moment of the reaction at the chosen support is zero. This is the standard, efficient method.

A non-uniform rod has its centre of mass at an unknown position, which becomes one of the unknowns to find. Given the reactions or the balance point, you take moments to locate the centre of mass, or to find the weight of the rod. These problems test the same principles but require careful, complete moment equations.

Throughout, consistency is key: mark distances from a fixed reference, give each moment its sense, and check that both the force and moment conditions are satisfied. A good check is that taking moments about a second, different point gives a consistent answer — a quick way to catch an error.

R_A + R_B = \text{total weight}, \qquad \sum M_{A} = 0 \Rightarrow R_B

Beam on two supports

Resolve vertically for the sum of reactions; take moments about one support to find the other.

Worked example

Reactions on a uniform beam

A uniform beam $AB$ of length $4\ \text{m}$ and weight $120\ \text{N}$ rests on supports at $A$ and $B$ . A load of $60\ \text{N}$ hangs $1\ \text{m}$ from $A$ . Find the reactions at $A$ and $B$ .

Take moments about A
Anticlockwise ( $R_B$ at $4\ \text{m}$ ) equals clockwise (weight $120\ \text{N}$ at $2\ \text{m}$ , load $60\ \text{N}$ at $1\ \text{m}$ ): $4R_B = 120(2) + 60(1) = 300$ .
Reaction at B
$R_B = \dfrac{300}{4} = 75\ \text{N}$ .
$R_B = \frac{300}{4} = 75\ \text{N}$
Resolve vertically
$R_A + R_B = 120 + 60 = 180$ , so $R_A = 180 - 75 = 105\ \text{N}$ .

Result: $R_A = 105\ \text{N}$ and $R_B = 75\ \text{N}$ .

Typical mistakes

Placing a uniform rod's weight at one end rather than at its midpoint.
Measuring distances from different reference points within the same moment equation.

Active revision

A uniform rod $AB$ of length $6\ \text{m}$ and weight $80\ \text{N}$ rests horizontally on supports at $A$ and at a point $1\ \text{m}$ from $B$ . Find the reaction at each support.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Tilting and limiting equilibrium#

AdvancedDfE GCE Mathematics — momentsAQA 7357 (Mechanics)

A ladder against a wall

Fig. 4In limiting equilibrium, moments about the foot combine with friction at the ground to keep the ladder from slipping.

A body supported at two or more points is on the point of tilting when it is about to rotate about one of the supports. At that instant the reaction at the support it is lifting away from becomes zero, because that support is just about to lose contact. Setting the appropriate reaction to zero is the key modelling step in every tilting problem.

With one reaction set to zero, the body is effectively pivoting about the remaining support, and taking moments about that support gives the condition for tilting. This determines, for example, the greatest load that can be placed at the end of a plank before it tips, or how far a person can walk along an overhanging beam before it overbalances.

For a ladder resting against a wall, limiting equilibrium combines moments with friction. The ladder is in equilibrium under its weight, the normal reactions at the wall and ground, and friction at the ground (and possibly the wall). Taking moments about the foot of the ladder, together with resolving horizontally and vertically, and using

F = \mu R

at the point of slipping, gives the conditions for it to stay up.

These problems bring together the whole mechanics toolkit: forces as vectors, resolving, the friction model and moments. The reliable method is a complete diagram, then the three equilibrium equations (two resolutions and one moment), with the limiting condition (

F = \mu R

, or a reaction set to zero) supplying the extra relationship needed to solve for the critical value.

\text{On the point of tilting: reaction at the lifting support} = 0

The tilting condition

Set the reaction at the support about to lose contact to zero, then take moments about the other support.

Worked example

On the point of tilting

A uniform beam $AB$ of length $6\ \text{m}$ and weight $100\ \text{N}$ rests on supports at $C$ and $D$ , where $AC = 1\ \text{m}$ and $AD = 4\ \text{m}$ . Find the greatest weight $W$ that can be hung at $B$ before the beam tilts about $D$ .

Tilting condition
About to tilt about $D$ , so the reaction at $C$ is zero; take moments about $D$ .
Moments about D
Weight of beam ( $100\ \text{N}$ at the centre, $2\ \text{m}$ from $D$ ) is anticlockwise; $W$ at $B$ ( $2\ \text{m}$ from $D$ ) is clockwise: $W \times 2 = 100 \times 2$ .
$2W = 100 \times 2 = 200$
Solve
$W = \dfrac{200}{2} = 100\ \text{N}$ .

Result: The greatest weight is $W = 100\ \text{N}$ ; beyond this the beam tilts about $D$ .

Typical mistakes

Not realising that a reaction becomes zero at the point of tilting.
Omitting friction, or the weight at the ladder's midpoint, in a ladder problem.

Active revision

A uniform plank of length $5\ \text{m}$ and weight $200\ \text{N}$ rests on supports $1\ \text{m}$ from each end. Find the greatest weight that can be hung from one end without the plank tilting.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Moments

4 sections~11 min reading time3 competenciesLevel Standard 2 · Advanced 2

Moments

Key points

Moment of an angled force

Key points

Balancing a beam

Key points

Reactions on a uniform beam

Key points

On the point of tilting

Sources

Moments

Key points

Moment of an angled force

Key points

Balancing a beam

Key points

Reactions on a uniform beam

Key points

On the point of tilting

Sources