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AQA A-level Mathematics (7357) · Moments
Mini-Lesson

Moments

This mini-lesson covers AQA section S — Moments: the moment of a force (force × perpendicular distance), clockwise and anticlockwise moments, the two conditions for equilibrium of a rigid body, uniform and non-uniform rods, reactions at supports, and tilting.

A B R(A) R(B) 10g (centre) 6g 1 m 2 m
A uniform rod: its weight acts at the centre. Take moments about a support to eliminate that unknown reaction.

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. We use g = 9.8 m s⁻², and rods are rigid, with uniform rods having their weight at the centre. Press Start when you're ready.

Section S1 · moments

The moment of a force

A moment measures the turning effect of a force about a point.

moment = force × perpendicular distanceunits: newton metres (N m) · state the sense: clockwise or anticlockwise
  • The distance must be measured perpendicular to the line of action of the force.
  • If a force of F acts at an angle θ to the rod, at a distance d from the pivot, the moment is F d sin θ (only the component perpendicular to the rod turns it).
  • A force whose line of action passes through the pivot has zero moment — the perpendicular distance is 0.
Worked example

A force of 20 N acts perpendicular to a spanner, 0.75 m from the bolt.

Moment = 20 × 0.75 = 15 N m (clockwise, say)

Calculate

Moment of a force

1A force of 20 N acts perpendicular to a spanner at a distance of 0.75 m from the bolt. Calculate the moment about the bolt.
N m
Hint: Moment = force × perpendicular distance = 20 × 0.75.
Quick check

Perpendicular distance

?A force of 10 N acts at 30° to a rod, at a distance of 2 m from the pivot. What is the moment about the pivot?
Section S1 · equilibrium

Equilibrium of a rigid body

For a rigid body (a rod, a plank, a beam) to be in equilibrium, two conditions must hold:

  • 1. Resolve: the resultant force is zero (so total up = total down).
  • 2. Take moments: the total moment about any point is zero — so total clockwise = total anticlockwise.

Strategy that saves marks: take moments about a point where an unknown force acts. That force then has zero moment and vanishes from the equation, leaving one unknown.

Worked example — the seesaw

A child of mass 30 kg sits 2 m from the pivot. Where must a 40 kg child sit to balance?

Moments about the pivot: 30g × 2 = 40g × d

g cancels: 60 = 40d → d = 1.5 m on the other side.

Calculate

Balancing a seesaw

2A child of mass 30 kg sits 2 m from the pivot of a light seesaw. How far from the pivot, on the other side, must a 40 kg child sit to balance it?
m
Hint: Moments about the pivot: 30g × 2 = 40g × d. The g cancels: 60 = 40d.
Sort it

Clockwise, anticlockwise, or zero?

Tap a force description, then tap the moment it produces about the pivot.

🕐 Clockwise

🕘 Anticlockwise

0️⃣ Zero moment

Section S1 · rods

Rods, supports and reactions

A uniform rod has its weight at the centre. A non-uniform rod does not — finding where its weight acts is a standard exam question.

Worked example — a uniform rod on two supports (g = 9.8)

A uniform rod AB of length 4 m and mass 10 kg rests horizontally on supports at A and B. A particle of mass 6 kg is attached 1 m from A. Find the reactions.

Moments about A (this kills R(A) — its distance is zero):

R(B) × 4 = (10g × 2) + (6g × 1) = 20g + 6g = 26g

R(B) = 26g ÷ 4 = 6.5 × 9.8 = 63.7 N

Resolve vertically: R(A) + R(B) = 10g + 6g = 16g = 156.8 N

R(A) = 156.8 − 63.7 = 93.1 N

Check by taking moments about B: R(A) × 4 = 10g × 2 + 6g × 3 = 20g + 18g = 38g → R(A) = 9.5g = 93.1 ✓

Tilting: when a rod is on the point of tilting about one support, the reaction at the other support is zero. Set that reaction to 0 and take moments — that is the whole trick.

Calculate

Reaction at B

3A uniform rod AB of length 4 m and mass 10 kg rests on supports at A and B. A 6 kg particle is attached 1 m from A. (g = 9.8) Find the reaction at B, to 1 d.p.
N
Hint: Moments about A: R(B) × 4 = 10g × 2 + 6g × 1 = 26g. So R(B) = 26 × 9.8 ÷ 4.
Calculate

Reaction at A

4For the same rod, calculate the reaction at A, to 1 d.p.
N
Hint: Resolve vertically: R(A) + R(B) = (10 + 6)g = 156.8 N. So R(A) = 156.8 − 63.7.
Calculate

Non-uniform rod

5A non-uniform rod AB of length 3 m and weight 60 N rests on supports at A and B. The reactions are R(A) = 40 N and R(B) = 20 N. Find the distance d of the centre of mass from A.
m
Hint: Moments about A: R(B) × 3 = 60 × d, so 20 × 3 = 60d.
Quick check

Two conditions

?What are the conditions for a rigid body to be in equilibrium?
Match it

Moments in words

Tap a phrase on the left, then how it finishes on the right.

Statement
Completion
Quick check

On the point of tilting

?A plank rests on two supports and a person walks towards one end. The plank is on the point of tilting. What do you know?
Quick check

Units check

?What is the unit of a moment?
Section S1 · tilting

On the point of tilting

When a plank rests on two supports and a load moves outwards, the reaction at the far support shrinks. At the moment of tilting, that reaction is exactly zero.

Worked example

A uniform plank AB of length 8 m and mass 30 kg rests on supports at A and at C, where AC = 6 m. A man of mass 40 kg walks from A past C. How far beyond C can he go before the plank tilts?

At the point of tilting about C, the reaction at A is zero. Take moments about C:

The plank's weight acts at its centre, 4 m from A — that is 2 m on the A-side of C: anticlockwise moment = 30g × 2 = 60g

The man is x m beyond C: clockwise moment = 40g × x

Balance: 40g x = 60g → x = 60 ÷ 40 = 1.5 m beyond C (i.e. 7.5 m from A).

Why it works: setting the far reaction to zero removes one unknown, and taking moments about the support it tilts around removes the other. One equation, one unknown.

Calculate

Tilting point

6A uniform plank AB of length 8 m and mass 30 kg rests on supports at A and C, with AC = 6 m. A man of mass 40 kg walks beyond C. How far beyond C can he walk before the plank tilts?
m
Hint: At tilting, R(A) = 0. Moments about C: 40g × x = 30g × 2 (the plank's centre is 2 m on the other side of C).
Quick check

A couple

?Two equal and opposite forces act on a rod, but along different lines of action (a couple). What happens?
Recap

The big ideas to know

Moment = force × perpendicular distance (N m); a force at angle θ to the rod gives Fd sin θ

Zero moment: any force whose line of action passes through the pivot

Equilibrium of a rigid body: resultant force = 0 and total moment about any point = 0

Uniform rod: weight acts at the midpoint · Non-uniform: find where it acts by taking moments

Take moments about an unknown force to make it disappear from the equation

On the point of tilting: the reaction at the other support is zero

That is the whole of section S — Moments for AQA A-level Mathematics (7357). Press Finish to see your score.

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