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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

LearnYMA01WME02 · Mechanics 2Statics of rigid bodiesMoments and beams on supports

Statics of rigid bodies · Subtopic 1 of 2

Moments and Beams on Supports

A moment measures a force's turning effect: force × perpendicular distance from the pivot. A beam in equilibrium has zero resultant force and zero resultant moment about every point. Take moments about a support to remove its unknown reaction.

Edexcel YMA01 WME01Edexcel YMA01 WME02Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Calculate the moment of a force about a point.
Find the resultant moment of several forces.
Use the two conditions for a beam in equilibrium.
Find reactions at supports, with loads on the beam.

Pause and think

Why is it easier to open a door by pushing near the handle than near the hinges?

Section 2 of 14

Core idea

Start hereWhy spanners have long handles

Push near the nut and it won't budge; push at the end of a long spanner and it turns. Turning effect depends on force and distance. Balancing turning effects is how bridges and seesaws are designed.

The big idea

Moment = force × perpendicular distance. Rigid body in equilibrium: resultant force zero and clockwise moments = anticlockwise moments about any point.

1. Moments

A 6 m uniform beam of weight 200 N on two supports

Supports at 0.5 m and 5 m from the left end; an 80 N load 5.7 m from the left. The beam's weight acts at its middle (uniform).

M=Fd

$F$: force (N).
$d$: perpendicular distance from the pivot to the line of the force (m).

2. Beams on supports

Worked exampleBeam with a load

6 m uniform beam, 200 N; supports at A (0.5 m) and B (5 m); 80 N load at 5.7 m. Find reactions.

Goal: $R_A$ , $R_B$ .
Strategy: Moments about A removes $R_A$ .

$R_B(4.5)=200(2.5)+80(5.2)$
Weight at 3 m, 2.5 m from A; load 5.2 m from A.
$R_B=916/4.5=203.6\text{ N}$
Solve.
$R_A=280-203.6=76.4\text{ N}$
Vertical balance.

Answer

R_A=76.4

R_B=203.6

✓ CheckAbout B:

76.4(4.5)+80(0.7)=399.8\approx200(2)

✓.

Worked exampleFinding the centre of mass

A non-uniform 4 m rod, weight 60 N, rests on supports at its ends; left reaction 40 N.

Goal: Position of the weight.
Strategy: Moments about the right end.

$40(4)=60d$
Left reaction vs weight.
$d=\tfrac83\text{ m from the right}$
Solve.

Answer

1.33

m from the left end.

✓ CheckBigger reaction on the left, so the weight is nearer the left ✓.

TrickTake moments about an unknown Taking moments about a support removes its reaction.

Pause and think

Moment of 30 N at 2 m?

Summary

$Fd$ (perpendicular).
Forces and moments both balance.

Moment: Turning effect of a force.
Uniform: Weight acts at the centre.

HistoryGive me a place to stand. Archimedes worked out the law of the lever around

250

BC and is reported to have said: "Give me a place to stand, and I will move the Earth." With a long enough lever, a small force can balance a huge one.

Fun factYour forearm is a lever. Your biceps attaches only a few centimetres from your elbow, but your hand is about

35

cm away. To hold a

5

kg bag, the muscle must pull with a force several times larger than the bag's weight.

Section 3 of 14

Key formulae

Moment

M=F\times d

Equilibrium

\sum\text{up}=\sum\text{down},\qquad\sum\text{clockwise}=\sum\text{anticlockwise}

TrickMoments about the unknown. Take moments about the point where an unknown force acts: it has zero distance, so it disappears from the equation. Pick the support whose reaction you don't want yet.

Section 4 of 14

Worked examples

Example 1 · Off-centre support0 / 2 steps

A uniform plank

AB

6

20

kg, rests on supports at

A

and

C

with

AC=4

m. Find the reactions.

Example 2 · See-saw0 / 1 steps

30

kg child sits

2

m from the pivot. Where must a

40

kg child sit to balance?

Section 5 of 14

Explore the graph

Reactions on a loaded beam

Loading the interactive graph…

Try this

A $6$ m, $20$ kg uniform beam rests on supports at its ends. A $30$ kg load sits $p$ m from $A$ . The lines show $R_B$ (lower when $p$ is small) and $R_A$ .
Move the load towards $B$ . Which reaction grows?
Check the two reactions always add to $50g=490$ N.

The nearer support carries more of the load.

In the real worldCranes and counterweights. A tower crane has a heavy concrete counterweight on its short arm. Its moment balances the moment of the load on the long arm, so the tower isn't tipped over. The crane's computer limits how far out a heavy load can go.

Section 6 of 14

Common mistakes

IncorrectMoment

=

force

\times

length of the beam

Correct

\times

distance from the pivot

Measure from the point you take moments about.

IncorrectWeight of a uniform beam at an end

CorrectAt the midpoint

Uniform means evenly spread.

IncorrectEach support takes half the weight

CorrectOnly when symmetric

Off-centre supports share unequally.

IncorrectMixing clockwise and anticlockwise

CorrectSet them equal

Write two lists, then equate them.

I'll take moments about the middle

Two unknown reactions

we'll both be in the equation, then

moments about one of the supports. Lesson learned.

Section 7 of 14

Exam tip

Exam tipSay where you take moments. Write "Taking moments about

A

:" before the equation. It earns the method mark even if a distance is wrong, and makes your working easy to follow.

Section 8 of 14

Guided practice

SA-MOM-G011 markL1 FoundationCalculatorMoments

A force of

20

N acts perpendicular to a rod,

3

m from a pivot. Find its moment about the pivot.

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-MOM1-0011 markL1 FoundationNon-calculatorMoments

A force of

15

N acts at right angles to a spanner,

0.4

m from the bolt. Find its moment about the bolt.

Salam Academy Original Practice · Proposed mark scheme

SA-MOM-0032 marksL1 FoundationCalculatorMoments

A child of mass

30

kg sits

2

m from the pivot of a see-saw. How far from the pivot, on the other side, must a child of mass

40

kg sit to balance it?

Salam Academy Original Practice · Proposed mark scheme

SA-MOM1-0022 marksL1 FoundationNon-calculatorMoments

25

kg child sits

1.6

m from the pivot of a see-saw. How far from the pivot, on the other side, must a

40

kg child sit to balance?

Salam Academy Original Practice · Proposed mark scheme

SA-MOM-0012 marksL1 FoundationCalculatorBeams

A uniform rod of length

4

m and mass

10

kg rests horizontally on supports at its ends. Find the reaction at each support.

Salam Academy Original Practice · Proposed mark scheme

SA-MOM1-0042 marksL1 FoundationNon-calculatorMoments

About a point

O

, a

10

N force has a clockwise moment with distance

2

m and an

8

N force has an anticlockwise moment with distance

3

m. Find the size of the resultant moment.

Salam Academy Original Practice · Proposed mark scheme

SA-MOM-0024 marksL2 StandardCalculatorBeams

A uniform plank

AB

of length

6

m and mass

20

kg rests horizontally on supports at

A

and

C

, where

AC=4

m. Find the reactions at

C

and at

A

Salam Academy Original Practice · Proposed mark scheme

SA-MOM1-0033 marksL2 StandardCalculatorBeams

A uniform rod

AB

of length

5

m and mass

12

kg rests horizontally on supports at

A

and

B

. A

30

kg mass hangs

1

m from

A

. Find the reaction at

B

Salam Academy Original Practice · Proposed mark scheme

SA-MOM-0075 marksL3 ExamCalculatorBeams

A uniform beam

AB

of length

4

m and mass

12

kg hangs horizontally from two vertical ropes, one at

A

and one at

C

, where

CB=1

m. A load of mass

8

kg is attached at

B

. Find the tension in each rope, to 3 significant figures.

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-MOM-0095 marksL2 StandardCalculatorBeams

A uniform beam

AB

of length

6

m and mass

20

kg rests horizontally on supports at

A

and

C

, where

AC=4

m. A child of mass

30

kg stands on the beam at

D

, where

AD=1

m. Find the reactions at

A

and at

C

. (

g=9.8

)

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Moment $=Fd$ (perpendicular distance).
Beam in equilibrium: forces balance and moments balance.
Uniform: weight at the midpoint.
Take moments about an unknown to remove it.

Section 12 of 14

Quick quiz

Q1 · SA-MOM-Q021 markL1 FoundationCalculatorMoments

The units of a moment are

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-MOM1-0051 markL1 FoundationCalculatorMoments

The moment of a force about a point on its line of action is

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: IAL Mathematics, unit WME01 (Mechanics 1)
Where: M1 · 6 Moments
Expected: The moment of a force about a point; equilibrium of a rigid body (rod or beam) under parallel coplanar forces, using resolving and taking moments.
Calculator: Calculators are permitted, but exact answers (surds, fractions, π) are expected where the question asks.
Typical style: "A uniform plank rests on two supports… find the reaction at each support."
Open in: YMA01 · WME01 YMA01 · WME02 9MA0 · Paper 3

Section 14 of 14

Past paper questions

1 question (10 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

WME01/01 · January 2025Question 410 marks
A uniform rod held by two ropes, with a particle added and about to tilt.
Question paper, page 12 Mark scheme

All YMA01 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel YMA01, 2018 specification (official specification).