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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

LearnYMA01WME03 · Mechanics 3Statics of rigid bodiesRigid bodies in equilibrium

Statics of rigid bodies · Subtopic 1 of 1

Equilibrium of a Rigid Body

A rigid body is in equilibrium when the resultant force is zero (resolve in two directions) and the total moment about any point is zero. Taking moments about the point where unknown forces act eliminates them. In limiting equilibrium, friction $F=\mu R$ .

Cambridge 9231 Paper 3Edexcel YMA01 WME03≈ 45 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

State and use the conditions for equilibrium of a rigid body.
Solve ladder and rod problems with friction.
Decide whether a body slides or topples.

Pause and think

Why is a ladder more likely to slip when you climb near the top?

Section 2 of 14

Core idea

Start hereWill the ladder slip?

A ladder leaning on a wall is held by the floor's friction. Climb higher and the ladder becomes more likely to slide. Particles only need forces to balance; a rigid body must also not turn.

The big idea

A rigid body is in equilibrium when forces balance in two directions and moments balance about any point. Three equations, so up to three unknowns.

1. The conditions

\sum F_x=0,\quad\sum F_y=0,\quad\sum M_P=0

$P$: any point; choose one that removes unknowns.

Worked exampleLoaded rod

A uniform rod $AB$ , $4$ m, $60$ N, rests on supports at $A$ and $B$ with a $40$ N load $3$ m from $A$ . Find both reactions.

Goal: $R_A,R_B$ .
Strategy: Moments about $A$ , then resolve.

$4R_B=60(2)+40(3)=240$
Moments about $A$ ; $R_A$ vanishes.
$R_B=60$
Solve.
$R_A=100-60=40$
Resolve vertically.

Answer

R_A=40

R_B=60

✓ CheckMoments about

B

4R_A=60(2)+40(1)=160

✓.

2. Ladders

Least

\mu

for a uniform ladder

$R=W,\quad F=S$
Resolve vertically and horizontally (smooth wall reaction $S$ ).
$S\cdot2a\sin\theta=W\cdot a\cos\theta$
Moments about the foot (length $2a$ , angle $\theta$ to ground).
$S=\frac{W}{2\tan\theta}$
Rearrange.
$F\le\mu R\Rightarrow\mu\ge\frac{1}{2\tan\theta}$
Friction needed must be available.

Worked exampleLadder at 60°

Find the least $\mu$ at $60^\circ$ .

Goal: $\mu$ .
Strategy: Use the result.

$\tan60^\circ=\sqrt3$
Exact value.
$\mu=\frac1{2\sqrt3}$
Substitute into $\mu\ge\frac1{2\tan\theta}$ .

Answer

\mu\approx0.289

✓ CheckSteeper ladders need less friction ✓.

Strategy

Drawing diagram…

Pick the equations to minimise algebra.

Common misconceptionPerpendicular distance Moment is force × perpendicular distance from the pivot to the line of action. Draw it.

Key takeawaySlide or topple On a tilting plane a block slides when

\tan\alpha=\mu

and topples when

\tan\alpha=\frac{\text{half-width}}{\text{height of }G}

; whichever angle is smaller happens first.

Pause and think

Why can three non-parallel forces in equilibrium always be drawn meeting at a point?

Summary

Resolve twice, moments once.
Smooth surfaces: normal reaction only.
Limiting: $F=\mu R$ .

Rigid body: Object that does not deform.
Limiting equilibrium: On the point of moving; friction at its maximum.
Couple: Two equal, opposite, parallel forces: turning effect only.

HistoryStevin. Simon Stevin (1586) proved the triangle of forces with his "wreath of spheres" argument: a chain on a triangular prism can't move by itself.

Fun factThe 75° rule. Safety guides say a ladder should lean at about

75^\circ

(the "1 out for every 4 up" rule), a balance of slip risk and backward tipping.

Section 3 of 14

Key formulae

Equilibrium

\sum F_x=0,\ \sum F_y=0,\ \sum M_P=0

Friction

F\le\mu R;\ \text{limiting: }F=\mu R

TrickMoments where unknowns meet. Take moments about the point where the most unknown forces act: they vanish from that equation.

Section 4 of 14

Worked examples

Example 1 · Ladder0 / 3 steps

A uniform ladder of weight

W

and length

2a

rests against a smooth wall at

60^\circ

to rough ground. Find the least

\mu

Section 5 of 14

Explore the graph

Least μ against ladder angle

Loading the interactive graph…

Try this

$x$ is the angle with the ground; the curve is the least $\mu$ for equilibrium.
With friction $m$ , the ladder is safe where the curve is below the line.

$\mu_{\min}=\frac1{2\tan\theta}$ for a uniform ladder, smooth wall.

In the real worldCranes. Tower cranes carry counterweights so the total moment about the mast stays near zero whatever the load.

Section 6 of 14

Common mistakes

IncorrectFriction at a smooth wall

CorrectNormal reaction only

Smooth means no friction.

IncorrectUsing

\sin

for the horizontal distance

CorrectDraw the perpendicular distance

Moment

=

force

\times

perpendicular distance.

Incorrect

F=\mu R

when not limiting

Correct

F\le\mu R

Friction only reaches

\mu R

on the point of slipping.

Forces balanced

Moments: am I a joke to you?

Section 7 of 14

Exam tip

Exam tipDiagram first. Mark every force with its point of action: weight at the centre, normal reactions perpendicular to surfaces, friction along them.

Section 8 of 14

Guided practice

SA-FM-RB-G012 marksL1 FoundationNon-calculatorConditions

A uniform rod

AB

of length

4

m and weight

60

N rests horizontally on supports at

A

and

B

. Find the reaction at

B

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FM-RB-0012 marksL2 StandardNon-calculatorConditions

The same rod has a

40

N load at

3

m from

A

. Find the reaction at

B

Salam Academy Original Practice · Proposed mark scheme

SA-FM-RB-0023 marksL3 ExamCalculatorLadders

A uniform ladder rests against a smooth vertical wall at

60^\circ

to rough horizontal ground and is about to slip. Find

\mu

(3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

SA-FM-RB-0033 marksL3 ExamCalculatorLadders

A uniform ladder on rough ground (

\mu=0.5

) rests against a smooth wall. Find the smallest angle (degrees) with the ground for equilibrium.

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-FM-RB-0043 marksL4 ChallengingCalculatorConditions

A uniform cube rests on a rough plane (

\mu=0.8

) that is slowly tilted. Will it slide or topple first? Enter the angle (degrees) at which it first moves.

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Resolve twice, take moments once.
Moments about the busiest point.
Limiting: $F=\mu R$ .

Section 12 of 14

Quick quiz

Q1 · SA-FM-RB-Q011 markL1 FoundationNon-calculatorConditions

For a rigid body in equilibrium under coplanar forces you need

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: IAL Mathematics WME03 (M3)
Where: Statics of rigid bodies
Expected: Equilibrium of rigid bodies including rods, ladders and frameworks; limiting friction; three-force problems.
Calculator: Permitted; take g = 9.8 m s⁻² unless told otherwise.
Typical style: Structured questions that build from a diagram; give answers to 2 or 3 significant figures.
Open in: YMA01 · WME03

Section 14 of 14

Past paper questions

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