Statics and Equilibrium

This lesson covers statics — the study of bodies in equilibrium — as required by the Edexcel 9MA0 A-Level Mathematics specification. You need to solve problems where forces and moments are in balance.

Conditions for Equilibrium

A body is in static equilibrium when:

The resultant force is zero (no acceleration).
The resultant moment about any point is zero (no rotation).

For a particle (no size, so no rotation):

Only condition 1 applies.
Resolve forces in two perpendicular directions and set each component sum to zero.

For a rigid body (has size, can rotate):

Both conditions apply.
Resolve forces AND take moments.

Resolving Forces

To check equilibrium, resolve all forces into two perpendicular directions (usually horizontal and vertical).

Horizontal equilibrium: ΣFx = 0 Vertical equilibrium: ΣFy = 0

Example: Three forces act on a particle: 10 N horizontally to the right, 6 N at 60° above the horizontal to the left, and a force P vertically downwards.

Horizontal: 10 - 6 cos 60° = 0? → 10 - 3 = 7 ≠ 0.

This system is not in equilibrium unless there is an additional horizontal force. This illustrates that all forces must be accounted for.

Triangle of Forces

If exactly three forces act on a particle in equilibrium, they can be represented as the sides of a triangle — drawn in order, the forces form a closed triangle.

This can be used to find unknown forces using:

Trigonometry (sine rule, cosine rule)
Geometry

Lami's Theorem: If three concurrent forces are in equilibrium, then:

F₁/sin α₁ = F₂/sin α₂ = F₃/sin α₃

where α₁, α₂, α₃ are the angles opposite to the forces F₁, F₂, F₃ respectively (i.e. the angles between the other two forces).

Equilibrium on an Inclined Plane

A particle on a rough inclined plane (angle θ) is in equilibrium when:

Along the plane: friction balances the component of weight along the plane. Perpendicular to the plane: normal reaction balances the component of weight perpendicular to the plane.

If additional forces act (e.g. a horizontal force P), resolve all forces along and perpendicular to the plane.

Rigid Body Equilibrium: Combining Forces and Moments

For a rigid body in equilibrium:

Resolve horizontally: ΣFx = 0
Resolve vertically: ΣFy = 0
Take moments about a suitable point: ΣM = 0

This gives three equations, which can be used to find up to three unknowns.

Exam Tip: Choose the point for taking moments wisely. Pick a point where unknown forces act — this eliminates them from the moment equation.

Worked Example: Ladder Problem

A uniform ladder of length 6 m and mass 20 kg leans against a smooth vertical wall at angle θ to the horizontal floor. The floor is rough with μ = 0.4. Find the angle θ at which the ladder is about to slip.

Forces:

Weight: 20g downwards at the midpoint (3 m from the base).
Normal reaction from wall: Rw horizontally at the top.
Normal reaction from floor: Rf vertically upwards at the base.
Friction from floor: F horizontally at the base (towards the wall).

Resolving horizontally: F = Rw ... (1) Resolving vertically: Rf = 20g ... (2) At limiting equilibrium: F = μRf = 0.4 x 20g = 8g ... (3)

From (1) and (3): Rw = 8g.

Taking moments about the base: Rw x 6 sin θ = 20g x 3 cos θ 8g x 6 sin θ = 60g cos θ 48 sin θ = 60 cos θ tan θ = 60/48 = 5/4 θ = arctan(1.25) ≈ 51.3°

Hinged and Pivoted Bodies

If a rigid body is hinged at a point, the hinge exerts a reaction force. This reaction typically has both horizontal and vertical components (Hx and Hy).

Method:

Take moments about the hinge to find other unknown forces (eliminating Hx and Hy).
Resolve horizontally and vertically to find Hx and Hy.

Bodies Supported by Strings

If a rigid body is suspended by one or more strings:

The tension in each string acts along the string.
Take moments about a convenient point.
Resolve forces to find remaining unknowns.

Summary

A body in equilibrium has zero resultant force and zero resultant moment.
For a particle: resolve in two directions. For a rigid body: resolve in two directions and take moments.
Use the triangle of forces or Lami's Theorem for exactly three forces on a particle.
For ladder problems: resolve horizontally, vertically, and take moments about the base or top.
Choose pivot points wisely to simplify the algebra.

A-Level Deep Dive: Statics and Equilibrium

Spec mapping

Edexcel 9MA0-03 specification, Paper 3 — Statistics and Mechanics, sections 10–11 (Forces, friction and moments) covers the principle of moments in simple static contexts; understand and use the equilibrium of forces on a particle and on a rigid body, including the use of $\Sigma F = 0$ and $\Sigma M = 0$ (refer to the official specification document for exact wording). This is exclusively Year 2 content — Year 1 mechanics treats only particles in equilibrium under concurrent forces (so $\Sigma F = 0$ alone suffices). The Year 2 extension adds rigid bodies, where forces no longer act through a single point, so a second condition is required: the resultant moment about every point must vanish. The formula booklet provides no statics-specific results — the moment definition $M = Fd$ (or $M = Fl\sin\theta$ for an oblique force) and the friction law $F \leq \mu R$ must be deployed from memory.

Worked example with full mark scheme

Question (8 marks):

A uniform ladder $AB$ of mass $30$ kg and length $5$ m rests with end $A$ on rough horizontal ground (coefficient of friction $\mu$ ) and end $B$ against a smooth vertical wall. The ladder makes an angle of $60°$ with the ground. Modelling the ladder as a uniform rod, find the smallest value of $\mu$ for which the ladder is in equilibrium. Take $g = 9.8$ m s $^{-2}$ .

Solution with mark scheme:

Step 1 — draw the force diagram.

Forces on the ladder: weight $W = 30g$ N acting vertically down at the midpoint $M$ (distance $2.5$ m from $A$ ); normal reaction $R$ at the ground acting vertically up at $A$ ; frictional force $F$ at the ground acting horizontally toward the wall at $A$ ; normal reaction $N$ from the smooth wall acting horizontally away from the wall at $B$ . The wall is smooth, so there is no vertical friction at $B$ .

B1 — correct, fully labelled force diagram with all four forces and their points of application. Candidates who omit the friction $F$ or the wall reaction $N$ lose this immediately.

Step 2 — resolve vertically.

$\Sigma F_y = 0: \quad R - 30g = 0 \implies R = 30g = 294 \text{ N}$

M1 — resolving vertically; A1 — correct value of $R$ .

Step 3 — resolve horizontally.

$\Sigma F_x = 0: \quad F - N = 0 \implies F = N$

M1 — resolving horizontally and equating $F$ and $N$ .

Step 4 — take moments about $A$ .

Taking moments about $A$ eliminates $R$ and $F$ (both pass through $A$ , contributing zero moment). The weight acts at the midpoint, perpendicular distance from $A$ along the horizontal is $2.5\cos 60°$ . The wall reaction $N$ acts horizontally at $B$ , perpendicular distance from $A$ along the vertical is $5\sin 60°$ .

$\Sigma M_A = 0: \quad N \cdot 5\sin 60° - 30g \cdot 2.5\cos 60° = 0$

M1 — taking moments about a sensible point with correct lever arms.

$N = \dfrac{30g \cdot 2.5\cos 60°}{5\sin 60°} = \dfrac{30g \cdot 2.5 \cdot 0.5}{5 \cdot (\sqrt{3}/2)} = \dfrac{37.5g}{2.5\sqrt{3}} = \dfrac{15g}{\sqrt{3}} = 5g\sqrt{3}$

So $N = 5g\sqrt{3} \approx 84.87$ N.

A1 — correct value of $N$ .

Step 5 — apply the limiting friction condition.

For minimum $\mu$ , friction is at its limiting value: $F = \mu R$ . Since $F = N$ :

$\mu R = N \implies \mu = \dfrac{N}{R} = \dfrac{5g\sqrt{3}}{30g} = \dfrac{\sqrt{3}}{6}$

M1 — using $F = \mu R$ at the point of slipping; A1 — final value $\mu = \dfrac{\sqrt{3}}{6} \approx 0.289$ .

Total: 8 marks (B1 M4 A3, split as shown).

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A uniform rod $AB$ of mass $8$ kg and length $2$ m is freely hinged at $A$ to a vertical wall. The rod is held horizontal by a light inextensible string attached at $B$ and to a point $C$ on the wall directly above $A$ , such that the string makes an angle of $30°$ with the rod.

(a) Find the tension in the string. (3)

(b) Find the magnitude and direction of the reaction at the hinge. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO3.3) — taking moments about $A$ to eliminate the hinge reaction: $T \sin 30° \cdot 2 - 8g \cdot 1 = 0$ , recognising the perpendicular component of $T$ at $B$ .
A1 (AO1.1b) — correct equation $2T\sin 30° = 8g$ .
A1 (AO1.1b) — $T = \dfrac{8g}{2 \cdot 0.5} = 8g = 78.4$ N.

(b)

M1 (AO1.1b) — resolving horizontally and vertically at the hinge: horizontal component $X = T\cos 30°$ , vertical component $Y = 8g - T\sin 30°$ .
A1 (AO1.1b) — substituting to obtain $X = 8g\cos 30° = 4g\sqrt{3}$ , $Y = 8g - 8g \cdot 0.5 = 4g$ .
A1 (AO2.1) — magnitude $\sqrt{X^2 + Y^2} = \sqrt{48g^2 + 16g^2} = 8g$ N $\approx 78.4$ N; direction $\arctan(Y/X) = \arctan(1/\sqrt{3}) = 30°$ above horizontal.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. Statics questions on Paper 3 are dominated by AO1 (procedural mechanics) with one or two AO2/AO3 marks reserved for choosing the smartest pivot or interpreting a vector reaction in magnitude–direction form.

Synoptic links

Connects to:

Forces and Newton's laws (section 8 of 9MA0-03): statics is the limiting case of dynamics where acceleration is zero. The force-resolution discipline learnt in projectile and connected-particle problems carries over directly: choose axes, resolve, equate. The only new ingredient is the moment equation.
Moments (section 11): the principle $\Sigma M = 0$ about any point is the single new mathematical idea in rigid-body statics. Choosing the pivot smartly (through an unknown force) eliminates that force from the equation — a strategic skill that distinguishes A* candidates.
Friction (section 10): rigid-body statics problems typically pair with the friction inequality $F \leq \mu R$ . "Minimum $\mu$ " or "on the point of slipping" cues replace the inequality with the equality $F = \mu R$ , converting two unknowns into one.
Vectors (Paper 1, section 10): computing the moment of an oblique force as $F \cdot d \cdot \sin\theta$ is the magnitude of the cross product $\vec{r} \times \vec{F}$ . The 2-D scalar moment is the $\hat{k}$ -component of the full 3-D vector moment — the same idea that underpins angular momentum and torque in further mechanics.
Engineering structures (applied mathematics extensions): trusses, bridges and cantilevers are analysed by applying $\Sigma F = 0$ and $\Sigma M = 0$ to each rigid member or joint. The A-Level ladder problem is the entry point to the method of joints and the method of sections used in first-year engineering statics.

Mark-scheme literacy

Statics questions on 9MA0 Paper 3 split AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	65–75%	Drawing the force diagram, resolving in two perpendicular directions, taking moments, applying $F = \mu R$
AO2 (reasoning / interpretation)	15–25%	Choosing the pivot strategically, interpreting "smooth" or "rough" cues, expressing reactions in magnitude–direction form
AO3 (problem-solving / modelling)	5–15%	Reading a real-world setup (ladder, hinge, signpost) and deciding which modelling assumptions to invoke

Examiner-rewarded phrasing: "taking moments about $A$ to eliminate the unknown reaction"; "since the wall is smooth, there is no frictional component at $B$ "; "modelling the ladder as a uniform rod, weight acts at the midpoint"; "on the point of slipping, friction is limiting, so $F = \mu R$ ". Phrases that lose marks: failing to state the modelling assumption ("uniform rod" implies weight at centre — say so); writing $F = \mu R$ without justifying that friction is limiting; omitting units on a final reaction force.

A specific Edexcel pattern: questions ending "...find the magnitude and direction of the reaction at the hinge" demand both. Many candidates compute the components, then forget to combine them and quote a direction — losing the final A1.

Grade-band model answers

3-mark question

Question: A uniform plank $AB$ of mass $12$ kg and length $4$ m rests on two supports at $A$ and $B$ . Find the magnitudes of the reactions at $A$ and $B$ .

Grade C response (~150 words):

The plank is uniform, so its weight $12g = 117.6$ N acts at the midpoint $M$ , halfway between $A$ and $B$ . By symmetry, both supports carry equal reactions.

Vertically: $R_A + R_B = 12g$ .

By symmetry: $R_A = R_B$ , so each is $\dfrac{12g}{2} = 6g = 58.8$ N.

Statics and Equilibrium

Statics and Equilibrium

Conditions for Equilibrium

Resolving Forces

Triangle of Forces

Equilibrium on an Inclined Plane

Rigid Body Equilibrium: Combining Forces and Moments

Worked Example: Ladder Problem

Hinged and Pivoted Bodies

Bodies Supported by Strings

Summary

A-Level Deep Dive: Statics and Equilibrium

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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