Statics

This lesson covers statics at A-Level — the study of objects in equilibrium under the action of coplanar forces. Statics problems involve finding unknown forces, angles, or distances when a body is not accelerating.

Conditions for Equilibrium

A body is in equilibrium when:

The resultant force is zero: $\sum F_x = 0$ and $\sum F_y = 0$
The resultant moment about any point is zero: $\sum M = 0$

For a particle (all forces act at one point), only the force conditions are needed. For a rigid body, both force and moment conditions are needed.

Resolving Forces

To apply equilibrium conditions, resolve all forces into two perpendicular components (typically horizontal and vertical):

$\text{Horizontal: } \sum F_x = 0$ $\text{Vertical: } \sum F_y = 0$

Example: A particle is in equilibrium under three forces: 10 N acting horizontally to the right, a force $P$ at 30 degrees above the horizontal to the left, and a vertical force $Q$ downwards.

Horizontal: $10 - P\cos 30\degree = 0$ , so $P = \frac{10}{\cos 30\degree} = 11.5$ N

Vertical: $P\sin 30\degree - Q = 0$ , so $Q = 11.5 \times \sin 30\degree = 5.77$ N

Triangle of Forces

If three forces acting on a particle are in equilibrium, they can be represented as the sides of a triangle (drawn head to tail). This is known as the triangle of forces.

Conversely, if three forces form a closed triangle when drawn head to tail, the forces are in equilibrium.

This method is particularly useful when the forces act at known angles, as it allows the use of the sine rule and cosine rule to find unknowns.

Lami's Theorem

If three forces acting at a point are in equilibrium, and the angles between the forces are $\alpha$ , $\beta$ , and $\gamma$ (where $\alpha + \beta + \gamma = 360\degree$ ):

$\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}$

where $\alpha$ is the angle opposite $F_1$ , etc.

Exam Tip: Lami's theorem is a quick method for three-force equilibrium problems but only works when exactly three forces act at a point. The angles are the angles between the forces, not the angles the forces make with the horizontal.

Rigid Body Equilibrium

For a rigid body (e.g., a beam or ladder), take moments about a suitable point to set up equations.

Strategy:

Draw a clear force diagram showing all forces and their points of application.
Resolve horizontally: $\sum F_x = 0$ .
Resolve vertically: $\sum F_y = 0$ .
Take moments about a convenient point: $\sum M = 0$ .

Choose the pivot point wisely — selecting a point where unknown forces act eliminates those unknowns from the moment equation.

The Ladder Problem

A common statics problem involves a ladder leaning against a wall:

The ladder has weight $W$ acting at its centre (if uniform).
The ground is rough (friction $F$ acts horizontally at the base).
The wall is often smooth (no friction at the top).
Normal reactions act at both contact points.

Equilibrium equations:

Horizontal: $F = R_{\text{wall}}$
Vertical: $R_{\text{ground}} = W$
Moments about the base: $R_{\text{wall}} \times L\sin\theta = W \times \frac{L}{2}\cos\theta$

From the moments equation: $R_{\text{wall}} = \frac{W\cos\theta}{2\sin\theta} = \frac{W}{2\tan\theta}$

Summary

Equilibrium requires zero resultant force (in all directions) and zero resultant moment.
Resolve forces into horizontal and vertical components.
Use the triangle of forces or Lami's theorem for three-force particle equilibrium.
For rigid bodies, use both force resolution and moments.
Choose the pivot point to eliminate unknowns from the moment equation.
Ladder problems are a classic application combining friction, moments, and equilibrium.

Exam Tip: In statics problems, always start by drawing a complete force diagram. Label every force clearly. Then choose the most efficient combination of equations (resolve horizontally, resolve vertically, take moments) to solve for the unknowns. Avoid unnecessary equations — three unknowns require three equations.

A-Level Deep Dive: Statics

Spec mapping

AQA 7357 specification, Paper 3 — Mechanics, Sections U and V (Year 2) covers moments in simple static contexts (refer to the official specification document for exact wording).; "Solve problems involving rigid bodies in equilibrium under coplanar forces, including ladders, hinged rods and beams". Statics sits at the apex of the Mechanics strand: it requires confident handling of vectors (Section O), Newton's laws and resolved forces (Section P), friction (Section Q) and moments (the new Year 2 content of Section U). The defining new idea is that for a rigid body — as opposed to a particle — equilibrium demands two simultaneous conditions: $\sum \mathbf{F} = \mathbf{0}$ (no translation) and $\sum M = 0$ about any point (no rotation). The AQA formula booklet does not give moment formulae explicitly; the result $M = Fd$ (force times perpendicular distance) and the identity $M = Fr\sin\theta$ for a force at angle $\theta$ to a position vector of length $r$ must be memorised.

Worked example with full mark scheme

Question (8 marks):

A uniform ladder $AB$ of weight $W$ and length $2a$ rests with end $A$ on rough horizontal ground and end $B$ against a smooth vertical wall. The ladder is inclined at angle $\theta$ to the horizontal. The coefficient of friction between the ladder and the ground is $\mu$ . Show that, for the ladder to remain in equilibrium, $\mu \geq \dfrac{1}{2\tan\theta}$ , and hence find the smallest possible value of $\theta$ if $\mu = \tfrac{1}{4}$ .

Solution with mark scheme:

Step 1 — diagram and forces.

Forces acting on the ladder:

Weight $W$ downward at the midpoint $G$ (since the ladder is uniform, $G$ is at distance $a$ from $A$ along the rod).
Normal reaction $N$ from the ground at $A$ , vertically upward.
Friction $F$ at $A$ , horizontal, directed toward the wall (opposing impending slip away from the wall).
Normal reaction $S$ from the smooth wall at $B$ , horizontal, directed away from the wall.

B1 — correct free-body diagram with all four forces and directions labelled.

Step 2 — resolve horizontally and vertically.

Horizontal: $F = S$ . Vertical: $N = W$ .

M1 — both resolved equations stated correctly.

Step 3 — take moments about $A$ (eliminates $N$ and $F$ ).

Moments of $W$ : lever arm $= a\cos\theta$ (horizontal distance from $A$ to $G$ ), so moment $= Wa\cos\theta$ clockwise. Moments of $S$ : lever arm $= 2a\sin\theta$ (vertical height of $B$ above $A$ ), so moment $= S \cdot 2a\sin\theta$ anticlockwise.

Setting $\sum M_A = 0$ :

$S \cdot 2a\sin\theta = W \cdot a\cos\theta$

M1 — taking moments about a sensible pivot ( $A$ chosen to eliminate two unknowns). A1 — correct moment equation.

Step 4 — solve for $S$ .

$S = \dfrac{Wa\cos\theta}{2a\sin\theta} = \dfrac{W}{2\tan\theta}$

A1 — correct expression for $S$ .

Step 5 — apply the friction inequality.

For limiting equilibrium or any static state, $F \leq \mu N$ . Substituting $F = S$ and $N = W$ :

$S \leq \mu W \implies \dfrac{W}{2\tan\theta} \leq \mu W$

Dividing by $W$ (positive):

$\mu \geq \dfrac{1}{2\tan\theta}$

M1 — correct application of the friction limit $F \leq \mu N$ . A1 — printed inequality obtained, with the inequality direction justified.

Step 6 — substitute $\mu = \tfrac{1}{4}$ .

$\dfrac{1}{4} \geq \dfrac{1}{2\tan\theta} \implies 2\tan\theta \geq 4 \implies \tan\theta \geq 2$

So the minimum angle is $\theta = \arctan(2) \approx 63.43°$ .

A1 — correct minimum angle stated to a sensible degree of precision.

Total: 8 marks (B1 M3 A4).

Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): A uniform horizontal beam $AB$ of length $4\,\text{m}$ and mass $20\,\text{kg}$ is freely hinged at $A$ to a vertical wall. The beam is held horizontal by a light inextensible cable attached to the beam at $B$ and to the wall at a point $C$ above $A$ . The cable makes an angle of $30°$ with the beam. A particle of mass $5\,\text{kg}$ hangs from $B$ . Find the tension in the cable and the magnitude of the reaction at the hinge $A$ . Take $g = 9.8\,\text{m s}^{-2}$ .

Mark scheme decomposition by AO:

M1 (AO3.1b) — model the beam as a rigid rod with weight $20g$ acting at its midpoint, suspended particle $5g$ acting at $B$ , tension $T$ along cable at $30°$ to the beam, and hinge reaction with components $X$ (horizontal) and $Y$ (vertical) at $A$ .
M1 (AO1.1b) — take moments about $A$ to eliminate the hinge reaction: $T\sin 30° \cdot 4 = 20g \cdot 2 + 5g \cdot 4$ .
A1 (AO1.1b) — $2T = 40g + 20g = 60g$ , so $T = 30g = 294\,\text{N}$ (3 s.f.).
M1 (AO1.1b) — resolve horizontally: $X = T\cos 30°$ ; vertically: $Y + T\sin 30° = 25g$ , so $Y = 25g - 15g = 10g$ .
M1 (AO1.1b) — magnitude of hinge reaction: $R = \sqrt{X^2 + Y^2} = \sqrt{(T\cos 30°)^2 + (10g)^2}$ .
A1 (AO2.5) — $X = 30g \cdot \tfrac{\sqrt{3}}{2} = 15g\sqrt{3}$ , so $R = g\sqrt{(15\sqrt{3})^2 + 10^2} = g\sqrt{675 + 100} = g\sqrt{775} \approx 273\,\text{N}$ (3 s.f.).

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. Statics questions in AQA Paper 3 are AO1-heavy because the technique is procedural once the model is drawn correctly; the AO3 mark is reserved for the modelling step (correctly identifying which forces act where), and the AO2 mark for combining the components into a single vector magnitude with appropriate accuracy.

Synoptic links

Connects to:

Mechanics Section O — Vectors: every force in a statics problem is a 2-D vector. Resolving a tension along and perpendicular to a beam, and recombining hinge components into a magnitude $\sqrt{X^2 + Y^2}$ and direction $\arctan(Y/X)$ , is direct vector arithmetic. The $\mathbf{r} \times \mathbf{F}$ representation of a moment generalises naturally to the cross product in further mechanics.
Mechanics Section P — Forces and Newton's laws: the $\sum \mathbf{F} = \mathbf{0}$ condition for a rigid body is the static specialisation of Newton's second law $\mathbf{F} = m\mathbf{a}$ with $\mathbf{a} = \mathbf{0}$ . The same free-body diagrams and resolution techniques carry over directly from particle dynamics.
Mechanics Section Q — Friction: ladder problems are the canonical context where friction inequality $F \leq \mu N$ becomes the binding constraint. The transition from $F < \mu N$ (genuine static equilibrium) to $F = \mu N$ (limiting equilibrium, on the verge of slipping) is a key conceptual distinction examined repeatedly.
Pure Section L — Trigonometry: every statics calculation involves resolving forces using $\sin\theta$ and $\cos\theta$ . Exact-value trigonometry is rewarded in modelling questions where forces are at $30°$ , $45°$ or $60°$ — students who use $\sin 60° = \sqrt{3}/2$ rather than $0.866$ keep all later answers in exact form.
Engineering and physics application: statics is the foundational subject for civil and mechanical engineering. Truss analysis, beam bending and bridge design all begin with rigid-body equilibrium under coplanar forces. The "method of joints" and "method of sections" used to analyse pin-jointed trusses are direct extensions of the moment and resolution techniques studied here.

Mark-scheme literacy

Statics questions on AQA 7357 Paper 3 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–70%	Resolving forces correctly, taking moments about a sensible point, applying $F \leq \mu N$ , solving simultaneous linear equations
AO2 (reasoning / interpretation)	15–25%	Justifying the choice of pivot, interpreting "limiting equilibrium" or "smooth wall" correctly, presenting answers with sensible accuracy and units
AO3 (modelling / problem-solving)	15–25%	Drawing the free-body diagram, deciding which idealisations apply (light rod, smooth contact, uniform beam), interpreting the physical situation in mathematical form

Examiner-rewarded phrasing: "taking moments about $A$ to eliminate the hinge reaction"; "by symmetry, the centre of mass lies at the midpoint of the beam"; "since the wall is smooth, the reaction at $B$ is horizontal"; "for limiting equilibrium, $F = \mu N$ , and so…". Phrases that lose marks: omitting units on a force (" $T = 294$ " rather than " $T = 294\,\text{N}$ "); rounding mid-calculation rather than at the end; using $\geq$ when the question asks for the minimum value (which is an equality at the boundary).

A specific AQA pattern to watch: questions phrased "show that the ladder will not slip" expect a clean inequality argument: derive an expression for the required friction coefficient, then compare with the given $\mu$ . Conversely, "find the smallest angle such that…" expects an equality at the boundary of limiting equilibrium. Reading the command word precisely — "show", "find", "determine the minimum" — flags which logical structure is wanted.

Grade-band model answers

3-mark question

Question: A uniform plank of length $3\,\text{m}$ and weight $120\,\text{N}$ rests horizontally on two supports, one at each end. Find the magnitudes of the reactions at the two supports.

Grade C response (~150 words):

By symmetry, since the plank is uniform and the supports are at the ends, the weight is shared equally between them. Each reaction is therefore $\tfrac{120}{2} = 60\,\text{N}$ .

Examiner commentary: Full marks (3/3). The candidate has spotted the symmetry argument and applied it correctly. While a more formal answer would set up the moment equation explicitly, the symmetry argument is acceptable here because the configuration is genuinely symmetric. A Grade B answer would normally also state $\sum F_y = 0$ to confirm $R_1 + R_2 = 120$ as a check.

Grade A response (~190 words):*

Let $R_A$ and $R_B$ be the reactions at the supports at the ends $A$ and $B$ respectively. The plank is uniform, so its weight $W = 120\,\text{N}$ acts at the midpoint $G$ , which is $1.5\,\text{m}$ from each end.

Taking moments about $A$ :

$R_B \cdot 3 = 120 \cdot 1.5 \implies R_B = 60\,\text{N}$

Resolving vertically:

$R_A + R_B = 120 \implies R_A = 60\,\text{N}$

Examiner commentary: Full marks (3/3). The candidate sets up the model formally with named reactions, identifies the location of the centre of mass explicitly using the "uniform" assumption, takes moments about a chosen pivot, and verifies the result by resolving vertically. The dual derivation (moments + resolution) demonstrates examiner-aware redundancy checking. This is the kind of presentation discipline that pays off on longer questions where one slip would otherwise be unrecoverable.

Statics

Statics

Conditions for Equilibrium

Resolving Forces

Triangle of Forces

Lami's Theorem

Rigid Body Equilibrium

The Ladder Problem

Summary

A-Level Deep Dive: Statics

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

6-mark question

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