A block of weight 40\,N sits on a slope at 30^. Find the component of weight down the slope. [2 marks]

The normal reaction on a block is 50\,N and μ = 0.3. Find the maximum friction. [2 marks]

CCEA CCEA 2019-style practice: A uniform rod AB of length 4\,m and weight 60\,N rests horizontally on supports at A and at a point C, where C is 3\,m from A. Find the reactions at the two supports.

The weight 60\,N acts at the midpoint, 2\,m from A. Take moments about A to find the reaction R_C at C: R_C × 3 = 60 × 2 = 120, so R_C = 40\,N. Resolve vertically for the reaction R_A at A: R_A + R_C = 60, so R_A = 60 - 40 = 20\,N. Markers reward the weight at the midpoint, taking moments about A, the reaction at C, and resolving vertically for the reaction at A.

Northern IrelandMathsSyllabus dot point

How do friction and moments govern the equilibrium and motion of bodies, including on inclined planes?

Resolving forces in two dimensions, the friction model with the coefficient of friction, motion and equilibrium on an inclined plane, and the moment of a force with the conditions for the equilibrium of a rigid body.

A CCEA A-Level Mathematics answer on resolving forces in two dimensions, the friction model with the coefficient of friction and limiting friction, motion and equilibrium on an inclined plane, and the moment of a force with the conditions for the equilibrium of a rigid body.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

CCEA wants you to resolve forces in two dimensions, apply the friction model with the coefficient of friction, analyse motion and equilibrium on an inclined plane, and use the moment of a force with the conditions for the equilibrium of a rigid body. This is the more demanding mechanics of A2 2, where bodies have size and friction matters.

The answer

Resolving forces and friction

The inclined plane

A body on the slope stays in equilibrium if the friction can balance $mg\sin\theta$ ; it slides if $mg\sin\theta$ exceeds the maximum friction.

Moments and rigid-body equilibrium

Worked example: equilibrium on a slope

Find the friction holding a block on a slope

A $10\,\text{kg}$ block rests in equilibrium on a rough slope inclined at $20^\circ$ . Find the friction force acting on it. Take $g = 9.8\,\text{m s}^{-2}$ .

step Resolve the weight

The component down the slope is

mg\sin\theta = 10 \times 9.8 \times \sin 20^\circ = 98 \times 0.342 = 33.5\,\text{N}.

step Apply equilibrium along the slope

In equilibrium the friction up the slope balances this component:

F = mg\sin\theta = 33.5\,\text{N}.

step Interpret

The friction is $33.5\,\text{N}$ up the slope. This is the actual friction needed for equilibrium, which must not exceed the maximum $\mu mg\cos\theta$ ; if it would, the block slides.

Examples in context

Example 1. A ladder against a wall. A ladder in equilibrium needs both force balance and moment balance; taking moments about the foot finds the wall reaction. Rigid-body equilibrium with moments is exactly the ladder-safety calculation.

Example 2. The steepest slope before sliding. A box stays put until the slope angle reaches $\tan^{-1}\mu$ , where $mg\sin\theta$ first exceeds $\mu mg\cos\theta$ . The friction model predicts the critical angle at which an object begins to slide.

Try this

Q1. A block of weight $40\,\text{N}$ sits on a slope at $30^\circ$ . Find the component of weight down the slope. [2 marks]

Cue. $40\sin 30^\circ = 20\,\text{N}$ .

Q2. The normal reaction on a block is $50\,\text{N}$ and $\mu = 0.3$ . Find the maximum friction. [2 marks]

Cue. $F_{\max} = \mu R = 0.3 \times 50 = 15\,\text{N}$ .

Q3. State the two conditions for a rigid body to be in equilibrium. [2 marks]

Cue. Zero resultant force and zero resultant moment about any point.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20217 marksA block of mass

5\,\text{kg}

rests on a rough horizontal surface with coefficient of friction

0.4

. A horizontal force

P

is applied. Find the value of

P

at which the block is on the point of moving. Take

g = 9.8\,\text{m s}^{-2}

Show worked answer →

Vertically, the normal reaction balances the weight: $R = mg = 5 \times 9.8 = 49\,\text{N}$ .

At the point of moving, friction is limiting: $F = \mu R = 0.4 \times 49 = 19.6\,\text{N}$ .

Horizontally, the block is on the point of moving when the applied force equals the limiting friction:

$P = F = 19.6\,\text{N}.$

Markers reward the normal reaction, the limiting-friction formula, and equating $P$ to the maximum friction.

CCEA 20196 marksA uniform rod

AB

of length

4\,\text{m}

and weight

60\,\text{N}

rests horizontally on supports at

A

and at a point

C

, where

C

3\,\text{m}

from

A

. Find the reactions at the two supports.

Show worked answer →

The weight $60\,\text{N}$ acts at the midpoint, $2\,\text{m}$ from $A$ .

Take moments about $A$ to find the reaction $R_C$ at $C$ :

$R_C \times 3 = 60 \times 2 = 120,$ so $R_C = 40\,\text{N}$ .

Resolve vertically for the reaction $R_A$ at $A$ :

$R_A + R_C = 60,$ so $R_A = 60 - 40 = 20\,\text{N}$ .

Markers reward the weight at the midpoint, taking moments about $A$ , the reaction at $C$ , and resolving vertically for the reaction at $A$ .

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)