What is the friction model?

A body stays at rest while the driving force is at most the maximum friction μ R; once it exceeds μ R, the body accelerates with friction at its maximum.

A 10\,kg block is on a slope at 20^. Find the component of its weight down the slope (take g = 9.8\,m s^-2). [2 marks]

A force of 8\,N acts at a perpendicular distance of 0.5\,m from a pivot. Find the moment. [1 mark]

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

WJEC WJEC A2 style-style practice: A block of mass 5\,kg rests on a rough plane inclined at 30^ to the horizontal. The coefficient of friction is 0.4. Determine whether the block remains at rest. Take g = 9.8\,m s^-2.

Resolve the weight along and perpendicular to the plane, then compare the down-slope force with the maximum friction. Component down the slope: mgsin 30^ = 5(9.8)(0.5) = 24.5\,N. Normal reaction: R = mgcos 30^ = 5(9.8)(0.866) = 42.4\,N. Maximum friction: μ R = 0.4 × 42.4 = 17.0\,N. Since the down-slope force 24.5\,N exceeds the maximum friction 17.0\,N, the block does not remain at rest and slides down. Markers reward resolving the weight into components, finding the normal reaction from the perpendicular direction, comparing the down-slope force with the limiting friction, and the correct conclusion that the block slides.

WJEC WJEC A2 style-style practice: A uniform rod AB of length 4\,m and weight 60\,N rests horizontally on supports at A and at a point 1\,m from B. Find the reaction at the support near B.

Take moments about the support at A to eliminate the reaction there, using the weight at the centre. The weight 60\,N acts at the centre, 2\,m from A. The support near B is 3\,m from A (since it is 1\,m from B). Let the reaction near B be R_B. Taking moments about A (clockwise equals anticlockwise): R_B × 3 = 60 × 2. R_B = 1203 = 40\,N. Markers reward taking moments about a sensible point (A) to remove one unknown, placing the weight at the centre of the uniform rod, and the correct reaction of 40\,N. Forgetting that a uniform rod's weight acts at its midpoint is the standard error.

WalesMathsSyllabus dot point

How do we resolve forces on an inclined plane, model friction, and take moments to analyse equilibrium?

Resolving forces on inclined planes, the friction model and the coefficient of friction, connected particles, and moments and the equilibrium of rigid bodies.

A focused answer to WJEC A2 Unit 4 forces, covering resolving forces on inclined planes, the friction model and the coefficient of friction, connected particles, and taking moments for the equilibrium of rigid bodies.

Generated by Claude Opus 4.814 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

WJEC wants you to resolve forces on an inclined plane, to use the friction model $F \le \mu R$ with the coefficient of friction, to analyse connected particles (including over a pulley with friction or on a slope), and to take moments for the equilibrium of rigid bodies such as rods and beams. This is the A2 mechanics core, combining with calculus kinematics in longer questions.

The answer

Resolving forces on an inclined plane

On a slope, the natural directions are along and perpendicular to the plane, so resolve the weight into these.

The friction model

A body stays at rest while the driving force is at most the maximum friction $\mu R$ ; once it exceeds $\mu R$ , the body accelerates with friction at its maximum.

Connected particles

The same principles as AS apply: write $F = ma$ for each particle, with a common acceleration and (for a light string over a smooth pulley) a common tension. On a slope, include the $mg\sin\theta$ component and friction in the equation for the body on the incline.

Moments and rigid-body equilibrium

A moment measures the turning effect of a force.

Take moments to find a reaction

A uniform beam $AB$ of length $6\,\text{m}$ and weight $200\,\text{N}$ rests on supports at $A$ and $B$ . A load of $100\,\text{N}$ hangs $2\,\text{m}$ from $A$ . Find the reaction at $B$ .

step Place the forces

The beam's weight $200\,\text{N}$ acts at the centre, $3\,\text{m}$ from $A$ . The load $100\,\text{N}$ is $2\,\text{m}$ from $A$ . The reaction $R_B$ acts at $B$ , $6\,\text{m}$ from $A$ .

step Take moments about A

Taking moments about $A$ removes the reaction at $A$ : $R_B \times 6 = 200 \times 3 + 100 \times 2$ .

step Evaluate the right-hand side

$R_B \times 6 = 600 + 200 = 800$ .

step Solve

$R_B = \dfrac{800}{6} = 133\,\text{N}$ (to three significant figures).

Examples in context

Example 1. The angle of friction. A block is on the point of sliding down a rough slope when the down-slope force equals the maximum friction: $mg\sin\theta = \mu mg\cos\theta$ , so $\tan\theta = \mu$ . This angle, where slipping just begins, is the angle of friction, and it depends only on $\mu$ , not the mass. Resolving on the incline derives this neat result.

Example 2. A ladder against a wall. A uniform ladder leaning on a smooth wall and rough floor stays up only if the friction at the base provides enough horizontal force. Taking moments about the base, plus resolving horizontally and vertically, gives three equations for the reactions and the minimum $\mu$ . Moments and force balance together solve the classic ladder problem.

Try this

Q1. A $10\,\text{kg}$ block is on a slope at $20^{\circ}$ . Find the component of its weight down the slope (take $g = 9.8\,\text{m s}^{-2}$ ). [2 marks]

Cue. $mg\sin 20^{\circ} = 10(9.8)(0.342) = 33.5\,\text{N}$ .

Q2. A force of $8\,\text{N}$ acts at a perpendicular distance of $0.5\,\text{m}$ from a pivot. Find the moment. [1 mark]

Cue. Moment $= 8 \times 0.5 = 4\,\text{N m}$ .

Q3. On a slope, the normal reaction is $50\,\text{N}$ and $\mu = 0.3$ . Find the maximum friction. [2 marks]

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

Exam-style practice questions

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

WJEC A2 style6 marksA block of mass

5\,\text{kg}

rests on a rough plane inclined at

30^{\circ}

to the horizontal. The coefficient of friction is

0.4

. Determine whether the block remains at rest. Take

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

Show worked answer →

Resolve the weight along and perpendicular to the plane, then compare the down-slope force with the maximum friction.

Component down the slope: $mg\sin 30^{\circ} = 5(9.8)(0.5) = 24.5\,\text{N}$ .

Normal reaction: $R = mg\cos 30^{\circ} = 5(9.8)(0.866) = 42.4\,\text{N}$ .

Maximum friction: $\mu R = 0.4 \times 42.4 = 17.0\,\text{N}$ .

Since the down-slope force $24.5\,\text{N}$ exceeds the maximum friction $17.0\,\text{N}$ , the block does not remain at rest and slides down.

Markers reward resolving the weight into components, finding the normal reaction from the perpendicular direction, comparing the down-slope force with the limiting friction, and the correct conclusion that the block slides.

WJEC A2 style5 marksA uniform rod

AB

of length

4\,\text{m}

and weight

60\,\text{N}

rests horizontally on supports at

A

and at a point

1\,\text{m}

from

B

. Find the reaction at the support near

B

Show worked answer →

Take moments about the support at A to eliminate the reaction there, using the weight at the centre.

The weight $60\,\text{N}$ acts at the centre, $2\,\text{m}$ from $A$ . The support near $B$ is $3\,\text{m}$ from $A$ (since it is $1\,\text{m}$ from $B$ ).

Let the reaction near $B$ be $R_B$ . Taking moments about $A$ (clockwise equals anticlockwise):

$R_B \times 3 = 60 \times 2$ .

$R_B = \dfrac{120}{3} = 40\,\text{N}$ .

Markers reward taking moments about a sensible point (A) to remove one unknown, placing the weight at the centre of the uniform rod, and the correct reaction of $40\,\text{N}$ . Forgetting that a uniform rod's weight acts at its midpoint is the standard error.

Related dot points

Sources & how we know this

WJEC GCE AS/A Level Mathematics specification (from 2017) — WJEC (2017)