Friction

This lesson covers friction as required by the Edexcel 9MA0 A-Level Mathematics specification. You need to understand the friction model, the coefficient of friction μ, and how to apply friction in equilibrium and motion problems.

The Friction Model

Friction is a force that opposes the motion (or attempted motion) of an object along a surface. It acts along the surface at the point of contact.

Key Properties

Friction always acts in the opposite direction to the motion (or to the direction in which the object would move if friction were absent).
The magnitude of friction depends on the normal reaction R and the coefficient of friction μ.
Friction is self-adjusting up to a maximum value.

The Laws of Friction

When the object is stationary or on the point of moving (limiting equilibrium):

F ≤ μR

where F is the frictional force, R is the normal reaction, and μ is the coefficient of friction.

When friction is at its maximum (limiting friction):

F = μR

This occurs when the object is on the point of sliding (limiting equilibrium) or when the object is already sliding.

When the object is in motion:

F = μR

The friction force equals μR while the object is sliding (the kinetic friction model used at A-Level).

The Coefficient of Friction

The coefficient of friction μ (mu) is a dimensionless number that depends on the two surfaces in contact.

0 ≤ μ
μ = 0 means a smooth surface (no friction).
Typical values: wood on wood ≈ 0.2-0.5; rubber on concrete ≈ 0.6-0.8.

Friction on a Horizontal Surface

Example: A 10 kg box is on a rough horizontal surface with μ = 0.4. A horizontal force of 50 N is applied.

Normal reaction: R = mg = 10(9.8) = 98 N Maximum friction: F_max = μR = 0.4 x 98 = 39.2 N

Since the applied force (50 N) > F_max (39.2 N), the box moves.

Resultant force = 50 - 39.2 = 10.8 N Acceleration: a = 10.8 / 10 = 1.08 m s⁻²

Friction on an Inclined Plane

On a plane inclined at angle θ to the horizontal:

Weight component down the slope = mg sin θ
Weight component perpendicular to the slope = mg cos θ
Normal reaction: R = mg cos θ
Maximum friction: F = μR = μmg cos θ

Object Sliding Down the Slope

Friction acts up the slope (opposing the downward motion):

mg sin θ - μmg cos θ = ma a = g(sin θ - μ cos θ)

Object Being Pushed Up the Slope

Friction acts down the slope (opposing the upward motion):

Applied force - mg sin θ - μmg cos θ = ma

Limiting Equilibrium on a Slope

An object is on the point of sliding down a slope when the friction is at its maximum and the net force along the slope is zero:

mg sin θ = μmg cos θ tan θ = μ

This gives the angle of friction — the steepest angle at which the object remains in equilibrium.

Exam Tip: If tan θ < μ, the object remains stationary (friction is not at its maximum). If tan θ > μ, the object slides. If tan θ = μ, the object is in limiting equilibrium.

Problems with Non-Horizontal Applied Forces

When a force is applied at an angle to the surface, it has:

A component along the surface (helps or opposes motion).
A component perpendicular to the surface (changes the normal reaction).

Example: A force P is applied at angle α above the horizontal to a box on a rough horizontal surface.

Resolving vertically: R + P sin α = mg, so R = mg - P sin α. Resolving horizontally: P cos α - F = ma (where F = μR for sliding).

Note: P sin α reduces the normal reaction, which in turn reduces the maximum friction. This is why pulling at an angle can be more efficient than pushing horizontally.

Worked Example: Two-Stage Problem

A 5 kg block is pushed along a rough surface (μ = 0.3) by a horizontal force of 25 N for 4 seconds, then the force is removed. Find how far the block travels after the force is removed.

Stage 1 (force applied): R = 5g = 49 N, F = 0.3 x 49 = 14.7 N. Resultant = 25 - 14.7 = 10.3 N. a = 10.3/5 = 2.06 m s⁻². v at end of Stage 1: v = 0 + 2.06(4) = 8.24 m s⁻¹.

Stage 2 (no applied force, friction decelerates): a = -14.7/5 = -2.94 m s⁻². v² = u² + 2as: 0 = 8.24² + 2(-2.94)s. s = 67.9/5.88 ≈ 11.5 m.

Summary

Friction opposes motion and satisfies F ≤ μR (with equality at limiting friction or during sliding).
On a slope: R = mg cos θ, and tan θ = μ at limiting equilibrium.
Non-horizontal forces change the normal reaction and therefore the friction.
Always draw a force diagram and resolve in two perpendicular directions.

A-Level Deep Dive: Friction

Spec mapping

Edexcel 9MA0-03 specification, Mechanics section 10 — Forces and friction covers the concept of a frictional force; understand and use the coefficient of friction; understand and use $F \leq \mu R$ , with $F = \mu R$ when on the point of moving (refer to the official specification document for exact wording). Friction sits at the heart of Year 2 Mechanics on Paper 3. It interacts directly with section 7 (Forces and Newton's laws), section 8 (Moments) when modelling rough hinges and ladders, and section 9 (Kinematics) whenever a particle decelerates on a rough surface. The Edexcel formula booklet does not state $F = \mu R$ — the relationship and the inequality must be quoted from memory, with the equality justified by the phrase "on the point of slipping".

Worked example with full mark scheme

Question (8 marks): A particle of mass $4\,\text{kg}$ rests on a rough plane inclined at $30°$ to the horizontal. The coefficient of friction between the particle and the plane is $\mu$ . A force of magnitude $P\,\text{N}$ is applied to the particle, parallel to and up the line of greatest slope. The particle is on the point of moving up the plane when $P = 30$ .

(a) Find the value of $\mu$ , giving your answer to 2 decimal places. (6)

(b) The force $P$ is now removed. Determine, with justification, whether the particle remains in equilibrium. (2)

Solution with mark scheme:

(a) Step 1 — resolve perpendicular to the plane.

Let $R$ be the normal reaction. With $g = 9.8\,\text{m s}^{-2}$ :

$R = 4g\cos 30° = 4 \times 9.8 \times \dfrac{\sqrt{3}}{2} = 19.6\sqrt{3} \approx 33.95\,\text{N}$

M1 — resolving perpendicular to the slope and using $\cos 30°$ correctly. Common error: writing $R = 4g$ (treating the slope as horizontal) loses M1 immediately.

A1 — correct value of $R$ .

Step 2 — resolve parallel to the plane (up positive).

Since the particle is on the point of moving up, friction acts down the plane and $F = \mu R$ .

$P - 4g\sin 30° - \mu R = 0$

M1 — equation of equilibrium parallel to the slope, with friction acting down the slope. The direction of $F$ is the load-bearing reasoning step: down because the impending motion is up.

Step 3 — substitute and solve.

$30 - 4 \times 9.8 \times 0.5 - \mu \times 19.6\sqrt{3} = 0$ $30 - 19.6 - 19.6\sqrt{3}\,\mu = 0$ $\mu = \dfrac{10.4}{19.6\sqrt{3}} = \dfrac{10.4}{33.95} \approx 0.306$

M1 — substituting $F = \mu R$ at the limiting case. A1 — correct algebraic rearrangement. A1 — $\mu \approx 0.31$ to 2 d.p.

(b) With $P = 0$ , the component of weight down the slope is $4g\sin 30° = 19.6\,\text{N}$ . Maximum available friction is $\mu R = 0.306 \times 33.95 \approx 10.4\,\text{N}$ .

Since $19.6 > 10.4$ , friction cannot balance the weight component; the particle slides down.

M1 — comparing weight component down the slope with maximum friction $\mu R$ . A1 — correct conclusion with numerical justification.

Total: 8 marks (M4 A4).

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A box of mass $8\,\text{kg}$ is pulled along rough horizontal ground by a light rope inclined at $20°$ above the horizontal. The tension in the rope is $T\,\text{N}$ . The coefficient of friction between the box and the ground is $0.4$ . The box accelerates at $0.5\,\text{m s}^{-2}$ .

(a) Show that $T(\cos 20° + 0.4 \sin 20°) = 0.4 \times 8g + 8 \times 0.5$ . (4)

(b) Hence find $T$ to 3 s.f. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO3.1b) — resolving vertically: $R + T\sin 20° = 8g$ , modelling the vertical component of tension lifting the box.
M1 (AO1.1b) — substituting $F = \mu R = 0.4(8g - T\sin 20°)$ for kinetic friction at the limiting/sliding value.
M1 (AO1.1b) — Newton's second law parallel to the ground: $T\cos 20° - F = 8 \times 0.5$ .
A1 (AO2.1) — eliminating $R$ and $F$ to reach the printed result.

(b)

M1 (AO1.1b) — substituting $g = 9.8$ and evaluating both sides.
A1 (AO1.1b) — $T \approx 27.0\,\text{N}$ (3 s.f.).

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This question is procedurally rich but structured: AO3 is earned at the modelling step where vertical resolution is needed because the tension has a vertical component.

Synoptic links

Connects to:

Section 7 — Forces and Newton's laws: friction is one of the three contact forces (with normal reaction and tension) that combine in $F = ma$ problems. Every rough-surface kinematics question routes through Newton II.
Section 7.4 — Resolution of forces: computing $R$ on a slope demands $\cos\theta$ and $\sin\theta$ resolution. Sign errors here propagate fatally into the friction calculation.
Section 7.6 — Statics of a particle: "on the point of slipping" problems are equilibrium problems with the extra constraint $F = \mu R$ . The angle of friction $\lambda = \arctan\mu$ unifies the analysis.
Section 8 — Moments (Year 2): rough ladder problems combine friction at the foot with the moment equation about the foot or the wall. The classic "ladder slips" question is friction + moments synoptic.
Modelling assumptions: "rough" enables the friction model; "smooth" forces $\mu = 0$ . Engineering applications — vehicle braking, conveyor belts, climbing — all reduce to the same Coulomb-friction inequality.

Mark-scheme literacy

Friction questions on 9MA0-03 split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Resolving correctly, applying $F = \mu R$ at the limit, executing Newton II
AO2 (reasoning / interpretation)	20–30%	Justifying friction direction, stating "on the point of slipping" implies equality, comparing weight component to maximum friction
AO3 (problem-solving / modelling)	15–25%	Choosing axes, identifying when vertical/horizontal vs along/perpendicular is the better resolution, recognising rope-angle effects

Examiner-rewarded phrasing: "since the particle is on the point of slipping, $F = \mu R$ "; "friction opposes the direction of impending motion"; "resolving perpendicular to the plane gives $R = mg\cos\theta$ ". Phrases that lose marks: "friction equals $\mu R$ " (without justifying the equality); writing $R = mg$ on a slope; stating a friction direction without referencing motion or impending motion.

A specific pattern: when a rope is inclined upward, the vertical component of tension reduces $R$ , which reduces $\mu R$ . Forgetting this and using $R = mg$ is the most common single mark-loss on Paper 3 friction questions involving inclined tension.

Grade-band model answers

3-mark question

Question: A block of mass $5\,\text{kg}$ rests on a rough horizontal surface with coefficient of friction $0.3$ . A horizontal force of $10\,\text{N}$ is applied. Determine whether the block moves.

Grade C response (~150 words):

The normal reaction is $R = 5g = 49\,\text{N}$ .

Maximum friction is $\mu R = 0.3 \times 49 = 14.7\,\text{N}$ .

The applied force is $10\,\text{N}$ , which is less than $14.7\,\text{N}$ , so the block does not move.

Friction

Friction

The Friction Model

Key Properties

The Laws of Friction

The Coefficient of Friction

Friction on a Horizontal Surface

Friction on an Inclined Plane

Object Sliding Down the Slope

Object Being Pushed Up the Slope

Limiting Equilibrium on a Slope

Problems with Non-Horizontal Applied Forces

Worked Example: Two-Stage Problem

Summary

A-Level Deep Dive: Friction

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

More in Mathematics