Friction

This lesson covers the mechanics of friction at A-Level. Friction is a contact force that opposes the tendency of surfaces to slide relative to one another. Understanding the friction model is essential for solving problems involving objects on rough surfaces.

The Friction Model

The frictional force $F$ between two surfaces obeys the inequality:

$F \leq \mu R$

where:

$F$ is the frictional force
$\mu$ is the coefficient of friction (a constant that depends on the surfaces in contact)
$R$ is the normal reaction force between the surfaces

Two Regimes

Situation	Friction	Condition
Static (not moving or about to move)	$F \leq \mu R$	Friction takes whatever value is needed to prevent motion, up to the limiting value
Limiting equilibrium (about to move)	$F = \mu R$	Friction is at its maximum value
Dynamic (sliding)	$F = \mu R$	Friction equals the limiting value (A-Level model assumes static and dynamic coefficients are equal)

Exam Tip: Friction is not always equal to $\mu R$ . It equals $\mu R$ only when the object is sliding or on the point of sliding (limiting equilibrium). If the object is stationary and not about to move, friction is less than $\mu R$ and must be calculated from the equilibrium conditions.

Direction of Friction

Friction always acts in the direction that opposes the motion or the tendency to move:

If the object is moving, friction acts in the opposite direction to the velocity.
If the object is stationary, friction acts in the direction that prevents the tendency to move.

Friction on a Horizontal Surface

Example: A 10 kg block rests on a rough horizontal surface with $\mu = 0.4$ . A horizontal force of 30 N is applied. Does the block move? If so, find the acceleration.

$R = mg = 10 \times 9.8 = 98$ N

Maximum friction: $F_{\max} = \mu R = 0.4 \times 98 = 39.2$ N

Since the applied force (30 N) < $F_{\max}$ (39.2 N), the block does not move. Friction = 30 N (equal and opposite to the applied force).

If the applied force were 50 N:

$50 > 39.2$ , so the block moves.

$F_{\text{net}} = 50 - 39.2 = 10.8$ N

$a = \frac{F_{\text{net}}}{m} = \frac{10.8}{10} = 1.08$ m/s $^2$

Friction on an Inclined Plane

On a rough inclined plane at angle $\theta$ :

Weight component down the slope: $mg\sin\theta$
Normal reaction: $R = mg\cos\theta$
Maximum friction: $F_{\max} = \mu R = \mu mg\cos\theta$

Condition for Sliding Down the Plane

The object will slide down the plane if:

$mg\sin\theta > \mu mg\cos\theta$ $\tan\theta > \mu$

The angle of friction $\lambda$ is defined by $\tan\lambda = \mu$ . The object slides when $\theta > \lambda$ .

Limiting Equilibrium on a Slope

When the object is on the point of sliding down:

$mg\sin\theta = \mu mg\cos\theta$ $\mu = \tan\theta$

This gives a method for finding $\mu$ experimentally.

Problems Involving Both Friction and Applied Forces

When an applied force $P$ acts on an object on a rough surface:

Draw a force diagram.
Resolve perpendicular to the surface to find $R$ (which may depend on $P$ if $P$ has a perpendicular component).
Calculate $F = \mu R$ (if sliding or limiting).
Resolve parallel to the surface and apply $F = ma$ .

Exam Tip: If a force is applied at an angle to the surface, it will change the normal reaction. For example, pushing downwards at an angle increases $R$ (and hence increases friction), while pulling upwards at an angle decreases $R$ (and hence decreases friction). Always resolve perpendicular to the surface first.

Summary

Friction opposes the tendency to move: $F \leq \mu R$ .
At limiting equilibrium or when sliding: $F = \mu R$ .
On an incline: $R = mg\cos\theta$ , weight component down slope = $mg\sin\theta$ .
Object slides if $\tan\theta > \mu$ .
Always resolve perpendicular to the surface first to find $R$ , then apply friction.
Friction is only at its maximum when the object is moving or about to move.

Exam Tip: A very common exam mistake is to write $F = \mu R$ without checking whether the object is actually sliding or on the point of sliding. If the object is in static equilibrium, $F < \mu R$ and you must find $F$ from the equilibrium conditions.

A-Level Deep Dive: Friction

Spec mapping

AQA 7357 specification, Paper 3 — Mechanics, Section U: Forces and Newton's Laws (Year 2 content) covers the concept of a friction force and its effect on motion. Use the model $F \leq \mu R$ to solve problems involving rough contact (refer to the official specification document for exact wording). Friction sits at the synoptic heart of A-Level Mechanics: it is examined explicitly in Paper 3 but draws on every prior mechanics topic — vector resolution, Newton's three laws, equilibrium of a particle, kinematics under variable acceleration — and feeds forward into rigid-body work in further-mathematics specifications. The AQA formula booklet does not list $F \leq \mu R$ explicitly because it is treated as a defining model assumption rather than a derived formula; you must recall it and state the equality condition (point of slipping) yourself. Friction questions on Paper 3 typically carry between 8 and 14 marks of a 100-mark paper, but the fingerprints of the friction model — opposing direction, bounded magnitude, threshold behaviour — appear far more widely.

Worked example with full mark scheme

Question (8 marks):

A particle of mass $5\,\text{kg}$ 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 directed up the line of greatest slope.

(a) Given that the particle is on the point of slipping down the plane, find the value of $P$ in terms of $\mu$ and $g$ , and hence find $P$ when $\mu = 0.2$ and $g = 9.8\,\text{m s}^{-2}$ . (5)

(b) Given instead that $P = 40\,\text{N}$ and the particle accelerates up the plane at $0.6\,\text{m s}^{-2}$ , find the value of $\mu$ . (3)

Solution with mark scheme:

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

Taking axes along and perpendicular to the slope, the weight $5g$ acts vertically downwards. Its perpendicular component (into the plane) is $5g\cos 30°$ and its parallel component (down the slope) is $5g\sin 30°$ .

$R = 5g\cos 30°$

M1 — resolution of weight perpendicular to the slope, with the cosine of the slope angle. A persistent error is to write $R = 5g\sin 30°$ — confusing parallel and perpendicular components. The geometry to anchor: the normal to the plane is closest to the vertical, so the perpendicular component is the cosine component.

A1 — correct expression $R = 5g\cos 30° = \tfrac{5g\sqrt{3}}{2}$ .

Step 2 — identify the direction of friction at the point of slipping down.

Because the particle is on the point of slipping down the plane, motion would be down the slope, so friction acts up the slope. At the threshold, $F = \mu R$ (the equality holds because the particle is on the limit of slipping; below this threshold $F < \mu R$ ).

M1 — recognising friction acts up the slope (opposing impending motion) and writing $F = \mu R$ (not $F \leq \mu R$ ) because the particle is on the point of slipping.

Step 3 — resolve parallel to the plane.

Up-slope forces: $P$ and $F = \mu R = 5\mu g\cos 30°$ . Down-slope force: $5g\sin 30°$ . Equilibrium parallel to the plane:

$P + \mu \cdot 5g\cos 30° = 5g\sin 30°$

$P = 5g\sin 30° - 5\mu g\cos 30° = 5g\left(\tfrac{1}{2} - \tfrac{\mu\sqrt{3}}{2}\right)$

M1 — equation of equilibrium parallel to the slope, with friction added to $P$ on the up-slope side (since both oppose the impending down-slope motion).

Step 4 — substitute $\mu = 0.2$ and $g = 9.8$ .

$P = 5 \cdot 9.8 \cdot \left(0.5 - 0.2 \cdot \tfrac{\sqrt{3}}{2}\right) = 49 \cdot (0.5 - 0.1732) = 49 \cdot 0.3268 \approx 16.0\,\text{N}$

A1 — $P \approx 16.0\,\text{N}$ (to 3 s.f.).

(b) Step 1 — same perpendicular resolution. $R = 5g\cos 30° \approx 42.44\,\text{N}$ .

Step 2 — friction now opposes upward motion. Because the particle accelerates up the plane, friction acts down the plane and is fully developed: $F = \mu R$ .

Newton's second law parallel to the plane (positive up the slope):

$P - 5g\sin 30° - \mu \cdot 5g\cos 30° = 5 \cdot 0.6$

M1 — Newton II equation with friction subtracted from $P$ on the up-slope side, mass-times-acceleration on the right.

Step 3 — substitute and solve for $\mu$ .

$40 - 24.5 - \mu(42.44) = 3$ $12.5 = 42.44\mu$ $\mu \approx 0.295$

M1 A1 — correct rearrangement and answer to a sensible accuracy. Examiners accept $0.29$ , $0.30$ , or $0.295$ to 2–3 s.f.

Total: 8 marks (M5 A3, split as shown).

The two parts together expose the single defining feature of friction problems: the direction of friction depends on the direction of impending or actual motion, not on the direction of the applied force. Many candidates resolve correctly but write the friction term with the wrong sign — and lose every accuracy mark downstream.

Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): A box of mass $8\,\text{kg}$ is pulled along rough horizontal ground by a light inextensible rope inclined at $25°$ above the horizontal. The coefficient of friction between the box and the ground is $0.4$ , and the tension in the rope is $T\,\text{N}$ .

(a) Given that the box is on the point of slipping, find $T$ . (4)

(b) The rope is then changed so that it pulls horizontally with the same magnitude $T$ . Show that the box accelerates, and find the acceleration. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO3.1b) — modelling: rope tension resolved into horizontal $T\cos 25°$ and vertical (upward) $T\sin 25°$ components; recognising that the upward component reduces the normal reaction.
A1 (AO1.1b) — $R = 8g - T\sin 25°$ stated with correct sign.
M1 (AO1.1b) — at the point of slipping, $F = \mu R$ , and horizontal equilibrium gives $T\cos 25° = 0.4(8g - T\sin 25°)$ .
A1 (AO1.1b) — solving for $T$ : $T(\cos 25° + 0.4\sin 25°) = 0.4 \cdot 8g$ , giving $T = \dfrac{0.4 \cdot 8 \cdot 9.8}{\cos 25° + 0.4 \sin 25°} \approx 28.5\,\text{N}$ .

(b)

M1 (AO1.1b) — with horizontal rope, $R = 8g$ (no vertical lift), so maximum static friction is $\mu R = 0.4 \cdot 8 \cdot 9.8 = 31.36\,\text{N}$ . Since $T \approx 28.5 < 31.36$ , you might expect no motion — but the rope is now pulling and the box was previously sliding (or the question implies kinetic friction continues). Resolve and apply Newton II: $T - \mu \cdot 8g = 8a$ .
A1 (AO2.4) — $28.5 - 31.36 = 8a$ gives $a \approx -0.36\,\text{m s}^{-2}$ , i.e. the box decelerates. Stating the negative sign and interpreting it physically earns the AO2 mark.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This question is genuinely synoptic in the AQA sense — it tests the modelling decision (does upward rope tension reduce $R$ ? yes), the mathematical execution (resolve, equate, solve), and the interpretive judgement (negative acceleration means deceleration, not motion in the opposite direction).

Synoptic links

Connects to:

Section R — Forces and Newton's Laws (Year 1 introduction): the very definition of $R$ as the normal contact force is set up in Year 1 mechanics. Friction questions exploit the fact that $R$ is not always equal to $mg$ — it depends on what other forces have vertical components. Every friction problem starts with the perpendicular-to-surface equation, and that equation is genuinely Year 1 content under stress.
Section S — Kinematics and section T — Moments (Year 2): when friction problems involve a rigid body on the point of toppling versus sliding, you must compare two failure modes. The body slides when $P > \mu R$ and topples when the line of action of the resultant falls outside the base. AQA Year 2 questions ask "which happens first?" — pure synoptic friction-plus-moments work.
Section V — Projectiles, momentum, work-energy (Year 2): the work-energy theorem with friction reads $\tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2 = -\mu R d$ for sliding distance $d$ . Friction is the canonical non-conservative force; the energy lost to friction equals $\mu R d$ and reappears as heat. Energy-method friction problems are standard Year 2 fare.
Section P — Quantities and units (Year 1): $\mu$ is dimensionless — the ratio of two forces. Spotting that a candidate answer " $\mu = 4.2\,\text{N}$ " must be wrong on dimensional grounds is a free sanity check.
Connected particles (pulleys, strings): when two masses are connected over a rough pulley (or when one slides on a rough surface while the other hangs vertically), each mass needs its own equation of motion with friction acting on the surface contact. A* candidates write both equations and add to eliminate tension — the same Newton-II-for-systems technique used in connected-particle problems without friction.

Mark-scheme literacy

Friction questions on AQA 7357 Paper 3 split AO marks broadly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Resolving forces, applying Newton II, computing $F = \mu R$ , manipulating the resulting equations
AO2 (reasoning / interpretation)	25–35%	Correctly identifying friction direction; justifying when $F = \mu R$ versus $F \leq \mu R$ ; interpreting negative answers
AO3 (problem-solving / modelling)	10–25%	Choosing which model assumptions apply (smooth pulley, light rope, particle on point of slipping); critiquing the model

Examiner-rewarded phrasing: "since the particle is on the point of slipping, $F = \mu R$ (with equality)"; "friction acts up the slope, opposing the impending motion down the slope"; "modelling the box as a particle, so we ignore rotational effects". Phrases that lose marks: writing $F = \mu R$ without justifying the equality (when the particle is in static equilibrium below the threshold, $F < \mu R$ and equating overestimates friction); omitting "rough" or "smooth" when stating contact assumptions in modelling answers; failing to specify the sense of friction (up the slope, down the slope) before resolving.

A specific AQA pattern to watch: questions phrased "find the least value of $P$ such that ..." or "find the greatest value of $\mu$ such that ..." are signalling that the threshold $F = \mu R$ is the operating condition. The word "least" or "greatest" pins down which direction friction is acting. Misreading these adverbs flips the sign of the friction term and loses the entire question.

Grade-band model answers

3-mark question

Question: A particle of mass $2\,\text{kg}$ rests on rough horizontal ground. The coefficient of friction is $0.3$ . A horizontal force of $P\,\text{N}$ is applied. Find the maximum value of $P$ for which the particle remains in equilibrium. Take $g = 9.8\,\text{m s}^{-2}$ .

Grade C response (~150 words):

The normal reaction is $R = 2g = 2 \times 9.8 = 19.6\,\text{N}$ .

At the maximum $P$ , the particle is on the point of slipping, so $F = \mu R = 0.3 \times 19.6 = 5.88\,\text{N}$ .

For equilibrium, $P = F$ , so $P_{\max} = 5.88\,\text{N}$ .

Friction

Friction

The Friction Model

Two Regimes

Direction of Friction

Friction on a Horizontal Surface

Friction on an Inclined Plane

Condition for Sliding Down the Plane

Limiting Equilibrium on a Slope

Problems Involving Both Friction and Applied Forces

Summary

A-Level Deep Dive: Friction

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

More in Mathematics