OCR A Level Mathematics A: Friction and rough surfaces (H240) – Revision Notes

These are condensed notes on friction, references 3.03r to 3.03v of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026). The study guide explains each idea in full with longer examples. Once the notes feel familiar, check yourself with the practice questions, the course hub, the checklist and the free diagnostics.

The first reference (3.03r) is Stage 1 content shared with AS. References 3.03s to 3.03v are A Level only. The specification excludes knowledge of the angle of friction. Take g = 9.8 m s⁻².

Key ideas and results

Ref	Idea	Result
3.03r (Stage 1)	Friction is a contact force along the surface that opposes sliding or the tendency to slide	If its value is given, treat it as any other force in F = ma
3.03s (A Level only)	Contact force = normal component + frictional component	Magnitude S = √(R² + F²)
3.03t (A Level only)	Coefficient of friction μ, no units	F ≤ μR
3.03t	Limiting friction	F = μR, on the point of moving
3.03u (A Level only)	Static and limiting equilibrium	Resultant is zero in two perpendicular directions
3.03v (A Level only)	Motion on rough surfaces	F = μR against motion; then F = ma

Must-know distinctions

Situation	Friction F	How to find it
At rest, not about to move	F < μR	From equilibrium
At rest, about to move	F = μR	Limiting equilibrium
Sliding	F = μR	Direction opposes motion
Smooth contact	F = 0	Model of no friction

• Normal force R is found from resolving perpendicular to the surface. It is not always mg.
• Friction F is found from resolving along the surface, or from μR when limiting or sliding.
• Contact force magnitude needs both R and F, joined by Pythagoras.

Method: any friction problem

1. Draw a diagram with weight, R, F, and any applied forces. Show the direction of motion or the tendency to move.
2. Resolve perpendicular to the surface to find R.
3. Decide the state: at rest, limiting, or sliding.
4. Write F = μR if limiting or sliding. Otherwise leave F unknown and solve from equilibrium.
5. Resolve along the surface: sum of forces = 0 if in equilibrium, or = ma if moving.
6. Check the answer: is F ≤ μR? Is R positive?

Formulas for the surface types

Setting	R	Friction or acceleration
Horizontal, no vertical forces besides weight	mg	Maximum friction μmg
Horizontal, force P at angle θ above	mg − P sin θ	Pull: R reduces
Horizontal, force P at angle θ below	mg + P sin θ	Push: R increases
Slope at angle α, at rest, no other force	mg cos α	F = mg sin α, stays at rest if tan α ≤ μ
Slope, sliding down	mg cos α	a = g sin α − μg cos α
Slope, sliding up	mg cos α	Deceleration g sin α + μg cos α
Horizontal, no driving force	mg	Deceleration μg

Contact is lost when R = 0.

Worked reminders

Limiting equilibrium with a pull. A 12 kg box on ground with μ = 0.4 is pulled at 25° above the horizontal and is about to slide.

R = 117.6 − T sin 25°,   T cos 25° = 0.4R
T = 47.04 ÷ (cos 25° + 0.4 sin 25°) = 43.7 N

Range of P up a slope. An 8 kg block on a 30° slope, μ = 0.3, with R = 67.9 N, mg sin 30° = 39.2 N and μR = 20.4 N.

Least P:    39.2 − 20.4 = 18.8 N   (about to slide down)
Greatest P: 39.2 + 20.4 = 59.6 N   (about to slide up)

Stopping on level ground. Projected at 7 m s⁻¹, μ = 0.2: deceleration 0.2 × 9.8 = 1.96 m s⁻², so s = 49 ÷ 3.92 = 12.5 m.

Down a slope. At 20°, μ = 0.45: a = 9.8(sin 20° − 0.45 cos 20°) = −0.792 m s⁻². It decelerates. Since tan 20° < 0.45, it stays at rest when it stops.

More worked reminders

Contact force (3.03s). Normal component 45 N, frictional component 18 N: S = √(45² + 18²) = √2349 = 48.5 N.

Friction given as a vector (3.03r). A 2.5 kg particle on a horizontal surface has an applied force (11i + 7j) N. Friction has magnitude 5 N and acts in the direction of −(4i + 3j), so it is (−4i − 3j) N.

Resultant = (7i + 4j) N
a = (2.8i + 1.6j) m s⁻²,   |a| = √10.4 = 3.22 m s⁻²

Two stages (3.03v). A 5 kg body on rough horizontal ground, μ = 0.2, is pulled by a horizontal 30 N force for 3 s from rest, then the force is removed.

Stage 1: a = (30 − 0.2 × 49) ÷ 5 = 4.04 m s⁻²,  v = 12.12 m s⁻¹
Stage 2: deceleration = 0.2 × 9.8 = 1.96 m s⁻²
         s = 12.12² ÷ (2 × 1.96) = 37.5 m

Method: will it move?

1. Find R, then the maximum friction μR.
2. Find the force that friction would have to balance. On a slope this is mg sin α plus or minus any applied force along the slope. On the flat it is the horizontal part of the applied force.
3. If that force is no greater than μR, the body stays at rest and F equals that force.
4. If it is greater, the body moves, F = μR, and the resultant is the difference.
5. State the conclusion in words, with both numbers shown.

Method: stopping and what happens next

1. Find the deceleration from friction (and the weight component, on a slope).
2. Use v² = u² + 2as with v = 0, or v = u + at, for the distance or time.
3. On a slope, compare tan α with μ. If tan α ≤ μ, the body stays at rest. If tan α > μ, it slides back down.

Quick self-test

1. The normal force is 30 N and μ = 0.5. What is the greatest possible friction force?
2. The normal component of a contact force is 24 N and the frictional component is 7 N. Find the magnitude of the contact force.
3. A 50 kg trunk is on rough horizontal ground, μ = 0.3. Find the maximum friction.
4. A block rests on a rough plane with μ = 0.5. What is the greatest inclination at which it can stay at rest? Give your answer to 3 s.f.
5. A 10 kg block slides down a plane at 30° with μ = 0.2. Find its acceleration to 3 s.f.
6. The same block is projected up the same plane. Find its deceleration to 3 s.f.
7. A 15 kg block on rough horizontal ground is pulled by a horizontal 60 N force and accelerates at 1.5 m s⁻². Find μ to 3 s.f.
8. A block is projected at 10 m s⁻¹ along rough horizontal ground with μ = 0.5. How far does it slide, to 3 s.f.?
9. A 3 kg block on a horizontal surface is pulled by a 20 N force at 40° above the horizontal. Find the normal force to 3 s.f., assuming it stays in contact.
10. A 5 kg block on rough horizontal ground, μ = 0.3, is pulled by a 25 N force at 30° above the horizontal. Does it move?

Answers

1. Maximum F = 0.5 × 30 = 15 N.
2. S = √(24² + 7²) = √625 = 25 N.
3. R = 50 × 9.8 = 490 N, so maximum F = 0.3 × 490 = 147 N.
4. tan α = μ, so α = tan⁻¹ 0.5 = 26.6°.
5. a = 9.8(sin 30° − 0.2 cos 30°) = 3.20 m s⁻².
6. Deceleration = 9.8(sin 30° + 0.2 cos 30°) = 6.60 m s⁻².
7. F = 60 − 15 × 1.5 = 37.5 N, R = 147 N, μ = 37.5 ÷ 147 = 0.255.
8. Deceleration = 0.5 × 9.8 = 4.9, s = 100 ÷ 9.8 = 10.2 m.
9. R = 3 × 9.8 − 20 sin 40° = 29.4 − 12.86 = 16.5 N.
10. R = 49 − 12.5 = 36.5 N, maximum F = 10.95 N. The horizontal pull is 25 cos 30° = 21.65 N, which is greater, so it moves.

Where marks are usually lost

• Using F = μR for a stationary body that is not close to slipping.
• Taking R = mg on a slope or with an angled rope.
• Putting friction in the wrong direction after the motion reverses.
• Adding R and F instead of using Pythagoras for the contact force.
• Rounding g, sine or cosine values early and losing accuracy in the answer.
• Not saying whether the body stays at rest, after finding that it stops.
• Writing a bare number with no comparison of F and μR when asked to show that a body does not move.
• Mixing up the least and greatest force: the least force has friction up the slope at its maximum.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. These notes follow references 3.03r to 3.03v.