Mathematics 9709 · syllabus 4.1
Cambridge A Level Mathematics 9709 — Forces and equilibrium topical past papers
Resolving forces into components, finding resultants, and using friction and normal contact forces to solve problems about particles in equilibrium.
- Questions
- 157
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 45
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What Forces and equilibrium covers
Mechanics in 9709 is Paper 4, and section 4.1 is about particles whose forces balance. You identify the forces acting, such as weight, the normal contact force, friction, tension, thrust and any applied force, draw them on a diagram, and resolve them into components. In equilibrium the components in any direction sum to zero, so resolving in two perpendicular directions gives two equations to solve.
A contact force between surfaces has two parts: the normal reaction R and the friction F. A smooth contact has no friction. On a rough surface F ≤ μR, with F = μR when the particle is in limiting equilibrium, on the point of slipping. Problems include particles on inclined planes, held by strings at angles, or pushed by a force at an angle to a slope, where the direction of friction depends on which way the particle is about to move.
Paper 4 treats every body as a particle, so moments and centres of mass are not required; those topics now belong to Further Mathematics. Answers come from calculation rather than scale drawing, and g is taken as 10 m s⁻². The same resolving underpins Newton's laws of motion, where the forces no longer balance.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 4.1 Forces and equilibrium
What you should be able to do
- Draw a clear force diagram showing every force on a particle
- Resolve forces into components and find the magnitude and direction of a resultant
- Use the conditions for equilibrium to find unknown forces or angles
- Use F ≤ μR, with F = μR in limiting equilibrium, on horizontal and inclined surfaces
- Apply Newton's third law to pairs of contact forces
Key ideas
- In equilibrium the sum of the components of the forces in any direction is zero
- Components X and Y have a resultant of magnitude √(X² + Y²), at an angle tan⁻¹(Y/X) to the direction of X
- Friction: F ≤ μR, with F = μR in limiting equilibrium
- On a slope at angle α, the weight mg has a component mg sin α down the slope and mg cos α perpendicular to it
- Newton's third law: the forces two bodies exert on each other are equal in size and opposite in direction
How Forces and equilibrium has been examined
From 2022 to 2026, Forces and equilibrium was on Paper 4 · Mechanics in 11 of 11 exam series, worth about 18% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 4 · Mechanics | 45 | 18% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 6.
Filed in Momentum under: Forces & Equilibrium.
A worked example
A box of mass 4 kg rests on a rough plane inclined at 30° to the horizontal, where the coefficient of friction is 0.2. A force of P N acts on the box up the plane, parallel to a line of greatest slope. Find the least value of P for which the box remains at rest.
- The weight is 4 × 10 = 40 N. At the least P the box is about to slide down, so friction acts up the plane and F = μR.
- Resolve perpendicular to the plane: R = 40 cos 30° = 34.64 N.
- So F = 0.2 × 34.64 = 6.928 N.
- Resolve along the plane: P + F = 40 sin 30° = 20.
- P = 20 − 6.928 = 13.07.
Answer: P = 13.1 N (3 s.f.)
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Taking the normal reaction as mg on a slope, or when another force has a vertical component: resolve perpendicular to the surface to find R each time.
- Setting F = μR when the particle is not on the point of slipping: friction only reaches μR in limiting equilibrium.
- Putting friction in the wrong direction: it opposes the motion the particle would otherwise have, so a particle about to slide up a slope has friction acting down it.
- Swapping sin and cos when resolving the weight on a slope: the component along a slope at angle α is mg sin α.
Exam technique
- Draw a large force diagram, and label every force and angle, before writing any equation.
- Resolve parallel and perpendicular to an inclined plane, which keeps the unknown normal reaction in one equation.
- Use g = 10 m s⁻² and give answers to 3 significant figures unless told otherwise.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.