AQA A-Level Mathematics: R: Forces and Newton's laws (7357) – Revision Notes

For full explanations and worked examples, read the Forces and Newton’s laws study guide first. These notes are for the final weeks.

They cover Section R: Forces and Newton’s laws (R1 to R6) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. The specification places Section R in Paper 2. A calculator is required in every 7357 paper. Take g = 9.8 m s⁻² unless told otherwise.

Links: practice questions · course hub · checklist · free 10-minute diagnostics · Quantities and units revision notes · Vectors revision notes · Kinematics revision notes · Moments revision notes.

Specification checklist

Ref	You must be able to
R1	Explain what a force is; use Newton’s first law
R2	Use F = ma in a straight line, with perpendicular forces, 2D vectors, or forces you resolve
R3	Use weight W = mg; motion under gravity; use g to the accuracy given
R4	Use Newton’s third law; equilibrium; smooth pulleys and connected particles; resolve in 2D
R5	Add forces; find resultants; dynamics in a plane
R6	Use F ≤ μR; coefficient of friction; rough surfaces; limiting friction and statics

Formulae to recall

Appendix B of the specification lists these as formulae you must be able to recall.

Result	Formula	Notes
Weight	W = mg	W in N, m in kg
Newton’s second law	F = ma	F is the resultant force
Friction	F ≤ μR	R is the normal reaction

Other results you use constantly (they follow from the above):

Situation	Result
Component of force P at angle θ to a direction	P cos θ along, P sin θ across
Particle on smooth plane at angle α	a = g sin α down the plane
Normal reaction on a plane at α (no other perpendicular forces)	R = mg cos α
On the point of slipping on a plane at α (weight, R and friction only)	μ = tan α
Equilibrium	resultant = 0 in every direction

The three laws in one line each

• First law: no resultant force ⇔ at rest or constant velocity.
• Second law: resultant force = mass × acceleration, in the same direction.
• Third law: A on B is equal and opposite to B on A; the pair act on different bodies.

About g (R3)

The specification says g may be assumed constant and is used to varying degrees of accuracy. It is not a universal constant: it depends on location. The inverse square law is not required. Use the value in the question, and round answers to match it.

Method: F = ma problems

1. Draw a clear diagram. Mark weight, normal reaction, tension, friction, driving force and resistances.
2. Mark the acceleration with a double arrow.
3. Resolve perpendicular to the motion: the total there is zero. This usually gives R.
4. Resolve along the motion: resultant = ma.
5. If friction acts and the body moves or is about to move, use F = μR.
6. Solve, then check the direction and size make sense.

Method: connected particles and pulleys

1. Model: string light and inextensible, pulley smooth → same tension on both sides, same size of acceleration.
2. Write F = ma for each particle in its own direction of motion.
3. Add the equations to remove T, find a, then substitute back for T.
4. For a rigid tow bar, find a from the whole system, then T from the rear body alone.
5. When a string goes slack or breaks, the tension becomes zero: start a new stage with new equations.

Method: equilibrium of a particle

1. Choose two perpendicular directions (often horizontal and vertical, or along and across a slope).
2. Resolve every force in each direction.
3. Set each total to zero and solve the pair of equations.
4. With forces as vectors, add the vectors and set the sum to 0.

Method: friction decisions

1. Find R, then the maximum friction μR.
2. Compare it with the force trying to cause motion.
3. Smaller than μR → stays at rest, friction equals that force.
4. Equal → limiting equilibrium, on the point of moving.
5. Larger → it moves; friction = μR, against the motion.

Worked reminders

A. Push downwards at an angle on a rough floor. A crate of mass 10 kg is pushed by a force of 50 N at 30° below the horizontal; μ = 0.3.

R = 10g + 50 sin 30° = 98 + 25 = 123 N   (push adds to R)
F_max = 0.3 × 123 = 36.9 N
Forward: 50 cos 30° = 43.30 N > 36.9 N → moves
a = (43.30 − 36.9) ÷ 10 = 0.640 m s⁻²

B. Vertical motion on a rope. Resultant = T − mg (up positive). Moving at constant speed → T = mg, whether going up or down.

C. Force on a smooth pulley. When the two string sections both hang vertically, the string pulls down on the pulley with 2T.

Must-know distinctions

• Mass vs weight: kg vs N; W = mg.
• Resultant force vs a single force: only the resultant goes in F = ma.
• Smooth vs rough: smooth means no friction (μ = 0).
• F = μR vs F ≤ μR: equality only when moving or on the point of moving.
• Equilibrium vs constant velocity: both have zero resultant; “at rest” is a special case.
• Third-law pair vs balanced forces: a third-law pair acts on two bodies; balanced forces act on one.
• Tension vs thrust: a string can only pull; a rod can pull (tension) or push (thrust).
• Light, inextensible, smooth pulley: each assumption has a job (see Overarching themes).

Quick self-test

1. Find the acceleration of a 3 kg particle acted on by a resultant force (6i − 9j) N, and its magnitude.
2. Find the weight of a 2.4 kg mass using g = 9.81 m s⁻².
3. A 12 kg block rests on rough horizontal ground, μ = 0.25. A horizontal force of 20 N acts on it. Find the friction force.
4. The force on the block in Q3 is raised to 36 N. Find its acceleration.
5. A particle slides down a smooth plane inclined at 20°. Find its acceleration.
6. A car of mass 900 kg has driving force 2000 N and total resistance 650 N on a level road. Find its acceleration.
7. A lift of mass 400 kg accelerates downwards at 1.2 m s⁻². Find the tension in the cable.
8. Particles of mass 4 kg and 6 kg hang either side of a smooth pulley on a light inextensible string. Find the acceleration and the tension.
9. A block on a rough plane inclined at 15° is on the point of slipping. Find μ.
10. Forces of 5 N along i and 12 N along j act on a particle. Find the magnitude of the resultant and its angle with i.
11. A van moves at constant speed with a driving force of 300 N. What is the total resistance? Which law?
12. A horizontal pull of 40 N gives an 8 kg block on rough horizontal ground an acceleration of 2 m s⁻². Find μ.

Answers

1. a = (2i − 3j) m s⁻²; |a| = √13 = 3.61 m s⁻².
2. W = 2.4 × 9.81 = 23.5 N.
3. μR = 0.25 × 117.6 = 29.4 N > 20 N, so it stays at rest and F = 20 N.
4. a = (36 − 29.4) ÷ 12 = 0.55 m s⁻².
5. a = 9.8 sin 20° = 3.35 m s⁻².
6. a = (2000 − 650) ÷ 900 = 1.5 m s⁻².
7. 400g − T = 400 × 1.2, so T = 400 × 8.6 = 3440 N.
8. a = (6 − 4)g ÷ 10 = 1.96 m s⁻²; T = 4(9.8 + 1.96) = 47.0 N.
9. μ = tan 15° = 0.268.
10. √(5² + 12²) = 13 N; tan⁻¹(12/5) = 67.4°.
11. 300 N, by Newton’s first law (zero resultant).
12. Friction = 40 − 8 × 2 = 24 N; R = 78.4 N; μ = 24 ÷ 78.4 = 0.306.

Where marks are usually lost

• Writing R = mg on a slope, or when a pulling or pushing force has a vertical component.
• Using F = μR for a body that is at rest and not about to move (Q3 above).
• Leaving out a force on the diagram, most often friction or the normal reaction, and then missing it in the equation.
• Using a single force, not the resultant, in F = ma.
• Sign errors when the body decelerates: keep one positive direction and let a come out negative.
• In pulley problems, using one particle’s weight in the other particle’s equation.
• Forgetting that tension drops to zero once a string breaks or goes slack.
• Resolving with sin where cos is needed: mark the angle on the diagram first.
• Giving an acceleration without its direction when the question asks for it, or a force without units.
• Rounding intermediate values (such as R) too early, so the final answer is out by more than the last figure.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for A-level exams June 2018 onwards: Section 3.19 R: Forces and Newton’s laws (R1 to R6), with Appendix B: mathematical formulae and identities.