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

This guide teaches 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 lists Section R under Paper 2, alongside vectors, quantities and units, kinematics and moments. A calculator is required in every 7357 paper. Take g = 9.8 m s⁻² throughout unless a question says otherwise, and give answers to 3 significant figures where they are not exact.

Use this guide with the Forces and Newton’s laws revision notes and the Forces and Newton’s laws practice questions. The course hub is at /boards/aqa/a-level/mathematics/, the printable checklist at /checklists/aqa/a-level/mathematics/, and the free 10-minute diagnostics will show you where to start.

What Section R covers

Ref	What you must be able to do
R1	Understand the concept of a force; use Newton’s first law
R2	Use Newton’s second law for motion in a straight line (forces in two perpendicular directions, or simple 2D vectors), including when forces must be resolved (2 dimensions only)
R3	Use weight and motion in a straight line under gravity; g and its value to varying degrees of accuracy
R4	Use Newton’s third law; equilibrium of a particle; smooth pulleys and connected particles; resolving forces in 2 dimensions; equilibrium under coplanar forces
R5	Add forces; find resultant forces; dynamics for motion in a plane
R6	Use the F ≤ μR model for friction; coefficient of friction; motion on a rough surface; limiting friction and statics

Appendix B of the specification lists three results here that you must be able to recall: weight = mass × g, friction: F ≤ μR, and Newton’s second law in the form F = ma. Units (N, kg, m s⁻²) are covered in Quantities and units in mechanics, and adding force vectors in the Vectors study guide.

R1: forces and Newton’s first law

A force is a push or a pull. It has size and direction, so it is a vector, measured in newtons (N). The forces you meet most often are:

• weight, W = mg, acting vertically down;
• normal reaction, R, from a surface, at right angles to it;
• friction, F, along a rough surface, opposing motion or the tendency to move;
• tension (pull) or thrust (push) in a string or rod;
• driving force and resistances for vehicles.

Newton’s first law: a body stays at rest, or moves with constant velocity, unless a resultant force acts on it. So “at rest” or “constant velocity” both mean the resultant force is zero: the forces balance in every direction.

Worked example 1. A boat moves in a straight line at a constant 4 m s⁻¹. The engine gives a forward thrust of 650 N. Water resistance is 480 N and air resistance is R N. Find R.

Constant velocity → resultant force = 0 (first law)
650 = 480 + R
R = 170 N

The speed is not needed: any constant velocity means balanced forces.

R2 and R3: Newton’s second law and weight

Newton’s second law: the resultant force on a body equals mass × acceleration, F = ma, and the acceleration is in the direction of the resultant. Draw every force, choose a positive direction, then write “resultant force in that direction = ma”.

For 2D force vectors, a = F ÷ m component by component.

Worked example 2. Forces (5i + 2j) N and (3i − 8j) N act on a particle of mass 4 kg. Find its acceleration and the magnitude of the acceleration.

F = (5 + 3)i + (2 − 8)j = (8i − 6j) N
a = F ÷ m = (8i − 6j) ÷ 4 = (2i − 1.5j) m s⁻²
|a| = √(2² + 1.5²) = √6.25 = 2.5 m s⁻²

Weight and g (R3). Weight is the force of gravity on a body: W = mg. The specification says g may be assumed constant, and its value is used to varying degrees of accuracy, such as 9.8 or 9.81 m s⁻². It also says g is not a universal constant: it depends on location. The inverse square law for gravitation is not required. Use exactly the value of g the question gives, and don’t give answers more precisely than g allows.

Worked example 3. A ball of mass 0.5 kg falls vertically. Air resistance is a constant 1.4 N. Find its acceleration.

Down positive: resultant = mg − 1.4 = 0.5 × 9.8 − 1.4 = 3.5 N
a = 3.5 ÷ 0.5 = 7 m s⁻² downwards

Resolving (R2 extended). When a force acts at an angle θ to your chosen direction, its component in that direction is F cos θ, and its component at right angles is F sin θ. Resolve along the motion to use F = ma, and perpendicular to the motion, where the acceleration is zero, to find a normal reaction.

Worked example 4. A sledge of mass 12 kg is pulled across smooth horizontal ice by a rope at 25° above the horizontal. The tension is 60 N. Find the acceleration and the normal reaction.

Horizontally (F = ma):   60 cos 25° = 12a
                         a = 54.378... ÷ 12 = 4.53 m s⁻²
Vertically (no motion):  R + 60 sin 25° = 12 × 9.8
                         R = 117.6 − 25.357... = 92.2 N

The rope lifts part of the weight, so R is less than mg.

R4: Newton’s third law, connected particles and equilibrium

Newton’s third law: if body A exerts a force on body B, then B exerts a force on A that is equal in size and opposite in direction. The pair act on different bodies, so never put both in one F = ma equation.

Worked example 5. A man of mass 70 kg stands in a lift of mass 500 kg. The lift accelerates upwards at 0.6 m s⁻². Find (a) the tension in the lift cable, (b) the force the man exerts on the floor.

(a) Lift and man together, up positive:
    T − 570 × 9.8 = 570 × 0.6
    T = 570 × 10.4 = 5928 N
(b) Man alone: R − 70 × 9.8 = 70 × 0.6
    R = 70 × 10.4 = 728 N (floor on man)
    Third law: the man pushes down on the floor with 728 N

Treating lift and man as one body removes the internal force pair.

Connected particles. For particles joined by a light inextensible string, the tension is the same throughout the string and the particles share the same size of acceleration. Over a smooth pulley the tension is the same on both sides. Write a separate F = ma equation for each particle, then solve together. These modelling words (light, inextensible, smooth) are explained in Overarching themes.

Worked example 6. Particle A, mass 3 kg, lies on a smooth horizontal table. A light inextensible string joins A, over a smooth pulley at the table edge, to particle B, mass 2 kg, which hangs freely. The system is released from rest. Find the acceleration and the tension.

A (horizontal):  T = 3a
B (down):        2g − T = 2a
Add:             2g = 5a  →  a = 19.6 ÷ 5 = 3.92 m s⁻²
                 T = 3 × 3.92 = 11.76 = 11.8 N
Check with B:    19.6 − 11.76 = 7.84 = 2 × 3.92 ✓

Worked example 7. A car of mass 1200 kg tows a trailer of mass 400 kg along a straight horizontal road using a light rigid tow bar. The driving force is 2600 N. Resistances are 500 N on the car and 300 N on the trailer. Find the acceleration and the force in the tow bar.

Whole system:  2600 − 500 − 300 = 1600a  →  a = 1.125 m s⁻²
Trailer:       T − 300 = 400 × 1.125
               T = 750 N (tension)

Equilibrium of a particle. The resultant force is zero, so resolve in two perpendicular directions and set each total to zero.

Worked example 8. A particle of mass 5 kg hangs from two light strings. One makes 30° with the horizontal and the other 45° with the horizontal, on opposite sides. Find the tensions.

Horizontal:  T₁ cos 30° = T₂ cos 45°
Vertical:    T₁ sin 30° + T₂ sin 45° = 5 × 9.8 = 49
Solve:       T₁ = 35.9 N,  T₂ = 43.9 N

Quick check: the steeper string carries more weight, so T₂ > T₁.

R5: resultant forces and motion in a plane

To add forces given by magnitude and direction, resolve each into perpendicular components, add them, then rebuild the resultant with Pythagoras and tan. Then a = resultant ÷ m, in the same direction.

Worked example 9. Three horizontal forces act on a particle of mass 3 kg: 12 N along i, 9 N along j, and 15 N at 135° anticlockwise from i. Find the resultant and the acceleration.

i:  12 + 15 cos 135° = 1.393... N
j:  9 + 15 sin 135°  = 19.606... N
|resultant| = √(1.393² + 19.606²) = 19.7 N
direction: tan⁻¹(19.606 ÷ 1.393) = 85.9° anticlockwise from i
|a| = 19.656... ÷ 3 = 6.55 m s⁻², in the same direction

R6: friction

On a rough surface the friction force F satisfies F ≤ μR, where R is the normal reaction and μ is the coefficient of friction. Friction is only as large as it needs to be, up to its maximum μR.

• If the body is at rest and not about to move, F is whatever balances the other forces, and F < μR.
• If the body is on the point of moving (limiting equilibrium), F = μR.
• If the body is moving, F = μR, opposing the motion.

A smooth surface is the model μ = 0.

Worked example 10. A box of mass 5 kg rests on rough horizontal ground, μ = 0.4. A horizontal force of 15 N acts on it. Does it move?

R = 5 × 9.8 = 49 N;  maximum friction = 0.4 × 49 = 19.6 N
15 < 19.6, so the box stays at rest and F = 15 N (not 19.6 N)

Worked example 11 (limiting friction and statics). A box of mass 20 kg is on rough horizontal ground, μ = 0.35. It is pulled by a rope at 20° above the horizontal and is on the point of moving. Find the tension P.

Vertical:    R = 20 × 9.8 − P sin 20° = 196 − 0.342P
Horizontal:  P cos 20° = F = μR (limiting)
             0.9397P = 0.35(196 − 0.342P)
             P(0.9397 + 0.1197) = 68.6
             P = 64.8 N

Worked example 12 (rough inclined plane). A block of mass 4 kg is released from rest on a rough plane inclined at 30°, μ = 0.25. Find its acceleration.

Perpendicular:  R = 4g cos 30° = 33.948... N
Max friction:   F = 0.25 × 33.948 = 8.487... N
Down the slope: 4g sin 30° = 19.6 N > 8.49 N, so it slides
F = ma:         19.6 − 8.487 = 4a  →  a = 2.78 m s⁻² down the plane

Common errors

• Using a single force instead of the resultant in F = ma.
• Writing R = mg when another force has a vertical component, or on a slope (R = mg cos θ there).
• Setting F = μR when the body is at rest and not on the point of moving.
• Swapping sin and cos when resolving: draw the angle and check which side it is next to.
• Writing a single F = ma equation for a pulley system whose particles move in different directions. Write one equation per particle.
• Quoting more figures than the value of g supports.

Next steps

• Condensed recall: Forces and Newton’s laws revision notes.
• Test yourself: Forces and Newton’s laws practice questions.
• Force vectors: Vectors practice questions.
• Related topics: Kinematics and Moments.
• Find your gaps: all free 10-minute diagnostics.
• Course hub: /boards/aqa/a-level/mathematics/. Checklist: /checklists/aqa/a-level/mathematics/.

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 A: mathematical notation (mechanics) and Appendix B: mathematical formulae and identities.