A-Level Maths forces: Newton's laws, friction, inclined planes and moments

Where forces sit in each specification

The key ideas

Draw every force: weight mg acting down, normal reaction R perpendicular to the surface, friction F along the surface opposing motion (or the tendency to move), tension T along a string, and any applied force. Then resolve in two perpendicular directions. For an object on a slope at angle θ, resolve parallel and perpendicular to the slope: the weight has components mg sin θ down the slope and mg cos θ into the slope.

In equilibrium the resultant force is zero, so the components in each direction balance. If the object accelerates, the resultant in the direction of motion equals ma. Friction can take any value up to μR; it reaches μR only when the object is sliding or on the point of slipping ('limiting equilibrium').

For connected particles, write F = ma for each particle separately, using the same tension and the same size of acceleration for both when the string is light and inextensible. For moments (Edexcel M1), the moment of a force about a point is force × perpendicular distance. In equilibrium, clockwise moments equal anticlockwise moments about any point, and the forces balance too.

Worked example 1 (g = 10): a 5 kg box is released on a rough slope at 30° with μ = 0.2

1. Perpendicular to the slope: R = 5 × 10 × cos 30° = 43.30 N.
2. Maximum friction: μR = 0.2 × 43.30 = 8.66 N.
3. Weight component down the slope: 5 × 10 × sin 30° = 25 N. This is more than 8.66 N, so the box slides and F = 8.66 N acts up the slope.
4. Newton's second law down the slope: 25 - 8.66 = 5a.
5. So a = 16.34 ÷ 5 = 3.27 m s^-2 (3 s.f.).

Worked example 2 (g = 10): a 3 kg block on a smooth table is joined by a string over a pulley to a hanging 2 kg mass

1. For the block (horizontal): T = 3a.
2. For the hanging mass (down positive): 2 × 10 - T = 2a, so 20 - T = 2a.
3. Add the equations to eliminate T: 20 = 5a, so a = 4 m s^-2.
4. Then T = 3 × 4 = 12 N.
5. Check with the hanging mass: 20 - 12 = 8 = 2 × 4. Correct.

Worked example 3 (Edexcel M1 only, g = 9.8): a uniform 4 m rod of mass 10 kg rests on supports at A and at C, 3 m from A

1. The weight 10 × 9.8 = 98 N acts at the centre, 2 m from A.
2. Take moments about A to remove the unknown force there: R(C) × 3 = 98 × 2.
3. So R(C) = 196 ÷ 3 = 65.3 N.
4. Resolve vertically: R(A) + R(C) = 98, so R(A) = 98 - 65.3 = 32.7 N.

Common mistakes that cost marks

• Leaving out a force, usually friction or the normal reaction.
• Swapping sin and cos when resolving on a slope. The component along the slope is mg sin θ.
• Using F = μR when the object is not moving and not on the point of slipping.
• Writing one equation for the whole system of connected particles and then using it to find tension.
• Putting the normal reaction equal to mg on a slope or when another force has a vertical component.
• In moments (Edexcel), using the distance along a rod instead of the perpendicular distance.

Exam technique and how a tutor helps

Always draw a large, labelled force diagram, even if the question does not ask for one. Write each resolving equation with a heading such as 'Resolving perpendicular to the plane' so the examiner can follow your method. If there is a choice of where to take moments, choose a point that an unknown force passes through.

In one-to-one lessons a tutor starts by having the student draw the force diagram and say each force aloud before writing an equation, because almost all mistakes in this topic begin with a wrong diagram. Then the student works through mixed questions with slopes, pulleys and limiting friction from their own board's past papers, with the tutor checking every resolution on the shared whiteboard.

Self-check: can you do these?

• A 4 kg particle rests on a rough horizontal floor with μ = 0.3 (g = 10). What horizontal force will just make it slip? (Answer: 12 N)
• A 2 kg particle is pulled up a smooth slope at 30° by a force of 16 N parallel to the slope (g = 10). Find its acceleration. (Answer: 3 m s^-2)
• Two particles of 5 kg and 3 kg hang over a smooth pulley (g = 10). Find the acceleration. (Answer: 2.5 m s^-2)
• What is the component of the weight perpendicular to a slope of angle θ? (Answer: mg cos θ)
• Is 'moments' examined in Cambridge 9709? (Answer: no, it is in Edexcel IAL M1)