OCR A Level Mathematics A: Two-dimensional equilibrium, connected particles and pulleys (H240) – Revision Notes

These notes condense outcomes 3.03l to 3.03q of the OCR A Level Mathematics A specification (H240, Version 3.1, August 2026, for first assessment in 2018). Only 3.03n is Stage 1 (common with AS); everything else here is Stage 2, so it is A Level only. For full explanations and worked examples, read the study guide. Test yourself with the practice questions. The course page is OCR A Level Mathematics, the printable list is the checklist, and the free diagnostics help you decide what to revise first. Take g = 9.8 m s⁻² unless told otherwise. All surfaces on this page are smooth.

Key results at a glance

Idea	Result
Force F inclined at θ to a line	Part along the line F cos θ; part at right angles F sin θ
Resultant of forces	Add i-parts and j-parts: R = (ΣX)i + (ΣY)j
Magnitude and direction	√(Rx² + Ry²); tan α =
Equilibrium (3.03m)	Resultant is zero; resolved parts sum to zero in any direction
Weight on a plane at angle α	mg sin α down the plane, mg cos α into it
Newton’s third law (3.03l)	Equal in size and opposite in direction, on different bodies
Smooth pulley, light string	Same tension throughout; same acceleration size for both particles
Force on a pulley	Vector sum of the two string tensions
Vector dynamics (3.03q)	F = ma in both components; integrate if F depends on t

Resultants (3.03p)

Velocity shows the direction of motion. Acceleration points the way the resultant force points.

Method:

1. Resolve every force into i and j parts, taking care over signs.
2. Add the parts to get Rx and Ry.
3. Sketch R, then find |R| and its angle.

Reminder: for two forces P and Q with angle φ between them, resolve along P. The parallel part is P + Q cos φ and the perpendicular part is Q sin φ. Solutions use calculation, not scale drawing.

Equilibrium (3.03m, 3.03n)

Method:

1. Draw a diagram with every force on the particle: weight, tensions, reactions, applied forces.
2. Choose two perpendicular directions that remove as many unknowns as possible.
3. Write “resolve →” or “resolve ↑” (or “parallel to the plane”) and set each total to zero.
4. Solve, then check by resolving in a third direction.

A polygon of forces closes when the forces are in equilibrium. For three forces this gives a triangle, which you can solve with the sine and cosine rules. A found value that is negative means the force acts opposite to the way you drew it.

Inclined planes (3.03o)

• Resolve parallel and perpendicular to the plane.
• A force P acting up the plane at angle β above the plane has parallel part P cos β and perpendicular part P sin β, which lifts the particle off the surface.
• A horizontal force on a plane at angle α has parallel part P cos α and perpendicular part P sin α, which presses the particle into the plane.
• Contact is lost when the normal reaction R reaches zero.

Connected particles (3.03l)

• Two bodies in contact: the force of the first on the second equals the force of the second on the first. They are opposite and act on different bodies, so they do not cancel on one body.
• A taut string pulls both ends inwards with the same tension.
• A system with no relative motion between its parts can be treated as a single particle.
• Method: whole system first for acceleration, then one body alone for an internal force.
• An angled force changes the normal reaction. It does not change the horizontal equation except through its horizontal part.

Pulleys and strings (3.03o, 3.03q)

Word in the question	What it gives you
Light string	Tension is the same along it
Inextensible	Both particles have the same speed and acceleration size
Smooth pulley or peg	Tension is the same on both sides
Light pulley	No mass, so no extra equation for the pulley

Method:

1. Draw each particle separately. Mark the tension, the weight, any reaction, and the direction of the acceleration.
2. Write F = ma for each particle, with the direction of motion positive.
3. Add the equations to remove T, or substitute.
4. Find T last, from the simplest particle.

Force on the pulley, for tensions T in two string parts with angle φ between them: magnitude 2T cos(φ/2), along the bisector of φ. For a vertical string, φ = 0 and the force is 2T.

Worked reminder: a 2 kg particle on a smooth table is joined by a string, passing over a pulley on the table’s rim, to a hanging 3 kg particle.

3g − T = 3a,   T = 2a
3g = 5a  →  a = 5.88 m s⁻²,  T = 11.8 N

Vector dynamics (3.03q)

• Constant force: a = F/m is constant. Use the vector forms of v = u + at and s = ut + ½at².
• Force depending on time: a = F/m, then v = ∫a dt with a constant vector from the initial velocity, then r = ∫v dt with a constant vector from the initial position.
• Speed is the magnitude of v. Direction of motion is the direction of v.
• A force given as magnitude and direction must be resolved before you use F = ma componentwise.

Worked reminder: forces (9i − 6j) N and (−3i + 12j) N act on a 3 kg particle that starts at rest.

F = (6i + 6j) N,  a = (2i + 2j) m s⁻²
After 4 s:  s = ½at² = 16i + 16j,  |s| = 22.6 m

Must-know distinctions

• Weight (acts on the body, downwards) and normal reaction (acts on the body, perpendicular to the surface) are not a Newton’s third law pair. The pair for the reaction is the force the body exerts on the surface.
• Tension is the same in a light string over a smooth pulley, but the force on the pulley is not T. It is the vector sum of two tensions.
• Equilibrium means zero resultant. It does not mean zero velocity: constant velocity is also equilibrium.
• Resultant is a single force. Components are a pair of perpendicular parts.

Quick self-test

1. Find the resultant of (3i + 4j) N and (5i − j) N, with its magnitude.
2. Find the horizontal and vertical components of a 20 N force at 30° above the horizontal.
3. A 10 kg particle sits on a smooth plane inclined at 20°. Find the component of its weight down the plane and into the plane.
4. Perpendicular forces of 8 N and 6 N act on a particle that is in equilibrium with a third force. Find the size of the third force and its angle to the 8 N force.
5. A smooth pulley carries a light inextensible string with 2 kg and 3 kg particles hanging at its ends. Find the acceleration and tension.
6. Find the force the string exerts on the pulley in question 5.
7. A force (12i − 9j) N acts on a 3 kg particle. Find the magnitude of its acceleration.
8. A particle starts from rest with acceleration 2t i m s⁻². Find its velocity at t = 3.
9. A 6 kg particle is on a smooth plane at 30°, joined by a string over a pulley at the top to a 5 kg particle hanging. Find the acceleration and tension.
10. What does “inextensible” tell you about the two particles?

Answers

1. (8i + 3j) N; magnitude √73 = 8.54 N.
2. Horizontal 17.3 N, vertical 10.0 N.
3. Down the plane 10g sin 20° = 33.5 N; into the plane 10g cos 20° = 92.1 N.
4. Taking the 8 N force along +x and the 6 N force along +y, the third force is (−8i − 6j) N. Its size is 10 N and it makes an angle of 143.1° with the 8 N force (36.9° below the negative x-axis).
5. a = g/5 = 1.96 m s⁻²; T = 2(g + a) = 23.5 N.
6. 2T = 47.0 N, vertically downwards on the pulley.
7. a = (4i − 3j) m s⁻², so |a| = 5 m s⁻².
8. v = t² i, so at t = 3 the velocity is 9i m s⁻¹.
9. 5g − T = 5a and T − 6g sin 30° = 6a give 11a = 49 − 29.4, so a = 1.78 m s⁻²; T = 5(g − a) = 40.1 N.
10. While the string is taut, both particles have the same speed and acceleration size.

Where marks are usually lost

• Mixing sin and cos on an inclined plane.
• Omitting the vertical component of an angled force from the normal reaction.
• Writing one equation for two particles together and getting signs wrong. Draw each separately.
• Leaving a negative acceleration unexplained instead of using it to reverse the assumed direction.
• Finding the force on a pulley as 2T when the strings are not parallel.
• Giving an angle without saying what it is measured from.
• Forgetting the constant of integration, or using the starting position as the starting velocity.
• Rounding g or intermediate values early, which pulls the final answer off.

Official syllabus

OCR, A Level Mathematics A (H240), Version 3.1 (August 2026), first assessment 2018: section 3.03, Forces and Newton’s Laws.