A Level Maths 9709 · Paper 4 Mechanics · 4.1 · ~6 min read
Forces and equilibrium
Identifying the forces on a particle, resolving them into components, using the conditions for equilibrium, finding resultants, and modelling friction with F ≤ μR, including limiting equilibrium on rough horizontal and inclined surfaces.
Types of forceForce diagramsResolvingResultant forceEquilibrium of a particleFriction & μLimiting equilibriumInclined planes
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end
In 30 seconds
- Draw the force diagram first: weight mg down, normal reaction R perpendicular to the surface, friction F along the surface opposing (possible) motion, and any tensions or applied forces.
- Resolve in two perpendicular directions. On a slope, use along and perpendicular to the slope.
- Equilibrium: the components in every direction sum to zero. That gives two equations for two unknowns.
- Friction: F ≤ μR. F = μR only when the particle is moving or about to move (limiting equilibrium).
By the end you can
- Identify the forces acting on a particle and draw a clear diagram.
- Find the resultant of several forces by resolving.
- Use the conditions for equilibrium of a particle in two dimensions.
- Model friction with F ≤ μR and use limiting friction, including on inclined planes.
1Forces and resolving
The language of mechanics
| Term | Meaning in a model |
|---|---|
| Particle | an object whose size is ignored; all forces act at one point |
| Light (string, rod) | its mass is ignored, so tension is the same throughout |
| Inextensible | its length doesn't change, so connected particles share an acceleration |
| Smooth | no friction |
| Rough | friction acts, up to a limit μR |
- Use g = 10 m s⁻² in this paper, unless the question says otherwise. (9702 Physics uses 9.81.)
- Resolving: a force P at angle θ to a direction has component P cos θ along it and P sin θ perpendicular to it.
Resultant: X = ΣP cos θ, Y = ΣP sin θ ; ǀRǀ = √(X² + Y²), tan φ = Y/X
resolve every force, add, then recombineWorked example
Forces of 15 N, P N and Q N act at a point at angles 0°, 120° and 250° to the positive x-axis, and are in equilibrium. Find P and Q.
- x: 15 + P cos 120° + Q cos 250° = 0 M1
- y: P sin 120° + Q sin 250° = 0 M1
- solving: P = 18.4, Q = 17.0 A1 A1
Common mistakeMixing sin and cos. The component next to the angle uses cos. Draw the angle on your diagram before resolving.
2Friction
F ≤ μR
Coefficient of friction μ — the ratio of limiting friction to normal reaction. A smooth surface has μ = 0.
- Limiting equilibrium: "on the point of moving" or "about to slip" ⇒ F = μR.
- Not yet limiting: F is just whatever is needed for equilibrium, and F < μR.
- Normal reaction is not always mg. A pull at an angle above the horizontal reduces R; a push down increases it.
Worked example
A box of mass 5 kg rests on rough horizontal ground. A force of 30 N at 20° above the horizontal makes it just about to move. Find μ.
- vertical: R + 30 sin 20° = 50 ⇒ R = 39.74 N M1 A1
- horizontal: F = 30 cos 20° = 28.19 N B1
- limiting: μ = F/R = 0.709 M1 A1
Common mistakeWriting R = mg when another force has a vertical component. The 2024 Paper 4 report singled this out.
3Inclined planes
Along and perpendicular
Along the slope: mg sin α ; perpendicular: R = mg cos α (if no other perpendicular forces)
α is the angle of the slopeWorked example
A particle of mass 2 kg is on a rough plane inclined at 25°, with μ = 0.3. (a) Show it would slide down if released. (b) Find the least force P, acting up the slope, that keeps it at rest. (c) Find the force P up the slope that makes it about to move up.
- (a) down-slope component 20 sin 25° = 8.45 N; maximum friction 0.3 × 20 cos 25° = 5.44 N; 8.45 > 5.44 so it slides M1 A1
- (b) friction acts up the slope: P + 5.44 = 8.45 ⇒ P = 3.01 N M1 A1
- (c) friction acts down the slope: P = 8.45 + 5.44 = 13.9 N M1 A1
- Which way does friction act? Opposite to the motion that would happen. In (b) the particle tends to slide down; in (c) it tends to move up. The two answers give the range of P for equilibrium.
Common mistakeFriction pointing the wrong way, usually because the direction of possible motion wasn't thought about first.
Linking questionThe forces here produce accelerations in Newton's laws of motion; work against friction is in Energy, work and power; and 3D vectors are in Vectors.
Test yourself — cover the answers first
Quick check
- What does "light" tell you about a string?
- A 4 kg mass is on a smooth slope at 30°. Find the component of weight down the slope.
- When is F = μR?
- What is the normal reaction on a 3 kg block on a horizontal floor with no other vertical forces?
- Find the resultant of 6 N east and 8 N north.
Exam-style
- A 4 kg particle rests on a rough plane inclined at 30° and is on the point of slipping down. Find μ.
- A sledge of mass 12 kg is pulled at constant speed along rough horizontal ground by a rope at 30° above the horizontal with tension 50 N. Find μ.
- A particle hangs in equilibrium from two strings at 30° and 60° to the horizontal. Its weight is 20 N. Find the tensions.
Show answers
- its mass is ignored, so the tension is the same throughout
- 40 sin 30° = 20 N
- when moving or on the point of moving
- 30 N
- 10 N at 53.1° north of east
- 40 sin 30° = μ × 40 cos 30° ⇒ μ = tan 30° = 0.577
- R = 120 − 25 = 95; F = 50 cos 30° = 43.3; μ = 0.456
- T₁ cos 30° = T₂ cos 60°; T₁ sin 30° + T₂ sin 60° = 20 ⇒ T₁ = 10 N, T₂ = 17.3 N
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