OxfordAQA International AS and A-level Mathematics: Statics: forces, friction and equilibrium of a particle (9660)

This guide covers the first part of section M2.3 (Statics and forces) of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2: force diagrams, friction, forces as vectors, resolving, and the equilibrium of a particle. The specification states this content is part of International A-level Unit M2 (Mechanics) and is not examined at International AS. Moments, rigid bodies in equilibrium and centres of mass are the rest of M2.3 and are taught separately.

Related pages: revision notes, practice questions, the M1 forces study guide, the course hub, the printable checklist and the free diagnostics.

What you must be able to do

Specification content	You must be able to
Force diagrams	Identify the forces present and label them clearly
Friction, limiting friction, F ≤ μR	Derive and work with inequalities from the relationship
Forces as vectors, resultants	Model forces as vectors; find the resultant of forces at a point; resolve forces
Equilibrium	Know the resultant is zero; find unknown forces on bodies at rest

Throughout, g = 9.8 m s⁻², the value the specification gives for M1.3. Forces are in newtons (N).

1. Force diagrams and modelling

Draw a large diagram before any algebra. Show only the forces acting on the body you are studying, as arrows from the particle.

Force	Direction
Weight, W = mg	Vertically downwards
Normal reaction, R	Perpendicular to the contact surface, away from it
Tension, T	Along the string, pulling the particle towards the string
Thrust	Along a rod, pushing the particle away from the rod
Friction, F	Along the surface, opposing the sliding or the tendency to slide
Applied force	As described: label its size and its angle

Do not draw components and the original force together without marking which is which. Either draw the force, or draw its components with the force crossed out.

The specification expects you to comment on modelling words. “Particle” means the size and rotation of the body are ignored. “Smooth” means there is no friction, so the only contact force is R. “Light” means the string or rod has no mass, so the tension is the same at both ends of a straight string. “Inextensible” means the string does not stretch.

A particle on a slope of angle α to the horizontal has these arrows:

weight mg      : straight down, from the particle
normal R       : at 90° to the slope, away from it
friction F     : along the slope, opposing the tendency to slide
pull or push   : along the slope, or at the stated angle

Mark the angle α between the slope and the horizontal, then check that the angle between the weight and the perpendicular to the slope is also α.

2. Forces as vectors

A force in two dimensions can be written as Xi + Yj, with i horizontal and j vertical (or any perpendicular pair). The resultant of several forces acting at a point is their vector sum: add the i-components, then add the j-components.

For a resultant Xi + Yj:

• magnitude = √(X² + Y²)
• direction: the acute angle with the i-axis is arctan(|Y|/|X|); say which axis and which way the angle is measured.

Example 1. Three forces act on a particle: F₁ = (6i − 2j) N, F₂ = (−i + 5j) N and F₃ = (−2i + 3j) N. Find the resultant, its magnitude and its direction.

R = (6 − 1 − 2)i + (−2 + 5 + 3)j = 3i + 6j
|R| = √(9 + 36) = √45 = 3√5 ≈ 6.71 N
angle with i: arctan(6/3) = arctan 2 = 63.4° above the i direction

If a fourth force F₄ makes the particle stay at rest, F₄ = −R = (−3i − 6j) N.

Forces given by size and direction. A force of magnitude F at angle θ to the i-axis has components F cos θ along i and F sin θ along j. Use the angle to whichever axis is given, and check the sign in each quadrant.

Example 2. Forces of 10 N at 0°, 14 N at 50° and 8 N at 130° act at a point, all angles measured anticlockwise from the positive x-axis. Find the resultant.

Force	x-component	y-component
10 N at 0°	10	0
14 N at 50°	14 cos 50° = 8.999	14 sin 50° = 10.725
8 N at 130°	8 cos 130° = −5.142	8 sin 130° = 6.128
Sum	13.857	16.853

|R| = √(13.857² + 16.853²) = 21.8 N (3 s.f.)
angle = arctan(16.853 / 13.857) = 50.6° above the positive x-axis

3. Resolving and equilibrium

A particle is in equilibrium when the resultant force on it is zero. In components, the forces sum to zero in every direction you choose. For a particle at rest, that is the whole of the condition.

Method

1. Draw the force diagram with every force labelled.
2. Choose two perpendicular directions. Pick them so that the most unknown forces lie along them.
3. Resolve in each direction: forces one way = forces the other way.
4. Solve the two equations.
5. Answer the question, with units and a sensible rounding.

On an incline of angle α, resolve along and perpendicular to the slope. The weight has component mg sin α down the slope and mg cos α into the slope.

Example 3. A 3 kg particle hangs from two light strings attached to a horizontal ceiling. The strings make 35° and 50° with the vertical, on opposite sides. Find the tensions T₁ and T₂.

horizontal:  T₁ sin 35° = T₂ sin 50°
vertical:    T₁ cos 35° + T₂ cos 50° = 3 × 9.8 = 29.4

From the first equation T₂ = T₁ sin 35° / sin 50° = 0.7488 T₁. Substitute:

T₁ (cos 35° + 0.7488 cos 50°) = 29.4
T₁ (0.8192 + 0.4813) = 29.4
T₁ = 22.6 N,  T₂ = 16.9 N

Example 4. A 7 kg particle is held at rest on a smooth plane inclined at 34° by a horizontal force H. Find H and the normal reaction R.

Resolve along the slope (smooth, so no friction) and perpendicular to it.

along slope:       H cos 34° = 7g sin 34°   →  H = 7g tan 34° = 46.3 N
perpendicular:     R = 7g cos 34° + H sin 34°
                   R = 56.87 + 25.87 = 82.7 N

Check: R cos 34° = 68.6 = 7g, the weight, as the vertical resolution requires.

4. Friction

Friction acts along the surface and opposes the motion, or the tendency to move. Its size is limited:

F ≤ μR

where μ is the coefficient of friction. When F = μR the friction is limiting: the particle is on the point of slipping, but still in equilibrium. When the particle is at rest and F < μR, friction takes whatever value is needed, found by resolving. So there are three situations.

Situation	What to use
At rest, not about to move	Resolve to find F, then check F ≤ μR
Limiting equilibrium (about to slip)	F = μR, and the direction of F opposes the slip
Finding a range of values	Write F ≤ μR with F from the resolving equation; solve the inequality

Example 5. A crate of mass 11 kg is on rough horizontal ground with μ = 0.28. A rope pulls it at 32° above the horizontal and the crate is on the point of slipping. Find the tension T.

vertical:    R = 11g − T sin 32° = 107.8 − T sin 32°
horizontal:  T cos 32° = μR = 0.28 (107.8 − T sin 32°)
T (cos 32° + 0.28 sin 32°) = 30.184
T (0.8480 + 0.1484) = 30.184
T = 30.3 N

Then R = 107.8 − 30.29 sin 32° = 91.7 N and F = 25.7 N. Notice R is smaller than the weight: an upward pull reduces R, and so reduces the friction available.

Example 6 (range of a force). A 10 kg particle rests on a rough plane inclined at α with tan α = 3/4, so sin α = 3/5 and cos α = 4/5. The coefficient of friction is 0.5. A force P acts up the slope along the line of greatest slope. Find the values of P for which the particle stays at rest.

perpendicular:  R = 10g cos α = 78.4
maximum friction: μR = 0.5 × 78.4 = 39.2
component of weight down the slope: 10g sin α = 58.8

If P is small, the particle tends to slide down, so friction acts up the slope: P + F = 58.8 with F ≤ 39.2, giving P ≥ 19.6. If P is large, it tends to slide up, so friction acts down the slope: P = 58.8 + F with F ≤ 39.2, giving P ≤ 98.

19.6 N ≤ P ≤ 98 N.

Deriving a condition. With no other force, a particle on a rough slope at rest has F = mg sin α and R = mg cos α. The condition F ≤ μR gives mg sin α ≤ μ mg cos α, so μ ≥ tan α. For the plane in Example 6, a particle left alone would stay at rest only if μ ≥ 0.75. With μ = 0.5 it would not, so P ≥ 19.6 N is needed to hold it.

Common errors

• Putting F = μR when the particle is not about to slip. Use F = μR only for limiting equilibrium. Otherwise F is found from the resolving equations.
• Taking R = mg on a slope, or when a rope or push has a vertical component. Always resolve perpendicular to the surface.
• Drawing friction in the wrong direction. Friction opposes the tendency to slide. For a range question, draw it both ways.
• Mixing sin and cos on an incline. The weight component along the slope is mg sin α, with α the angle of the slope to the horizontal.
• Giving a single value when the question asks for a range, or reversing an inequality when dividing by a negative.
• Writing a direction without a reference axis. “63.4°” alone is incomplete; say from which axis and which way.
• Using g = 9.8 for part of a question and 10 for the rest. Keep one value throughout.

Next steps

Test yourself on the practice questions, then condense with the revision notes. For friction on a sliding particle and Newton’s laws, see the M1 forces guide. Use the diagnostics to find weak spots, and tick off the topic on the checklist.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section M2.3 Statics and forces, published by OxfordAQA.