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

Condensed notes for the first part of section M2.3 (Statics and forces) of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2. The specification states this content is part of International A-level Unit M2 (Mechanics), not examined at International AS. For full explanations, use the study guide. Then try the practice questions. Moments, rigid bodies and centres of mass are not covered here. See also the course hub, the checklist and the free diagnostics.

Key facts

Item	Result
Weight	W = mg, downwards, with g = 9.8 m s⁻²
Normal reaction R	Perpendicular to the surface, away from it
Friction	Along the surface, opposing (the tendency to) motion; F ≤ μR
Limiting friction	F = μR, particle on the point of slipping
Components of F at angle θ to an axis	F cos θ along the axis, F sin θ perpendicular to it
Resultant of Xi + Yj	Magnitude √(X² + Y²); angle with i is arctan(Y/X), with signs checked
Equilibrium of a particle	Resultant force = 0, so the components sum to zero in every direction
Weight on a slope of angle α	mg sin α along the slope (down), mg cos α into the slope
Rest on a rough slope, no other force	μ ≥ tan α

Modelling words

Word	Meaning
Particle	Size and rotation ignored; all forces act at one point
Smooth	No friction; the only contact force is R
Rough	Friction can act; F ≤ μR
Light	No mass; tension equal at both ends of a straight string
Inextensible	Length of string does not change

Method for any equilibrium problem

1. Draw a large force diagram. Label every force with its size or a letter.
2. Mark friction opposing the tendency to move. If unsure, draw it one way and let a negative answer tell you.
3. Choose two perpendicular directions, usually along and perpendicular to the surface.
4. Resolve in each direction.
5. If the particle is at rest, not limiting, solve for F and then check F ≤ μR.
6. If limiting, substitute F = μR.
7. Give units. Round only the final answer, normally to 3 significant figures.

Vector reminders

To find a resultant, add i-parts and j-parts separately. To hold a particle in equilibrium against known forces, add the force that is the negative of their resultant.

Reminder A. Three forces (4i + 3j) N, (−7i + j) N and (pi + qj) N keep a particle at rest.

i: 4 − 7 + p = 0  →  p = 3
j: 3 + 1 + q = 0  →  q = −4
magnitude of third force = √(9 + 16) = 5 N

Reminder B. A 22 N force points 28° above the horizontal. Components: 22 cos 28° = 19.4 N horizontally and 22 sin 28° = 10.3 N vertically.

Friction: three situations

Situation	What to do
Particle at rest, not about to move	F comes from the resolving equation. Check F ≤ μR
About to slip (limiting equilibrium)	F = μR, direction against the slip
Range of values (of a force or of μ)	Put the resolving expression for F into F ≤ μR and solve the inequality

For a range of an applied force, there are two limits. One has friction acting one way (the particle on the point of sliding one way), and the other has friction acting the other way. Solve both.

Reminder C (rough slope). A particle of mass 2 kg is at rest on a rough plane inclined at 20°, with no other force. Friction needed is F = 2g sin 20° = 6.70 N. Normal reaction R = 2g cos 20° = 18.4 N. So μ must be at least F/R = tan 20° = 0.364.

Reminder D (horizontal floor). A block has R = 50 N and μ = 0.3, so the greatest friction is 15 N. A horizontal push of 12 N gives F = 12 N, not 15 N. The block stays at rest. A push of 15 N is the limiting case. A push above 15 N moves the block.

Reminder E (slanting force). A pull at an angle above the horizontal reduces R (R = mg − T sin θ), so it reduces the friction available. A push at an angle below the horizontal increases R.

Reminder F (range for a pulling force). A 6 kg particle is on a rough plane at 30° with μ = 0.2. A force P acts up the line of greatest slope.

R = 6g cos 30° = 50.92,   greatest friction = 0.2 × 50.92 = 10.18
weight component down the slope = 6g sin 30° = 29.4
lower limit (friction up the slope):   P + 10.18 = 29.4  →  P = 19.2
upper limit (friction down the slope): P = 29.4 + 10.18  →  P = 39.6

The particle stays at rest for 19.2 N ≤ P ≤ 39.6 N. Say what each end of the range means in words: at the lower end it is about to slide down, at the upper end it is about to slide up.

Smooth surfaces and strings

• On a smooth plane inclined at α, a force P parallel to the slope that holds a particle of mass m satisfies P = mg sin α. Then R = mg cos α.
• A horizontal force H holding the particle on a smooth plane gives H = mg tan α and R = mg / cos α.
• For a particle hung from two strings, resolve horizontally and vertically. The horizontal components balance. The vertical components together support the weight.
• In a symmetrical set-up with each string at angle β to the vertical, 2T cos β = mg.

Must-know distinctions

Do not confuse	Difference
F = μR and F ≤ μR	The first is for limiting equilibrium only
Weight and normal reaction	Equal only when R and the weight are the sole vertical forces
Sin and cos on a slope	Weight along the slope is mg sin α, into the slope mg cos α
Resultant and equilibrant	The equilibrant is the force that cancels the resultant: equal size, opposite direction
One value and a range	A range needs two limiting cases, and the answer is an inequality

Quick self-test

1. Find the weight of a 7.5 kg parcel.
2. A 22 N force is directed at 28° to the horizontal. State its horizontal and vertical components to 3 s.f.
3. Find the magnitude of the resultant of (2i − 5j) N and (−6i + j) N.
4. A 2 kg particle rests on a rough plane at 20° to the horizontal with no other force. Find the friction force.
5. A surface has R = 40 N and μ = 0.45. What is the limiting friction?
6. A block with R = 50 N and μ = 0.3 is pushed horizontally with 12 N. What is the friction force, and does the block move?
7. A particle of mass 3 kg is held on a smooth plane at 40° by a force parallel to the slope. Find the force and the normal reaction.
8. A particle is at rest under three forces, two of which are perpendicular, of 5 N and 12 N. Find the magnitude of the third.
9. State why F = μR cannot be used for every particle on a rough surface.
10. A 4 kg particle hangs from two light strings, each at 30° to the vertical. Find the tension in each.

Answers

1. 7.5 × 9.8 = 73.5 N.
2. 19.4 N and 10.3 N (22 cos 28° = 19.42; 22 sin 28° = 10.33).
3. Resultant = −4i − 4j, magnitude √32 = 5.66 N (4√2).
4. 2 × 9.8 × sin 20° = 6.70 N.
5. 0.45 × 40 = 18 N.
6. Limiting friction is 15 N, and the push is smaller, so F = 12 N and the block stays at rest.
7. Force = 3g sin 40° = 18.9 N; R = 3g cos 40° = 22.5 N.
8. √(25 + 144) = 13 N, opposite in direction to the resultant of the other two.
9. Friction only reaches μR when the particle is on the point of slipping. Otherwise F is smaller and is found by resolving.
10. 2T cos 30° = 39.2, so T = 39.2 / (2 × 0.8660) = 22.6 N.

Where marks are usually lost

• Using R = mg on a slope or when a force has a vertical component.
• Putting F = μR in a question where the particle is not about to slip.
• Leaving out one of the two limits in a range question, so the answer is one-sided.
• Friction drawn in the wrong direction and then not checked against the sign of the answer.
• Components swapped: cos where the angle is to the other axis.
• A direction given as an angle with no axis or sense.
• Rounding g to 10 part-way through, or rounding early and losing accuracy in the final answer.
• A diagram that shows a force and its components together, so it is unclear what is acting.
• No units, or a magnitude given as a negative number.

Official syllabus

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