Forces and equilibrium

AS Level Mathematics · Mechanics. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Start every question with a clear diagram of the forces on the particle: weight, tension, any applied force, and the contact force from a surface. The contact force is treated as two parts. The normal reaction R acts at right angles to the surface, and friction F acts along it, against the direction of sliding.

A force is a vector, so it can be split into two perpendicular components. A particle is in equilibrium when the forces balance. Then the components add up to zero in every direction, and you can write one equation for each of two perpendicular directions.

A smooth surface is a model with no friction at all. On a rough surface, friction grows only as large as is needed, up to a limit.

Rules to remember

Common mistake

Students write R = mg every time. If a force pulls partly upwards, or the surface is sloping, R is different. Always find R by resolving at right angles to the surface. Also, F = μR holds only when the object is about to slide.

Worked example

A block of mass 2 kg rests on a rough horizontal surface. A string pulls it with a force of 10 N at 30° above the horizontal, and the block is about to slide. Find the coefficient of friction, to 3 significant figures. (g = 10 m s−2)

  1. Vertically: R + 10 sin 30° = 2 × 10, so R = 20 − 5 = 15 N.
  2. Horizontally: F = 10 cos 30° = 8.66 N.
  3. Limiting equilibrium: μ = F ÷ R = 8.66 ÷ 15.

Answer: μ = 0.577

Practice questions

Try each one, then open the answer.

1. A particle is in equilibrium under three forces: 5 N in the positive x-direction, 12 N in the positive y-direction, and a force F. Find the magnitude of F.

  1. A
    7 N
  2. B
    17 N
  3. C
    13 N
  4. D
    √119 N
Show answer

Answer: C. For equilibrium F must balance the resultant of the other two. That resultant has magnitude √(52 + 122) = 13 N, so F = 13 N, directed opposite to it.

2. Two forces of 6 N and 8 N act on a particle at right angles to each other. Find the magnitude of the resultant and the angle it makes with the 8 N force.

  1. A
    10 N at 53.1° to the 8 N force
  2. B
    10 N at 36.9° to the 8 N force
  3. C
    14 N along the bisector
  4. D
    2 N at 36.9° to the 8 N force
Show answer

Answer: B. Magnitude = √(62 + 82) = 10 N. The angle with the 8 N force satisfies tan θ = 6/8, so θ = 36.9°. Forces are vectors, so 6 + 8 = 14 is wrong.

3. A block of mass 5 kg rests in limiting equilibrium on a rough plane inclined at 30° to the horizontal. Find the coefficient of friction, to 3 significant figures. (g = 10 m s−2)

  1. A
    0.577
  2. B
    0.5
  3. C
    0.866
  4. D
    1.73
Show answer

Answer: A. Along the plane: F = 50 sin 30° = 25 N. Perpendicular: R = 50 cos 30° = 43.3 N. Limiting, so μ = F/R = 25/43.3 = 0.577 (= tan 30°). Using R = 50 N gives the wrong 0.5.

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