All subjects · Mathematics · WME01 (M1)

Equilibrium of a particle

A particle is in equilibrium when the resultant force on it is zero. You find unknown forces by resolving in two perpendicular directions, or by adding force vectors.

Conditions

In equilibrium the forces add to zero. In practice: resolve in two perpendicular directions and set each total to zero. With i and j vectors, the i parts add to 0 and the j parts add to 0.

Choosing directions

Resolve horizontally and vertically for strings and weights. Resolving perpendicular to an unknown force removes it from that equation, which can save a step.

Triangle of forces

Three forces in equilibrium form a closed triangle when drawn nose to tail. The sine rule or right-angled trigonometry on this triangle is a quick alternative for three-force problems.

Worked example

A 2 kg particle hangs from two strings making 30° and 60° with the horizontal. Find the tensions.

  1. Horizontally: T_1 cos 30° = T_2 cos 60°.
  2. Vertically: T_1 sin 30° + T_2 sin 60° = 2g.
  3. Solving: T_1 = 9.80 N and T_2 = 17.0 N.

Answer: T_1 = 9.80 N, T_2 = 17.0 N

Key idea: Equilibrium: resolve in two perpendicular directions and set each sum to zero (or add the vectors to 0).

Check yourself

  1. A particle is in equilibrium under forces (3i - 2j) N, (-5i + 4j) N and F. Find |F|, to 3 significant figures.
    Show answer

    2.83 N. F = 2i - 2j, so |F| = sqrt(8) = 2.83 N.

  2. Using the worked example, find the tension in the string at 60° to the horizontal, to 3 significant figures.
    Show answer

    17.0 N. T_2 = 17.0 N.

  3. A 5 kg particle hangs on a string and is held by a horizontal force P so that the string makes 20° with the vertical. Find P, to 3 significant figures.
    Show answer

    17.8 N. Vertically: T cos 20° = 5g. Horizontally: T sin 20° = P. So P = 5g tan 20° = 17.8 N.

  4. For the same particle, find the tension in the string, to 3 significant figures.
    Show answer

    52.1 N. T = 5g/cos 20° = 52.1 N.

  5. A smooth ring of mass m kg is threaded on a light string whose ends are fixed at the same level. Each part of the string makes 40° with the horizontal and the tension is 12 N. Find m, to 3 significant figures.
    Show answer

    1.57 kg. Smooth ring: the tension is the same in both parts. Vertically: 2 × 12 sin 40° = mg, so m = 1.57 kg.

  6. Forces of 6 N due east and 8 N due north act on a particle. Find the magnitude of the third force that keeps it in equilibrium.
    Show answer

    10 N. The third force balances the resultant of the other two: sqrt(36 + 64) = 10 N.

  7. What is the condition for a particle to be in equilibrium?
    • No forces act on it
    • It is at rest
    • The resultant force on it is zero
    • Its weight equals the normal reaction
    Show answer

    The resultant force on it is zero. Equilibrium means zero resultant force; the particle may be at rest or moving at constant velocity.

Practise with instant feedback

Sources

Last updated 7 November 2026. Spotted a mistake? Tell us.