All subjects · Mathematics · WME02 Mechanics 2

Moments of forces at an angle

Edexcel IAL Mathematics · WME02 Mechanics 2

In M1, forces were perpendicular to the rod. In M2 they can act at any angle, so you need the perpendicular distance from the pivot to the line of action, or the perpendicular component of the force.

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Two equivalent methods

The moment of a force F about a point P is F × perpendicular distance from P to the line of action of F.

Equivalently, resolve F into components along and perpendicular to the rod: the component along the rod passes through P and has no moment; the perpendicular component F sin θ, at distance d along the rod, gives moment F sin θ × d.

Weights on inclined rods

A uniform rod of length 2a inclined at θ to the horizontal: its weight acts vertically at the midpoint, so the perpendicular distance from the lower end to the line of action is a cos θ. A horizontal force at the top end has perpendicular distance 2a sin θ from the bottom.

Vector forces

For a force F = Xi + Yj acting at the point (x, y), the anticlockwise moment about O is xY - yX. This is useful for laminas with forces given as vectors.

Worked example

A rod AB of length 4 m has a force of 10 N applied at B at 30° to the rod. Find its moment about A.

  1. Perpendicular component: 10 sin 30° = 5 N.
  2. Moment = 5 × 4 = 20 N m.
  3. (Or: perpendicular distance from A to the line of action = 4 sin 30° = 2 m, and 10 × 2 = 20 N m.)

Answer: 20 N m

Key idea: Moment = force × perpendicular distance = (perpendicular component) × distance along the rod. Components along the rod have no moment.

Check yourself

  1. A force of 25 N acts at the end of a rod 3 m long, at 60° to the rod. Find its moment about the other end, to 3 significant figures.
    Show answer

    65.0 N m. 25 sin 60° × 3 = 65.0 N m.

  2. A force (3i + 4j) N acts at the point with position vector (2i + j) m. Find its anticlockwise moment about O.
    Show answer

    5 N m. xY - yX = 2 × 4 - 1 × 3 = 5 N m.

  3. A uniform rod of length 6 m and weight 50 N is inclined at 40° to the horizontal with its lower end on the ground. Find the moment of its weight about the lower end, to 3 significant figures.
    Show answer

    115 N m. Perpendicular distance = 3 cos 40°, moment = 50 × 3 cos 40° = 115 N m.

  4. A horizontal rod of length 2 m is pivoted at one end. A string at the other end makes 30° with the rod. The string must provide a moment of 30 N m about the pivot. Find the tension.
    Show answer

    30 N. T sin 30° × 2 = 30, so T = 30 N.

  5. A ladder of length 5 m makes 60° with the horizontal. A horizontal force of 40 N acts at its top. Find the moment of this force about the foot of the ladder, to 3 significant figures.
    Show answer

    173 N m. Perpendicular distance = 5 sin 60°, moment = 173 N m.

  6. A force acts at the end B of a rod AB, directed along the rod. What is its moment about A?
    • Force × length of the rod
    • Half of force × length
    • Zero, because its line of action passes through A
    • It depends on the weight of the rod
    Show answer

    Zero, because its line of action passes through A. The perpendicular distance from A to the line of action is zero.

  7. A spanner is turned by a force of fixed size at its end. At what angle to the spanner should the force act to give the largest moment?
    • 90°
    • 45°
    • 0°
    • 180°
    Show answer

    90°. F sin θ × d is greatest when sin θ = 1.

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Last updated 8 November 2026. Spotted a mistake? Tell us.