Turning effects of forces

Definition. Moment $M = F \times d_\perp$ where $d_\perp$ is perpendicular distance from the PIVOT to the LINE OF ACTION.

Units. N·m (not joules — same units but different concept).

Sign convention. Choose clockwise or anticlockwise as positive.

Force at angle. If force makes angle $\theta$ with the lever, take component perpendicular to lever: $M = F d \sin\theta$.

Example. 25 N at $30°$ to a spanner, 0.30 m from pivot. $M = 25 \times 0.30 \times \sin 30° = 3.75$ N·m.

Cambridge tip. Mark scheme rewards labelling perpendicular distance clearly.

Couple. Two equal, opposite, parallel forces that don't share a line of action.

Properties.

• Net FORCE = 0 (forces cancel).
• Net MOMENT (torque) ≠ 0.
• Produces ROTATION without translation.

Torque of couple. $\tau = F \times d$ where $d$ is perpendicular distance between the two forces.

Examples. Steering wheel, screwdriver, tap.

Cambridge tip. State that net force is zero — common mark scheme requirement.

Statement. For a body in equilibrium, the sum of CLOCKWISE moments about any point = sum of ANTICLOCKWISE moments about the same point.

Two conditions for full equilibrium.

1. Net FORCE = 0 (translational).
2. Net MOMENT = 0 about ANY point (rotational).

Method.

1. Choose a pivot (often eliminate an unknown).
2. List all forces and their perpendicular distances.
3. Sum clockwise = sum anticlockwise.
4. Solve.

Uniform body. Weight acts at the CENTRE OF MASS (geometric centre for uniform).

Example. 1.0 m beam, weight 20 N, pivot at centre. 30 N at 0.20 m left of pivot. Find $W$ at 0.30 m right to balance.

• Beam weight gives no moment (at pivot).
• $30 \times 0.20 = W \times 0.30 \Rightarrow W = 20$ N.

Cambridge tip. Choose pivot through an unknown force to eliminate it from moment equation.