Moments and equilibriumEdexcel International A Level Physics: Revision notes

The moment of a force about a point is the force multiplied by the perpendicular distance from the point to the line of action of the force:

$\text{moment} = Fx$

The unit is the newton metre (N m). A moment is a turning effect: it is clockwise or anticlockwise about the chosen axis. A spanner of length $0.25\ \text{m}$ with a perpendicular force of $40\ \text{N}$ gives a moment of $40 \times 0.25 = 10\ \text{N m}$.

This is why a longer handle makes it easier to turn a nut: for the same moment, a larger $x$ needs a smaller $F$.

Only the perpendicular distance counts. If a force $F$ acts at an angle $\theta$ to a rod of length $L$ pivoted at one end, the perpendicular distance from the pivot to the line of action is $L\sin\theta$, so:

$\text{moment} = F L \sin\theta$

Equivalently, only the component of the force perpendicular to the rod, $F\sin\theta$, produces a moment. For $40\ \text{N}$ at $60^\circ$ to a $0.25\ \text{m}$ handle: $40 \times 0.25 \times \sin 60^\circ = 8.7\ \text{N m}$.

The centre of gravity of an object is the point through which the whole weight of the object may be considered to act. For a uniform object, whose mass is evenly distributed, it lies at the geometrical centre, so the weight of a uniform rod or plank acts at its midpoint.

This lets the weight of an extended body be treated as a single force when taking moments. An object suspended from a point comes to rest with its centre of gravity vertically below the point.

The principle of moments states that for a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.

A body in equilibrium has no resultant force and no resultant moment. So the upward forces equal the downward forces, and the moments balance about every point. You choose the point. Taking moments about a point where an unknown force acts removes that force from the moments equation, which makes the algebra shorter.

A uniform plank of length $4.0\ \text{m}$ and weight $200\ \text{N}$ rests on supports A and B at its ends. A $600\ \text{N}$ child stands $1.0\ \text{m}$ from A.

Moments about A: $F_B \times 4.0 = 200 \times 2.0 + 600 \times 1.0 = 1000\ \text{N m}$, so $F_B = 250\ \text{N}$.

Force balance: $F_A + F_B = 800\ \text{N}$, so $F_A = 550\ \text{N}$.

Check by taking moments about B: $F_A \times 4.0 = 200 \times 2.0 + 600 \times 3.0 = 2200$, so $F_A = 550\ \text{N}$.