Moments and levers

Physics · Senior school physics · about 7 min

A spanner, a seesaw and a wheelbarrow all work the same way: a force applied further from a pivot has a bigger turning effect. That turning effect is a moment, and balancing moments is the basis of statics in engineering.

The key idea

The moment of a force about a pivot is the force times its perpendicular distance from the pivot:

M = F \times d

An object is balanced when the clockwise moments equal the anticlockwise moments:

\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}}

A horizontal lever arm with a round pivot at its left end, resting on a triangular support. A downward force arrow at the right end. A dimension under the arm from the pivot to the force. A curved arrow near the pivot shows the clockwise turning effect. — A moment is the force times its perpendicular distance from the pivot.

Try it yourself

Move the weights along the lever and find the balance point.

Perpendicular distance matters

The distance in $M = F d$ is measured at right angles to the force's line of action. Pushing on a door at an angle, only the part of your push that's perpendicular to the door turns it, which is why pushing straight on is easiest.

Top view of a door, drawn as a straight horizontal bar with a hinge at its left end O. At the right end, a dark red force arrow pushes on the door at an angle, with the angle marked between the arrow and the door. A thin dashed dark blue arrow from the same point shows the part of the force at right angles to the door. A dimension under the door from O to the force. — Only the part of the force at right angles to the arm turns it.

Levers trade force for distance

A long handle lets a small force balance a large one: if your effort is $5$ times further from the pivot than the load, you need only $\tfrac{1}{5}$ of the force. The catch is that your hand has to move $5$ times further. Levers don't create energy; they change how it's applied.

Worked example: balancing a seesaw

A $30$ kg child sits $1.5$ m from the pivot of a seesaw. Where must a $45$ kg child sit on the other side to balance it?

A seesaw: a long horizontal plank balanced on a triangular pivot at its centre. A rectangular block on the left and a larger rectangular block on the right, the right one closer to the pivot. Dimensions under the plank from the pivot to each block's centre. — The seesaw in this example.

Show the solution

The weights are $30g$ and $45g$ , but $g$ cancels, so work with the masses.

30 \times 1.5 = 45 \times d

d = \frac{45}{45} = 1.0 \text{ m}

The heavier child sits closer to the pivot, $1.0$ m away.

Common mistakes

Measuring the distance along the lever instead of perpendicular to the force

Use the perpendicular distance. For an angled force, use $F d \sin\theta$ .

Forgetting the weight of the lever itself

A uniform beam's weight acts at its centre and has its own moment about the pivot.

Mixing units like grams and kilograms on each side

Keep the same units on both sides; if you use masses instead of weights, $g$ must cancel from both.

Check your understanding

1. A $20$ N force acts $0.5$ m from a pivot, at right angles. What is its moment?

2. Why is a long spanner easier to use than a short one?

3. A $600$ N load sits $0.4$ m from a pivot. What force $1.2$ m from the pivot on the other side balances it?

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