The Principle of Moments

curriculum-key-fact
If an object is balanced, the total clockwise moment about a point is equal to the total anticlockwise moment about the same point. This is called the Principle of Moments. Total clockwise moment = Total anticlockwise moment.
Question

The diagram below shows two masses balanced on a level beam.

The diagram below shows two masses balanced on a level beam.

How far is the 10 N weight from the pivot?

The beam is balanced and so from the Principal of Moments we know that:

Total clockwise moment about the pivot = Total anticlockwise moment about the pivot.

Calculate each individual moment first:

Anticlockwise moment

Perpendicular distance from the pivot = d m.

Force F = 10 N.

Anticlockwise moment = F x d = 10 N x d m = 10d Nm.

Clockwise moment

Perpendicular distance from the pivot = 1 m.

Force F = 20 N.

Clockwise moment = F x d = 20 N x 1 m = 20 Nm.

Total clockwise moment = Total anticlockwise moment

10d = 20.

{d}=20 Nm ÷ 10 N

d = 2 m

The 20 N weight is 2 m from the pivot.

Question

A parent and child are at opposite sides of a playground see-saw. The parent sits 0.8 m from the pivot. The child sits 2.4 m from the pivot and weighs 250 N.

Calculate the weight of the parent if the see-saw is balanced.

The see-saw is balanced and so from the Principal of Moments we know that:

Total clockwise moment about the pivot = Total anticlockwise moment about the pivot

Anticlockwise moment

The anticlockwise moment is the child's moment = Fd.

Perpendicular distance of child from the pivot = 2.4 m.

Force F = 250 N.

Anticlockwise moment = 250 N × 2.4 m = 600 Nm.

Clockwise moment

The clockwise moment is the parent's moment = Fd.

Perpendicular distance of adult from the pivot = 0.8 m.

Force F = F N.

Clockwise moment = F N x 0.8 m = 0.8F Nm.

Total clockwise moment = Total anticlockwise moment.

0.8 F = 600.

F = 600 Nm ÷ 0.8 m.

F = 750 N.

The parent’s weight equals 750 N.