Moments

The idea

The moment of a force about a point is force $\times$ perpendicular distance. A rigid body is in equilibrium when the resultant force is zero and the moments about any point balance. A uniform rod's weight acts at its middle.

Draw the rod with every force and its distance from one end.
Take moments about a point where an unknown force acts, to remove it.
Resolve vertically for the other unknown.

Watch out Use the perpendicular distance from the pivot to the line of the force.

See it worked

Worked example

A uniform rod of length $4\text{ m}$ and mass $10\text{ kg}$ rests on supports at its ends. Find each reaction.

Weight $10g = 98\text{ N}$ acts at the middle.
By symmetry the reactions are equal.
Each is $49\text{ N}$ .

Answer $49\text{ N}$ each

Your turn

Question 1Non-calculator3 marks

A uniform beam $AB$ of length $6\text{ m}$ and weight $120\text{ N}$ rests on supports at $A$ and $B$ . A person of weight $60\text{ N}$ stands $2\text{ m}$ from $A$ . Find the reaction at $B$ .

Question 2Non-calculator3 marks

For the same beam, find the reaction at $A$ .

Question 3Non-calculator3 marks

A child of mass $30\text{ kg}$ sits $2\text{ m}$ from the pivot of a seesaw. What mass sitting $1.5\text{ m}$ from the pivot on the other side balances it?

In past papers

31 official questions in 31 papers since 2018. The latest:

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