Ladders and rods in equilibrium

The idea

A ladder against a wall is a moments problem with several forces: weight at the middle, reactions from the ground and the wall, and friction where a surface is rough. Resolve in two directions and take moments about the foot.

Draw every force: weight, both normal reactions and any friction.
Resolve horizontally and vertically.
Take moments about the foot of the ladder; in limiting equilibrium $F = \mu R$ .

Watch out A smooth wall gives no friction: its reaction is horizontal only.

See it worked

Worked example

A uniform ladder of weight $W$ rests at $60^\circ$ to rough ground against a smooth wall. Find the friction at the ground.

Moments about the foot (length $2a$ ): $S \times 2a\sin 60^\circ = W \times a\cos 60^\circ$ .
$S = \frac{W}{2\tan 60^\circ} = 0.289W$ .
Horizontally, friction $= S = 0.289W$ .

Answer $0.289W$

Your turn

Question 1Calculator allowed3 marks

A uniform ladder rests against a smooth wall on rough ground, at $60^\circ$ to the ground, about to slip. Find $\mu$ to 3 significant figures.

Question 2Calculator allowed3 marks

A uniform ladder of weight $200\text{ N}$ rests against a smooth wall at $70^\circ$ to the ground. Find the reaction from the wall to 3 significant figures.

Question 3Non-calculator3 marks

For the same ladder (weight $200\text{ N}$ , smooth wall), find the normal reaction from the ground.

In past papers

15 official questions in 15 papers since 2018. The latest:

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