Question

A uniform 80.0 N ladder 4.0 m long is placed against a frictionless wall with its base situated 2.0 from the wall. Find the forces exerted by the wall and by the ground on the ladder.

Answer 1

Let normal reaction at endpoints be $N_1, N_2$

Force of friction acting between ground and wall is f.

The weight of the ladder will act at the mid-point of the ladder.

Let angle made by the wall with ground is $\theta, cos\theta = \frac{2}{4}= \frac{1}{2}$

Then for equilibrium in the horizontal and vertical direction

$f = N_2$ (1)

$N_1 = mg = 80N$ (2)

Taking torque about point on ground,

$N_2 lsin\theta = mg\frac{l}{2} cos\theta \implies N_2 = 23.1N$