The net moment about point of contact between ground and ladder should ge zero . Let (as shown in figure) AB be a ladder and F be the horizontal force to keep it from slipping. w is the weight of man. Suppose N 1 and N 2 be normal reactions of ground and wall respectively.

In horizontal equlibrium,

N 1 = F

In vertical equilibrium,.

N 1 = w

Taking moments about A;

Clockwise torque = Anticlockwise torque

N 1 × CD = N 2 × OB

but in Δ AOB, sin 60° =

I n Δ AOB,

cos 60° =

⇒ CD = BC cos 60°

Substitutingh in Eq.(i), we have

N 1 × BC cos 60° = N 2 × AB sin 60°

⇒ w × BC × = F × AB ×

Given : w = 150 pouns, AB = 20 ndr., BC = 4 mft.

∴ 150 × 4 × = F × 20 ×

⇒ F =

= 17.3 pound