A weightless ladder 20 ft long rests against a frictionless

Home » Mechanics » System of Particles and Rotational Motion » Question

A weightless ladder 20 ft long rests against a frictionless wall at an angle of 60º from the horizontal. A 150 pound man is 4 ft from the top of the ladder. A horizontal force is needed to keep it from slipping. Choose the correct magnitude of the force from the following

1. 175 lb
2. 100 lb
3. 120 lb
4. 69.2 lb

Correct Option: D

AB is the ladder, let F be the horizontal force and W is the weigth of man. Let N₁ and N₂ be normal reactions of ground and wall, respectively. Then for vertical equilibrium
W = N₁ .....(1)
For horizontal equilibrium, N₂ = F .....(2)
Taking moments about A,
N₂(AB sin60°) – W(AC cos 60°) = 0 ......(3)
Using (2) and AB = 20 ft, BC = 4 ft, we get

F =		20 ×	√3		= W		20 ×	1		pound
			2					2

⇒ F =	8W × 2	=	4W	=	150 × 4	pound
	20√3		5√3		5√3

= 40√3 = 40 × 1.73 = 69.2 pound