Question

A uniform ladder of length 1 and weight W=50N rests against a smooth, vertical wall. If the coefficient of static friction between ladder and ground is µ=0.5, find the minimum angle ѳ min between the ladder and ground such that the ladder will not slip: A.45∘ B.53 ∘ C. 60∘ D. 37∘

Answer 1

Take moments about the base. With W at the midpoint, the horizontal reaction at the wall is
H = (W/2) cotθ.

Horizontal equilibrium at the base gives friction f = H, and static friction limit is f ≤ µN = µW. So
(W/2) cotθ ≤ µW → cotθ ≤ 2µ → tanθ ≥ 1/(2µ).

For µ = 0.5: tanθ ≥ 1 → θ ≥ arctan(1) = 45°.

So the minimum angle is 45° (choice A).