A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is μs = 0.495, determine the smallest angle the ladder can make with the floor without slipping.

45.3°

Explanation:

Given,

Length of ladder = l

Weight of ladder = w

Coefficient of friction = μs = 0.495

Smallest angle the ladder makes = θ

If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,

N1 = mg and

N2 = μmg

Moment at a point A in the clockwise direction is

N2 Lsinθ - mg.(L/2).cosθ = 0

μmgLsinθ - mg.(L/2).cosθ = 0

μmgLsinθ = mg.(L/2).cosθ

μsinθ = cosθ/2

sin θ / cos θ = 1 / 2μ

Tan θ = 1 / 2μ

Substituting the value of μ = 0.495, we have

Tan θ = 1 / 2 * 0.495

Tan θ = 1 / 0.99

Tan θ = 1.01

θ = tan^-1(1.01)

θ = 45.3°

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