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Question:

In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with the horizontal floor. The coefficient of friction between the wall and the ladder is μ₁ and that between the floor and the ladder is μ₂. The normal reaction of the wall on the ladder is N₁ and that of the floor is N₂. If the ladder is about to slip, then

μ₁≠0 , μ₂≠0 and N₂=mg/(1+μ₁μ₂)

μ₁=0 , μ₂≠0 and N₂tanθ=mg/2

μ₁≠0 , μ₂=0 and N₁tanθ=mg/2

μ₁=0 , μ₂≠0 and N₁tanθ=mg/2

Solution:

From the condition of equilibrium of bodies:
ΣFx=0 → -N₁+μ₂N₂=0 → N₁=μ₂N₂
Also ΣFy=0 → N₂+μ₁N₁-mg=0 → N₂+μ₁μ₂N₂-mg=0 → N₂(1+μ₁μ₂)=mg → N₂=mg/(1+μ₁μ₂)
Applying torque equation about corner (left) point on the floor:
mgl/2cosθ=N₁lsinθ+μ₁N₁lcosθ