JEE Advanced | 2014FrictionMedium
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
Options
A.μ1 = 0, μ2 ≠ 0 and N2 tan θ = 
B.μ1 ≠ 0, μ2 = 0 and N1 tan θ =
C.μ1 ≠ 0, μ2 ≠ 0 and N2 = 
D.μ1 = 0, μ2 ≠ 0 and N1 tan θ =
Solution
μ2 can never be zero for maximum equilibrium.
When μ1 = 0 we have
N1 = μ2N2 .....(i)
N2 = m2g .....(ii)
τB = 0 ⇒ mg
cosθ = N1L sin θ ⇒ N1 =
⇒ N1 tan θ =
When μ1 ≠ 0 we have
μ1N1 + N2 = mg .....(i)
μ2N2 = N1 .....(ii)
⇒ N2 =
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