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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 μ 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

A
μ1=0μ20 and N2 tan θ=mg/2
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B
μ10μ2=0 and N1 tan θ=mg/2
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C
μ10μ20 and N2=mg1+μ1μ2
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D
μ1=0μ20 and N1 tan θ=mg/2
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Solution

The correct option is D μ1=0μ20 and N1 tan θ=mg/2
Consider of translational equilibrium
N1=μ2N2 (i)
N2+μ1N1=Mg (ii)

Solving N N2=mg1+μ1μ2
N1=μ2mg1+μ1μ2
Applying torque equation about corner (left) point ont he floor
mg l2 cos θ=N1l sin θ+μ1N1l cos θ
Solving tan θ=1μ1μ22μ2

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