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Question

# A ladder of mass m is leaning against a wall. lt 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

A
μ1=0;μ20 and N2tanθ=mg2
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B
μ10;μ2=0 and N1tanθ=mg2
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C
μ10;μ20 and N2=mg1+μ1μ2
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D
μ1=0;μ20 and N1tanθ=mg2
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Solution

## The correct options are C μ1≠0;μ2≠0 and N2=mg1+μ1μ2 D μ1=0;μ2≠0 and N1tanθ=mg2Balancing forces in horizontal direction,N1=μN2Balancing forces in vertical direction,N2+μN1=mgSolving we get,N2=mg1+μ1μ2N1=μ2mg1+μ1μ2Applying rotational equilibrium,mgl2cosθ=N1lsinθ+μN1lcosθ Solving, tanθ=1−μ1μ22μ2If μ1=0,μ2≠0tanθ=12μ2N1=μ2mgor N1tanθ=mg2

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