Q.
In 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 isμ2. The normal reaction of the wall on the ladder is N1 and that of the floor is N2. If the ladder is about toslip, then
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a
μ1=0μ2≠0 and N2tanθ=mg2
b
μ1≠0μ2=0 and N1tanθ=mg2
c
μ1≠0μ2≠0 and N2=mg1+μ1μ2
d
μ1=0μ2≠0 and N1tanθ=mg2
answer is C.
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Detailed Solution
If μ1=0,μ2≠0t1=0, balancing torques about AN1ℓsinθ=mgcosθ2; N1tanθ=mg2 If μ1≠0,μ2≠0T2=0, equilibrium cannot be attained. If μ1≠0,μ2≠0N1=t2=μ2N2;N2+t1=mgN2+μ1N1=mg; N2+μ1μ2N2=mg; N2=mg1+μ1μ2
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