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Engineering
Physics
Equilibrium
Static and Kinetic Friction
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  N2tanθ=mg2

 µ1≠0,µ2≠0  and  N1=mg1+µ1µ2

 µ1≠0,µ2=0  and  N1tanθ=mg2

 µ1=0,µ2≠0  and  N1tanθ=mg2

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Solution

µ2N2 = N1

µ1N1 + N2 = mg

mg L2 cos θ = N1L sin θ + µ1N1 cos θ

if µ1 = 0,                        N2 = mg

                                     N1 = mg2 cot θ

 ⇒N1tanθ=mg2

if          µ2 = 0, N1 = 0

           N2 = mg

 ⇒   not possible

 µ1≠0, µ2≠0

N2 = mg1+ μ1μ2

 

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