Laws of Motion

As the rod is about to slip, wall and floor exert limiting friction on the ladder.

Case 1 : If $\mu$₁ = 0, $\sum {{{\overrightarrow \tau }_A} = \overrightarrow 0 } $

$mg\left( {{l \over 2}\cos \theta } \right) = {N_1}(l\sin \theta )$

${N_1} = {{mg\cot \theta } \over 2}$

or ${N_1}\tan \theta = {{mg} \over 2}$

and ${N_2} = mg$

Case 2 : If $\mu$₂ = 0, N₁ remain unbalanced and rod can never be in equilibrium.

Case 3 : If $\mu$₁ $\ne$ 0, $\mu$₂ $\ne$ 0

${N_1} = {f_2} = {\mu _2}{N_2}$

${N_2} + {f_1} = mg$

or ${N_2} + {\mu _1}{N_1} = mg$

or ${N_2} + {\mu _1}({\mu _2}{N_2}) = mg$

or ${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$

The forces acting on the block are F = 1 N towards the left, weight mg = 0.1 $\times$ 10 = 1 N downwards, normal force N, and the frictional force f. Resolve F and mg along and perpendicular to the inclined plane.

When $\theta$ = 45$^\circ$, the net force that brings the block down is

${F_d} = mg\sin \theta - F\cos \theta = {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 0$.

Thus, the block is stationary if $\theta = 45^\circ $. When $\theta > 45^\circ $, the force ${F_d} > 0$, and hence the block has a tendency to move down. Thus, the frictional force f acts on the block upwards i.e., towards Q.