Two identical ladders are arranged as shown in the figure. Mass of the block is $m$ and the mass of each ladder is $M$ . The length of each of the ladder is $L$ . The system is in equilibrium. What is the magnitude of frictional force acting at $A$ or $B$ ?

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Drawing force diagrams of the rod, we have

This is the equilibrium of coplanar forces, hence using the equations

$\sum F_{x} = 0, \sum F_{y} = 0$ and

Net moment about point O = 0

we have the equations,

$N + N_{2} = M g +$ $\frac{m g}{2}$ ... (i)

$N_{1} = f$ ...(ii)

$N = N_{2} + M g +$ $\frac{m g}{2}$ ... (iii)

and $M g$ $\frac{L}{2}$ $\cos θ + f L \sin θ = N L \cos θ$ ... (iv)

Solving these four equations, we have

$f = (\frac{M + m}{2}) g cot θ$

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