**Problem:**
Two identical ladders, each of mass M and length L are resting on the rough horizontal surface as shown in the figure. A block of mass m hangs from P. If the system is in equilibrium, find the direction and magnitude of frictional force acting at A and B.
(IIT JEE 2005)

**Solution:**
Let the block of mass $m$ hangs from the point P by a string attached to the hinges of the two ladders.

Newton's second law gives the tension $T$ in the string as $T=mg$. By symmetry, string tension pulls down each ladder by a force \begin{align} T/2=mg/2.\nonumber \end{align} By Newton's third law, the reaction forces acting on the two ladders at the hinge point P are equal and opposite. These are shown by $R_1$ and $R_2$ in the figure. By symmetry, the normal reaction $N$ at A and B are equal, the friction forces $f$ at A and B are equal in magnitude but opposite in direction. Another force acting on both the ladders is their weight $Mg$ which pass through their centre of mass. In equilibrium, the resultant forces on the two ladders are zero i.e., \begin{align} \label{xsb:eqn:1} &N+R_2-Mg-mg/2=0,\\ \label{xsb:eqn:2} &f=R_1,\\ \label{xsb:eqn:3} &N-R_2-Mg-mg/2=0. \end{align} The above equations give $R_2=0$ (as expected!) and $N=(M+m/2)g$. In equilibrium, the net torque about any point is zero. Thus, the torque about the point P for the left ladder is \begin{align} \label{xsb:eqn:4} Mg(L/2)\cos\theta+fL\sin\theta-NL\cos\theta=0. \end{align} Substitute $N=(M+m/2)g$ in above equation and simplify to get \begin{align} f=\left(\tfrac{M+m}{2}\right)g\cot\theta.\nonumber \end{align}

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