**Problem:**
In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle $\theta$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $\mu_1$ and that between the floor and the ladder is $\mu_2$. The normal reaction of the wall on the ladder is $N_1$ and that of the floor is $N_2$. If the ladder is about to slip, then
(IIT JEE 2014)

**Solution:**
The forces on the rod are its weight $mg$, normal reactions $N_1$, $N_2$, and frictional forces $f_1$ and $f_2$.

When the rod is about to slip, frictional forces attain their maximum values \begin{align} &f_1=\mu_1N_1,\\ &f_2=\mu_2 N_2, \end{align} and their directions are as shown in the figure. Newton's second law in the horizontal and the vertical directions gives \begin{align} &N_1=f_2,\\ &mg=N_2+f_1. \end{align} Eliminate $f_1$, $f_2$, and $N_1$ from above equations to get \begin{align} N_2=\frac{mg}{1+\mu_1\mu_2}. \end{align} Torque on the rod about any point is zero. The torque about A is \begin{align} mg(l/2)\cos\theta-N_1 l\sin\theta-\mu_1 N_1 l\cos\theta=0. \end{align} Simplify above equation to get \begin{align} N_1(\tan\theta+\mu_1)={mg}/{2}. \end{align}

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