Q:

Why we can't take torque equation for the mid point of the ladder in the question **Source IIT JEE 2005 PHYSICS**

Question (Source IIT JEE 2005 Physics)
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 magnitude and the direction of frictional force at A and B.
Attempt
For the equilibrium of the whole system:
$ΣFy=0$
So,$$N=\frac{(2M+m)g}{2}$$
And for rotational equilibrium
$$mg\frac{lcos(\theta)}{2} +N(\frac{lcos(\theta)}{2}-f\frac{lsin\theta}{2}=0$$ This what I have written for either rod's rotational equilibrium about the mid point of the rod since the rod is in equilibrium.
Main Difficulty
When I approached my teacher for the answer, mine was wrong and then my teacher tried to explain why it should not be correct for the rotational equilibrium equation for the rod, I was not able to understand. This step is my main difficulty or obstacle. Please help me in clearing this doubt.

Thank you for helping me.

A:

The left ladder presses the right ladder at point P. This is a force you need to include as a torque unless you select P as your axis of rotation. 

