Q.
An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If $AB = BC$, and the angle made by $AB$ with downward vertical is $\theta$, then :
Solution:
Let mass of one rod is $m$.
Balancing torque about hinge point.
$mg \left(C_{1}P\right) = mg \left(C_{2}N\right) $
$ mg \left(\frac{L}{2} \sin\theta\right) = mg \left(\frac{L}{2} \cos\theta - L \sin\theta\right) $
$ \Rightarrow \frac{3}{2} mg L \sin\theta = \frac{mgL}{2} \cos\theta $
$ \Rightarrow \tan \theta = \frac{1}{3} $
