Falling Rod

Suppose the minimum value of $\mu$ occurs at $\theta$ . And at the moment I pause the person.

So the FBD will be like 581

We write the torque equation about the intersecting point of F and mg line of force.

Since there is no motion at contact point of rod and floor. Then, $Fr × \bigg( \frac{l}{2 \sin \theta} +\frac{l \sin \theta}{2} \bigg) - N × \frac{l \cos \theta}{2} = 0$

Note that moment of $F$ and $mg$ becomes zero.

Compare with $Fr = {\mu} N$ ${\mu} =\frac{ cos \theta sin \theta}{1+ sin² {\theta}}$ Min value of the expression gives the minimum value of $\mu$ $\displaystyle{{ \mu }_{ min }=\cfrac { 1 } { 2\sqrt { 2 } } }$

some people might not have done like kushal ( finding intersection of mg and f ) but it does'nt matter , ultimately mg will get cancelled $$ but you must get rid of moment of $F$ doint that and writing the conditions, of translational and rotational equilibrium you will get $\mu$ as $\frac {1}{\sqrt8}$ which is the required answer :) !