Explanation

(d) As the block remains stationary therefore
For translatory equilibrium
$\sum_{}^{}F_{x} = 0\ \therefore F = N$
and $\sum_{}^{}{F_{y} = 0\ \therefore f = mg}$
For rotational equilibrium $\sum_{}^{}{\tau = 0}$
By taking the torque of different forces about point 0
$\overrightarrow{\tau_{F}} + \overrightarrow{\tau_{f}} + \overrightarrow{\tau_{N}} + \overrightarrow{\tau_{mg}} = 0$
As $F$ and $mg$ passing through point $O$
$\therefore\overrightarrow{\tau_{1}} + \overrightarrow{\tau_{N}} = 0$
As $\overrightarrow{\tau_{f}} \neq 0\ \ \therefore\overrightarrow{\tau_{N}} \neq 0$ and torque by friction and normal reaction will be in opposite direction