(d) The cubical block is in equilibrium
For translational equilibrium
(a) `SigmaF_x = 0 rArr F =N`
(b) `SigmaF_y = 0 rArr f = mg`
For Rotational Equilibrium
`Sigma tau_c = 0`
Where `tau_c` = torque about c.m.
Torque created by frictional force (f) about C =fxxa in
clockwish direction.
There should be another torque which should counter this torque . The normal reacton N on the block acts as
shown. This will create a torque Nxxb in the anticlockwise direction.
Such that fxxa = Nxxb.