(N/A) Yes,a body can be in partial equilibrium. This means it may be in translational equilibrium but not in rotational equilibrium,or it may be in rotational equilibrium but not in translational equilibrium.
Case $(a)$: Rotational equilibrium but not translational equilibrium.
Consider a light rod of negligible mass $AB$ of length $2a$. Two parallel forces,both equal in magnitude $F$,are applied perpendicular to the rod at ends $A$ and $B$ in the same direction as shown in figure $(a)$.
Let $C$ be the midpoint of $AB$,so $CA = CB = a$.
The net force on the rod is $\sum \vec{F} = F + F = 2F \neq 0$. Thus,it is not in translational equilibrium.
The net torque about the midpoint $C$ is $\tau = (F \times a) - (F \times a) = 0$. Thus,the rod is in rotational equilibrium.
Case $(b)$: Translational equilibrium but not rotational equilibrium.
Consider the same rod $AB$ of length $2a$. Two equal and opposite forces $\vec{F}$ are applied perpendicular to the rod at ends $A$ and $B$ as shown in figure $(b)$.
The net force on the rod is $\sum \vec{F} = F - F = 0$. Thus,the rod is in translational equilibrium.
The net torque about the midpoint $C$ is $\tau = (F \times a) + (F \times a) = 2Fa \neq 0$. Since the torques are in the same direction (both cause anti-clockwise rotation),the rod is not in rotational equilibrium.