A body may be in partial equilibrium, means it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium.

Consider a light rod of negligible mass $\mathrm{AB}$ and at the its two ends of which two parallel forces both equal in magnitude are applied perpendicular to the rod as shown in above figure.

$\mathrm{C}$ be the midpoint of $\mathrm{AB}$.

$\therefore \mathrm{CA}=\mathrm{CB}=a$ assumed.

The moment of the forces at $\mathrm{A}$ and $\mathrm{B}$ will both be equal in magnitude $(a \mathrm{~F})$, but opposite in sense as shown. The net moment on the rod will be zero. The system will be in rotational equilibrium,

but it will not be in translational equilibrium : $\sum \overrightarrow{\mathrm{F}} \neq 0$

Now consider a light rod of negligible mass and length $2 a$. Its two ends $\mathrm{A}$ and $\mathrm{B}$ of which two equal and opposite forces $\overrightarrow{\mathrm{F}}$ applied perpendicular to the rod. $\mathrm{C}$ be the midpoint of $\mathrm{AB}$. $\therefore \mathrm{AC}=\mathrm{BC}=a$

Hence, both the torque on the rod in the same direction, cause anti-clockwise rotation of the rod. So, the total force on the body is zero, so the body is in translational equilibrium, but it is not in rotational equilibrium due to torque.

If rod is fixed at $\mathrm{C}$, then it undergoes pure rotational motion.