A rigid body is said to be in mechanical equilibrium, if both the linear momentum and angular momentum are not changing with time, the body has neither linear acceleration nor angular acceleration.

$(i)$ Translational equilibrium :

If the total force means the vector sum of the forces on the rigid body is zero, then

$\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}+\ldots+\overrightarrow{\mathrm{F}}_{n}=0$

$\therefore\sum_{i=1}^{n} \overrightarrow{\mathrm{F}}_{i}=0 \text { where } i=1,2,3, \ldots, n$

$\ldots \text { (1) }$

If the total force on the body is zero, then the total linear momentum of the body does not change with time.

Equation $(1)$ gives the condition for the translational equilibrium of the rigid body.

$(ii)$ Condition for the rotational equilibrium :

If the total torque means the vector sum of the torques on the rigid body is zero, then

The rotational equilibrium condition is independent of the location of the origin.

Equation $(1)$ is equivalent to three scalar equations as under, they corresponds to

$\sum_{i=1}^{n} \overrightarrow{\mathrm{F}_{i x}}=0, \sum_{i=1}^{n} \overrightarrow{\mathrm{F}_{i y}}=0 \text { and } \sum_{i=1}^{n} \overrightarrow{\mathrm{F}_{i z}}=0 \ldots \text { (3) }$

where $i=1,2,3, \ldots, n$ and $\mathrm{F}_{i x}, \mathrm{~F}_{i y}$ and $\mathrm{F}_{i z}$ are respectively the $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ components of the forces $\overrightarrow{\mathrm{F}}_{i}$.

Equation (2) is equivalent to three scalar equations

$\sum_{i=1}^{n} \overrightarrow{\tau_{i x}}=0, \sum_{i=1}^{n} \overrightarrow{\tau_{i y}}=0 \text { and } \sum_{i=1}^{n} \overrightarrow{\tau_{i z}}=0 \ldots$

where $i=1,2,3, \ldots, n$ and $\tau_{i x}, \tau_{i y}$ and $\tau_{i z}$ are respectively the $X$, $Y$ and $Z$ component of the torque $\vec{\tau}_{i}$