Question

In a stable equilibrium for completely submerged bodies, what is the relation between forces? Given $F_B=\text{Buoyant force},W=\text{weight of body}$

• A. $F_B=W$ and the centre of buoyancy is below the centre of gravity.
• B. $F_B=W$ and the centre of buoyancy is above the centre of gravity.
• C. $F_B<W$
• D. None of the above mentioned statements are true.

Answer 1

The correct option is B $F_B=W$ and the centre of buoyancy is above the centre of gravity.

For stable equilbrium $F_{\text{net}}=0,{\tau }_{\text{net}}=0$ and in case of small angular displacements, a restoring torque must act on the body, which would tend to bring it back to the original position.
$\Rightarrow F_B=W$ and they will form a couple


In case of small angular displacements, net torque due to couple of $F_B\&W$ will tend to bring the body to equilbrium position.This is possible only if $B$ lies above $\left(G\right)$.

Case $1$. If ${\tau }_{\text{ext}}$ tends to displace the body in clockwise direction through small angle $\left(d\theta \right)$, in this case the net torque due to couple $\left(F_B\&W\right)$ will tend to rotate the body in anti-clockwise direction. Hence, It will restore the body back to its original equilibrium position.


Case $2$.If ${\tau }_{\text{ext}}$ tends to displace the body in anti-clockwise direction through small angle $\left(d\theta \right)$, in this case the net torque due to couple $\left(F_B\&W\right)$ will tend to rotate the body in clockwise direction. Hence, It will restore the body back to its original equilibrium position.


Therefore for stable equilibrium, $B$ should always lie above $G$.