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When a cylinder is placed on a fluid, it is partially submerged and attains equilibrium at that state.

In equilibrium, $$F_B=mg$$ where $F_B$ is the buoyant force and $mg$ is the weight of the cylinder.

But we know that $F_B=m_{\text{liquid displaced}}\ g$. Thus, $ m_{\text{liquid displaced}}\ g=mg$

But in the above case, $m_{\text{liquid displaced}}\neq m$ as only a part of $m$ is inside the fluid.

During partial submersion, how can we write the equation for equilibrium state?

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    $\begingroup$ Your equation is correct. The mass of the displaced liquid is equal to the mass of the entire cylinder (and larger than the mass of the portion of the cylinder which is below water). $\endgroup$ Apr 4, 2021 at 19:08

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There are two equal and opposite forces on the cylinder, $w = m_{cyl}g$, and $F_B$.

$F_B$ is the force from pressure that water exerts on the cylinder. Since the pressure increases with depth, forces on the bottom are greater, and these are directed upward.

These bouyancy forces are the same, no matter what object with the same size and shape is subject to them. In particular, just enough water to fill the hole is subject to the same bouyancy force.

$$F_B = m_{water}g$$

So $m_{cyl}g = m_{water}g$. We know the cylinder has a larger volume because some of it stick up above the water. So the cylinder is less dense.

This means if you had a truncated cylinder the same size and shape as the water that filled the hole, it would not be in equilibrium fully submerged in the same position.

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