A hemispherical portion of radius ' R ' is removed from the bottom of a cylinder of radius ' R '. The volume of the remaining cylinder is ' V ' and its mass is ' M &

Question

A hemispherical portion of radius ' R ' is removed from the bottom of a cylinder of radius ' R '. The volume of the remaining cylinder is ' V ' and its mass is ' M '. It is suspended by a string in a liquid of density ' ρ ' where it stays vertical. The upper surface of the cylinder is at a depth ' h ' below the liquid surface. The force on the bottom of the cylinder is

TestHub · Accepted Answer

The net upward force exerted by the liquid on the cylinder is the buoyant force, which is given by Archimedes' principle as:

$F_{B} = V ρ g$

This buoyant force is the resultant of the forces exerted by the liquid on all surfaces of the cylinder. Since the horizontal forces cancel out due to symmetry, the buoyant force is the difference between the upward force on the bottom surface and the downward force on the top surface:

$F_{B} = F_{b o tt o m} - F_{t o p}$

The top surface of the cylinder is flat, has an area $A = π R^{2}$ , and is at a depth $h$ . The downward force on the top surface is:

$F_{t o p} = P_{t o p} A = (ρ g h) (π R^{2})$

Substituting $F_{t o p}$ and $F_{B}$ into the force balance equation, we get the upward force on the bottom surface:

$F_{b o tt o m} = F_{B} + F_{t o p}$

$F_{b o tt o m} = V ρ g + ρ g hπ R^{2}$

$F_{b o tt o m} = ρ g (V + π R^{2} h)$

Answer: ( $ρ g (V + π R^{2} h)$ )

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