Solution: Maximum Buoyant Force on a Valve
To find the mass required to keep the block in equilibrium, we must calculate the maximum possible upward force exerted by the fluid. This maximum occurs at a specific water level where the lift is optimized against the downward pressure.
1. The Logic of Maximum Force
The fluid exerts an upward pressure on the lower faces and a downward pressure on the upper faces. The net upward force is maximized when the water level is high enough to create maximum pressure at the bottom, but not so high that it covers too much of the top (which would push the block down).
For the diagram above net force due to water would be upward and equal to weight of the jutted out portions. Any more water added would create an extra downward force, thereby reducing the upward force.: The width of the dry block at the surface equals the width of the hole ($a$).
2. Calculating the Net Force
We use the “missing flux” method. The force is the weight of the displaced fluid volume ($V_{sub}$) minus the force lost because the hole is exposed to air (pressure correction).
Submerged Volume ($V_{sub}$):
The total volume of the prism minus the top “dry” tip and the bottom “exposed” tip. Since both tips have base width $a$ (and thus height $a/2$), their combined area is $a^2/2$.
$$ V_{sub} = b \left( c^2 – \frac{a^2}{2} \right) $$
Pressure Correction:
We must subtract the force that would act on the hole area if it were sealed. The depth at the hole is $H = \sqrt{2}c – a$.
$$ F_{hole} = P \times Area = (\rho g [\sqrt{2}c – a]) (ab) $$
3. Final Derivation
Subtracting the hole correction from the buoyant force: $$ F_{net} = \rho g b \left( c^2 – \frac{a^2}{2} \right) – \rho g b a (\sqrt{2}c – a) $$ Expanding and rearranging terms reveals a perfect square: $$ F_{net} = \rho g b \left( c^2 – \sqrt{2}ac + \frac{a^2}{2} \right) = \rho g b \left( c – \frac{a}{\sqrt{2}} \right)^2 $$
The mass must exceed the maximum lifting force: $$ m > \rho b \left( c – \frac{a}{\sqrt{2}} \right)^2 $$