For an object floating at rest, buoyancy balances its weight. The displaced liquid volume need not equal the object's entire volume. Keeping these two volumes separate is the central step in this original foundation exercise.
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Define the model
A uniform object has volume 0.003 m3 and density 800 kg/m3. It floats in water of density 1000 kg/m3. Ignore air buoyancy and surface-tension effects. Take g = 9.8 m/s2.
The object has mass 2.4 kg and weight 23.52 N. At rest, the water must provide the same upward force.
Solve for displaced volume
Using buoyancy = water density × g × displaced volume gives a displaced volume of 0.0024 m3. Its fraction of the whole volume is 0.0024/0.003 = 0.80.
Thus 80% of the object's volume is submerged under these assumptions. For a nonuniform shape, this is a volume fraction, not automatically 80% of its height.
Check the fully submerged comparison
If held completely underwater, the object would displace 0.003 m3 and experience buoyancy of 29.4 N. This exceeds its weight, so an additional downward force would be needed to hold it still in this ideal model.
Do not use the full volume in the original floating calculation and then claim the forces balance. Draw the actual state and identify the displaced volume first. This is a teaching problem, not an official Olympiad question or an experimental instruction.
Continue reading: British Physics Olympiad Mechanics: Momentum vs Kinetic Energy.
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Official information and further reading: Source 1
