Reject a floating prediction when required displacement is impossible
A freely floating object in vertical equilibrium has buoyancy equal to weight: ρfluidgVsub=mg. Therefore required submerged volume is m/ρfluid. For a homogeneous solid, the submerged fraction equals its average density divided by fluid density when that ratio is below one. For a hollow sealed object, use total mass and external displaced volume, not the density of its wall material alone.
Learning outcome: Compare payload4 and8 kg at fluid density1000. Record the required versus available displacement before accepting any draft. Then compare densities800 and1200 at fixed payload. Explain the special full-immersion equality case and the stability questions not answered.
Starting recap
Review density and unit conversion, perpendicular force components, static force balance, work-energy, mass conservation and projectile motion.
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A sealed box has base mass4 kg, maximum displacement0.010 m³ and horizontal area0.05 m². In waterρ1000, a4 kg payload gives total8 kg, required displacement0.008 m³ and draft0.16 m. With8 kg payload, total12 kg requires0.012 m³, exceeding capacity; at full immersion buoyancy100 N is less than weight120 N.
Open the experimentPredict
Can a change of fluid density rescue a floating balance that failed at the same payload?
Experiment
Compare payload4 and8 kg at fluid density1000. Record the required versus available displacement before accepting any draft. Then compare densities800 and1200 at fixed payload. Explain the special full-immersion equality case and the stability questions not answered.
Test a floating equilibrium before accepting its draft
Worked teaching model · synthetic values · Original sealed rigid box base mass4 kg, external maximum displaced volume0.010 m³, horizontal area0.05 m²,height0.20 m,g10. Vertical force balance only; no flooding, stability, capsize or real vessel safety model. Required volume above capacity rejects floating equilibrium.
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Feasibility before draft: required displaced volume 0.008 m³ versus available0.010 m³. The vertical balance fits; rotational stability and real safety are not evaluated.
| Quantity | Value | Unit |
|---|---|---|
| Total mass | 8 | kg |
| Required displaced volume for balance | 0.008 | m³ |
| Maximum available displaced volume | 0.01 | m³ |
| Maximum supportable mass in this model | 10 | kg |
| Floating balance feasible | 1 | 1 yes,0 no |
| Net upward force at full immersion | 20 | N |
| Equilibrium draft | 0.16 | m |
| Fraction of external volume submerged | 0.8 | ratio |
Accessible plot coordinates
| coordinate | value |
|---|
A vertical force-balanced displacement fits within the supplied external volume; stability is not established.
For vertical equilibrium ρfluidgVsub=mg. If the object is fully immersed first, buoyancy saturates at the full external volume. Do not hide a failed float prediction by clipping the required volume to its maximum. Neutral full immersion has no positive freeboard and is not a safety margin.
A vertical force-balanced displacement fits within the supplied external volume; stability is not established.
For vertical equilibrium ρfluidgVsub=mg. If the object is fully immersed first, buoyancy saturates at the full external volume. Do not hide a failed float prediction by clipping the required volume to its maximum. Neutral full immersion has no positive freeboard and is not a safety margin.
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Why can a hollow metal object float even though the metal itself is denser than water?
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