Derive buoyancy from displaced liquid

A sinking object still receives an upward force

Predict the forces on a sealed cube held entirely underwater. If the cube is denser than the liquid, does the liquid stop pushing upward? Sinking describes the imbalance between forces, not the absence of buoyancy.

The investigation uses a rigid cube with volume 0.001 m³, which is 1 litre, and height 0.1 m. You choose its average density and the surrounding liquid density. Initially its top is 0.05 m below the surface, so the whole cube is submerged. The liquid is uniform and at rest; teaching g is 10 m/s². The diagram may show a possible floating equilibrium, but the force bars always describe this fully submerged starting test. Their labels distinguish these two configurations.

Build buoyancy from pressure differences

The cube’s bottom is deeper than its top, so the bottom pressure is greater. Pressure on the bottom pushes upward; pressure on the top pushes downward. Opposite side forces cancel in this symmetric arrangement. The remaining pressure force is upward.

For a cube with horizontal face area A and height H, the bottom-to-top pressure difference is ρ_liquid gH. Multiplying by A gives B = ρ_liquid gAH = ρ_liquid gV_displaced. B is buoyant force in newtons. The density is that of the liquid, and V_displaced is the volume occupied in the liquid by the object. A wholly submerged sealed cube displaces its full external volume.

This is the weight of the liquid that would occupy that volume. It is not necessarily the object’s weight. The OpenStax buoyancy reference states this displaced-fluid relation. The current lesson applies it to a cube with separate pressure and weight accounting.

Check the same force two ways

At liquid density 1000 kg/m³, full-submersion buoyancy is 1000 × 10 × 0.001 = 10 N upward. The cube’s face area is 0.01 m². With surface pressure 100 kPa, its upper face at depth 0.05 m has pressure 100500 Pa and downward force 1005 N. The lower face at depth 0.15 m has pressure 101500 Pa and upward force 1015 N. Subtracting gives the same 10 N upward.

The large common pressure contribution cancels between opposite faces. The buoyant force is this pressure-force difference. Adding another 10 N after already summing the pressure forces would count the same effect twice. Object weight is a separate gravitational force and must still be included.

For object density 500 kg/m³, mass is density times volume: 500 × 0.001 = 0.5 kg. Weight is 0.5 × 10 = 5 N downward. The fully submerged hydrostatic-plus-weight force is 10 − 5 = +5 N, with upward positive. It has an upward tendency when released; a motion history is not supplied by this static calculation.

Change object density without changing displaced volume

Keep liquid density at 1000 kg/m³ and compare object densities 500, 1000 and 1500 kg/m³. All three rigid cubes have the same external volume, so fully submerged buoyancy is 10 N in every case. Their weights are 5, 10 and 15 N. The force differences are +5, 0 and −5 N respectively.

Capture these cases. Explain why a sinking cube can have a substantial buoyant force that simply fails to balance its weight. Increasing object density at fixed external volume changes mass and weight; it does not change how much liquid is displaced while the object stays fully submerged.

Now change the liquid density while holding the object density fixed. Denser liquid gives a larger buoyant force for the same displaced volume. At liquid density 1500 kg/m³, the full-submersion force is 15 N. The object’s mass and weight remain unchanged.

Know which geometry the formula uses

For partial submersion, use only the volume below the liquid surface. Using the whole object volume would overestimate liquid buoyancy. The next lesson determines that fraction from equilibrium rather than assuming it.

For a fully submerged rigid object in this uniform incompressible liquid, moving deeper increases both top and bottom pressure by the same amount. Their difference and buoyancy remain unchanged. This conclusion can fail for a compressible object whose volume changes or a fluid whose density varies with depth. The model also omits air buoyancy, surface tension and floor contact.

The pressure ledger predicts a hydrostatic force balance. It does not include drag, fluid inertia during motion, terminal speed, settling time or the behavior of a leaking or deforming object. Do not divide the displayed force by mass and claim a measured acceleration history without an appropriate motion model.

Transfer to another volume

A rigid sealed object of volume 0.002 m³ and mass 2.4 kg is fully submerged in liquid of density 900 kg/m³. Use g = 10 m/s². Find buoyancy, weight and the signed vertical force difference. Predict whether pushing it deeper changes buoyancy under the stated constant-density, fixed-volume assumptions.

Check the buoyant force transfer

Buoyancy is 900 × 10 × 0.002 = 18 N upward. Weight is 2.4 × 10 = 24 N downward. The upward-positive difference is −6 N, a downward tendency. Greater depth does not change the displaced volume or liquid density in this model, so buoyancy remains 18 N while fully submerged.