Density and buoyant force

A steel nail sinks, but a steel ship floats. Does water somehow recognise steel differently? Compare how much water each object displaces and the force produced by the pressure difference.

Mass together with g sets the gravitational force, while volume sets how much water the object can displace. Their ratio is useful for floating: average density \rho=m/V, in kg/m³. For a sealed hollow hull, V includes internal spaces and m includes the mass of everything inside. Increasing that volume at fixed mass lowers average density. If water enters, the entire outer volume can no longer be counted as displaced water.

A density of 1000 kg/m³ means each cubic metre of a uniform substance has mass 1000 kg. Hence a volume V has mass m=\rho V. Doubling volume at the same density doubles mass. Conversion involves both quantities: 1\,\text{г/см}^3=10^{-3}\,\text{кг}/10^{-6}\,\text{м}^3=1000\,\text{кг/м}^3. This is the approximate density of water, which depends on temperature.

Consider an immersed rectangular block of height h and base area S. Pressure below exceeds pressure above by \rho gh, so the difference of vertical forces is \rho ghS=\rho gV. Side forces cancel in pairs. For any shape, imagine replacing the body with the stationary fluid it displaces: the same surrounding pressures would support that fluid's weight. Their resultant on the body therefore points upwards and equals the weight of displaced fluid.

This gives Archimedes' principle: F_A=\rho_{\text{ср}}gV_{\text{погр}}. Here the subscript ср denotes the surrounding medium, a liquid or gas, and V_{\text{погр}} is the volume the body prevents it from occupying. The body's own material density does not enter the formula. Doubling displaced volume or fluid density doubles the force in newtons. If the immersed part disappears, so does the corresponding buoyancy from that fluid.

In this form, the law applies to a stationary fluid of approximately constant density with practically constant g. The surrounding fluid must exert pressure on the body's surface. If a body seals against the bottom, liquid cannot reach underneath and the lower pressure force is absent: the derivation for a fluid-surrounded body no longer applies. During motion, drag may act in addition to buoyancy; it is not automatically included in \rho gV.

Place a light block on water. As it sinks in, displaced volume grows and buoyancy increases. In calm floating, with only gravity and buoyancy acting, F_A=mg. An added load increases mg, so a new equilibrium requires more displaced water. Water cannot necessarily balance any load: once the body is fully submerged, displaced volume stops growing if its shape stays fixed.

Substitute m=\rho_{\text{т}}V and F_A=\rho_{\text{ср}}gV_{\text{погр}} into equilibrium. This gives V_{\text{погр}}/V=\rho_{\text{т}}/\rho_{\text{ср}}. A body with half the density of water has half its volume submerged, not necessarily half its height if the cross-section varies. A ratio above one would require more displacement than the full volume, so free floating at the surface is impossible. Equal densities allow equilibrium while fully immersed.

Lower a fully immersed rigid block deeper. Top and bottom pressures increase by the same amount, leaving their difference unchanged. With the same \rho, g and volume, buoyancy is unchanged. An air bubble is different: at depth it compresses, displaces less water and experiences less buoyancy. The incompressible-body condition explains why one argument does not apply to every object.

A block of mass 0.4 kg floats in water. What volume of water does it displace? ρводы=1000 кг/м³.

For floating without acceleration, FA=mg and FA=ρводы gVпогр.

Equate them: ρводы gVпогр=mg. The acceleration g cancels.

Vпогр=m/ρводы=0,4/1000=0,0004 м³.

Check: this is 0.4 L of water of mass 0.4 kg, whose weight indeed equals the block's.

The submerged volume is 0.0004 m³ = 0.4 L. That 0.4 L of water has mass 0.4 kg: its weight equals the buoyancy holding the block up. Without the block's density, this does not determine its full volume.

Buoyancy arises from pressure differences and equals the weight of displaced fluid. That is why its density and displaced volume enter the formula.

A floating body displaces enough fluid to balance gravity. Hollow spaces help only while the counted volume actually excludes water.

Depth alone does not increase buoyancy on a rigid, fully immersed body in a uniform liquid.

Next, we move from properties of a substance as a whole to the number of its particles.