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The centre of buoyancy is the point through which the resultant hydrostatic buoyant force acts. In a uniform fluid under uniform gravitational acceleration, it is the volume centroid of the displaced fluid. For a floating body its position generally changes when the body tilts, because the displaced volume changes shape.

For a small heel of a floating body in a uniform fluid, the metacentre is the intersection of the new vertical buoyancy line with the original vertical through the centre of buoyancy. With the centre of mass a signed distance $BG$ above the original centre of buoyancy, the metacentric height is $GM=I_{\mathrm{wp}}/V_d-BG$. Here $I_{\mathrm{wp}}$ is the waterplane second moment of area about the heel axis and $V_d$ the displaced volume. Positive $GM$ gives a restoring small-angle torque and negative $GM$ gives instability. The formula describes local static stability, not all large-angle behaviour or effects of free liquid surfaces inside the body.

Take a body floating at rest in a homogeneous fluid, with uniform gravity, no tether or other support, and negligible air buoyancy. Imagine replacing the submerged part by the same volume of the surrounding fluid. That imaginary fluid parcel would be at rest: the surrounding hydrostatic pressure would balance its gravitational weight. The actual body experiences the same pressure traction over the submerged boundary, so the resulting buoyant force is upward with magnitudeThe body's vertical force balance therefore gives Archimedes' principle in the requested floating case:The argument determines displaced submerged volume, not necessarily the total volume of a partly emerged body.

The pressure resultant acts through the centre of buoyancy, the centre of mass of the imaginary displaced fluid. Let $G$ be the body's centre of gravity and $B_c$ the centre of buoyancy. The two remaining external forces are equal and opposite vertical forces. If their lines of action were distinct, they would form a nonzero couple and the body would turn. Rotational equilibrium consequently requiresThis is a condition for equilibrium, not sufficient evidence of its stability. Tilting changes the displaced shape and hence the position of the centre of buoyancy; the resulting torque determines whether the body returns towards its original orientation. A centre of gravity above the centre of buoyancy does not by itself settle that floating-body stability question.

Archimedes' principle states that the resultant hydrostatic pressure force on a body equals the weight of the fluid displaced, directed opposite to gravity. Let $V_d$ be the displaced fluid region, with outward normal $\hat{\mathbf{n}}$ from the body into the fluid. Since the fluid pressure force is inward,Hydrostatic balance is $\nabla p={\rho }_f\mathbf{g}$, with downward gravitational acceleration $\mathbf{g}$. Thereforefor uniform fluid density and gravity. Uniform atmospheric pressure gives no net force on a closed surface. For a floating body, the same displaced-volume construction includes the free surface; any significant air buoyancy can be included as a second displaced fluid.

A freely floating body of mass $m$ satisfies $mg={\rho }_{f}g{V}_d$, so its displaced volume is $m/{\rho }_f$. A homogeneous body of density ${\rho }_b<{\rho }_f$ floats with submerged fraction ${\rho }_b/{\rho }_f$; if ${\rho }_b>{\rho }_f$, hydrostatic buoyancy alone cannot support it. The buoyant force acts through the centre of buoyancy, the centroid of the displaced volume for constant density, while weight acts through the centre of mass. Their relative positions determine rotational equilibrium and stability. For a fully submerged body, a centre of mass below the centre of buoyancy gives a restoring torque; a floating body's changing displaced shape additionally requires the metacentric height criterion. Surface tension, externally driven flow and contact forces can alter the simple force balance.