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In a uniform gravitational field, the centre of gravity is the point $G$ through which the resultant gravitational weight acts. It coincides with the center of mass:Indeed the sum of the elemental gravitational torques about any origin isReplacing the distributed gravitational forces by the resultant at $G$ therefore preserves both total force and torque. A strongly nonuniform field need not permit the same identification.

For equilibrium, all external forces and torques must balance. A freely suspended rigid body supported at a pivot requires $G$ on the vertical through that pivot. With $G$ below the pivot, a small angular displacement raises its gravitational potential energy and gives a restoring torque; with $G$ above it, the equilibrium is unstable. A body supported on a horizontal base can remain at rest under its weight only if the vertical projection of $G$ lies within the region that the support forces can sustain. At an edge the overturning torque becomes decisive.

For motion, the center-of-mass equation isThe centre moves as a particle carrying the total mass under the resultant external force, while rotation about it is governed by the external torque and angular momentum. This separates translation from rotation; it does not determine a rigid body's full orientation from its centre's trajectory alone. In uniform gravity and negligible air resistance, a thrown rotating body has a parabolic centre-of-mass path even while its parts follow more complicated paths.

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.