= Solution
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 magnitude
$$
B=\rho_f gV_{\rm displaced}.
$$
The body's vertical <force> balance therefore gives <Archimedes' principle> in the requested floating case:
$$
\boxed{mg=\rho_f gV_{\rm displaced}.}
$$
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 requires
$$
\boxed{G\text{ and }B_c\text{ lie on the same vertical line}.}
$$
This 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.
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