= Floating equilibrium
{title2=$V_{\mathrm{displaced}}/V_{\mathrm{body}}=\rho_{\mathrm{body}}/\rho_f$}
A freely floating body at rest has no supporting contact or applied holding <force>. Neglecting <surface tension> and the <buoyancy> of air, its upward <buoyancy> balances its downward <force> from <gravity>:
$$
F_b=mg.
$$
For a body of uniform <mass density> $\rho_{\mathrm{body}}$ and total <volume> $V_{\mathrm{body}}$, its <mass> is $m=\rho_{\mathrm{body}}V_{\mathrm{body}}$. Substitution into <Archimedes' principle> gives
$$
\frac{V_{\mathrm{displaced}}}{V_{\mathrm{body}}}
=\frac{\rho_{\mathrm{body}}}{\rho_f}.
$$
Partial immersion is possible when $\rho_{\mathrm{body}}<\rho_f$. A nonuniform body uses its mean <mass density> $m/V_{\mathrm{body}}$ instead; for a hollow body this <volume> must count the enclosed space that excludes <water>. This explains how the amount of displaced <water> can support a vessel whose walls alone are denser than <water>. https://openstax.org/books/college-physics-2e/pages/11-7-archimedes-principle[OpenStax, College Physics, §11.7: floating fraction and mean density]{ref}
For the $600\ \mathrm{kg\,m^{-3}}$ <cube> in <water> of <mass density> $1000\ \mathrm{kg\,m^{-3}}$, the displaced fraction is $600/1000=0.60$: sixty percent of its <volume> lies below the surface. With a horizontal top and bottom, the immersed depth $d$ satisfies $V_{\mathrm{displaced}}=Ad$, so
$$
V_{\mathrm{displaced}}=0.000600\ \mathrm{m^3},\qquad
d=\frac{0.000600}{0.010}=0.060\ \mathrm m.
$$
Its remaining height is $0.040\ \mathrm m$, and the <buoyancy> is $1000\times9.81\times0.000600=5.886\ \mathrm N$, exactly balancing its downward <force> from <gravity>.
\Image[/buoyancy-floating-and-sinking.png]
{title=Floating, neutral buoyancy and sinking for three equal-size cubes}
{description=All three <cubes> have side $0.10\ \mathrm m$ in <water> with stipulated <mass density> $1000\ \mathrm{kg\,m^{-3}}$ and <gravitational acceleration> $9.81\ \mathrm{m\,s^{-2}}$. Left: the $600\ \mathrm{kg\,m^{-3}}$ body floats with $0.060\ \mathrm m$ immersed and equal upward and downward <forces> of $5.886\ \mathrm N$. Middle: a fully immersed $1000\ \mathrm{kg\,m^{-3}}$ body has <neutral buoyancy>, with both <forces> equal to $9.81\ \mathrm N$. Right: a fully immersed $1200\ \mathrm{kg\,m^{-3}}$ body has downward <force> from <gravity> $11.772\ \mathrm N$ against upward <buoyancy> $9.81\ \mathrm N$, so release from rest gives an initial net downward <force> $1.962\ \mathrm N$.}
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