= Archimedes' principle
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{title2=$F_b=\rho_fgV_{\mathrm{displaced}}$}
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<Archimedes' principle> states that <buoyancy> equals the downward <gravitational weight>, or weight, of the displaced <fluid>. In a resting <fluid> of constant <mass density> $\rho_f$ under uniform <gravitational acceleration> $g$, its magnitude is
$$
F_b=\rho_fgV_{\mathrm{displaced}}.
$$
Here $V_{\mathrm{displaced}}$ is the <volume> from which the body excludes the <fluid>: the whole external <volume> for a fully immersed solid body, and just the immersed <volume> for a partly immersed one. The body's own <mass> does not enter this <buoyancy> formula. https://openstax.org/books/university-physics-volume-1/pages/14-4-archimedes-principle-and-buoyancy[OpenStax, University Physics, §14.4: Archimedes' principle]{ref}
A short proof outline for a fully immersed body is to imagine replacing it by exactly the <fluid> it displaces. The surrounding <hydrostatic pressure> is unchanged. That replacement <fluid> is at rest, so its upward <pressure> <force> must balance its downward <force> from <gravity>. The original body receives the same inward <pressure> <forces> on the same closed surface, hence the same upward <buoyancy>.
Equivalently, let $\Omega$ be the displaced region, $\partial\Omega$ its closed surface and $\mathbf e_z$ the upward unit direction. The <divergence theorem>, applied component by component to the <surface integral>, and <hydrostatic pressure> give
$$
\mathbf F_b=-\int_{\partial\Omega}p\mathbf n\,dA
=-\int_\Omega\nabla p\,dV
=\rho_fgV_{\mathrm{displaced}}\mathbf e_z.
$$
This also explains why horizontal <forces> cancel even for an irregular shape. A spatially constant surface <pressure> contributes zero to the closed-surface <integral>. If the <mass density> varies with depth, replace $\rho_fV_{\mathrm{displaced}}$ by the <mass> $\int_\Omega\rho_f\,dV$ of the displaced <fluid>.
The closed-surface argument assumes that the <fluid> applies <pressure> over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no <fluid> beneath it; then the lower-face <pressure> <force> used in the proof is absent, and a supporting contact <force> must be treated separately. For partial immersion, the usual <buoyancy> formula assumes the exposed surface receives the common atmospheric <pressure>, neglects the much smaller <buoyancy> from the air, and neglects <surface tension>.
For the $0.10\ \mathrm m$ <cube> in the stipulated <water>, full immersion displaces $0.001\ \mathrm{m^3}$, so $F_b=1000\times9.81\times0.001=9.81\ \mathrm N$. If its uniform <mass density> is $600\ \mathrm{kg\,m^{-3}}$, its <mass> is $m=600\times0.001=0.600\ \mathrm{kg}$ and its downward <force> from <gravity> is $mg=5.886\ \mathrm N$. To hold it fully immersed at rest requires an additional downward <force>
$$
F_{\mathrm{hold}}=F_b-mg=9.81-5.886=3.924\ \mathrm N.
$$
On release from rest it initially moves upward; its eventual <floating equilibrium> has a smaller displaced <volume>. A same-size <cube> of uniform <mass density> $1200\ \mathrm{kg\,m^{-3}}$ has <mass> $1.200\ \mathrm{kg}$ and downward <force> from <gravity> $11.772\ \mathrm N$. Full immersion still gives $9.81\ \mathrm N$ of <buoyancy>, leaving an initial net downward <force> of $11.772-9.81=1.962\ \mathrm N$. Subsequent motion also involves <drag>.
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