= Hydrostatic pressure
{title2=$dp/dz=-\rho_fg$}
{wiki}
<Hydrostatic pressure> is the <pressure> distribution in a resting <fluid>, balancing its <body force> from <gravity>. Take $z$ upward and a uniform downward <gravitational field> $\mathbf g=-g\mathbf e_z$, where $g>0$. The vertical <pressure gradient> satisfies
$$
\frac{dp}{dz}=-\rho_f g.
$$
A short derivation balances the vertical <forces> on a thin horizontal layer of <area> $A$ and thickness $\Delta z$: $p(z)A-p(z+\Delta z)A-\rho_fgA\Delta z=0$. Dividing by $A\Delta z$ and taking the <derivative> gives the equation above. For constant <mass density> $\rho_f$, <integration> from the surface at $z_s$ gives
$$
p(z)=p_{\mathrm{surface}}+\rho_fg(z_s-z)
=p_{\mathrm{surface}}+\rho_fgh,\qquad h=z_s-z.
$$
Thus <pressure> increases with depth $h$. For the stipulated <water> with <mass density> $1000\ \mathrm{kg\,m^{-3}}$ and <gravitational acceleration> $9.81\ \mathrm{m\,s^{-2}}$, depths $0.20\ \mathrm m$ and $0.30\ \mathrm m$ give <gauge pressures> $1962\ \mathrm{Pa}$ and $2943\ \mathrm{Pa}$. Their difference produces the <buoyancy> of the example <cube>. https://www.feynmanlectures.caltech.edu/II_40.html#Ch40-S1[Feynman Lectures, II §40-1: hydrostatics]{ref}
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