So far we have looked at the pressure and gravitational forces on a fluid element. What happens when we submerge a solid in the fluid?

Here we have the same pressure forces on top and bottom of a control volume of solid, as well as the gravitational force pulling downward on the solid; however, what is different is that we "subtracted" some fluid in order to put the solid in there.

**Archimedes' Principle** states that the buoyant force, the
force resulting from the "subtracted" fluid (whose volume is
$V_f$), is equal to the weight of the displaced fluid *in the
direction opposite to gravity*:

$\vec{F_B} = -\rho_f V_f \vec{g}$

This force acts as a body force, going through the center of
mass of the body, and therefore can be thought of as a *density
correction* for the gravitational force (if the solid is
completely submersed so that $V_f = V_s$):

$F_g = (\rho_s - \rho_f) g V_s$

Derive the pressure field equation

Use Archimedes' principle to calculate buoyant forces on (partially) immersed objects

There are two forces acting on the block: the weight of the solid itself acting downward, the buoyant force (given as the weight of the displaced fluid). These forces must be equal if the block is not moving:

$\displaystyle{W = \rho_s g V_s = F_B = \rho_f g V_f}$

If we define $f$ as the fraction of the block that is under fluid, then $V_f = fV_s$ so:

$\displaystyle{\rho_s g V_s = \rho_f g (fV_s)}$

which can be rearranged to give:

$\displaystyle{\rho_s = f\rho_f}$