LTR: Archimedes' Principle for Buoyancy Force

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:

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