---
title: "Archimedes' principle"
firstLetter: "A"
publishDate: 2021-04-10
categories:
- Fluid statics
- Fluid mechanics
- Physics

date: 2021-04-10T15:43:45+02:00
draft: false
markup: pandoc
---

# Archimedes' principle

Many objects float when placed on a liquid,
but some float higher than others,
and some do not float at all, sinking instead.
**Archimedes' principle** balances the forces,
and predicts how much of a body is submerged,
and how much is non-submerged.

In truth, there is no real distinction between
the submerged and non-submerged parts,
since the latter is surrounded by another fluid (air),
which has a pressure and thus affects it.
The right thing to do is treat the entire body as being
submerged in a fluid with varying properties.

Let us consider a volume $V$ completely submerged in such a fluid.
This volume will experience a downward force due to gravity, given by:

$$\begin{aligned}
    \va{F}_g
    = \int_V \va{g} \rho_\mathrm{b} \dd{V}
\end{aligned}$$

Where $\va{g}$ is the gravitational field,
and $\rho_\mathrm{b}$ is the density of the body.
Meanwhile, the pressure $p$ of the surrounding fluid exerts a force
on the surface $S$ of $V$:

$$\begin{aligned}
    \va{F}_p
    = - \oint_S p \dd{\va{S}}
\end{aligned}$$

We rewrite this using Gauss' theorem,
and replace $\nabla p$ by demanding
[hydrostatic equilibrium](/know/concept/hydrostatic-pressure/):

$$\begin{aligned}
    \va{F}_p
    = - \int_V \nabla p \dd{V}
    = - \int_V \va{g} \rho_\mathrm{f} \dd{V}
\end{aligned}$$

For the body to be at rest, we require $\va{F}_g + \va{F}_p = 0$.
Concretely, the equilibrium condition is:

$$\begin{aligned}
    \boxed{
        \int_V \va{g} (\rho_\mathrm{b} - \rho_\mathrm{f}) \dd{V}
        = 0
    }
\end{aligned}$$

It is commonly assumed that $\va{g}$ has a constant direction
and magnitude $\mathrm{g}$ everywhere.
If we also assume that $\rho_\mathrm{b}$ and $\rho_\mathrm{f}$ are constant,
and only integrate over the "submerged" part, we find:

$$\begin{aligned}
    0
    = \mathrm{g} (\rho_\mathrm{b} - \rho_\mathrm{f}) V
    = \mathrm{g} (m_\mathrm{b} - m_\mathrm{f})
\end{aligned}$$

In other words, the mass $m_\mathrm{b}$ of the submerged portion $V$ of the body,
is equal to the mass $m_\mathrm{f}$ of the fluid it displaces.
This is the best-known version of Archimedes' principle.

Note that if $\rho_\mathrm{b} > \rho_\mathrm{f}$, then,
even if the entire body is submerged,
the displaced mass $m_\mathrm{f} < m_\mathrm{b}$,
and the object will continue to sink.



## References
1.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.