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---
title: "Archimedes' principle"
sort_title: "Archimedes' principle"
date: 2021-04-10
categories:
- Fluid statics
- Fluid mechanics
- Physics
layout: "concept"
---

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 entire surface $$S$$ of $$V$$:

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

Where we have used the divergence theorem.
Assuming [hydrostatic equilibrium](/know/concept/hydrostatic-pressure/),
we replace $$\nabla p$$,
leading to the definition of the **buoyant force**:

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

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 $$\va{g}$$ is constant everywhere, with magnitude $$\mathrm{g}$$.
If we also assume that $$\rho_\mathrm{f}$$ is constant on the "submerged" side,
and zero on the "non-submerged" side, we find:

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

In other words, the mass $$m_\mathrm{b}$$ of the entire 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 the displaced mass $$m_\mathrm{f} < m_\mathrm{b}$$
even if the entire body is submerged,
and the object will therefore continue to sink.



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