From 71b9e1aa3050dd492761973ad4be73c6d65e7eb1 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 10 Apr 2021 19:58:09 +0200
Subject: Expand knowledge base

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+---
+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.
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