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Archimedes' Principle (Law)

Archimedes' Principle states that

When a floating body of mass $M$ is in equilibrium with a fluid of constant density, then it displaces a mass of fluid $M_d$ equal to its own mass; $M_d = M$.

Archimedes' principle can be justified via arguments using some elementary classical mechanics. We use a Cartesian coordinate system oriented such that the $z$-axis is normal to the surface of the fluid.

Let $\mathbf{g}$ be The Gravitational Field (taken to be a constant) and let $\Omega$ denote the submerged region of the body. To obtain the net force of buoyancy $\mathbf{F}_B$ acting on the object, we integrate the pressure $p$ over the boundary of this region

$\displaystyle \mathbf{F}_B = \int_{\partial\Omega}{-p\mathbf{n}\,dS} $
Where $\mathbf{n}$ is the outward pointing normal to the boundary of $\Omega$. The negative sign is there because pressure points in the direction of the inward normal. It is a consequence of Stokes' theorem that for a differentiable scalar field $f$ and for any $\Omega\subset\mathbb{R}^3$ a compact three-manifold with boundary, we have

$\displaystyle \int_{\partial\Omega}{f\mathbf{n}\,dS} = \int_\Omega{\nabla f\,dV} $
therefore we can write

$\displaystyle \mathbf{F}_B = -\int_\Omega{\nabla p\, dV} $
Now, it turns out that $\nabla p = \rho_f\mathbf{g}$ where $\rho_f$ is the volume density of the fluid. Here is why. Imagine a cubical element of fluid whose height is $\Delta z$, whose top and bottom surface area is $\Delta A$ (in the $x-y$ plane), and whose mass is $\Delta m$. Let us consider the forces acting on the bottom surface of this fluid element. Let the z-coordinate of its bottom surface be $z$. Then, there is an upward force equal to $p(z)\Delta A\mathbf{e}_z$ on its bottom surface and a downward force of $-p(z + \Delta z)\Delta A\mathbf{e}_z + \Delta m\mathbf{g}$. These forces must balance so that we have

$\displaystyle p(z)\Delta A = p(z + \Delta z)\Delta A - \Delta m\vert\mathbf{g}\vert $
a simple manipulation of this equation along with dividing by $\Delta z$ gives

$\displaystyle \frac{p(z + \Delta z) - p(z)}{\Delta z} = \frac{\Delta m}{\Delta ... ...A\Delta z}{\Delta A\Delta z}\vert\mathbf{g}\vert = \rho_f \vert\mathbf{g}\vert $
taking the limit $\Delta z \to 0$ gives

$\displaystyle \frac{\partial p}{\partial z} = \rho_f\vert\mathbf{g}\vert $
Similar arguments for the $x$ and $y$ directions yield

$\displaystyle \frac{\partial p}{\partial x} = \frac{\partial p}{\partial y} = 0 $
putting this all together we obtain $\nabla p = \rho_f\mathbf{g}$ as desired. Substituting this into the integral expression for the buoyant force obtained above using Stokes' theorem, we have

$\displaystyle \mathbf{F}_B = -\int_\Omega{\rho_f\mathbf{g}\, dV} = -\rho_f\mathbf{g}\int_\Omega{dV} = -\rho_f\mathbf{g}$Vol$\displaystyle (\Omega) $
where we can pull $\rho_f$ and $\mathbf{g}$ outside of the integral since they are assumed to be constant. But notice that $\rho_f$Vol$(\Omega)$ is equal to $M_d$, the mass of the displaced fluid so that

$\displaystyle \mathbf{F}_B = -M_d\mathbf{g} $
But by Newton's second law, the buoyant force must balance the weight of the object which is given by $M \mathbf{g}$. It follows from the above expression for the buoyant force that

$\displaystyle M_d = M $
which is precisely the statement of Archimedes' Principle.



"Archimedes' Principle" is owned by joshsamani.

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Cross-references: volume, field, scalar, Stokes theorem, boundary, object, The Gravitational Field, system, classical mechanics, equilibrium, mass

This is version 3 of Archimedes' Principle, born on 2008-04-27, modified 2008-04-28.
Object id is 279, canonical name is ArchimedesPrinciple.
Accessed 838 times total.

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Physics Classification: 47.85.Dh (Hydrodynamics, hydraulics, hydrostatics)
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