ArchimedesPrinciple 2008-04-27 21:15:11 2008-04-27 22:56:27 Law % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Archimedes' Principle states that \emph{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 \[ \mathbf{F}_B = \int_{\partial\Omega}{-p\mathbf{n}\,dS} \] 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 \emph{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 \[ \int_{\partial\Omega}{f\mathbf{n}\,dS} = \int_\Omega{\nabla f\,dV} \] therefore we can write \[ \mathbf{F}_B = -\int_\Omega{\nabla p\, dV} \] 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 \[ p(z)\Delta A = p(z + \Delta z)\Delta A - \Delta m|\mathbf{g}| \] a simple manipulation of this equation along with dividing by $\Delta z$ gives \[ \frac{p(z + \Delta z) - p(z)}{\Delta z} = \frac{\Delta m}{\Delta A\Delta z}|\mathbf{g}| = \frac{\rho_f \Delta A\Delta z}{\Delta A\Delta z}|\mathbf{g}| = \rho_f |\mathbf{g}| \] taking the limit $\Delta z \to 0$ gives \[ \frac{\partial p}{\partial z} = \rho_f|\mathbf{g}| \] Similar arguments for the $x$ and $y$ directions yield \[ \frac{\partial p}{\partial x} = \frac{\partial p}{\partial y} = 0 \] 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 \[ \mathbf{F}_B = -\int_\Omega{\rho_f\mathbf{g}\, dV} = -\rho_f\mathbf{g}\int_\Omega{dV}- \rho_f\mathbf{g}\text{Vol}(\Omega) \] where we can pull $\rho_f$ and $\mathbf{g}$ outside of the integral since they are assumed to be constant. But notice that $\rho_f\text{Vol}(\Omega)$ is equal to $M_d$, the mass of the displaced fluid so that \[ \mathbf{F}_B = -M_d\mathbf{g} \] 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 \[ M_d = M \] which is precisely the statement of Archimedes' Principle.