PlanetPhysics/Archimedes Principle

Archimedes' Principle states that

When a floating body of \htmladdnormallink{mass {http://planetphysics.us/encyclopedia/CosmologicalConstant.html} $M$ 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 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 </math>p(z)\Delta A\mathbf{e}_z$onitsbottomsurfaceandadownwardforceof$-p(z + \Delta z)\Delta A\mathbf{e}_z + \Delta m\mathbf{g}$.Theseforcesmustbalancesothatwehave<math>p(z)\Delta A=p(z+\Delta z)\Delta A-\Delta m|\mathbf {g} |$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 $\mathbf {F} _{B}=-\int _{\Omega }{\rho _{f}\mathbf {g} \,dV}=-\rho _{f}\mathbf {g} \int _{\Omega }{dV}=-\rho _{f}\mathbf {g} =Vol=(\Omega )$where we can pull $\rho _{f}$ and $\mathbf {g}$ outside of the integral since they are assumed to be constant. But notice that $\rho _{f}=Vol=(\Omega )$ 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.