PlanetPhysics/Archimedes Principle

Archimedes' Principle states that

When a floating body of \htmladdnormallink{mass {http://planetphysics.us/encyclopedia/CosmologicalConstant.html} $M$ is in [[../InertialSystemOfCoordinates/|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 [[../NewtonianMechanics/|classical mechanics]]. We use a Cartesian coordinate [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]] oriented such that the $z$-axis is normal to the surface of the fluid.

Let $𝐠$ be [[../GravitationalField/|The Gravitational Field]] (taken to be a constant) and let $\mathrm{\Omega }$ denote the submerged region of the body. To obtain the net [[../Thrust/|force]] of buoyancy ${𝐅}_B$ acting on the [[../TrivialGroupoid/|object]], we integrate the pressure $p$ over the [[../GenericityInOpenSystems/|boundary]] of this region ${𝐅}_B=\underset{\partial \mathrm{\Omega }}{\int }-p𝐧dS$Where $𝐧$ is the outward pointing normal to the boundary of $\mathrm{\Omega }$. The negative sign is there because pressure points in the direction of the inward normal. It is a consequence of [[../StokesTheorem/|Stokes' theorem]] that for a differentiable [[../Vectors/|scalar]] [[../CosmologicalConstant2/|field]] $f$ and for any $\mathrm{\Omega }\subset {ℝ}^3$ a compact three-manifold with boundary, we have $\underset{\partial \mathrm{\Omega }}{\int }f𝐧dS=\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }fdV$therefore we can write ${𝐅}_B=-\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }pdV$Now, it turns out that $\mathrm{\nabla }p={\rho }_f𝐠$ where ${\rho }_f$ is the [[../Volume/|volume]] density of the fluid. Here is why. Imagine a cubical element of fluid whose height is $\mathrm{\Delta }z$, whose top and bottom surface area is $\mathrm{\Delta }A$ (in the $x-y$ plane), and whose mass is $\mathrm{\Delta }m$. 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}$.Theseforcesmustbalancesothatwehave<math>p(z)\mathrm{\Delta }A=p(z+\mathrm{\Delta }z)\mathrm{\Delta }A-\mathrm{\Delta }m|𝐠|$a simple manipulation of this equation along with dividing by $\mathrm{\Delta }z$ gives $\frac{p(z+\mathrm{\Delta }z)-p(z)}{\mathrm{\Delta }z}=\frac{\mathrm{\Delta }m}{\mathrm{\Delta }A\mathrm{\Delta }z}|𝐠|=\frac{{\rho }_f\mathrm{\Delta }A\mathrm{\Delta }z}{\mathrm{\Delta }A\mathrm{\Delta }z}|𝐠|={\rho }_f|𝐠|$taking the limit $\mathrm{\Delta }z\to 0$ gives $\frac{\partial p}{\partial z}={\rho }_f|𝐠|$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 $\mathrm{\nabla }p={\rho }_f𝐠$ as desired. Substituting this into the integral expression for the buoyant force obtained above using Stokes' theorem, we have ${𝐅}_B=-\underset{\mathrm{\Omega }}{\int }{\rho }_f𝐠dV=-{\rho }_f𝐠\underset{\mathrm{\Omega }}{\int }dV=-{\rho }_f𝐠=Vol=(\mathrm{\Omega })$where we can pull ${\rho }_f$ and $𝐠$ outside of the integral since they are assumed to be constant. But notice that ${\rho }_f=Vol=(\mathrm{\Omega })$ is equal to $M_d$, the mass of the displaced fluid so that ${𝐅}_B=-M_d𝐠$But by Newton's second law, the buoyant force must balance the weight of the object which is given by $M𝐠$. It follows from the above expression for the buoyant force that $M_d=M$which is precisely the statement of Archimedes' Principle.