Archimedes' Principle

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.



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As of this snapshot date, this entry was owned by joshsamani.