Short Derivation of Archie's Law

 "captain's log: supplemental"

It's been quite a while since I wrote the previous post about Archie, his wife, a crown etc.
While I was writing that, I did not want to derive Archimedes' law on my own, so I looked around on the net, to maybe find someone who already posted that, sadly, I couldn't.

Also, this morning my wife and I had an argument about this law, and guess what, she was right, I was wrong... Isn't it weird these are the arguments my wife and I have?!? I mean, seriously, how nerdy can you get? 

So for the interest of completeness (which is actually a mathematical axiom, but let's leave that be for now), I give you a short derivation of Archie's law:

Deriving the law (I AM THE LAW)

Let us consider an infinitesimal volume element of a material with density \(\rho\) , and incident area and height \(A,h\) respectively.
So now let's suppose the material is lighter than the medium surrounding it we get a force equation that looks like this:
\[\Sigma F=A\cdot P_{up} -A\cdot P_{down}-Mg\]
Where the force equivalent (sum of all forces) is positive.

Now, let's wrap the forces that are the result of pressure and name it the Buoyancy force thus:
\[F_{Buoyancy}=A\cdot\left(P_{up}-P_{down}\right)\]
and with a little massage we get:
\[A\cdot\left(P_{up}-P_{down}\right)=A\cdot h \left(\frac{P_{up}-P_{down}}{h}\right)=-V\nabla P\]

So, now, let's take a small detour to understand who \(\nabla P\) is shall we?

suppose we are dealing with an element that is filled with the same material as the medium it's in, in that case we get a mechanical equilibrium and \(\Sigma F\) is simply zero. and so we get:
\[Mg=F_{Buoyancy}=-V\nabla P\]
and breaking up the \(Mg\) element we get:
\[V\rho_{medium}\cdot g=-V\nabla P\Rightarrow \nabla P = -\rho_{medium}\cdot g\]
and so we get quite simply:
\[F_{Buoyancy}=V\cdot \rho_{medium}\cdot g\]

Proper usage

What's useful with this representation, is that we get the force outright, thus we can make force calculations directly.

An important thing to understand is that the buoyancy force is directed not UPWARDS (which is a common misconception) but against the direction of the pressure gradient . which is quite different. that means for example that in a spinning tube of air, the buoyancy force will be mostly inwards as the pressure gradient is directed outwards (in cylindrical coordinates).

That's it.

this time it was really short.