Physically interpreting Archimedes Law I understand that Archimedes law is "the buoyant force is equal to the weight of the fluid displaced".
It is often derived from $F=\rho Vg$, where $\rho = m/V$ and the formula is rearranged. From this derivation, I can clearly see how mathematically Archimedes' law works - but from this I'm left wondering: is Archimedes' law just a mathematical coincidence?
Or is there a physical way to interpret why the weight of the fluid displaced would equal to the force that an object is pushed up on? I would prefer a more physical way to interpret it if possible. As an example, Lorentz came up with the equations Einstein ended up using for Special Relativity, but Lorentz could not derive the physical meaning behind his transformations (I recall).
I feel like there is a physical meaning to Archimedes' law beyond just some mathematical manipulation but I don't know what. Is there one?
 A: Imagine we take the submerged (or floating) object away and fill the volume $V$ that it displaced with liquid instead.
If the buoyancy force were greater than the weight of the volume $V$ of liquid then that volume of liquid would rise. If the buoyancy force were less than the weight of the volume $V$ of liquid then that volume of liquid would sink. In order to have hydrostatic equilibrium, the buoyancy force on the volume $V$ must be exactly equal to the weight of an equivalent volume of liquid.
A: Archimedes' law is a law of physics, so of course "there is a physical meaning" - the meaning is that bodies in liquid experience buoyant force and the force is of same magnitude but opposite direction than gravity force on liquid that would be there if the body was not.
What you are after seems to be an explanation or derivation of Archimedes' law from something more familiar. This can be done without using much mathematics: Archimedes buoyant force is obviously sum of all pressure forces of water elements acting on the surface of the body touching the liquid. This sum is determined by shape and position of the body in the liquid and has to be the same as if the body below the liquid level was made of the same liquid. Since such a liquid body would be at rest, the buoyant force magnitude has to be the same as weight of the liquid body (but direction is opposite so in case of liquid body these two cancel each other).
A: It can be derived from the hydrostatic pressure of the fluids. Althoug I'm not aware of the derivation for general shaped bodies, one can easily calculate the force for a box. (Assuming the box is completly in the fluid.)
$$F_\textrm{upward}=A(p_\textrm{lower}-p_\textrm{upper})$$
where $F_\textrm{upward}$ is the upward force, $A$ is the area of the horizontal surfaces of the box, $p_\textrm{lower}$ is the pressure at the lower side of the box (deeper in the fluid) and $p_\textrm{upper}$is the pressure at the upper side of the box.
