Visualizing Archimedes’ Principle

Question

Today we were shown the derivation of Archimedes’ Principle in this way. Consider a cube of fluid inside the container. On the upper side, it has a pressure $P_1$, and on the lower side, it has a pressure $P_1+\rho gh$ since it must be in equilibrium. Now, replace the cube of fluid with a cube of other material. It will experience the same net force, pgh, minus its weight, so it will float or not depending whether its density is higher or lower than the fluid’s.

Although the maths involved are very simple, I have trouble visualizing that this new cube will have the same pressures acting on it. We were told that these pressures don’t depend on the cube of fluid itself, but rather on its surroundings. It doesn’t look intuitive at all to me. Isn’t the pressure difference precisely derived from the fact that there is a block of mass under its weight that must be in equilibrium? Maybe it could be argued that, if you compare points at the same height at the container with the block and the one without it, they should be the same. Thus the pressure under the new block must be the same as the one acting on the block of fluid. But still, this doesn’t offer a meaningful physical interpretation.

Answer 1

I have trouble visualizing that this new cube will have the same pressures acting on it. We were told that these pressures don’t depend on the cube of fluid itself, but rather on its surroundings.

This is true. The pressure is generated by the surrounding fluid (present in a gravitational field) itself. Note that the equation you included $P=P_0+\rho gh$ is a simple statement of the fact that the pressure at point in the fluid depth $h$ is the pressure on the surface of the fluid (atmospheric pressure) added to the pressure caused by the layers of fluid above the point $h$. This is no way dependent on the object placed in the fluid. Even if you removed the object from the fluid, the pressure that was acting on the bottom of the object is the same as the pressure in the fluid at the same point where the bottom of the object was previously.

Isn’t the pressure difference precisely derived from the fact that there is a block of mass under its weight that must be in equilibrium?

It has nothing to do with the block. The pressures at the top and bottom of the object are there even if the block is removed. Gravitational force pulling the fluid down causes different pressures at different depths. Archimedes principle states that the buoyant force on an object is identical to the weight of the fluid displaced by the object. So $${\bf F_B-F_W=0}\therefore\rho V{\bf g}-{\bf F_W}=0$$ if the object is in equilibrium, and the quantities we are concerned with is the density $\rho$ of the fluid and the volume $V$ of displaced fluid (${\bf F_W}$ is the weight of the object).

And because $\rho=\frac{m_f}{V}$ then $\rho V{\bf g}=(m_f\ \bf g)$ is the weight of the displaced fluid.

Maybe it could be argued that, if you compare points at the same height at the container with the block and the one without it, they should be the same. Thus the pressure under the new block must be the same as the one acting on the block of fluid.

Exactly!

But still, this doesn’t offer a meaningful physical interpretation.

It does. It just takes some time to appreciate, and when you progress in your studies of fluid/Newtonian mechanics, you'll see that it is indeed physically meaningful.