Hydrostatic Pressure - Theoretical Scenario

AI Thread Summary
The discussion centers on understanding hydrostatic pressure and the conditions under which a container of water will collapse onto a piston. It is clarified that the pressure exerted by the water and the walls of the container is not necessarily equal, especially when the system is not in equilibrium. The potential energy of the system plays a crucial role in determining whether the container will collapse; if the water's potential energy increases with a small movement, the system will not spontaneously change. Additionally, in scenarios with a massless container, the water's pressure can lead to unexpected movements, as the container may not exert downward pressure on the water. Ultimately, the dynamics of pressure and potential energy dictate whether the container will hold itself up or collapse onto the piston.
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I have come to understand the basics of Pascal's Principle/Hydrostatic pressure, but there is one scenario I am looking for help with.

If I have a piston at the bottom of a chamber of water, and I want to push the piston up to lift the water, I understand that I need to exert a greater PSI on the Piston than the PSI the water is exerting on the surface of the Piston.

However, let's say that (as indicated by the attached picture) I have a Piston that is attached to the ground, and the container of water is "free standing"... meaning the container of water can collapse onto the Piston (thereby displacing the water upwards).

How do I know if the container would indeed fall on the Piston and thereby displace the water?

Aren't the walls (and thereby the Piston) exerting the same PSI on the water as the water is exerting on the walls (and Piston)? If so, who wins?

Intuitively, I want to say that no matter what... in any scenario such as I pictured... the container of water will fall on top of the Piston every time... but my intuition has proved wrong many times before.

Is it really possible that the PSI of the water at the bottom of the container could hold the container up and prevent it from collapsing onto the Piston?

Are there scenarios where the container of water will hold itself up? Or will it collapse every time?

Thanks for any help. This is one where I don't really know how to even begin doing the calculation.
 

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Here is a response I received to the same question/example on another forum. It helps a little, but am still hoping someone can be more detailed.

"Are there scenarios where the container of water will hold itself up? Or will it collapse every time?"

Yes, there are. Let's take the limit where there is no overlap around the piston, so all the water is above the piston. Also imagine the container to be massless. If the container were to lower itself...*none* of the water would go down with it. In fact some water would be raised up into the thin neck. Since overall the water would be higher, this position has more potential energy. It won't move into that position spontaneously.

In general you can examine what the potential energy of the system before and after a small movement to see if it would be spontaneous.

In your image, if the area of the water that is "around" or "outside" the piston is larger than the area in the neck, it will collapse.

In cases where you're working with small quantities of water, the mass of glass, metal, or even plastic container will tend to overwhelm the mass of the water, and will overcome the water pressure. For a massless container, it may well move in ways that intuition does not expect.

"Aren't the walls (and thereby the Piston) exerting the same PSI on the water as the water is exerting on the walls (and Piston)? If so, who wins?"

No, the walls and the water may not be pushing the same. They are *if* it is in equilibrium. Otherwise there are net forces and the container will be accelerated. For instance, if there is no overlap, the a massless container will not be pushing down on the water, but the water will be pushing up (if there is water in the neck), and the container will accelerate upward.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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