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Pascal's principle states that pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid.

How is the pressure transmitted in the case where the fluid is not enclosed?

For instance, pushing a block of wood to a beaker with water such that the area of the bottom of the block of wood is does not enclose the mouth of the beaker.

How do we solve for the force on the walls of the beaker?

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I'm not an expert in fluid physics, but after a bit of thought, I believe the following can give clues to your answer.

If the volume of fluid is not enclosed, it means any pressure applied to it that exceeds atmospheric pressure (as in, any amount at all by the block) will just make it flow to the parts of the beaker where the fluid is free to move.

The force applied to walls of a beaker will be a function of the amount of fluid that is over that particular point on the wall - hence why dams are constructed thicker at their bottoms than at their tops. The water at the top of the volume will be at zero pressure relative to the atmosphere (at depth almost 0 the wall can have thickness almost zero and still hold the water). The water at the bottom of the beaker will have a column of water over it - the size of which will be affected by how much volume the block is currently occupying. I believe every 10 or so meters of water equals 1 atmosphere or about 100Kpa.

I'd need to review my textbooks to be able to do real calculations - and currently don't have time for it, but hopefully these are some good starting points for your investigation.

Good luck!

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