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Water pressure under a cube

My question is: when a cube is in perfect contact with the base of a beaker containing fluid, and there is no fluid under the cube, will there be an upthrust on the cube?
If yes, then how do we derive out the upthrust expression?
If no, then wouldn't Archimedes principle be violated as he mentioned that so long as there is fluid displaced, there will be upthrust?


Answer:

After a great deal of thought I have the following comments.
(a) this is a theoretical question because no surface is completely flat. Even on a polished metal cube there will be imperfections and bumps and pits on the surface. This means that two surfaces are never in close contact over the whole of their touching surfaces
(b) there will always be some water molecules that can "seep" into the gaps between these bumps

However if we do assume that they can touch over their whole surface then we have some interesting results.

It certainly does appear that there will be no upthrust. If you draw a force diagram there will be no vertical force due to liquid below the cube since there is no liquid below the cube. Think about a situation where the cube is actually stuck to the base of the container. There is no upthrust in this case.

An extension of this is to think of a hollow cube. If we follow the above argument it mans that there is no upthrust on the hollow cube and therefore it would not float up to the surface in spite of it having an average density less than that of water.

 
 
 
© Keith Gibbs 2013