Physics exercise: Pressure in this Fresh Water Tank

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The discussion centers on a physics exercise involving a sealed fresh water tank filled with air and water, creating a trapped air pocket. Participants seek clarification on applying momentum and understanding fluid statics, particularly regarding the force at the center of pressure on the tank's plate. There is a need for clearer diagrams and a full problem description, including the coordinate system used. The air compression due to the water being pumped in is highlighted as a key factor in determining pressure changes. Overall, the focus is on accurately visualizing and solving the fluid dynamics scenario presented.
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Homework Statement
The manhole cover of the fresh water tank is screwed on and the tank is filled to the indicated level. Determine the pressure at the bolts.
Relevant Equations
dF=P*dA ; H=550mm ; Varignon's Theorem
I did some of the exercise, but I don't know how to apply the momentum :'(

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Can you explain this some more. The lid is put on the tank (making a seal). The tank is initially filled with air, then water is pumped in, trapping a pocket of air between the water surface and the lid. Is that the scenario?

Or looking at your solution attempt a bit more, it looks like a fluid statics problem where you find the force acting at the center of pressure on the plate. Either way, I think you need clearer diagramming of where/how this plate is oriented on the tank.
 
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Not sure I understand the diagrams. Pls supply the full description of the problem and state your coordinate system.
 
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erobz said:
Can you explain this some more. The lid is put on the tank (making a seal). The tank is initially filled with air, then water is pumped in, trapping a pocket of air between the water surface and the lid. Is that the scenario?
I think that's what's happening. The air is compressed, so the solution is to find the change in pressure given the corresponding change in volume.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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