Solving Fluid Static Problem with Parallelepipeds

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SUMMARY

The discussion focuses on solving a fluid static problem involving two parallelepipeds combined to form an "L" shaped solid filled with water. The key equation referenced is Pascal's principle, which states that pressure is uniform throughout a fluid. The user initially calculated the pressure against the wall as \( P = \rho g h \) and the force as \( |F| = \rho g h \times S \), where \( S \) is the wall's cross-sectional area. However, this approach was incorrect, indicating a misunderstanding of the equilibrium conditions in the isolated system.

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Homework Statement



There are 2 parallelepiped with dimension x,l,h those were united to create a bigger solid with an "L" form ( see the picture ). The solid is isolated and full of water.
What's the force that must be done against the top wall, to make the system in equilibrium?

Here's the image:
l1t2ad.jpg


Homework Equations



Pascal's principle: everywhere the pressure's the same
Pressure at the bottom: $P=P_ext+\rho gh$

The Attempt at a Solution



The external pressure is 0 because the system is isolated; so I thought that the pressure against the wall may be $P=\rho g h$ and the Force would have been ( modular ) $|F|= \rho g h \times S $ where S is the section of the wall.. Unlucky this is not the right answer..

May someone help me?
 
Last edited:
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I think I know this stuff and would try to help but the problem is not clear to me as stated. If you can make an attempt at being more clear I will give it another shot.
 

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