Pressure difference between the top and bottom surfaces of this cube

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The discussion focuses on understanding the pressure difference between the top and bottom surfaces of a cube submerged in a fluid. It clarifies that ρ represents density, not pressure, and that ρl refers to the density of the liquid filling the cube. The pressure within the liquid increases with depth, which is crucial for calculating the pressure difference. The formula discussed relates to the gravitational force acting on the liquid column above the cube. Overall, the pressure difference is influenced by the depth and density of the fluid surrounding the cube.
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Homework Statement
We consider the case where the wagon is at rest in profile view (below). Determine the pressure difference deltarho = rho under - delta on between the pressure under the inner surface of the cube and the pressure on the upper surface of the cube.
Relevant Equations
rho = F/A
Hi all,

I think I have to take the formula for pression. rho = F/A -> rho = ma/A but not sure, how to handle it here.

Image 23.06.23 at 17.09.jpeg


I thought, it should be rho under - rho above = pl*L - pc *a

Solution is: pl *g *a
 
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ρ stands for density, not pressure. ρl would be the density of the liquid that fills the wagon. (I presume that's what's going on here.)

How does pressure relate to the depth below the surface of a fluid?
 
It's the length from beginning of the cube till ground?
 
The pressure within the liquid depends on the depth below the surface. Pressure increases with depth.

Read this: Static Fluid Pressure
 
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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