The Pressure at the Bottom of a Container

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Homework Help Overview

The discussion revolves around the pressure at the bottom of fluid-filled containers, specifically comparing cone-shaped and cylindrical containers. The original poster attempts to understand why the pressure calculations yield different results despite having the same height and fluid density.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster presents a mathematical approach to calculate pressure using the volume of a cone and compares it to that of a cylinder, leading to a discrepancy in results. Some participants question the assumptions made regarding the forces acting on the container walls and the implications of fluid pressure direction.

Discussion Status

Participants are actively engaging with the original poster's reasoning, providing insights about the forces involved and questioning the assumptions made. There is no explicit consensus yet, but the discussion is exploring important aspects of fluid mechanics.

Contextual Notes

There is an indication that the original poster may not have fully considered the effects of container shape on pressure distribution and the role of container walls in fluid dynamics.

mlb2358
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Homework Statement


My physics textbook states that the pressure at some point in a fluid filled container a distance h below the fluid air interface is dependent only upon h, the density of the fluid, and the acceleration due to gravity. If this is the case, then the pressure at the bottom of a cone shaped container should be the same as a cylindrical container as long as the height is the same and they are filled with fluids of equivalent density. However, when I try to show this mathematically, it doesn't seem to work out.


Homework Equations


P = Po + ρgh

P = F/A

The Attempt at a Solution


The volume of a right cone is: V = ∏r2h/3. The force due to the liquid in a container of this shape is F = ρgV = ρg(∏r2h/3). The pressure is F/A, so P = ρg(∏r2h/3)/∏r2 = ρgh/3, however a similar analysis for a cylinder leads to the conclusion that P = ρgh. This indicates that the pressure at the bottom of the two containers is different, even though the height below the surface of the air fluid interface is the same. I am not sure what I am missing here.
 
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You are missing the fact that there is a force from the walls of the container (the liquid acts with a force on the wall, so apply Newton's third law).

Also, if your argument was correct and you turned the cylinder tip down, the pressure at the tip would be infinite.
 
Orodruin said:
You are missing the fact that there is a force from the walls of the container (the liquid acts with a force on the wall, so apply Newton's third law).

Specifically, the fluid pressure acts perpendicular to the surface of whatever container the fluid is in.

Also, if your argument was correct and you turned the cylinder tip down, the pressure at the tip would be infinite.

I think the cone has the tip, rather than the cylinder.
 
SteamKing said:
I think the cone has the tip, rather than the cylinder.

Of course, it seems I was not yet fully awake when I wrote that :-p
 

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