The Pressure at the Bottom of a Container

[mlb2358]

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 = P_o + 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.

 

[Orodruin]

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.

 

[SteamKing]

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.

 

[Orodruin]

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 :tongue: