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## Main Question or Discussion Point

So, I know that according to everything online water pressure is independent from everything except gravity, height, and density of fluid. The situation I'm talking about is slight different.

Imagine a standard water bottle that angles inwards near the top. Now imagine a point on the

I mean, think about the way the equation cancels out.

P = F/A

P = MG/A

P = Density x Volume x Gravity / Area

In the column-of-water scenario, the volume of the column (m^3) cancels out with the area (m^2), leaving behind just M, aka height. But in my scenario, the volume and the area wouldn't cancel out to produce height because the volume isn't just a function of length, width, and height: it's irregular.

I can add some diagrams later if this is a confusing scenario to envision.

All I really need to know is, am I correct in assuming that here, the shape of the bottle is not irrelevant?

Imagine a standard water bottle that angles inwards near the top. Now imagine a point on the

*side*of the water bottle, near the on the slanted part. I know that what matters is the height of a "column" of water above your point, but you can't make a column of water here -- once the bottle starts slanting inwards, going straight up would mean you hit the slanted side of the bottle.I mean, think about the way the equation cancels out.

P = F/A

P = MG/A

P = Density x Volume x Gravity / Area

In the column-of-water scenario, the volume of the column (m^3) cancels out with the area (m^2), leaving behind just M, aka height. But in my scenario, the volume and the area wouldn't cancel out to produce height because the volume isn't just a function of length, width, and height: it's irregular.

I can add some diagrams later if this is a confusing scenario to envision.

All I really need to know is, am I correct in assuming that here, the shape of the bottle is not irrelevant?