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Why does pressure not depend on the shape of the container?

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  1. Oct 7, 2015 #1
    Hey there. I'm trying having a bit of trouble wrapping my head around this.

    1. The problem statement, all variables and given/known data
    That pressure is equal at all the same height no matter the shape of the container.

    2. Relevant equations
    [itex]P = \frac{F}{A}[/itex]
    [itex]P = ρgh[/itex]

    3. The attempt at a solution

    I'm trying to understand this from my physics class, but I can't seem to get a hold on it. Why does pressure in water only depend on the height? I can understand that pressure at a depth is caused by the weight of the water on top of it, as shown by [itex]P = \frac{F}{A}[/itex]

    Now, I know that the following is also true. [itex]P = ρgh[/itex] Now, this equation shows directly that it is related to height, and it can also be rearranged to [itex]P = \frac{F}{A}[/itex]. However, this is where to start to run into issues.

    Let's say we have a cylindrical container of fluid, like so.

    OG3v99D.jpg

    This cylinder, filled with fluid, will have a pressure on the bottom. Now, let's assume we have a conular container, with the same height, and equal base area, like so.

    S6zXEWI.png

    From my knowledge of [itex]P = ρgh[/itex], I know that the pressures are equal at the base. However, I can't begin to understand why. There is a lower volume of water in the cone, and therefore a lower mass. This means that there's a lower force pressing down from the water than in the cylinder. Because of this, shouldn't there be two different gauge pressures? What's the reasoning for there being equal gauge pressures? I can't begin to figure it out for the life of me.

    Thanks so much.
     
  2. jcsd
  3. Oct 7, 2015 #2

    andrewkirk

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    Consider a column of water whose base is near the outside of the cone. It is short, so the base only has a small weight of water above it. Now think about the top cc of that column. On one side of that cc the container wall is pushing on it. What is pushing on the other side, and how much water is above that?

    You're right though, it is unintuitive. I use siphoning regularly to put grey water from my washing machine onto my garden, and every time I do it I can't quite believe it is happening.
     
  4. Oct 7, 2015 #3
    I'm sorry, I don't quite understand what you mean. What's a cc?
     
  5. Oct 7, 2015 #4

    SteamKing

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    cc = cubic centimeter
     
  6. Oct 7, 2015 #5
    Ollen,

    The pressure would be the same even if you had a narrow cylinder like a capillary tube. All the equation is saying is the pressure at a certain dept ONLY depends on the depth of the water at a depth your interested in. And also, the pressure is the same in all directions at a certain depth. Does that help?
     
  7. Oct 7, 2015 #6
    Why? I can't figure that out.

    Now knowing what a cc is, pushing on the other side, would it be the other walls?
     
  8. Oct 7, 2015 #7

    andrewkirk

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    No. Remember that we are talking about a thin column of water whose base is close to the wall. So where that column hits the wall, the side of the column opposite to the side that is against the wall cannot be against another wall. If its not pushing against (and being pushed by) a wall, then it's pushing against .... what?
     
  9. Oct 7, 2015 #8
    Is it the atmosphere?
     
  10. Oct 7, 2015 #9

    andrewkirk

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    No. That would mean that the column of water nearer the wall is taller than the column of water next to it but further from the wall, which would mean the surface of the water is not horizontal. Draw a picture, showing the column, hitting the sloping container wall at the top, then draw a column next to it, which will hit the container wall higher up.
     
  11. Oct 7, 2015 #10
    Got it. That makes more sense.

    Thanks!
     
  12. Oct 7, 2015 #11
    In your truncated cone example, fluid pressure acts perpendicular to the slanted wall, and the slanted wall pushes back on the fluid. Since this pressure force is perpendicular to the slanted wall, there is a component of this force in the downward direction. So the slanted wall is exerting a downward force on the fluid. This turns out to exactly compensate for the missing weight.

    Chet
     
    Last edited: Oct 7, 2015
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