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Pressure at surface and bottom of pool

  1. Aug 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Two pools A and B have exactly the same depth, but A has 10 times the surface area, both at the top and at the bottom. Find the ratio of the total pressure (a) at the top surface of A to that at the top surface of B
    (b)at the bottom of A to that at the bottom of B.

    2. Relevant equations
    P=F/a or P=pgh

    3. The attempt at a solution
    For a, since they're at the surface they both have a pressure = atmospheric pressure so the ratio is 1?
    For b, since they're at the same depth, they have the same pressure because even if the area of A is 10 times the area of B, that means there's a force of about 10 times as big acting on the area as well so the pressure is the same. Is my reasoning correct? I don't see how the area affects part a? I mean the force and area at the surface for both will change in a way that their pressure will equal the atmospheric pressure. Do I need to consider atmospheric pressure in b as well? Even then, the ratio would be 1?
     
  2. jcsd
  3. Aug 3, 2016 #2

    Doc Al

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    You are correct. Surface area has nothing to do with pressure. (If they asked for total force, not pressure, then area would count for something.)
     
  4. Aug 3, 2016 #3
    Thank you!

    Just one more thing. We have a cube immersed fully in a liquid where the density of the liquid is p1 and the density of the cube is po. The cube is linked to the bottom of the container of the liquid by a rod. If p1 > po then the body would want to float so the net force would be upwards hence the rod would be under tension. If p1 < po then the body would want to sink so the net force would be downwards hence the rod will be under compression. If p1=po then it's neutrally stable, so it will move wherever, hence the rod will be under compression or will it be under nothing? Am I right with these 3?
     
  5. Aug 3, 2016 #4

    Doc Al

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    Assuming the rod is thin enough, then you are correct. When p1 = po, the buoyant force cancels the object's weight, so it doesn't add any tension to the rod.
     
  6. Aug 4, 2016 #5
    Yes that makes sense, so it won't be under any compression or tension? Thank you!

    I was just worried about this one:

    If p1 < po then the body would want to sink so the net force would be downwards hence the rod will be under compression.
     
  7. Aug 4, 2016 #6

    haruspex

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    That is correct too.
     
  8. Aug 4, 2016 #7
    Thank you Doc Al and Haruspex for your help :)
     
  9. Feb 11, 2017 #8
    Hi again!

    This particular question that I asked a few months ago (in reply number 3):

    "We have a cube immersed fully in a liquid where the density of the liquid is p1 and the density of the cube is po. The cube is linked to the bottom of the container of the liquid by a rod. If p1 > po then the body would want to float so the net force would be upwards hence the rod would be under tension. If p1 < po then the body would want to sink so the net force would be downwards hence the rod will be under compression."

    Wouldn't the rod be under compression if the body wanted to float since the rod is trying to keep the body down?
    And wouldn't it be under tension if the body wanted to sink?
     
  10. Feb 11, 2017 #9

    Doc Al

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    Just the opposite. If the body wanted to float it would be stretching the rod; if it wanted to sink, it would be compressing the rod.
     
  11. Feb 11, 2017 #10
    How will it be stretching the rod?
     
  12. Feb 11, 2017 #11

    Doc Al

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    When the cube pulls up on the rod, it stretches the rod creating tension. (You could replace the rod with a rope.)
     
  13. Feb 12, 2017 #12
    I can't seem to get my head around this, wouldn't the rod be pulling the rod when it would want to sink and pushing against the rod when it would want to float?
    The rod is what is keeping the cube to the bottom of the liquid so it would push on the rod when it would want to float? :frown:
     
  14. Feb 12, 2017 #13

    Doc Al

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    Forget about the fluid for a minute. Remove the liquid. You have a cube sitting on top of a rod. Is the rod compressed or stretched?
     
  15. Feb 12, 2017 #14

    haruspex

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    Maybe you are confusing yourself by thinking in terms of what the cube "wants".
    What the cube does is a consequence of the forces acting on it. If it does not move then those forces must be in balance.
    The pressure in the liquid is greater at greater depths, so it pushes more strongly on the underside of the cube than on top. That is what creates the buoyant force. When the liquid is dense the buoyant force is strong. If it exceeds the gravitational force on the cube then to balance the buoyant force the rod must pull down on the cube.
     
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