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Different pressure in water

  1. Apr 21, 2012 #1
    I would like to know if pressure is different in water due to the difference of gravity (like 1/d²) ? If yes, why a cone shape with a helicoid like the drawing show don't turn in water without torque ? I think the vertical is closer to the center than the horizontal surface so the pressure is greater ?
     

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  3. Apr 21, 2012 #2

    mfb

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    No. The pressure in gases and fluids (in equilibrium and without movement) can be calculated via the mass (more precise: area density) of the material above it.

    I don't see your argument why this shape should rotate in any way. The problem is symmetric under rotation.
     
  4. Apr 21, 2012 #3

    HallsofIvy

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    If you mean "is the pressure different at different depths" then the answer is "yes". The pressure acting on a one square m surface is equal to the weight of water above it (plus the weight of atmosphere above the water) no matter how the square is oriented.
     
  5. Apr 23, 2012 #4
    No, at a depth give, the pressure decrease with "x" axe because we increase the distance ? (it's the hypotenuse), d2>d1 so the pressure is lower at d2, no ?

    If the pressure change with "x" torques are differents (vertical surface / horizontal surface). And if pressure don't change with "x" why ?
     
    Last edited: Apr 23, 2012
  6. Apr 23, 2012 #5
    In practice, the earth is so huge compared to any structure you could build that d1 = d2, and therefore the same gravity is experienced at the same depth. The water pressure experienced at a point is not just a function of the gravity at that point, rather it is a function of all the water above that point crushing down, and the gravity that that water experiences at all the points above. Again, because the earth is so large, for most systems, gravity is effectively constant. The differing pressures at different depths in the ocean is not because gravity is changing, rather because you have more water piled on top of you at greater depths.
     
  7. Apr 23, 2012 #6
    I can understand d1=d2 approximation, but we can imagine a mass smaller than Earth, in this case, the pressure is different at d2 ? If yes, with a cone and a helicoid on it like the drawing show, why the cone don't turn with torque ? When I think with a fixed pressure or a gravity pressure identical at same height ok the cone don't turn but here I see differents pressures.
     
  8. Apr 24, 2012 #7

    mfb

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    Your system would still be invariant under rotation.
    It is ugly to calculate forces on each individual face of your body, if you can use a simple symmetry like this. It does not rotate, even if its size is comparable to its distance to the center of mass of earth.
     
  9. Apr 24, 2012 #8
    I can understand it's difficult to calculate. I think with a full cone and I retrieve small part for have a small helicoid. With a fixed pressure you apply the same pressure at each point and we know this cone don't rotate. With gravity like 1/d² the pressure in the vertical surface is bigger than the horizontal surface, in this case why the cone don't rotate (even the torque is very small) ?
     
  10. Apr 26, 2012 #9
    I add an image for show how I see the pressure in water due to the gravity. Each line is where pressure is constant. How can I calculate the torque on the cone ? Is there a simple equation for the surface ?
     

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  11. Apr 26, 2012 #10

    mfb

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    Integrate pressure x (radial vector) over the whole surface, where x denotes the cross product. This is ugly to do if you don't use nice integral properties*, but should be possible somehow. And the result will be 0.

    *these lead to the symmetry argument, which leads to 0.
     
  12. Apr 26, 2012 #11
    I'm sure the result is 0. But I would like to know why and calculate allow to give some information for understand. How can I have the equation of surfaces with a simple example because it's not necessary to calculate all the surface I think ?
     
  13. Apr 26, 2012 #12

    russ_watters

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    The simple example is a small surface and the equation is simply pa-pa=0
     
  14. Apr 26, 2012 #13
    yeah ;) ok but it's not like this I can understand how a surface where the pressure is lower can cancel the torque from a surface where the pressure is greater. Imagine with a fixed pressure, radius and surface equilibrate for cancel the torques. But here, the pressure is not the same... it's strange for me.
     
  15. Apr 28, 2012 #14
    I don't find the solution with a cone, it's too difficult for me, but I think it's the same problem with a parallelepipede like the drawing show. I remove some material for obtain a slope (grey color). In this case, I think it's easier to see where is the problem. For me, the green surface give a torque which must be compensated by the red surface. With a static pressure or with the approximation for gravity it's ok for me: no torque. But if I take the exact gravity (1/d² and the curve) in this case I don't understand how torques can be compensated ?
     

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    Last edited: Apr 28, 2012
  16. Apr 28, 2012 #15

    mfb

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    Imagine another object right in the region where you removed the material. Together, you have a simple block, with no torque. This means that the torque of your block is opposite to the torque applied on the part which you removed.
    But this part is just a parallelepiped with two equal sides.
     
  17. Apr 28, 2012 #16
    This problem is not comparable to the cone, the cone like I drawn has one surface more close to the center all the time, the another surface is from more center to external radius, I don't know how torques can be cancelled. I would like to calculate it but I don't know how to start with equations of surfaces.
     

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    Last edited: Apr 28, 2012
  18. Apr 29, 2012 #17
    @mfb: I said the torque is 0 for your example but only if I don't take the change of density in water. Now, with the fact that gravity is in 1/d² and the density change in the same time. If I take your shape I don't have a torque to 0 because the pressure change like 1/d3 and when I integrate the pressure I have 1/d² and it's not possible to cancel the forces with 1/d² it's not linear. But maybe something is wrong with my calculation.
     
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