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Why won't pressure on an irregular shape make it move?

  1. Feb 8, 2009 #1
    This is a very simple doubt, since pressure is force per unit area, if I place an object with say one side twice the area of the other underwater ( or even in the atmosphere ) wouldn't there be twice as much force on one side? ( the one with larger area ) and shouldn't that make the object move in the direction of that force?
  2. jcsd
  3. Feb 8, 2009 #2


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    I would love to know what shape you have that has more cross sectional area on one side than the other (as force is directional, cross sectional area is in essence what matters).
  4. Feb 8, 2009 #3
    lets say a wedge, with one edge a bit flat
    Last edited: Feb 8, 2009
  5. Feb 8, 2009 #4
    or imagine a cone with the top part chopped off and imagine the hole to be solid.
  6. Feb 8, 2009 #5


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    Staff: Mentor

    A wedge and a cone don't have more cross sectional area on one side than the other...

    The sloped side is still exposed to the pressure!

    Draw yourself a diagram: use a right triangle triangle sitting on the floor of a tank. Pressure acts perpendicular to each surface, so calculate the magnitude and direction of the forces on the sloped and vertical sides...then calculate the horizontal component of the force on the sloped side....
  7. Feb 8, 2009 #6


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    As Russ said, the sloped region is still subject to the pressure. Just because it isn't perfectly parallel to the opposite side doesn't mean that the net force is unbalanced. The exercise with the right triangle that Russ mentions is a good way to show this.
  8. Feb 8, 2009 #7
    I don't know if this helps, but imagine an air-filled PVC pipe that is capped at each end.
    Let's say that the pipe is 3-ft long with an inside diameter of 3-inches.

    Go to your favorite swimming pool and force the pipe down as far as practical and then release.

    Ver. 1) If held horizontally, then released, the air-filled PVC pipe will rise to the surface horizontally.

    Ver. 2) If held vertically, then released, the pipe during ascent will tend to rotate until gaining a complete horizontal configuration. If the water depth is too shallow, the conversion does not significantly alter until complete ascent. If the water depth is deep, the conversion to a horizontal position occurs under water before reaching the surface.
  9. Feb 8, 2009 #8
    Now, here is where it gets interesting:

    Let's change the shape of the tube from being an elongated circle, to an elongated equilateral triangle.

    Back to the swimming pool....

    The end result of rotation to a horizontal position remains the same, but now a new factor is introduced, which causes an additional rotation around it's long axis, such to where one of the flat sides eventually points down, the end result being that 2-sides are now rising to and then above the surface.
  10. Feb 8, 2009 #9
    With this information in hand, one might ask if it is possible to have both a vertical and lateral movement, at the same time, of a specifically shaped buoyant object underneath a body of water with no other internal/external forces applied.

    The answer is yes. It's easy.

    Imagine a basketball. Now, affix with strong glue a Plexiglas plate on the top of the basketball that is "wedge shape"... say, higher on the left side and sloping to flat on the right.

    Now, introduce weighted materials such to weight balance the left-vs-right Plexiglas wedge.
    Then, affix a reasonable weight to the center bottom of the basketball so that it doesn't flip over when submerged in the swimming pool yet can still rise from buoyancy.

    There you have it!
    During ascent, the ball will move upward by virtue of buoyancy, and laterally by virtue of wedge dynamics against the upper water.

    It can be much easier than this, but this explains the potential.
    Last edited: Feb 8, 2009
  11. Feb 9, 2009 #10
    thank you for the replies, I think i've got it now.
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