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Why should buoyancy occur?

  1. Aug 12, 2009 #1
    I guess it makes sense that the buoyant force of an object is equal to the weight of the displaced fluid, but why should this be the case? Why should that displaced liquid provide an upward force on the object?

    It seems apparent that as long as there is some part of the object above the surface, the result of the pressure on the submerged portion would net an upward force. But what about a completely submerged body? The pressure is now applied to every surface of the body (including the previously un-submerged surfaces), so why the net upward force? Is there some application of a conservation law that I am missing?
     
  2. jcsd
  3. Aug 12, 2009 #2

    Doc Al

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    The pressure is not the same everywhere. The pressure is greater on the lower part of the body--since fluid pressure increases with depth--leading to a net upward force
     
  4. Aug 12, 2009 #3
    OK, i thought that may be the case, but then geometry of the body would affect its buoyancy, not just its weight.
     
  5. Aug 12, 2009 #4
    The pressure applied to the low side is larger than the high side pressure. Remember the hydrostatic pressure, for the case of fluids exposed to a gravitational force:

    70d6a72229f1c9e7b9be66465519621c.png

    On the low side, the value of h is larger.
     
    Last edited by a moderator: Apr 27, 2017
  6. Aug 12, 2009 #5

    Doc Al

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    No. It turns out that the shape is irrelevant. Any object totally submerged will have a buoyant force equal to the weight of the displaced fluid, regardless of shape.
     
  7. Aug 12, 2009 #6
    Suppose i reconstruct the object so that it protrudes deeper than before, therefore having a larger pressure at the bottom. Now to keep the same density, those lower regions would have less surface area, so the increase in pressure would result in a lower force. How can i show that this trade off is exactly 1-to-1?

    Basically, how can i show that all these factors disappear, and leave me with only volume?
     
  8. Aug 12, 2009 #7
    I know it doesnt, and shouldnt matter what the shape of the object is, but i am really digging deep for a rigorous explanation.
     
  9. Aug 12, 2009 #8

    Doc Al

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    Read this explanation: http://hyperphysics.phy-astr.gsu.edu/Hbase/pbuoy.html" [Broken]. Note that the argument does not depend on the shape of the object.
     
    Last edited by a moderator: May 4, 2017
  10. Aug 12, 2009 #9
    Do you know multivariable calculus? If you do it is a piece of cake, if you don't you will need a heuristic explanation which might not be satisfying.
     
  11. Aug 12, 2009 #10
    I did a bit of thinking about 2 cylinders. Each the same volume and weight, but one twice as long as the other. The long one would have a pressure on its lower surface twice that of the shorter one, but half the surface area, therefore the same upward force. I suppose i would need some multivariable calculus to show that this applies to any general shape. I'd have to brush up on it. It's been a while, and im not sure if i know it well enough to solve this problem.
     
  12. Aug 12, 2009 #11
    If you mean the fact that the buoyant force is the weight of the displaced fluid, you're looking for the derivation of an (mg) term, not a volume. There is an article at this site that shows the derivation, on the following page:

    https://www.physicsforums.com/library.php?do=view_item&itemid=123

    The derivation on that page makes the volume a cube, which isn't necessary. To be more general I would prefer to select a cylinder whose base has any shape, base area A, height h.

    Pressure difference between top base and bottom base is rho gh.

    Buoyant force is = (pressure difference between the two bases)(base area) = rho ghA

    volume of displaced fluid = hA
    mass of displaced fluid = (density) (volume) = rho hA
    weight of displaced fluid = (mass)(g) = rho gha

    Therefore the buoyant force equals the weight of the displaced fluid.
     
    Last edited: Aug 12, 2009
  13. Aug 12, 2009 #12

    Doc Al

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    While you could use calculus to show that Archimedes' Principle holds for any shape, that's definitely the hard way to go. The argument given in the link I provided (and which is given in most textbooks) should convince you with no calculation--or calculus--whatsoever.
     
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