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Really basic buoyancy question

  1. Jul 15, 2008 #1
    Hi All,

    I'm having trouble understanding why exactly there is a net force upward when an object is submerged in a liquid. I thought the liquid exerts a force all around the object and thus cancels out...so how can there be a netforce from underneath the object?

    I think my understanding of this concept is off...help please!

    thanks!
     
  2. jcsd
  3. Jul 15, 2008 #2

    stewartcs

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    The magnitude of the pressure field varies with depth. So the bottom of the object has a higer upward force since it is at a greater depth. The top of the object has a lower downward force as compared to the bottom since it is at a shallower depth.

    Thus the net buoyant force is upward. This of course assumes a closed pressure field.

    CS
     
  4. Jul 15, 2008 #3
    Basically how it works is that when an object is under water, the bottom part of that object is deeper than the top part, even if it's only by a tiny bit. Since the deeper you go, the more pressure water exerts on you, the bottom part feels more pressure than the top part, therefore there is a net force up on the object.

    EDIT: And of course, I just used water arbitrarily. It can be any fluid. That's the reason why helium goes up in our atmosphere, it's lighter than our regular air.
     
  5. Jul 16, 2008 #4
    Another way of looking at it:

    When you place an object in a liquid (let's say a boat being placed in water), it displaces a certain volume of that liquid. If the volume of water that the boat displaces is equal to or greater than the mass of the boat itself, the boat will float. If you are not using water as your liquid, don't forget to consider the density of the liquid you ARE working with.
     
  6. Jul 16, 2008 #5
    pressure is a thermodynamical quantity so the origin of this force is thermodynamics.

    of all possible distrubitions, the distrubition that the object goes up has the minimum internal energy hence the origin of this force.

    proof: increasing the temperature will also increase this force?
     
  7. Jul 16, 2008 #6
    GREAT! thanks everyone. You have helped immensely! yay!
     
  8. Jul 16, 2008 #7
    I never thought of it this way...

    So if you have a hollow (closed, waterproof) tube of a reasonably light material and filled with air, and you put it underwater, does it come up faster if it's vertical and slower if it's horizontal?
     
  9. Jul 16, 2008 #8

    Andy Resnick

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    There's still going to be buoyancy from the difference in density between air and displaced water. But the horizontal tube will rise more slowly due to the increase in drag.
     
  10. Jul 16, 2008 #9

    stewartcs

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    Neglecting the drag...they would rise at the same speed. The area (if the cylinder was perfectly vertical) of the bottom is smaller than the total area compared to the horizontal position. So the net buoyant force acting upward would be the same since the depth would be different (i.e. greater pressure acting on a smaller area = smaller pressure acting on greater area). The force from the pressure field acting perpendicular to the cylinder would be zero (again assuming it is perfectly vertical).

    CS
     
  11. Jul 16, 2008 #10

    stewartcs

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    I can't type that fast Andy! :smile:
     
  12. Jul 16, 2008 #11
    ...and if you put a hair-thin sheet of material (lighter than water) underwater, and position it flat horizontally, then it should almost stay there (or at least rise very very slowly) because there is almost no pressure difference between the bottom and the top?

    PS Neglect imprecisions, assume it is perfectly flat.

    (well in that case, if near-zero thin, there is no volume displaced either...)
     
  13. Jul 16, 2008 #12

    Andy Resnick

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    The bouyancy force is simply [tex]\Delta mg[/tex]. For a rigid body, regardless of how the mass is distributed in space, the force will be the same. An exceedingly thin (rigid) bubble-sheet-thing will have the same upward force as a spherical bubble, if the volume is the same.
     
  14. Jul 16, 2008 #13

    Andy Resnick

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    Years mis-spent on chat rooms :)
     
  15. Jul 17, 2008 #14
    That's what I would have expected... But does the concept of pressure difference between top and bottom still hold?
     
  16. Jul 17, 2008 #15

    russ_watters

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    Yes, it does. Don't let "exceedingly thin" fool you - it still has a thickness and there is still a pressure difference.
     
  17. Jul 17, 2008 #16

    Andy Resnick

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    Pressure jumps across an interface can manifest themselves in different ways. Let's consider a neutrally bouyant volume of oil in water (or oil in a water-alcohol bath such that the density is matched). No matter what shape the oil is deformed into, it is still neutrally bouyant and there is no unbalanced force. However, the oil-water interface, being curved, is still associated with a pressure jump. The equilibrium shape of the oil drop is governed by the Laplace equation:

    [tex]\Delta P = -\sigma\kappa[/tex],

    Where [tex]\sigma[/tex] is the interfacial tension and [tex]\kappa[/tex] the interface curvature. If the oil is not neutrally bouyant, the pressure term has an additional component from the density difference (something like [tex]\Delta P =-\sigma\kappa + \Delta mg A[/tex].. it's too early in the morning to write it down properly)

    So our oil drop will assume a different equilbrium shape in the presence of gravity- the surface curvature will change to balance the gravitational component. In liquid bridges, the typical shape in an 'amphora":

    http://www.ulb.ac.be/polytech/mrc/research_en.html (near the bottom)

    Rigid bodies don't deform, so there is no balancing act between shape and bouyancy. Instead, there is simply an unbalanced force (P = F/Area), which results in motion.
     
  18. Jul 18, 2008 #17
    I am more puzzled now...

    If you manage to get a body that has exactly the same density as water, and therefore the weight of the water displaced is equal to the body's weight, according to Archimede's principle this body should float and neither fall or rise, right?

    But the body being extended in 3D, there is still a pressure difference from bottom to top. Doesn't this suggest it should move? Or does it just compensate for the gravity?
     
  19. Jul 18, 2008 #18

    russ_watters

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    Yes.
    It just compensates for gravity. Think about water itself, without an object placed in it - pressure varies with depth. Put a bubble aound some water and nothing changes: pressure still varies with depth.
     
  20. Jul 18, 2008 #19

    Andy Resnick

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    Another way to picture it is that a neutrally bouyant fluid will assume the shape of a sphere- not only given two immiscible fluids, but also blobs of fluid floating around the Space Station. The curvature is constant over the entire surface, thus the pressure jump across the surface is constant over the entire surface, so there are no unbalanced forces.

    Although the 'absolute' pressure varies in height, the pressure *jump* across the interface does not vary.
     
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