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Buoyancy of Oil in Water

  1. Jan 17, 2016 #1
    I have a hypothetical scenario I'd like to try understand, and I cannot find anything of the sort online nor in the forums.

    If I have a balloon of 1 m3 filled with oil, of a density of 850 kg/m3. Now I place this balloon in a body of water and tether it to a fixed point, how can I find the buoyancy force of this balloon that would 'pull' the tether?

    Is this purely an application of Archimedes' principle? Because if that is the case then obviously the force that is created by the ballon of oil is equal to 1 m3 of water that is being displaced. But this does not account for the fact that the oil will have the tendency to rise to the top of the body of water which in itself will cause additional force to act on the tether. Or is the wrong way to think about this problem?

    Also, would the depth of water cause additional exertion of force, say if it was 10m below the surface vs. 100m as we know that there is a pressure differential due to increasing weight of the body of water above the object.

    Any help is appreciated.
  2. jcsd
  3. Jan 17, 2016 #2

    Simon Bridge

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    Newtonmeter on the tether.

    Mathematically - Yes.

    ... no: the balloon will float unless you hold it under or the tether lifts it higher... it will only displace it's own volume if held under.
    (Assuming: the balloon material has negligible mass and volume.)

    Yes - where does oil get it's "tendency to rise" from?

    Archimedes principle is an emerging effect from the way the pressure of a fluid changes with depth (or height).
    It is the pressure gradient that provides the buoyancy force - not the absolute pressure.
    Last edited: Jan 17, 2016
  4. Jan 17, 2016 #3
    Yes, as Simon pointed out, this is an application of Archimede's principle.

    If the balloon is completely submerged (because the tether is short enough) and the buoyant force is greater than the weight force (which it would obviously be since the oil is less dense than water) then the tension in the tether would be the difference in the weight of the balloon filled with oil and the weight of the water displaced by the balloon. In order words, the tension in the tether in Newtons I believe would be 9.8*(1000kg - 850 kg) = 1470 N?
  5. Jan 17, 2016 #4

    Simon Bridge

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    Oh I was picturing it the other way up!
    ... well: 9.8(N/kg)*(1000kg - 850kg) = 1470N ... niggle: if I'm going to include units in the calculation that is ;)

    OP should verify the calculation.
  6. Jan 17, 2016 #5
    Thank you Simon & leright.

    I was thinking along the same lines in terms of the density difference between the water and the oil would contribute to the buoyancy force of the balloon.

    See the form of Archimedes' Principle I have observed only allows you to account for one fluid density, and then the volume of the body in the fluid.

    Can we extend the density to the 'difference between the densities'?
  7. Jan 17, 2016 #6
    So there are two forces acting on the balloon. There is the weight of the balloon and the bouyancy. The difference between these forces is the tension in the tether. W = 9.8*850 kg and Fb = 9.8*1000 kg. The difference is the net force on the balloon, or a net force upward (toward the surface of the water) of 9.8*150 = 1470 N.
  8. Jan 17, 2016 #7
    That makes sense, to reiterate:

    Fbuoyancy > Fballoon-weight -> Balloon will rise to surface
    Fbuoyancy < Fballoon-weight -> Balloon will fall to bottom.

    In the case of 1 m3 of oil in water, as you stated Fbuoyancy = 9.8*1000kg = 9800 N & Fballoon-weight = 9.8*850kg = 8330 N. Therefore, Fbuoyancy > Fballoon-weight = true -> rises to surface.

    If we had 1 m3 of concrete, 2500 kg/m3, (in a balloon) then we'd end up with Fballoon-weight > Fbuoyancy -> fall to bottom.
  9. Jan 17, 2016 #8
    You got it!
  10. Jan 17, 2016 #9
    Awesome, thank you for your help!

    One last thing, if I were to take my balloon 1000m under water, would there be a difference in the forces? or would they act the same. As I know that as you go deeper there is a pressure differential as Simon stated, would this apply more force onto the balloon?
  11. Jan 17, 2016 #10
    The pressure differential is the same at all depths. As I think Simon alluded to, you can theoretically do the same calculation of the buoyant force by integrating the pressure (rho*g*h) over the whole surface area of the balloon. The resulting force will be the same at all depths. This is not a real easy calculation though.

    Also, If there was no balloon what takes its place in the volume the balloon would have took up? Water! The water around that water must exert a force to "hold it up", which is equal to the weight of the water being "held up". That was Archimede's original reasoning I believe.
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