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Buoyancy Question

  1. Apr 28, 2008 #1
    This question was presented as one of those "think about it" questions in my physics class. Its not worth any points, but its driving me crazy! I can't figure it out. Here it is:

    Consider a pineapple sitting on a sensitive scale resting
    on the bottom of a large glass jar which is connected to a
    vacuum pump. The reading on the scale is 2.315 pounds
    with air in the jar. The air is now pumped out of the
    jar. Which of the following gives the best description of
    what happens to the scale reading and why.

    (a) The scale reading increases slightly because the
    buoyant force of the air is now gone.
    (b) The scale reading decreases slightly because the
    air used to be pushing down on the pineapple, but now
    it isn’t anymore.
    (c) Weight is caused by gravity and doesn’t have anything
    to do with the air, so the reading stays the same.
    (d) The reading decreases slightly because the scale
    reading used to include the weight of the air in the jar,
    but now the air is gone.
     
  2. jcsd
  3. Apr 28, 2008 #2

    stewartcs

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    To make sure I understand what you are asking...you are saying you have a pineapple inside of a glass jar, both of which are sitting on top of a sensitive scale...correct?

    Is the jar sealed?

    CS
     
  4. Apr 28, 2008 #3
    I believe both the pineapple and scale are in the jar. Also, I believe the jar is sealed (since it can hold a vacuum).

    Thanks for your quick response!
     
  5. Apr 28, 2008 #4

    mgb_phys

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    And assuming the scale didn't seal like a piston inside the jar you can ignore air pressure since it acts on the under side of the scale jusst as much as on the top.
    So it's just a question of boyancy - I would say 'a'
     
  6. Apr 29, 2008 #5
    mgb_phys - that is very true. However, I don't think we are given enough information to say that 'a' is the answer.

    If the scale is low-mass and high-volume (beach ball in water) and the pineapple is high-mass and low-volume (small stone in water), then the reading on the scale would be higher than normal, since the scale would be pushing up on the pineapple. If this was the case, the reading on the scale would actually decrease when the air is removed. This, of course, also depends on how much air was in the glass (how big the glass is). There are too many unknowns to say that 'a' is the answer, don't you think?

    Thanks again for your input! I love this forum - everyone is a genius.
     
  7. Apr 29, 2008 #6

    mgb_phys

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    You aren't normally expected to take into account the effect of bouyancy on the scale - but good point.
     
  8. Apr 29, 2008 #7
    To bouy or not to bouy

    First, I think two of the answer are just about the same thing said a little differently. (b&d) Secondly, the idea of buoyancy assumes the density of the thing that the fluid is pushing up is higher,(or should I say greater than), than the fluid's. In this case, it isn't. So no bouyancy. Third, the scale is inside the jar so the air pressure on the scale in the sealed jar is unaffected by the outside air :vacuum or not. But the problem states that the scale is very sensitive so the missing air that was formerly in the jar weighing down on the scale is now missing.
    Conclusion: the scale in the jar would register slightly less but of course the pineapple would not actually weigh less because it's mass did not change. And we know that its weight is supposed to be,(and is), proportional to its mass.
     
  9. Apr 29, 2008 #8
    Sorry, I meant to say the density of the fluid is greater than the object being bouyed up. And the object is therefore less dense than the fluid.
     
  10. Apr 29, 2008 #9

    Andy Resnick

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    I'm going with (c), only becasue the scale does not appear sensitive enough- if it was, I vote for (a). The atmospheric pressure has no effect, because the rigid bell jar separates the inside from outside. The scale simply measures applied force, which here is

    F = mg - [tex]\Delta\rho V g[/tex], where [tex]\Delta\rho[/tex] is the density difference between pineapple and air, and V the volume of the pineapple. Reducing the density of the air (via pressure) reduces the bouyancy force. Replacing the air with say, water, increases the bouyancy. Although in that case, 'm' would be different as well, as the scale would be reading some of the water weight.
     
  11. Apr 29, 2008 #10

    mgb_phys

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    Not if the water was above and below the pan of the scale it wouldn't
    - for the same reason that you don't need to worry about the weight of air on your kitchen scale.
     
  12. Apr 29, 2008 #11
    Let me repeat. There is no bouyancy force at all unless the density of the object, in this case the pineapple is less than the fluid that is inside of, this case air and then vacuum. Your equation above is subtracting from the weight of the pineapple as if there is a bouyancy force but there is not. If the fluid were more dense than the pineapple then the equation would be correct. Think for a minute about bouyancy itself. It is a force produced when the outside fluid is heavier than the object and tries to push itself under it, precisely because it is heavier. If the object is heavier than its equivalent volume in fluid then it is impossible for the fluid to push itself under it. Hence no upward force. We are then left with only the weight of the pineapple and the weight of the air in the jar on top of it. That being the case the answer is "d" since that air is now gone.
     
  13. Apr 29, 2008 #12

    mgb_phys

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    Yes there is - if you put a 1m^2 rock in water it doesn't float, but still receives an upthrust of 1000Kg and weighs 1000kg less
     
  14. Apr 29, 2008 #13
    Leaving out the effects of the air on the scale, bouyancy is caused by the pressure gradient in the air. Remove the air, decrease the pressure gradient, measured weight is more.
     
  15. Apr 29, 2008 #14
    You're right. Thanks for the correction. But since there is no fluid around to push up at that rate can't we now safely say there is no bouyancy?-- no force pushing up? Since in your example the water is still trying to push its way under the rock but if no water is there no upward force can be created.
     
  16. Apr 29, 2008 #15
    My professor is going to let us know tomorrow what the correct answer is. As for me, I've decided the correct answer must be (a). Here's why (I've pulled a lot of ideas from all of you so thanks!).

    (b) and (d) essentially say the same thing. Both attribute the increased scale reading to the weight of the air (rather, the air pressure). While the air pressure does exert a force, the force would push up on the scale face just as much as it pushes down.
    (c) is fundamentally wrong. Weight is a force caused by both gravity and mass. Since air has mass, it certainly is a factor that cannot be ignored.
    (a) must be the correct answer. I think I've done a fairly good quantitative analysis below.

    First of all, all solids in fluid will experience the buoyant force (fluids include both liquid and gas, so air is a fluid). The magnitude of the buoyant force always equals the weight of the fluid displaced by the object.

    First, we find the volume of the pineapple:
    Let's simplify the shape of the pineapple by saying it is a cylinder that is .4m tall and .2m in diameter. The volume of this cylinder would be about .05m^3.

    Next, we find the weight of the air displaced by the pineapple:
    The density of air is about 1.29 kg/m^3. So, we multiply the volume of the pineapple (.05) by the density of air (1.29) and get .0645.

    The scale in the problem reads to three significant digits, so the difference would be detected by the scale.

    Any thoughts?
     
  17. Apr 29, 2008 #16
    Wait a second. On second thought, I'm not so sure about your assertion about the rock. Granted, as the rock falls to the bottom it is falling through water that has pressure in all directions equal to the column of water over it(Bernoulli) but once the rock is flat on the sea bottom then there is no water beneath it and hence no water pressure. I doubt then it will have any bouyancy lying flat on the sea floor. It would be stuck like a sucker to the floor bottom. It would revert back to its weight in a vacuum.
     
  18. Apr 29, 2008 #17
    You forgot the column of air that was originally on top of the scale that is now gone. It weighs something too. Are you saying its too light to register on the scale?
     
  19. Apr 30, 2008 #18
    Again, the air does not effect the reading on the scale. Yes, the air has mass - and if the air was in a container on top of the scale, we would have to take the air's weight into account. However, the air is not in a container on top of the scale, the air is all around the scale. You could say that the scale is immersed in the air. Assuming that the scale is not sealed, the pressure from the air will effect the top of the scale's plate just as much as the bottom. I don't think the air makes a difference.
     
  20. Apr 30, 2008 #19
    The big problem is that the scale will also experience a buoyant force - now that just really complicates things...
     
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