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Classic buoyancy demo discussion on fluid dynamics

  1. May 4, 2007 #1
    Hello, I was attending a physics discussion on fluid dynamics (fairly basics concepts) and the classic buoyancy demo came up. (A test tube is inverted and allowed to be filled with enough water to float with the volume of air trapped in the tube, the test tube is inside a larger bottle which is then squeezed, compressing the gas and allowing the tube to sink)
    The Argument I became engaged in was the manner in which the sinking was occurring. I argued that the gas was compressed, causing a lower volume of gas, and thus a smaller amount of buoyant force. This then caused the test tube to sink. The teachers argument was that the amount of water displaced remained the same, however the density of the gas increased, causing the test tube to sink. The same demo was performed using a non-rigid container filled in a similar manner with a gas, and in this case it was agreed that the compresion of the gas was responsible. Was I right?
     
  2. jcsd
  3. May 5, 2007 #2

    Doc Al

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    I'd say that you were wrong and that the teacher's argument was incomplete. You can't consider the bubble by itself, since it's stuck inside a rigid tube--you must consider the tube as a whole.

    If the inverted test tube is beneath the water surface, it always displaces the same amount of water, so the buoyant force on the tube (pretend the open end is sealed) never changes. If it just floats, that means the buoyant force just balances the weight of the tube and its contents: water and bubble. (Another way of saying this is that the average density of the tube + contents just equals the density of water.)

    What happens when you squeeze the bottle is that you force more water into the tube. (The bubble gets smaller due to the increased water pressure, thus more water enters the tube.) Now the weight of the tube and contents is greater than the buoyant force, so it sinks. (The average density of the tube + contents has increased.)

    When a flexible container is used, the increased pressure does decrease the volume of water displaced and thus the buoyant force.
     
  4. May 5, 2007 #3
    Yes, you were. The increase of the density of the gas doesn't change its mass or its weight. It was, as you said, the volume of water displaced that changed.

    It is possible that the teacher and you were measuring the water displaced in different ways. For you (and for me) the water, including the water inside of the tube, was the outside water. For the teacher, perhaps, the outer water began at the lower end of the tube. In this case, the volume of displaced water did not changed, but the mass of the floater (sinker?) did change.

    Teachers are human and can be wrong sometimes. You cannot ask that they never make a mistake. Be tolerant as you want the others to be tolerant with you.
     
  5. May 5, 2007 #4

    Doc Al

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    I see your point. It certainly makes a difference how you interpret the amount of water "displaced". In calculating the buoyant force on something submerged, what counts is the volume of the thing--that determines the amount of fluid "displaced" as typically used in Archimedes's principle. Considering the tube as a whole, its volume doesn't change, so the water displaced remains the same. (Of course, if you measure the amount of water displaced by how the water level changes--perfectly reasonable--you'll get a different answer. But then you can't use the typical statement of Archimedes's principle that the buoyant force equals the weight of displaced fluid.)
     
  6. May 5, 2007 #5
    I can. For me, the exterior of the object starts at the level of the water inside the tube. When the air expands, it displaces the water. It works very well in my head and I reason very well in this way. Yours also works, of course. But it asks that the displaced water rests the same and that it is the mass of the object that changes. It is you that can't use Archimedes for the variation of boyancy.
    I do not think that one or the other way to see the problem is the "good one". I understand your way, and I could use it, but I prefer mine.
     
    Last edited: May 5, 2007
  7. May 5, 2007 #6

    Doc Al

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    I see what you're saying: You are taking the object to be empty tube+bubble. Excellent! (If that's what Dbjergaard was doing, then his reasoning is perfectly fine.)
    By my definition of "object" (tube, bubble, and enclosed water), the buoyant force is constant. But I like yours too! (I'll bet that the teacher was using a definition closer to mine--thus the confusion.)
     
  8. May 5, 2007 #7
    Yes. And maybe, for Dbjergaard and his teacher, there was also a confusion between the density of the air and the mean density of the tube contents.
     
    Last edited: May 5, 2007
  9. May 5, 2007 #8
    It wasn't a violent argument (at least from my perspective, I can see how a teacher could be upset if the student insisted he was wrong) I thought it could be argued both ways, and he was pretty sure that I was wrong and that he was arguing the correct way. I was just curious as to how other people visualized the problem. Now, what I'm curious about is the non-rigid case. Those of you who are arguing with the teacher with the test tube, and then siding with me in the non-rigid case (also the stance taken be the teacher) seem to be contradicting themselves. How in one case can the manner in which you calculate buoyant force be different from another? Isn't the study of physics a constant drive for consistancy in a multitude of situations?
     
  10. May 5, 2007 #9

    Doc Al

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    Once you've defined your object, you calculate buoyant force on the object exactly the same way. With the rigid tube--if you define your object as the tube, bubble, and enclosed water--then the volume of the object does not change; but with a flexible container, the volume may well change.

    But if you define your object as just container + bubble (which I presume, after reading lpfr's comments, is what you did as well), then the volume of the object will change, rigid container or not.

    I kind of prefer that second approach now. :smile:
     
  11. May 6, 2007 #10
    It just seems better to use the reasoning which is more widely applicable to more situations, thats just my opinion, and I'm by no means qualified to judge one way or the other.
     
  12. May 6, 2007 #11
    I think that the better reasoning is the one that you "feel". The second conclusion of this thread is that you must not reject too fast the others reasoning (a I also did). It can be also correct.
     
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