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Archimedes' buoyancy question

  1. Jan 4, 2016 #1
    Question: Balls A and B of equal mass are floating in a swimming pool, as shown below. Which will produce a greater buoyant force? (Image shows two circles with circle A larger than circle B)

    A. Ball A
    B. Ball B
    C. The forces will be equal
    D It is impossible to know without knowing the volume of each ball

    I think this question is poorly written. Based on the givens, I deduced that volume A > volume B, and thus density A has to be less than density of B in order for the two balls to have equal mass. So, the volume submerged in order for A to float has to be less than the volume submerged for B to float. Thus, it doesn't necessarily have to be ball A that will produce a greater buoyant force. Thoughts? Thanks.
     
  2. jcsd
  3. Jan 4, 2016 #2
    Hi Jmiz:

    Your analysis seems completely correct to me, up to
    although you didn't say explicitly your choice of A, B, C, D.

    I don't understand you comment:
    Did you pick "A"? If so I am surprised. I think the greater the submerged volume then the greater the buoyant force.

    ADDED
    I goofed. The buoyant force = the gravity force. Therefore "C".
    https://en.wikipedia.org/wiki/Buoyancy

    Regards,
    Buzz
     
  4. Jan 4, 2016 #3

    Merlin3189

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    What does it mean to float?
    Apart from any other properties or deductions, what is the essential of floating?
     
  5. Jan 4, 2016 #4
    Yes the answer was ball A. Kaplan answer key probably assumed that the volume submerged will be greater for ball A because it was drawn as larger...
     
  6. Jan 4, 2016 #5
    The essential of floating is when the buoyant force equals the weight.
     
  7. Jan 4, 2016 #6

    Merlin3189

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    And you are told the balls have equal mass - what does that say about their weight?

    Then you are asked about about the buoyant forces.

    (PS I just noticed your other post where you say, you looked up the answer (in Kaplan?) and ball A has the greater buoyant force. IS that what you said? Or have I misread it?)
     
  8. Jan 4, 2016 #7
    The answer key stated choice A. I picked choice C because it was the only one that made some what sense to me, but if there was a choice that said "it is impossible to know without knowing the density of each ball" I would have selected that.

    Given consideration of your information, the balls will have equal weight. So the buoyant forces resulting from each ball will be equal, which makes logical sense to me. However, Kaplan's answer key disagrees:

    "The buoyant force is equal to the weight of water displaced, which is quantitatively expressed as F = m (fluid displaced) g = p (fluid) V (fluid displaced) g
    The volume of displaced fluid is equal to the volume of the ball. The density of the fluid remains constant. Therefore, because ball A has a larger volume, it will displace more water and experience a larger buoyant force."

    The situation that their answer applies for I believe is not for this question's situation but for a setting that has both balls completely submerged in water, which the question stem does not explicitly state.
     
  9. Jan 4, 2016 #8

    Doc Al

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    Just curious, does the image show the balls floating on the surface (as the text implies) or submerged (as the answer key implies)?
     
  10. Jan 4, 2016 #9
    The image is just as I described: 2 circles nothing else one circle, labeled A, is larger than the second, labeled B. No bodies of water in picture.
     
  11. Jan 4, 2016 #10

    Doc Al

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    Strange. As you said, a poorly crafted question. If you ignore the diagram, which is meaningless if it doesn't show the water level, then the answer is clear. (And not what the answer key gives.)
     
  12. Jan 4, 2016 #11

    Merlin3189

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    Well, I don't know what to make of that book answer. As far as i can see, you would be absolutely correct with C. If they have equal weight and are floating, then the buoyant force must be the same and the volume displaced must be the same. All other factors are irrelevant.

    The only possibility that I can see is, they are asking, "what WOULD the buoyant forces be, IF the balls were pushed under until completely submerged" (rather than left floating.)

    The comment that, "The volume of displaced fluid is equal to the volume of the ball." is nonsense. It is one of the common mistakes that students make, to assume that the volume displaced is always equal to the volume of the object. This is only true when the object is completely submerged and this is normally when the object is sunk or barely floating.
    When an object is "floating" (even barely floating), the only thing you can say for sure, is that the weight of the object is equal to the weight of fluid displaced. Thence the weight of fluid of known density allows you to calculate the volume of fluid displaced and this must be the submerged volume of the object (, which must be less than or equal to the total volume of the object, else it would not float.)

    Your comment, "if there was a choice that said "it is impossible to know without knowing the density of each ball" I would have selected that." is effectively D. If you knew the volume of each ball, you would know the relative density and if you knew the density of each ball, you would know the relative volumes of the balls ( & NB, the total volumes of the balls, not the submerged volumes.) But neither would matter one jot! It is the weight (or mass) and the fact that they are floating that answers the question. The only volumes that would help you answer the question are the submerged volumes and neither D, nor your alternative formulation of knowing the density, will tell you that.
     
  13. Jan 4, 2016 #12

    Doc Al

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    Exactly.
     
  14. Jan 4, 2016 #13

    russ_watters

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    That's odd because the way you worded the question states that the body of water is in the picture ("floating in a swimming pool, as shown below"). You sure you didn't miss the surface and think it was a border of some kind?
     
  15. Jan 5, 2016 #14
    Thanks for your very helpful explanation!
     
  16. Jan 5, 2016 #15
    Thanks a lot to everyone who contributed! It seems clear to me that this was a mistake by Kaplan.
     
  17. Jan 5, 2016 #16
    @russ_watters
    You are right, of course, but looking for the problem in their MCAT book, the OP is also right. :)


    https://books.google.ca/books?id=aY-9AwAAQBAJ&pg=PA143&lpg=PA143&dq=Balls+A+and+B+of+equal+mass+are+floating+in+a+swimming+pool,+as+shown+below&source=bl&ots=zpB6VIsTxC&sig=3wYreTW6sHvRlzkrbR0uM6ZnLZw&hl=en&sa=X&ved=0ahUKEwjQy-amzZPKAhUH7R4KHWEcChsQ6AEIKjAC#v=onepage&q=Balls%20A%20and%20B%20of%20equal%20mass%20are%20floating%20in%20a%20swimming%20pool%2C%20as%20shown%20below&f=false [Broken]
     
    Last edited by a moderator: May 7, 2017
  18. Jan 5, 2016 #17
    Hi Jmiz:

    If both balls have a density greater than water, then both would sink. Since A is larger, it would displace more water, have greater buoyancy, and therefore sink slower

    Regards,
    Buzz
     
  19. Jan 5, 2016 #18
    I would think that the "bouyancy force" for two objects of identical mass are the same--even if one of the two is not floating--provided both are at rest.
     
  20. Jan 5, 2016 #19

    jbriggs444

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    Ignoring the corner case of neutral buoyancy, a mass that is not floating is either sinking or resting on the bottom. If it is sinking, it is not at rest. If it is resting on the bottom, it is not displacing its own weight.
     
  21. Jan 6, 2016 #20

    Merlin3189

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    I don't know what one can say about this. It appears to be a completely different concept of buoyancy from that I know (and since the OP asked about two FLOATING balls, it seems irrelevant here.)
    For me, buoyancy equals the weight of fluid displaced, which in turn equals g x the density of the liquid x the volume of liquid displaced.
    Nowhere does the mass of the object ever enter into consideration.
    For floating objects THEIR volume is also irrelevant.

    As for the 'at rest' bit, I prefer to think that the buoyant force is unaffected by movement and to consider dynamic effects separately, but that is a matter of choice.
     
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