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What would happen to the ball?

  1. Dec 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Assume that the beaker is half filled with Hg, the steel ball being lighter will float in it half-submerged. Now we pour water into the beaker, so that the beaker is completely filled with water on top, Hg in the bottom and steel ball floating between them.

    Now my question is what would happen to the relative position of the steel ball, will it go up, or down or just remain at its place?

    2. The attempt at a solution
    I think that the ball should go down, due to the force applied by the weight of the water
     
  2. jcsd
  3. Dec 31, 2016 #2

    Doc Al

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    What does Archimedes' principle have to say?
     
  4. Dec 31, 2016 #3
    Archimedes' principle says that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces and it acts in the upward direction.
     
  5. Dec 31, 2016 #4

    Doc Al

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    Good. And since the ball is floating, that buoyant force must equal what?
     
  6. Dec 31, 2016 #5

    haruspex

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    But the water is also pushing down on the mercury.
    Try to answer Doc Al's question, then write it as an equation for each of the two cases: mercury only and mercury+water.
     
  7. Jan 2, 2017 #6
    To the Weight of the ball + the weight of the water, i think.
     
  8. Jan 2, 2017 #7

    Doc Al

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    Archimedes' tells you that the buoyant force equals the weight of displaced fluid.

    But since the ball is floating, the net force on the ball must equal what?
     
  9. Jan 3, 2017 #8
    Weight of the Hg displaced.
     
  10. Jan 3, 2017 #9

    haruspex

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    The buoyant force always equals weight of fluid displaced, floating or not. Doc Al was asking what else that equals when the object is floating.
    But I don't think that will get you past your misunderstanding, because when the water is added you will still answer "weight of ball plus weight of water on the ball".
    What you are overlooking is that the buoyant force is the resultant of the pressures acting on the object in all directions. The upwards pressures underneath the object are greater than the downwards ones on top because of the greater depth, so the net force is upwards. When water is added, that adds downwards pressures on top of the ball, but it also presses down on the mercury around the ball. That increases the upwards pressures from the mercury under the ball.

    Archimedes' principle avoids having to think about all these different pressures. You can just consider removing the ball and putting something else in its place which would also be in equilibrium. If the ball is floating on mercury, removing the ball would create a void consisting of two spherical caps, one in the mercury and one in the air. To restore equilibrium, you could fill the lower cap with mercury and the upper with air. The weight of the ball must equal the sum of the weights of those two. Generally, we ignore the weight of the air displaced.

    Now, with the water added, what would you need to replace the ball by when the ball is removed?
     
  11. Jan 3, 2017 #10

    Doc Al

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    Here's what I was looking for: Given that the ball is floating, the net force on it must be zero. The buoyant force (upward) must equal the weight (downward).

    Your answer is correct for the initial situation, where the ball is floating in mercury alone. And that tells you that the weight of the ball must be equal to the weight of Hg displaced, which has a volume of half the ball.

    So, if you pour water on it now, can the Hg level still be at the half way mark of the ball? To figure it out, try assuming that it's true and see if things still make sense. (The ball's still floating and Archimedes' still holds.) If the mercury stayed at the half way mark after you added water, what must happen to the buoyant force per Archimedes'?

    (haruspex and I must be in sync! :)
     
  12. Jan 4, 2017 #11
    Okay, i get it now. Now we would need to replace the ball with mercury in the lower cap and water in the upper cap.
     
  13. Jan 4, 2017 #12
    If we now pour water on the arrangement, then water must push the ball down, due to its weight, so the mercury now can't be at the half way mark. But it also presses down on the mercury around the ball. That increases the upwards pressures from the mercury under the ball, as said by haruspex.
     
  14. Jan 4, 2017 #13

    Doc Al

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    I think you've got it. So answer your initial question: What happens to the relative position of the ball?
     
  15. Jan 4, 2017 #14
    It will sink a little bit, but will still float.
     
  16. Jan 4, 2017 #15

    Doc Al

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    Think it through. If it sunk a bit that would mean that more than half of the ball is submerged in Hg. The ball would therefore be displacing even more Hg than before. Is that possible?
     
  17. Jan 4, 2017 #16

    haruspex

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    Right.
    How does the total weight of that compare with the weight of mercury displaced when there was no water? What does that tell you about the volume of mercury displaced when the water is added?
     
  18. Jan 5, 2017 #17
    The total weight now is greater than that of the displaced mercury when there was no water, as now the half of the sphere is filled with water. But what does it tell us about the volume of mercury displaced after water is added?

    Maybe that now as the additional weight of the water presses down on the mercury, so the upward force on the ball increases and the now less amount of mercury is displaced. So the volume of mercury displaced becomes less.
    Is it correct?
     
    Last edited: Jan 5, 2017
  19. Jan 5, 2017 #18

    No, its not possible. As the water presses down on mercury, it will increase the total upward buoyant force, and the ball now would be less sunk in mercury than it was before the water was added.
    Okay. Am I right now?
     
  20. Jan 5, 2017 #19

    Doc Al

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    I think you've got it.
     
  21. Jan 5, 2017 #20

    haruspex

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    Yes.
    Without the water, the weight of mercury displaced equals weight of ball.
    With the water, weight of water displaced plus weight of mercury displaced still equals weight of ball.
    Since water weight > 0, weight of mercury displaced must have reduced.
     
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