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The weight with a fluid: center of mass or force at bottom?

  1. Dec 24, 2014 #1
    1. The problem statement, all variables and given/known data

    It is a question I asked to myself because I need to reply to another one. I imagined this case:

    http://imageshack.com/a/img912/7730/R964Vq.png [Broken]

    The step2 needs the energy E1, the step3 needs nothing because the gravity is 0 (I imagine the device between the Earth and the Moon), the air container is fixed, at step4 I recover E1, at step5 I recover an energy E2. At final I have -E1+E1+E2 = E2.

    2. Relevant equations

    --

    3. The attempt at a solution

    With a solid I'm agree to put the force of weight in the center of mass and if I do that the energy is conserved. But why I need to do that with a fluid ? The force from the weight is at the bottom of the container not in the center of mass. If I use the true force in the bottom of the container, I need to add an upward force ?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 24, 2014 #2

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    Wrong. You haven't lowered all the water to the original starting height.
     
  4. Dec 24, 2014 #3
    Sorry, I changed my message because it's not Fd it's an integration of F.

    And what ? the force of pressure is at the bottom of the container, no ? d is the same at step2 and at step4.
     
  5. Dec 24, 2014 #4

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    You have it correct, "NO."
     
  6. Dec 24, 2014 #5
    Could you explain please ?
     
  7. Dec 24, 2014 #6

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    What is the increment of weight, dw, for each increment of volume in your container? It's not zero at the top, is it?
     
  8. Dec 24, 2014 #7
    I don't understand the question, could you reformulate it please ?
     
  9. Dec 24, 2014 #8

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    Take nine duplicates of your "box." Arrange them on the table with the original. Place 1/10 of the water in each box. What's the mass of the water? What's the force the water exerts on the table? In terms of the original force.
     
  10. Dec 24, 2014 #9
    The same.
     
  11. Dec 24, 2014 #10

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    Now, how much work are you going to have to do to put all the water back into the original box?
     
  12. Dec 24, 2014 #11
    It depends how boxes are arranged. When the box move up or move down (in my cycle), nothing moves inside the container.
     
  13. Dec 24, 2014 #12

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    Last chance: blow it, and this thread gets locked, and you get labelled as a perpetual motion/over unity nut.
    How much work do you have to do to lift the water spread among ten boxes back into the original box?
     
  14. Dec 24, 2014 #13
    Weigth*h/2 but I'm not sure, with h the height of the container
     
    Last edited: Dec 24, 2014
  15. Dec 25, 2014 #14
    Someone could explain to me why the force F is apply in the center of mass and not at the wall ? Why the work at step 4 is not the same than in the step 2 ?
     
  16. Dec 25, 2014 #15

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    All right! Do it your silly way, BOTH directions. Don't use COM lifting, and shift to bottom of mass when lowering. Pick ONE OR THE OTHER.
     
  17. Dec 25, 2014 #16
    I understood ! if I take a lot of small parts, it's logical, thanks. I thought the force was the same for an altitude for one container with air or not. But it's not the same weight. So I can reply to my another question: if the weight is the same with a fixed bubble inside ?: the answer is no, the weight is lower. Now, with a bubble that is moving up, in one part I would say it's lower and in the other part when the bubble accelerates to the top the water accelerates to the bottom so the weight must be a little higher I don't know the sum. But it's ok for this question, thanks for your patience. I asked myself if I move down the container to the center of the Earth and place the container with air where g=0.
     
    Last edited: Dec 25, 2014
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