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Calculate the minimum work to extract a body of the water

  1. Apr 23, 2013 #1
    A hemisphere of 8 kg mass and 20 cm radius is at the bottom of a tank containing water. Find out the minimum work that an agent must to do to extract the hemisphere of the water.(g = 10 m/s2)
    prob_ener_pot_hid_01.jpg
    Please help me
     
  2. jcsd
  3. Apr 24, 2013 #2

    SteamKing

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    Draw a free body diagram of the sphere. It would also be helpful to know the depth of the water. Id the depth the same as the radius as the figure implies?
     
  4. Apr 24, 2013 #3

    rude man

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    Just posting to be notified of the progress of this thread, if any.
    I notice the density of the sphere is < density of water ....
     
  5. Apr 24, 2013 #4
    We assume that level of water is constant and inicially half of the semiesphera is inside of water.
     
  6. Apr 24, 2013 #5
    We assume that the density of the sphere is > density of water
     
  7. Apr 24, 2013 #6

    rude man

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    That's very difficult to do:
    Volume of sphere = 4/3 pi r^3 with r = 20cm → V = 33,510 cc
    Mass of sphere = m = 8kg = 8000 gm
    Therefore density of sphere = m/V = o.24 gm/cc whereas water density = 1 gm/cc.

    This whole problem looks ridiculous.

    EDIT: or maybe not: if we assume the sphere does not touch the bottom then this is actually a good problem.
     
    Last edited: Apr 24, 2013
  8. Apr 24, 2013 #7

    haruspex

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    You don't need to make any such assumption. We're not told why it is at the bottom of the tank. Maybe it was being held down somehow.
    On release, we could let it float and then lift it, but that would waste some of the potential energy. Assuming no losses, we can utilise the stored PE. That makes it a very easy calculation.
     
  9. Apr 24, 2013 #8

    haruspex

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    The depth of water cannot be constant. I think you mean the hemisphere (i.e. half the sphere were it complete) is fully immersed.
     
  10. Apr 25, 2013 #9
    You're right rude man. Density of body < density of water. Initially the force F points down to keep submerged hemisphere.

    I hope this image can help to unterstand better this problem.

    problem_min_work.jpg
     
    Last edited: Apr 25, 2013
  11. Apr 25, 2013 #10
    The level of the surface of water is constant and at beginning the hemisphere whole is inside of the water.
     
  12. Apr 25, 2013 #11
    Right haruspex.:smile:
    Could help me in that (calculation)
     
  13. Apr 25, 2013 #12

    rl.bhat

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    Net force on the hemisphere= dF = weight of the hemisphere - bouyancy force.
    Assume the density of hemisphere is greater than that of the water.
    The volume of the submerged part of the hemisphere is equal to V = pi/3 x h^2 x (2r - h)
    Bouyancy force = V x density of water.
    Total work done W = Int. dF x dh form H to zero.
     
    Last edited: Apr 25, 2013
  14. Apr 25, 2013 #13

    haruspex

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    To make the surface constant height we have to assume an infinitely wide reservoir. As the hemisphere rises, the space it occupied is filled with water. Where, in effect, does that water come from? (I'm asking for a specific height within the reservoir.)
     
  15. Apr 25, 2013 #14

    rl.bhat

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    In this problem, the surface of the water need not be at constant height. Bouyant force is independent of the size of the tank. If we assume the figure is correct, then the density of hemisphere is equal to the density of water. In that case it just submerses.
     
  16. Apr 25, 2013 #15

    haruspex

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    You do need to assume that. If the level drops as the hemisphere rises then the buoyant force decreases faster.
     
  17. Apr 25, 2013 #16

    rl.bhat

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    Since we are doing integration to find the total work done, the rate of change of depth does not matter. At any instant the buoyant force depends on the depth of the submersed portion of the hemisphere from the water surface.
     
  18. Apr 25, 2013 #17

    haruspex

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    Yes, but having raised the hemisphere a distance x, how much is still submerged? If the water level is constant it is 20cm-x, but if the tank has a finite extent the water level will have dropped and the submerged height will be less, reducing the buoyancy. As against that, it will not be necessary to raise the hemisphere 20cm to get it clear of the water. So it is not clear whether the work done will be more, the same or less. From a bit of algebra, looks to me like a falling water level requires more work to be done if, and only if, the density of the hemisphere exceeds half that of the water.
     
  19. Apr 25, 2013 #18

    rl.bhat

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    Yes. You are right.
     
  20. Apr 27, 2013 #19
    I've resolved thus. I've using a concept called "energy potencial of pression" wich is equal to pression of center gravity of body multiply for volumen of the body.
    Could you verify my calculation please.
    solved_problem.jpg
     
  21. Apr 27, 2013 #20

    haruspex

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    I get the same answer by what is probably an equivalent method. Since the water level is constant, the water that flows in to replace the hemisphere effectively comes from the surface of the reservoir. The average distance it falls in the process is 3R/8.
     
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