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Archimedes's principle of float

  1. May 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A swimming pool with 5 M length and 4 M width.A big layer of ice floats on the surface of the pool and above the layer of ice a rock with 40 KG mass and 5 R.D.,If the ice is melted and the rock fall, calculate the change in the height of water in the pool

    final answer (1.6 * 10^-3 m)

    2. Relevant equations
    F_b (upthrust force) = ρ_water * volume_immersed * gravity


    3. The attempt at a solution
    i tried by getting the volume of the displaced liquid as it equals the upthrust force and then divide it by the area to get the height of the displaced liquid but the answer is wrong. i think we should include the ice as it melted but it didn't give any info. about it though
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 2, 2014 #2
    This problem has nothing to do with Archimedes' Principle.

    But you are right the thickness of ice is assumed to be negligible.
     
  4. May 2, 2014 #3

    SammyS

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    Hello justaus3r. Welcome to PF !

    What is R.D. ?

    Relative Density ?


    Does the height of the water increase, or does it decrease ?
     
  5. May 2, 2014 #4

    SammyS

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    This problem has everything to do with Archimedes' Principle .
     
  6. May 2, 2014 #5
    Your calculation is almost correct. It correctly computes the level rise due to the rock on top of the ice. When the ice melts, the rock doesn't vanish in thin air. It goes to the bottom where it still affects the water level, but the amount of water level rise is different than before. Compute the difference.
     
  7. May 2, 2014 #6
    Yes, I see you are right now.
     
    Last edited: May 2, 2014
  8. May 2, 2014 #7
    yes R.D. is relative density = Density of material / density of water

    it increases
     
  9. May 2, 2014 #8
    archimedes' principle includes cases where opposing force(s) has higher value than upthrust force which leads to the sinking of the body
     
  10. May 2, 2014 #9
    thanks in advance for helping but unfortunatly i don't have the height of water in the pool
     
  11. May 2, 2014 #10

    SammyS

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    You don't need the height of water in the pool. you just need to find out how much it changes.
     
  12. May 2, 2014 #11
    Ok, as for the ice part should i include it or i can't bec. i don't have its area ?
     
  13. May 2, 2014 #12

    SammyS

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    Consider an ice cube floating in a glass of water.

    How much water does it displace?


    When it melts, what is the weight of the (liquid) water that the melted ice becomes?
     
  14. May 2, 2014 #13
    it displaces amount of water equals to the upthrust force acting on ice from surface of water

    I think it remains the same (am not sure though )
     
  15. May 2, 2014 #14

    SammyS

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    The ice cube displaces an amount of water equal to the weight of the ice cube.

    When the ice cube melts the weight of the water it yields is equal to the weight of the water it used to displace.
     
  16. May 2, 2014 #15
    their masses are equal but their volumes aren't equal, right ?
     
  17. May 2, 2014 #16
    Area = 20 m^(2)
    Before melting
    I think i solved it but correct me if i am wrong
    if the height increased due to floatation of that ice layer with the body equals 2 * 10 ^(-3) m
    since the body is floating therefore the M_body*g = ρ_water * vol._immersed * g
    vol_immersed = M_body / ρ_water = 40 / 1000 = 4 * 10 ^(-2) then V = A * h, h = V / A
    equals 4 * 10 ^ -2 / 20, therefore h = 2 * 10 ^(-3) m

    since the ice had melted and the body fall
    weight of water displaced = upthrust force
    M * g = volume of body * ρ_water *g
    M * g = 8 * 10 ^(-3) * 1000 * g
    M = 8 Kg
    volume of water = 8 / 1000 = 8 * 10(^-3) m^3
    height = 8 * 10 ^(-3) / 20 = 4 * 10 ^(-4) m


    change in height = 2 * 10^(-3) - 4 * 10^(-4) = 1.6 * 10 ^(-3) m
     
  18. May 2, 2014 #17

    ehild

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    Is the final height lower or higher then the initial height?

    The change of height is the final height - initial height. You subtracted the final height from the initial one, so got wrong sign for the change. Otherwise, it is correct.

    ehild
     
  19. May 3, 2014 #18
    I supposed that the change in height = height_big - height_small
     
    Last edited: May 3, 2014
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