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Floating iceberg

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data
    I think it should be pretty simple, but my result and that of the book are different:
    How much water does an iceberg displace (Its emerged part is ##V_i=100m^3##)


    3. The attempt at a solution
    knowing the density of sea water is ##d_w=1.03*10^3 kg/m^3## and that of ice ##d_i=0.92*10^3 kg/m^3## i can calculate the volume of the submerged part of the iceberg:

    ##\frac{d_i}{d_w}=0.89%## meaning the total volume of the iceberg is ##100:11=x:100##
    x=909. The submerged part is then 909-100=809m^3. now, since the iceberg floats, its weight and archimedes' force should be equal, then:

    ##m \cdot g= d_w \cdot g \cdot V_w## and so:
    ##809 \cdot 92=103 \cdot v_w## ##\Rightarrow## ##V_w=723 m^3##
    but according to the book it should be 1050m^3
    what's wrong in my reasoning?
    thank you in advance
     
  2. jcsd
  3. Jun 5, 2013 #2

    SteamKing

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    IMO, the book's answer is incorrect.
    I don't follow your ratio method.

    What I did was to say the x = submerged volume of the iceberg. Then I wrote an equation setting the weight of the iceberg = to the buoyancy of the iceberg. Solve for x.
     
  4. Jun 5, 2013 #3
    I don't understand your method. I get two incognitas:

    Weight of the iceberg: mass multiplied g, where the mass is volume divided density and x is the submerged volume. Buoyancy= g multiplied the volume of displaced water (which i don't know) multiplied the density of the water

    ##\frac{x+100}{d_{ice}} \cdot g=V_w \cdot g \cdot d_{w}##
    but i don't know both x and V_w
     
  5. Jun 5, 2013 #4

    Borek

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    How is displaced water different from submerged volume?
     
  6. Jun 5, 2013 #5

    haruspex

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    Because it gets subtracted from 1, effectively, the rounding error in truncating it to 0.89 becomes significant. You need to use a couple more digits of precision.
    The method looks ok to here, but as Borek points out this should also be the volume of water displaced. I don't understand what you did from here.
    Fwiw, I get 836 cu m.
     
  7. Jun 5, 2013 #6

    Borek

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    And you are not alone :wink:
     
  8. Jun 5, 2013 #7
    now everything's clear :) thank you!
     
  9. Jun 5, 2013 #8

    SteamKing

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    Let x = submerged volume of iceberg

    Total volume of iceberg = (x + 100) cu. m.

    Mass of iceberg = 920 kg/m^3 * (x + 100) m^3

    Since the iceberg is floating, the mass of the iceberg = mass of the displaced water
    (this is Archimedes Principle),

    Therefore,
    mass of displaced water = x * 1030

    equating displacement of iceberg to mass of iceberg,

    1030*x = 920 * (x + 100)

    Solve for x
     
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