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Basic Boyancy problem

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 73.4 kg person to be able to stand on it without getting his or her feet wet?

    (use 920kg/m^3 as density of ice and 1000 kg/m^3 for the density of freshwater)

    2. Relevant equations
    Archimedes Principle:
    Fg = Fb
    mg = pVg

    3. The attempt at a solution
    So my problem was I was really sick and I missed my couple first lectures on this topic. I've learned what I can from my textbook... but i think i'm missing a step or something in this problem.

    This is what I figured I should do:
    Since the buoyant force is equal to amount of liquid displaced, i plugged in the mass of the person and density of freshwater into Archimedes equation to solve for volume of water displaced.
    Vwater displaced = (73.4 kg) / (1000 kgm-3)
    Vwater displaced = 0.0734m3

    Than from here, I really haven't worked with this subject for long enough to understand what i can and cannot do. My next attempt would be to solve for the volume of ice by doing:
    pwaterVwater displaced = piceVice
    Vice = pwaterVwater displaced / pice
    Vice = (1000kgm-3)(0.0734m3) / (920kgm-3)
    Vice = 0.0798 m3

    Is this the correct answer? if not what should i have done?
    Last edited: Jan 29, 2009
  2. jcsd
  3. Jan 29, 2009 #2


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    Gold Member

    You've just worked out what volume 73.4 kilos of ice would be. So you know when the man steps on the ice and the ice sinks into the water due to his weight, it must displace0.0734m3 of water to support his weight without him getting his feet wet. So what you need to do is find what volume of ice has a volume of 0.0734m3 sticking out of the water when its floating there without the man.
  4. Jan 29, 2009 #3
    Alright... well I think I'm correct in saying something floats when its weight is equal to the weight of the displaced fluid (in this case fresh water). So I've figured out that the volume of ice floating above the water is 0.0734m3, than the total volume of the block of ice can equal 0.0734m3+ x, where x is the volume of the ice under water, and therefor also the volume of water displaced with just the ice floating.

    So with just the floating ice:

    piceVice = pwaterVwater
    (920 kgm3)(0.0734m3 + x) = (1000 kgm3)x
    (0.0734)(920)+(0.0734)(x) = 1000x
    67.528 = 1000x - 920x
    67.528 = x (1000-920)
    x = 0.8441m3 --> volume of ice under water

    Plug this value back into the Volume of ice, which was 0.0734m3 + x, therefore the total Volume of ice for a 73.4 kg man to stand on without getting his feet wet would be 0.9175m3 ?
  5. Jan 29, 2009 #4


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    Very good! :smile:
  6. Jan 29, 2009 #5
    Killer, thanks.
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