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Bouyancy of Ice in Water

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Two Identical glasses are filled to the same level. One is filled with fresh water (density=1 gm/cm^3) and the other with salt water (density=1.025g/cm^3) Into each glass a cube of ice (density= 0.92 gm/cm^3) one cm on each side is placed. What is the height of the ice above water level in each glass?


    2. Relevant equations
    Maybe Archimedes' principal? We never really went over this in class


    3. The attempt at a solution
    I honestly have no idea... Do we subtract the densities to get the cm of displacement? Like the ice in freshwater is .08 becase 1-.92=.08? I am truly lost.
     
  2. jcsd
  3. May 6, 2009 #2

    LowlyPion

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    Welcome to PF.

    That's pretty much it ... for plain water.

    Now what about the salt water?
     
  4. May 6, 2009 #3
    Well if that were the case for the freshwater I would assume it to be the same for the salt water. so 1.025-.92=.105? It just seems to simple for the final exam I suppose. But if this is correct can you offer any explanation why we simply subtract the densities?
     
  5. May 6, 2009 #4

    LowlyPion

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    Salt water is a little different. Your first result was a consequence of the fact that fresh water was given as a density of unity.

    Buoyancy is the amount of weight displaced isn't it? (I'm letting gravity cancel out here.) So all the cube needs to displace is .92g . The mass that the cube could displace and be even would be 1.025 g. So the percentage of the cube that will be poking out will be .92/1.025 subtracted from 1 won't it?
     
  6. May 6, 2009 #5
    Well that does give me an answer on the exam but I don't have an answer key and I don't really understand why I suppose... Maybe I just don't really understand bouyancy
     
  7. May 6, 2009 #6

    LowlyPion

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  8. May 6, 2009 #7

    LowlyPion

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    Here is a lecture that covers it in a little more detail:


    https://www.youtube.com/watch?v=ngABxM7jl0Q
     
  9. May 7, 2009 #8
    Ok so that makes a bit more sense. But it is the Mass of the object-the apparent mass of the object submerged=the density of water*volume of the object... So how does this relate to my problem? are we solving for the mass of the object submerged? giving us mass (which in this case would be .92) divided by the density of the saltwater*volume of the object (which is one)?
     
  10. May 7, 2009 #9

    LowlyPion

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    You are solving for the displacement of the water necessary to hold up the object. The cube weighs .92 grams as ice. How much liquid water does it need to displace in order to provide .92 grams of buoyant force? Once that amount of water is displaced, then it is in equilibrium and it is floating.
     
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