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Archimedes principle problem

  1. Dec 24, 2014 #1
    1. The problem statement, all variables and given/known data
    A small boat weighing 1000 N has a surface area of 3 m^2. It floats only 5 cm above the water level when in a fresh-water lake. How high out of the water will it ride in a salt-water lake? Assume the surface area of the boat does not change as it rises (salt water has a density of about 1.03 x 10^3 kg/m^3).

    2. Relevant equations
    B = pVg

    3. The attempt at a solution
    I can use the above equation to solve for the volume of the water displaced which will be the same as the volume of the boat. Unfortunately I am not sure were to go from there. I am not sure how to find the relationship between the height the boat rises in the water and it's volume. I also am not sure what to do with the information it gave me on the boats surface area.
  2. jcsd
  3. Dec 24, 2014 #2


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    The "surface area" of the boat should be understood as the area enclosed by the boat's outline at the water line. If this area is 3 m^2 and the boat rises by 5 cm, by how much will the submerged volume of the boat be reduced?
  4. Dec 24, 2014 #3


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    I'm sure it would be much clearer if they'd said the boat is in the shape of an upright rectangular box!
  5. Dec 24, 2014 #4
    So it's not the surface area of the entire boat just the part of the boat under the water?
  6. Dec 24, 2014 #5
    What is the displaced volume of salt water required to support the weight of the boat? What is the displaced volume of fresh water required to support the weight of the boat? What is the difference in displaced volume between salt water and fresh water?

  7. Dec 24, 2014 #6
    Volume of water displaced for fresh water = .102 m^3
    volume of water displaced for salt water = .0991 m^3
    difference in water displaced = .0029 m^3
  8. Dec 25, 2014 #7
    If the boat cross sectional area at the water line is 3 m^2, for the volume difference you calculated, how much higher will the boat float in salt water?

  9. Jan 5, 2015 #8
    I'm not sure that's the part I'm having trouble with
  10. Jan 5, 2015 #9


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    But it's the only step you have left to do.
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