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Fluid displacement

  1. Apr 8, 2008 #1
    [SOLVED] Fluid displacement

    This has been bugging me. Any help will be much appreciated. Thanks.

    1. The problem statement, all variables and given/known data

    A 355 mL soda can is 6.2 cm in diameter and has a mass of 20 g. Such a soda can half full of water is floating upright in water. What length of the can is above the water level?

    The answer is 5.22 cm but I can't reach that answer.

    2. Relevant equations

    Bouyant force = density(given by rho) * g * volume displaced

    Density of water = 1000 kg/ cubic meter

    3. The attempt at a solution

    mass of water inside the can: 355/2 mL * 1kg/L = 0.1775 kg
    mass of water inside the can plus the can itself = 0.1975

    From the bouyant force equation, mg = rho * g * volume displaced
    m = rho * volume displaced
    volume displaced = m/rho = .1975 kg /(1000 kg/(M^3)) = 0.0001975 M^3

    The area of the can's top is (3.1 cm)^2 * pi = 30.19 cm^2 = 0.003019 M^2

    volume = area * x where x is the depth to which the can has sunk into the water. Which I reason should also be the height of the part sticking out of the water since the can is half full.

    0.0001975 M^3 = 0.003019 M^2 * x
    x = 0.0654 M = 6.54 cm
    Last edited: Apr 9, 2008
  2. jcsd
  3. Apr 9, 2008 #2
    you almost got it. you found the immersion depth. how high is the can?
  4. Apr 9, 2008 #3
    Thanks for letting me know I was on the right track. With that hint, I solved it.
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