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Fluid dynamics - hydorstatic

  1. Dec 14, 2008 #1
    Hi everyone,

    I hope you can help me out with this question as my exam is tomorow.

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

    A 355 mL soda can with diameter=6.2 cm has a mass of 20g. If it is half full with water, and it is floating upright on water, what length of the can is abover the water level?

    2. Relevant equations
    [tex]\rho[/tex] = m/V

    F of buoyancy = [tex]\rho[/tex] of fluid*V of fluid displaced*g = [tex]\rho[/tex] of object*V of object*g

    Since V= Area*height, and the area is the same, then the Area cancel out and we are left with:
    [tex]\rho[/tex] of fluid*h of fluid displaced = [tex]\rho[/tex] of object*h of object

    Isolating for height of fluid displaced

    height of fluid displaced = [tex]\rho[/tex] of object*h of object/[tex]\rho[/tex] of fluid

    [tex]\rho[/tex] is density

    so height above water level = (H of object) - (H of fluid displaced)

    3. The attempt at a solution

    m of object = 0.02 kg + (355 mL/2)(1 g/1 mL)
    = 177.52 g

    density of object = m of object/ volume
    = 0.17752 kg / 0.000355 m^3
    = 500 kg/m^3

    h =V / ([tex]\pi[/tex] * radius^2)
    = 11.76 cm
    = 0.1176 m

    Now, subbing into the buoyancy eqation:

    height of fluid displaced = ([tex]\rho[/tex] of object) (h of object) / ([tex]\rho[/tex] of fluid)
    = (50)(0.1176 )/1000
    = 0.0588 m

    height above water level = (H of object) - (H of fluid displaced)
    = 0.1176 -0.0588
    = 5.88 cm

    The answer is 5.2 cm.
  2. jcsd
  3. Dec 14, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Recheck your arithmetic.
  4. Dec 14, 2008 #3

    God...Thank you sooo much. It took me a while to notice it actually. :)
  5. Dec 15, 2008 #4
    i got a similar problem with fluid dynamics , i got a test later in the week havent been prepped for it at all just been handed some sheets wondering if any one could help with an example i have got , as it will help me see what i need to do for future reference
    thanks in advance


    A cube of 0.25m length sides which has a weight of 50N is immersed in a tank of fluid .

    If the specific gravity of teh fluid is 0.8, calculate
    a) the density of the cube material
    b) the density of the fluid
    c) the resultant up-thrust of the cube


    50/0.8 = m
    m = 62.5 kg

    p=4000 kg/m3

    Thats all i can manage i havent really been taught this yet so any help would be appreciated
  6. Dec 16, 2008 #5

    I'll try to help with what I know.

    It might help you to know this:

    specific gravity of a fluid = density of fluid/density of water.
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