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Bouyant force and apparent weight

  1. Jan 12, 2009 #1
    1. The problem statement, all variables and given/known data

    This question is continued from a problem that I have already solved. This is what I know from the previous problem that will be helpful:

    r= 0.20 m
    roh of solid sphere= 850 kg/m^3
    mass sphere= 3.94 x 10 ^-5 kg

    I also have 2 apparent weights that I solved for when the tension in the system is
    a. 150 N ----> Normal force (apparent weight) Fn = -1.49 x 10^2
    b. 0 N -----> Fn = -3.86 x 10^-4 N

    (Hopefully those are correct)

    Now- the problem I came for help on:

    A room is partially filled with water with a sphere in it and the sphere is on a scale. What percentage of the sphere must be immersed in the water in order to achieve the same apparent weights as above (a and b).
    And- What percentage of the sphere must be immersed to achieve an apparent weight of zero?

    2. Relevant equations

    % immersed = V immersed/ V total = roh obj/roh fluid
    Bouyant force (Fb) = roh x V x g
    Fn = mg + ma (a = O)

    And from a full body diagram I have this:

    Fn + Fb - mg = ma (whch is zero)

    3. The attempt at a solution

    Well....I really am not sure what to do. One fact that I know and I think will help me is that if an object is say 90% underwater, then the roh of the object is 90% of the fluid....

    I need a kick in the right direction please.
  2. jcsd
  3. Jan 12, 2009 #2


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    Homework Helper

    Start with this, the 3 forces acting on the sphere. Fill in the detailed formula for each force.
  4. Jan 13, 2009 #3
    Ok- here's what I got, but Im not sure what this implies...

    mg + roh x V x g -mg = 0

    mg cancels and I end up with roh x V x g = 0
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