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Simple Problem about Air Pumped into a Balloon

  1. Dec 7, 2009 #1
    In the problem I am working a shipping crate is underneath the water, the dimensions of the crate are given. They tell you that when a balloon is inflated to a radius of 1.3m the crate starts to rise. What is the mass of the crate?

    FB = Wfluid
    P= F/A

    I calculated the buoyant force acting on the crate, and now I am trying to find the force generated in the balloon. To do this can I use the atmospheric pressure of 1.01 X 10^5, find the area of a sphere and then calculate force? Or does the pressure change even though the balloon is expanding?

    Last edited: Dec 7, 2009
  2. jcsd
  3. Dec 7, 2009 #2
    I think you may be making this harder than necessary. I would simply assume that you have achieved neutral buoyancy when the balloon is inflated. The volume of the balloon displaces enough water to equal the weight of the crate.
  4. Dec 7, 2009 #3
    ok so I calculated the mass of the water which would have been displaced... where can I go from there though? Where does the buoyant force that the water is exerting on the crate come into play?
  5. Dec 7, 2009 #4
    for the crate to rise: Buoyant force>=mass of balloon + mass of crate.

    The Mass of balloon negligible (assumption). I think you are done. The buoyant force just has to be equal to the mass of the crate. Naturally, as the balloon raises it will expand accelerating the rate of rise to the surface, but since water is incompressible, the mass at depth is the same as if the two were barely above water.
  6. Dec 7, 2009 #5
    Alright... so I just found the FB on the crate, as well as the FB on the balloon. Now I divided that by 9.8 and got a mass. (since mg = pvg and pvg is the sum of the buoyant forces)
    Does that sound right?
  7. Dec 7, 2009 #6
    Never mind :)
    I was feeling daring and imputed my answer, it was right

    Thanks for all your help!
  8. Dec 7, 2009 #7
    you bet, sometimes problems turn out to be a lot simpler than we dare imagine!
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