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Hot air balloon buoyancy homework

  1. Dec 2, 2004 #1
    A hot air balloon plus cargo has a mass of 291 kg and a volume of 809 m3. The balloon is floating at a constant height of 6.25 m above the ground. The density of the outside air may be assumed to be 1.29 kg/m3. What is the density of the hot air in the balloon?
    can someone help me with this problem? i know that i need to use the buoyant force but i'm not sure how to relate the height with the density
     
  2. jcsd
  3. Dec 2, 2004 #2

    Tide

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    I think the height is mentioned only to indicate the balloon is not tethered to the ground and the fact that the height is a constant means the balloon is not accelerating.
     
  4. Dec 2, 2004 #3
    oh ok. so how would you find the density? i tried 291/809 = 0.36kg/m^3 but that wasn't correct
     
  5. Dec 2, 2004 #4

    Tide

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    Equate the buoyant force to the weight of the balloon, cargo and the air (whose density you do not know) contained in that volume. The buoyant force is the weight of the displaced air and whose density you do know.
     
  6. Dec 2, 2004 #5
    so Fb = 1atm + (d)(9.8N/kg)(809m^3) = (1.29kg/m^3)(9.8N/kg)(809m^3) is that right?
     
  7. Dec 2, 2004 #6

    Tide

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    The buoyant force is

    [tex]F_b = 1.29 kg/m^3 \times 9.8 m/s^2 \times 809 m^3[/tex]

    and the weight of the hot air is

    [tex]W_{hot} = \rho_{hot} \times 9.8 m/s^2 \times 809 m^3[/tex]
     
  8. Dec 2, 2004 #7
    so the [tex]F_b = W_{hot} + W_{balloon}[/tex] ?

    [tex]W_{balloon} = (291kg)(9.8N/kg)[/tex]

    [tex]\rho_{hot} = 0.93kg/m^3[/tex]
     
    Last edited: Dec 2, 2004
  9. Dec 2, 2004 #8

    Tide

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    Way to go, Mike!!
     
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