1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bouyant force accelerates released balloon

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A small balloon filled with helium is released. What is it's initial upward acceleration? The only data given is ρ air = 1.29 kg/m^3 and ρ helium is .18 kg/m^3

    I assume the bouyant force ( FB ) = the weight of the air displaced by the balloon

    so BF = W(air) = ρ(air) * g * VolB = MassB * Accel

    and VolB = MassB / ρ(helium)


    2. Relevant equations

    so Accel = [ ρ(air) / ρ(he) ] * g = ( 1.29 / .18 ) * 9.81 = 70.3 m/s^2



    3. The attempt at a solution

    the answer in the book is 61 m/s^2, and I found that ρ(air) and ρ(he) change with temperature, but I have not been able to get the answer in the book. At ρ(air) 20degrees C, I get Acc = 65.4 m/s^2 which is still too high.
     
  2. jcsd
  3. Mar 30, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Don't forget the weight of the balloon (well, the helium inside) itself.
     
  4. Mar 30, 2013 #3
    Maybe something like BF = W(air) - W(he) = g * VolB( p(air) - p(he) ) = MassB * Acc

    then Acc = [ p(air) - p(he) / p(he) ] * g = [ (1.29 - .18) / .18 ] * 9.81 = 60.5 !!!!!!!!!!!!!!!

    You sir, are a freakin' genius :-)
     
  5. Mar 31, 2013 #4
    Finally, I see the concept: It's 2 problems in one. Net Force vertical .. and .. bouyant force vertical.

    Net Force = BF(air) - W(he) = W(air) - W(he) = p(air)*g*VolB - p(he)*g*VolB = MassB*Acc = p(he)*VolB*Acc

    then Acc = g * [ p(air) - p(he) ] / p(he) = 60.5 m/s^2 which is the correct answer.

    I wonder if this is also true of an object under water ? And we just ignore the gravitational attraction since it is only a very small part of the answer ??? Like I've been doing.
     
  6. Mar 31, 2013 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    For something filled with gas, under water, you can usually neglect the gravitational force on the gas, right.
     
  7. Mar 31, 2013 #6
    Yes. But when I needed to not ignore it, I didn't have a clue. It still seems a bit strange, but I finally realized that the gravitational force on the object was independent of it's volume and had to be considered separately. I might see this problem again with a heavy submarine .. and I'll be watching for it. Thanks MFB.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Bouyant force accelerates released balloon
  1. Bouyant Force (Replies: 5)

  2. Bouyant force (Replies: 3)

  3. Bouyant force help (Replies: 8)

  4. Bouyant Force Problem (Replies: 3)

Loading...