# Bouyant force accelerates released balloon

1. Mar 30, 2013

### johns123

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. Mar 30, 2013

### Staff: Mentor

Don't forget the weight of the balloon (well, the helium inside) itself.

3. Mar 30, 2013

### johns123

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 :-)

4. Mar 31, 2013

### johns123

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.

5. Mar 31, 2013

### Staff: Mentor

For something filled with gas, under water, you can usually neglect the gravitational force on the gas, right.

6. Mar 31, 2013

### johns123

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.