# Buoyant Force, How's my logic?

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1. Nov 6, 2014

### hanag

1. The problem statement, all variables and given/known data
A recreational (open) hot air balloon (i.e., Pinside is approximately Poutside) has a volume of 2107 m3 when fully inflated. The total weight of the balloon, basket, ballast and pilot is 1832.6 N (412 lbs). By how much must the density of the air in the balloon be smaller than that of the surrounding atmosphere in order to keep the balloon floating level near the ground?
2. Relevant equations
Buoyant force= volume x g x rho

3. The attempt at a solution
I got the right solution, but it was explained much differently in class (much more complicated), and I didn't even use all the values given in the problem. I'm not sure if my logic is correct.

I figured that the buoyant force must be equal to gravity because the balloon is not accelerating.
v x rho x g = m x g

2107 m^3 x rho x g= 1832.6 N
2107 m^3 x rho = 187 kg
difference in rho= 0.0888 kg/m^3

Does what I did make any sense at all? Thanks in advance!

2. Nov 6, 2014

### nasu

The logic is a little loose even though the answer is probably right.
You label the density by rho in the buoyant force. And then you change the meaning to a density difference. This is not good practice and is missing some steps.

3. Nov 6, 2014

### hanag

So how would I go about doing this the right way? I don't even know where to start.

4. Nov 8, 2014

### haruspex

Put in an unknown density for the cold outside air, $\rho_c$, and write the thing you want to find, the difference, as $\Delta \rho$. In terms of those write the density of the hot air. Then write an expression for the buoyant force in terms of those, and observe the cancellation of the unknown $\rho_c$.