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This is a conceptual question in a solution I am trying to understand.

Problem statement: I have a balloon with a volume of V $m^3$. The outside air temp is $K$ kelvin and mass to lift is $m$ kg.

I am to find the temperature inside the balloon to barely lift the given mass.

The formula used is $F_{b}=ρgV$

Solution:

I know that in order for the balloon to lift $F_b > F_g$ (subscript $b$ = buoyancy and subscript $g$ is gravity force)

Using $F_{b}=ρgV$ I tried taking the differences of the two air densities and setting it equal to the mass I need to lift. As seen here.

$(ρ_{air}-ρ_{balloon})Vg=mg$

Concern: Since $ρVg=F_{b}$ I understand this as there are THREE acting forces;

One inside the balloon and one outside the balloon(which is in the NEGATIVE y-direction(drag?)) and mg. I find this strange since the balloon is not moving.

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2 Answers

up vote 2 down vote accepted

The force balance

There are indeed 3 contributions to the nett force. You have gravity acting on the mass you want to lift so $F_{g,1}=mg$. However you are also lifting the air inside the balloon: $F_{g,2}=m_{balloon} g= \rho_{balloon}V g$. At the same time the mass of air is pushing down: $F_{g,3}=m_{air} g = \rho_{air}V g$.

So: $(\rho_{air}-\rho_{balloon})V g=m g$ which yields $$\rho_{balloon}=\rho_{air}-\frac{m}{V}$$

You can interpret the term with the difference in densities in two ways, you can view it as 1 force which becomes 0 for equal densities or you can view it as 2 forces which balance when densities are equal. The latter is the standard convention, but I think your confusing stems from the former interpretation.

The temperature inside your balloon

Your question doesn't completely stop there, because now you want to known to what temperature you should go. The simplest assumption to start with is to assume that air behaves as an ideal gas which means: $$ P M = \rho R T $$ where $P$ is the pressure, $M$ the molecular mass, $\rho$ the density, $R$ the gas constant and $T$ the temperature.

The gas in the balloon is heated, but it can still equilibrate pressure with the outside air because of the opening. Obviously the gas constant and the molecular mass don't change so we can find the density in the balloon as $$\rho_{balloon}= \frac{P M}{R T_{balloon}} $$

For the surrounding air we can do something similar: $$\rho_{air}= \frac{P M}{R T_{air}} $$

So our equation becomes $\frac{P M}{R T_{balloon}} =\frac{P M}{R T_{air}}-\frac{m}{V}$ which can be rewritten for the tempearture in the balloon to:

$$T_{balloon} =\left(\frac{1}{T_{air}}-\frac{mR}{PVM}\right)^{-1}$$

This equation nicely shows that an increase in mass to lift ($m$) will result in a higher temperature for the air in the balloon, but for example also that a higher pressure (nice weather) will result in a lower temperature necessary to lift the same mass on a bad weather day.

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These are the tree forces in your equation:

The gravitational force on the lifted weight: $mg$

The gravitational force on the balloon: $m_{balloon}g=\rho_{balloon}Vg$

The buooyancy force on the balloon: $\rho_{air}Vg$

For the balloon to lift, the sum of the first two must be equal or less than the third:

$$ mg+\rho_{balloon}Vg=\rho_{air}Vg $$

Which can be rewritten:

$$ mg=(\rho_{air}-\rho_{balloon})Vg $$

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