# Archimede's Principle-Hot air balloon floating

1. Homework Statement
A hot-air balloon plus cargo has a mass of 1890kg and a volume of 11430m^3. The balloon is floating at a constant height of 6.25m above ground. What is the density of hot air in the balllon?

2. Homework Equations :
The force of buoyancy is equal to F=(p_f)(g)(V)
where p_f is equal to the density of the fluid displaced by the object. The force of buoyancy is equal to the weight of the air displaced.

3. The Attempt at a Solution :
Im really stuck. I recognize that for the balloon to be floating at a constant height, the buoyancy force must equal the weight of the balloon.
That is,
(p_f)V=m (gravity cancels on both sides)
Solving this for the density of the air displaced, i get 0.165kg/m^3. Unless if this is the answer and i am really overthinking it, im not sure what aspect of this situation im missing.
I thought that the density found by equating the buoyancy force to the weight is the density of the air outside the balloon.

How does the height of the balloon factor into the equation/situation?
I thought about using the equation P_2=P_1+pgh
but i dont think the atmospheric pressure would vary at a height of 6.25 m.

Im confused.

## Answers and Replies

mgb_phys
Homework Helper
The weight of the ballon PLUS the hot air inside it must equal the weight of the air it is displacing.

The height (at least that close to the ground) doesn't have any effect, it might be there because there was another part to the question.

I dont think i understand.
I assumed the mass given was the mass of the balloon and cargo fully inflated? perhaps that was incorrect.

If i understood your hint correctly, i changed my equation to read:
(p_f)(g)(V)=(m_B+m_a)(g)
where m_B is the mass of the balloon and m_a is the mass of the air.
In this situation, i used a commonly accepted value for the density of air (1.29kg/m^3) and found the mass of hot air to be 12854.7, which gives a density of 1.12kg/m^3.

Is this what you meant

mgb_phys