Archimede's Principle-Hot air balloon floating

AI Thread Summary
To determine the density of hot air in a balloon floating at a constant height, the buoyancy force must equal the weight of the balloon and its cargo. The initial calculation yielded a density of 0.165 kg/m^3, but confusion arose regarding the role of height and whether the mass included the air inside the balloon. By adjusting the equation to account for the mass of both the balloon and the air, a new density of 1.12 kg/m^3 was calculated using a standard air density value. The discussion emphasizes that the height does not significantly affect the calculations at this altitude. Understanding the relationship between the balloon's mass, the displaced air, and the densities involved is crucial for solving the problem accurately.
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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, I am not sure what aspect of this situation I am 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 don't think the atmospheric pressure would vary at a height of 6.25 m.

Im confused.
 
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The weight of the balloon 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 don't 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
 
Yes, I assumed that the mass of the balloon didn't include the air in it - mainly because your assumption gave the wrong answer ;-)

Another way to think of it is that, the volume of air * difference in density = empty mass of the ballon.
 
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