The discussion revolves around determining the volume of helium required to lift a 340 kg balloon, given the densities of helium and air. Participants are exploring the principles of buoyancy and the relationship between the weight of the balloon, the weight of the displaced air, and the volume of helium needed.
There is ongoing exploration of various approaches to the problem, with some participants providing equations and others questioning the correctness of their calculations. While some have arrived at differing volume estimates, there is no explicit consensus on the correct answer, and participants continue to seek clarification on the underlying concepts.
Participants are navigating potential discrepancies between their calculations and textbook answers, leading to discussions about the assumptions made in their equations. There is also mention of confusion regarding the correct interpretation of buoyant force and the weights involved.
No, set the buoyant force equal to the weight of balloon + helium.pat666 said:so F_b=w_air+w_helium
No, the textbook is fine. For some reason you are setting the buoyant force equal to 340g. That's incorrect. (340g is the weight of the empty balloon.) Write an expression for the buoyant force in terms of the volume (what you call x).340g=1.28*9.81x+0.18*9.81x
that comes out to 232.877m^3 which is what someone previously said. maybe the answer in the textbook is wrong??
Why are you adding the weights of air and helium?pat666 said:w_air + w_helium = rho*g*v? if that's right what is the fluid on the right side of the equation?
It does! But the buoyant force must lift both the helium and the empty balloon.pat666 said:I thought that the buoyant force the helium had to provide would be equal to the force due to gravity so that it could float up?