Buoyancy and Archimedes' Principle (helium ballon carrying cargo), HELP

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SUMMARY

The discussion centers on calculating the maximum cargo weight a helium balloon can lift, given a radius of 7.35 meters and a balloon structure mass of 930 kg. The correct buoyant force equation is established as F(b) = (mass of helium + mass of cargo) × 9.8, with the density of air at 1.29 kg/m³. The volume of the balloon is calculated using the formula for the volume of a sphere, resulting in a volume of 1663.2 m³. The final cargo weight is determined to be 920 kg after accounting for the balloon's structure weight.

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confusedbyphysics
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I can not figure this out for the life of me! I know the answer is 920 kg, but I have to know how to get there so I can do my WebAssign problem!

Here's the problem:

" A spherical balloon has a radius of 7.35 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself."

So the buoyant force must = the weight of the helium plus the weight of the balloon

So F(b) = (mass of helium + mass of cargo) X 9.8

F(b) = Density of air X volume X 9.8

9.8 cancels out on both sides.

Density of air is 1.29 kg/m^3 so
1.29V = (mass of helium + mass of cargo)

mass of helium = density of helium (.179) X volume

so 1.29V - .179(V) = mass of cargo
V for a sphere = 4/3 (pi) (r)^3 = 1663.2

plug that in and I get 1847.8 for the mass, which is WRONG
strangely if I divide that by 2 I get the right answer though... 924, which rounded down would be 920, but I have no idea why id divide it by 2..

someone help please (i hope this makes some sense, I don't think I am making sense to myself, lol)
 
Physics news on Phys.org
You forgot to subtract off the 930 kg weight of the balloon skin and structure!
 
Oh crap, you're right

Wow I feel stupid. Thanks Halls of Ivy (I'm going to go hide now LOL)
 

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