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I have an assignment about the movie Up. Here is the question:

Critique the brief floating scene. Did the animators use enough balloons during this scene? What important variables, (which affect gases) did the animators seemingly fail to consider during the floating scenes as Carl's house is slowly rising after initial takeoff? What important variables, (which affect gases) did the animators fail to consider during the floating scenes as Carl traveled from the US to Venezuela? Explain why these factors should be considered.

For reference, we are supposed to assume that Carl's house was 125,000 pounds. Could someone please help me with this question? Thanks!

I'm thinking for the important variables about pressure, volume, and number of moles.

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closed as too localized by Chris White, Waffle's Crazy Peanut, David Z May 7 '13 at 5:58

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Hi Someone - this is a site for conceptual questions about physics, not general homework help. If you can edit your question to ask about the specific physics concept that is giving you trouble, I'll be happy to reopen it. See our FAQ and homework policy for more information. –  David Z May 7 '13 at 5:59
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1 Answer 1

You need to roughly estimate the volume of one balloon first. I would say it's about 4 liters, (that's how many bottles of milk I think I could fit in there) so 4 $dm^3$ which ammounts to $ 4\cdot 10^{-3} m^3$

The buoyancy force upward extered on one balloon is

$F_b = \rho_{air}V_{balloon}g$

where $\rho$ is the density of displaced liquid (here air) and $V$ is the volume of liquid the balloon displaces - so its own volume. I assume the balloon is inflated with helium. So, the gravitational pull downward on the balloon is

$F_g = \rho_{helium}V_{Balloon}g$

The net force pushing up one balloon is

$F_{up} = (\rho_{air}-\rho_{helium})V_{balloon}g$

Now the densities are repsectively $1.18 \frac{kg}{m^{3}}$ and $0.18 \frac{kg}{m^{3}}$ for air and helium. So

$F_{up} \simeq 1\cdot V_{balloon}g = 3.9 \cdot 10^{-2} N$

This net force up is the force the balloon actually pulls on the house with. The weight of the house is

$P = M_{house}g = 56250*9.81 N = 5.52\cdot 10^{5} N$

Where I have translated 125'000 pounds into kilograms. You would then roughly need 10 million balloons to lift up the house. You can easily replace the volume of the balloon with what you see fit. However, I think the number of balloons is not enough :)

For what is being neglected : air density changes as a function of height. Moreover, temperature changes will affect the volume of each balloon due to perfect gas law. The pressure of the air is also something that changes with height.

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Oh I'm so sorry, the balloon is supposed to be 3 feet in diameter, which would make its volume about 14.1 ft^3 –  Someone May 5 '13 at 15:45
    
I don't like to calculate with feet. 3 feet is about 0.92 meters. So the balloon volume is (spherical) $0.4 m^3$ which is a hundred times greater than what I had assumed. Since the force is linear in the volume, this means in the end you should need 100 times less baloons than I had calculated, so about 100 thousand. It's still a lot. –  Mathusalem May 5 '13 at 16:20
    
Yup. But what about the other parts of the question, such as the variables? –  Someone May 5 '13 at 16:24
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@Someone: It's better if you put some thought into it. This site is to help people learn from each other, and that doesn't work if your brain doesn't kick in. –  Mike Dunlavey May 5 '13 at 19:39
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