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I have the following situation in mind:

A big airtight bag of arbitrary shape with a person standing on it. The bag gets inflated with air to lift the person. Assuming that the bag is much larger than the persons footprint, how do I find the minimal overpressure in the bag that I need to lift the person of the ground?

I was thinking of just dividing the normal force of the standing person by the footprint area, but I am not sure on that approach $$F_n = 80×9.81 = 784\text{ N}$$ $$P_n = \frac{784}{0.2×0.3} = 13066\text{ Pa}$$

I have the feeling that the bag dimensions play a role as well, as intuitively I would say that to do this, a small bag would work better than a big bag, but again I'm not sure...

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Hello Bart, this is not a home-work question. But, the tag is appropriate. Also, please make use of TeX while mentioning Formulas... –  Waffle's Crazy Peanut Oct 12 '12 at 14:49
    
This is indeed not a homework problem, but for a DIY project. Do you mean that I should add the homework tag nonetheless? I was not aware of the availability of TeX, I will update the formula asap. –  Bart Arondson Oct 12 '12 at 14:51
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I could see that... It's a good question. But, Please see the definition of homework tag... It applies to this question also. BTW, Don't worry. I've suggested an edit to your question :-) –  Waffle's Crazy Peanut Oct 12 '12 at 14:54

2 Answers 2

up vote 3 down vote accepted

It is just as simple as you suggest. At the moment my feet are exerting a pressure on the ground of my weight divided by whatever the area of my shoes is and the pressure exerted by the ground on me is what keeps me stationary. Exactly the same applies to your air bag. once the air pressure is the same as the pressure you exert on the bag it will support you.

But there are a couple of extra things to consider. When you stand on the bag you will compress the air in it and you'll sink until the air is compressed enough to match the pressure of your shoes. So the initial pressure can be lower than your shoe pressure and the bag can still keep you off the ground.

You mention the bag size, it's probably easier to compress the gas a lot in a small bag than in a large bag, so a small bag would probably work better. There's nothing especially fundamental about this; it's just that a large bag allows more room for the air to move into as your feet compress it.

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OK, thanks for the answer. I'm still curious about the part with the bag size. Would you know an analytic relation between bag volume and pressure? Say I have a bag with 10L volume, how much air should I put in it to achieve the necessary pressure? –  Bart Arondson Oct 12 '12 at 22:13
    
John, I think you are wrong, in any realistic bag made of an elastic or non-elastic material, the surface tension of the bag caused by your feet sinking into the bag is much more likely to lift you up way before the psi times the area of your feet will lift you. Effectively the surface tension will distribute your weight over a much larger surface area. The only way to avoid surface tension is if the bag is completely perfectly and infinitely elastic in which case the pressure of the bag would never exceed atmospheric pressure and you would never be lifted anyway. –  FrankH Oct 13 '12 at 2:57
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Hmm, good point, I assumed the surface tension in whatever the bag was made from was negligable. The surface tension will have some effect, though I doubt it's the major effect. After all it's not like standing on a trampoline where the surface is secured at the edges. The bag material is free to respond to stress by changing shape. Anyhow that's a good point and I'd need to sit down and think about how big the effect would be for the sort of bags used for industrial lifting. –  John Rennie Oct 13 '12 at 6:51
    
@BartArondson: just the ideal gas equation of state $PV = nRT$ where $n$ is the number of moles of gas and $R$ is the ideal gas constant. Note that the temperature $T$ has to be in degrees Kelvin. This will be pretty close for air at the sorts of pressures you're using. –  John Rennie Oct 13 '12 at 7:00
    
This answer is only partially right. @FrankH makes a valid point by involving the surface tension in this. That, the shear stress in the wall of the bag and the boiler equations are somehow involved, I think... –  Bart Arondson Oct 21 '12 at 10:49

According to this website it is not recommended for the object being lifted to have only a small footprint on the inflating bag:

enter image description here

You can see that with a small (not recommended) footprint, the surface tension of the bag (which apparently is quite stiff) will contribute to the lifting of the object (the "sling effect" mentioned in the image). These bags have multiple layers of rubber and are reinforced with either strong synthetic fibers (aramid) or steel cables.

Your calculation of 13066 Pa is correct, but it will really be the upper limit to the true amount of pressure needed to lift the person. Surface tension of the bag material will effectively increase the area that is supplying the lift and thus a lower pressure will be suffice. The exact pressure needed is impossible to calculate without detailed knowledge about the bag material, it's properties and the height you want to lift the person.

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