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If a soda can could contain a vacuum, could it float?

  1. Jul 30, 2011 #1
    Say there was a theoretical soda can made of a super strong material that could hold up to the pressure of the atmosphere while weighing the same as a regular soda can. If everything was sucked out of the can, would it float in Earth's atmosphere?

    I think physics says it would float, but tell me what you think. Also, how much of a buoyancy force would a very light object with a vacuum in it exert? How strong would the force be if we could vacuum out a balloon and keep it taking up the same volume?
     
  2. jcsd
  3. Jul 30, 2011 #2
    The buoyancy force is from the fluid surrounding the object, and not exerted from the object itself. Also, buoyancy is independent from the weight/mass of object, but instead depends on the volume of the object and the density of the fluid around it. So a bowling ball and a balloon (of same volume) has the same buoyancy.

    To find out if the can would float:
    First, find the density of an empty can. Next, find the density of the air around it. If the density of the empty can is less than the air around it, the can will float around in mid air.

    I highly doubt it would, though.
     
  4. Jul 30, 2011 #3

    AlephZero

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    One way to get a handle on this is simplify the problem and think about a spherical can. That would be the way to use the smallest amount of material to resist the air pressure anyway.

    If the can has radius [itex]r[/itex], thickness [itex]t[/itex], and density of the can and the air are [itex]\rho_c[/itex] and [itex]\rho_a[/itex], the mass of the can is approximately [itex]4 \pi r^2 t \rho_c[/itex] and the mass of the displaced air is approximately [itex] 4 \pi r^3 \rho_a/3 [/itex].

    So the can will float if
    [tex]4 \pi r^2 t \rho_c \le 4 \pi r^3 \rho_a/3[/tex]
    or
    [tex] \frac {\rho_c}{\rho_a} \le \frac {r}{3t}[/tex]
    For an aluminum can, that would mean that r/t would have to be greater than about 6500. That's a VERY thin can!
     
  5. Jul 30, 2011 #4
    Sorry, but I just didn't get why you equated the mass of the evacuated air in the sphere with the mass of the spherical shell. Can you please explain? :(
     
  6. Jul 30, 2011 #5
    I'm pretty sure the weight difference between an "empty" can and a can full of air is not that much different.
     
  7. Jul 30, 2011 #6

    SpectraCat

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    That's what determines the buoyant force on an object suspended in a continuous medium. If the mass of the object is less than the mass of the air displaced by it, then there will be an upward force on it. So, for an evacuated shell (assuming a perfect vacuum), the total mass of the shell has to be less than the mass of the equivalent volume of air under the same conditions.
     
  8. Jul 31, 2011 #7
    for the can not to implode,it would have to be made of some strong material, and that would probably increase the mass. thus, the density would increase and i doubt it's density would still be less than that of air, so i think it wouldn't :)

    at least, that's how i see it :)
     
  9. Jul 31, 2011 #8
    Thanks for the replies. I wasn't sure on how to go about doing the math for this - I didn't need an explanation of buoyancy.
     
  10. Jul 31, 2011 #9

    mheslep

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    Or large. A can w/ 5 mil Al walls would need r = ~33" to have pos. buoyancy. But then of course such a can would be crushed by the surrounding air.
     
  11. Jul 31, 2011 #10

    uart

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    Ok, lets ignore the problem of strength and just look at the numbers (weight versus buoyancy force) for a typical soda can.

    In Australia the volume of a typical (Aluminum) soda can is 375mL and the typical mass of the can is 14.7 grams.

    The gravitational force is (F = mg) is approx 144 mN directed downwards.
    The buoyancy force (F = rho g V) is approx 4.8 mN directed upwards.

    So clearly the answer is, no it would not float, even if it were strong enough to not be crushed by air pressure. The percentage weight reduction is if fact only about 3%.

    Where g=9.8 m/s^2, V = 0.375/1000 m^3 and rho (density of air) is approximately 1.3 kg/m^3

    Ref.
    - http://www.aluminium-cans.com.au/Facts.html [Broken]
    - http://en.wikipedia.org/wiki/Beverage_can
     
    Last edited by a moderator: May 5, 2017
  12. Jul 31, 2011 #11
    Your understanding of buoyancy was flawed, hence your uncertainty of approaching the question. We explained it to you so that you would know what is going on.
     
  13. Aug 1, 2011 #12
    Blood, I understand buoyancy fine with the exception of what you pointed out (that objects don't exert the buoyancy force). If I was wondering why things float I would have made a thread with a question relating to that.

    To show off your undergrad knowledge, please try answering the question at hand instead of correcting a very simple mistake in wording. I'm not asking for a review of 7th grade physical science on how things float.
     
  14. Aug 1, 2011 #13

    ZapperZ

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    Then what EXACTLY are you asking? Please re-read the original post that you made, and tell me that your question hasn't been answered. You asked if your can would float in ATMOSPHERE! Yet, you dismissed the physics on what makes things float! What gives?

    A few members here have spent time and effort in addressing your question. Instead of trying to learn from the responses, you hurled insults in return. Either you clarify what exactly that you are looking for, and show a bit of appreciation to the help that you've been getting, or this thread is done.

    Zz.
     
  15. Aug 1, 2011 #14
    A soda can has a volume of 355 ml. Air has a density of 1.39 kg/m3. The mass of the displaced air is 1.39 kg/m3 x 3.55 x 10-5 m3 = 4.93 x 10-5 kg = 49.3 mg. The empty soda can weighs more than this, so no.
     
  16. Aug 1, 2011 #15

    uart

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    You slipped a decimal point there dickfore.

    1.39 kg/m3 x 3.55 x 10-4 m3 = 4.93 x 10-4 kg = 0.49 grams.

    BTW. That agrees with my answer in post #10 above, (0.49 grams corresponds to approx 4.8 mN force). It represents about about 3% of the cans weight.
     
  17. Aug 3, 2011 #16
    What I said was only directed at one person (Bloodthunder) who decided to ignore the question and tell me how buoyancy works, saying that my knowledge is "flawed." I meant to praise everyone else who took the time to answer the question that I asked. I'll make sure to directly quote the person to whom I'm speaking next time. Sorry if anybody felt that what I said was directed at you.
     
  18. Aug 3, 2011 #17
    And... how did I ignore the question? I told you one of the ways of calculating it, as compared to others who helped you calculate as well.

    Edit: Ah forget it. The question has already been answered anyway. Don't care anymore.
     
    Last edited: Aug 3, 2011
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