I'm having a discussion at the moment regarding the mass of $1\,{\rm kg}$ of feathers and $1\,{\rm kg}$ of steel.

The person I'm arguing with states that $1\,{\rm kg}$ of feathers will be lighter when weighed, compared to the $1\,{\rm kg}$ of steel, because the feathers are more buoyant.

She has done her calculations for the density of feathers and works out that $1\,{\rm kg}$ of feathers will displace $400\,{\rm L}$ of water. And because the feathers are more buoyant than the steel they would actually weigh less.

We are talking about the kilogram as a unit of mass, not weight.

The argument goes that if a scale is balanced perfectly in vacuum, it is not in air. The feathers, being more buoyant in air, would cause the scale to tip toward the steel.

I’m sure she is wrong, and even though buoyancy may be a factor, she calculates that the feathers would only have about half the weight in air that they do in vacuum.

So the question is, if equal masses of feathers and steel were balanced in a vacuum, would they still balance in $1\,{\rm atm}$? If not, what would be the difference in weight?

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    $\begingroup$ Do the test in water $\endgroup$ – user6760 Dec 20 '18 at 1:21
  • $\begingroup$ Then what does kilogram mean? $\endgroup$ – Mawg Dec 20 '18 at 7:13
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    $\begingroup$ You may want to distinguish weight (the gravitational force on an object) from apparent weight (the gravitational force minus the buoyant force). Weight is directly proportional to mass, but apparent weight depends on density and the properties of the surrounding fluid. $\endgroup$ – J. Murray Dec 20 '18 at 7:22
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    $\begingroup$ As @J.Murray mentions, the entire premise (or the wording) here is actually wrong. For 1 kg steel and 1 kg feathers, both the masses $m$ will be equal and the weights $w=mg$ will be equal, since $$m_1=m_2\quad \text{and}\quad w_1=m_1g=m_2g=w_2$$ But as is clear from the context, you don't really mean physical weight, you rather mean total force (or apparent weight, if you will, as J.Murray suggests). With a tipping scale you namely get a comparison of the total force on the objects and not a comparison of their weights. $\endgroup$ – Steeven Dec 20 '18 at 8:35
  • $\begingroup$ @Steeven. To be fair, the meaning of the word weight is just not standardized. There are plenty of people who use weight to mean force due to gravity, and there are plenty of people who use weight to mean what is read by a force scale. $\endgroup$ – march Dec 20 '18 at 21:11

Feathers are made from keratin, with a density of about 1.3 grams/cc. The net volume displaced by a kilogram of feathers is then 751 cc's. Steel has a density of 7.86 g/cc and a kilogram of it displaces 127 cc's.

Sea level air has a density of 0.0012 g/cc, so the buoyant force on 751 cc's of keratin is then 0.92 grams and the buoyant force on 127 cc's of steel is 0.155 grams.

This means that if we weigh the keratin in a vacuum, it will weigh 1 kilogram but in air it will weigh 999.08 grams. If we weigh the steel in a vacuum, it will weigh 1 kilogram but in air it will weigh 999.85 grams.

If we place the two bodies- keratin and steel, one kg of each- on a pivoting balance equidistant from the pivot in a vacuum, they will be in balance, but in air the keratin will be lighter by (999.85-999.08) grams or 0.77 grams.

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    $\begingroup$ A heap of feathers has smaller density than keratin: there are lots of voids at least between individual feathers. $\endgroup$ – Ruslan Dec 20 '18 at 6:48
  • $\begingroup$ Thank you sir, this is exactly the reply I expected and had hoped to get. $\endgroup$ – James Thorpe Dec 20 '18 at 7:11
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    $\begingroup$ @Ruslan In the atmosphere, those voids would be filled with neutrally-buoyant air. $\endgroup$ – Glurth Dec 20 '18 at 7:14
  • $\begingroup$ Though I understand it, I don’t like using kg as a unit for weight like you did in paragraph 3. $\endgroup$ – Dave Dec 20 '18 at 16:49
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    $\begingroup$ @dave, I was wondering when someone would call me out for that. I did it because I'm a recovering ex-engineer who has just embarked on the 12-step program. Step 4 is to learn the distinction between mass and weight and I'm not there yet. $\endgroup$ – niels nielsen Dec 20 '18 at 16:51

You can argue that her feathers contain a lot of air (never perfectly compressed) so that the volume is not true. You could also take your kilo of steel and make it into a hollow sphere (empty) of the same volume as the feathers (whatever that is) and then it would actually weigh less because you have no air inside. But you could be nice and let her make the point about a bigger volume being more bouyant, she is correct about this but ..... even volume is tricky.

  • $\begingroup$ The argument is not that buoyancy makes no difference, it is the amount of difference it makes that is in dispute. She is arguing that the buoyancy of the feathers is related to the density of feathers in the bag including the air within. This I do not agree with. She calculates that it’s buoyancy would make it weigh almost half of what it would weigh in a vacuum. I do not agree with either of these figures. $\endgroup$ – James Thorpe Dec 20 '18 at 3:03
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    $\begingroup$ She also calculates that a mass of one kilogram of feathers would displace 400 litres of water if immersed. I do not agree with that figure either. $\endgroup$ – James Thorpe Dec 20 '18 at 3:12
  • $\begingroup$ Yes good points. $\endgroup$ – PhysicsDave Dec 20 '18 at 5:25

Due to buoyancy one kilogram of hot air balloon indeed weighs less than one kilogram of steel!

  • $\begingroup$ It would be nice if you told us by how much, haha. $\endgroup$ – FGSUZ Dec 20 '18 at 0:45
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    $\begingroup$ This would be better as a comment $\endgroup$ – Aaron Stevens Dec 20 '18 at 1:22
  • $\begingroup$ I have used the above to elaborate on the question. Thanks. $\endgroup$ – James Thorpe Dec 20 '18 at 1:28
  • $\begingroup$ @AaronStevens I disagree that this should have been posted as a comment --- it's not a request for clarification or a suggestion for improvement to the question. (I'm glad it resulted in an improvement to the question anyway.) It's more like an answer, since it sets up a very clear example of a case where the buoyant force results in a different "weight" for two objects with the same mass, but it's so terse that it's not a terribly useful answer. It's fun, but here on Stack Exchange our relationship to fun is nuanced. $\endgroup$ – rob Dec 21 '18 at 4:03

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