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, 2018 at 1:21
  • $\begingroup$ Then what does kilogram mean? $\endgroup$ Dec 20, 2018 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, 2018 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, 2018 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, 2018 at 21:11

4 Answers 4


Feathers are made from keratin, with a density of about $1.3\ \mathrm{g/cm^3}$. The net volume displaced by a kilogram of feathers is then $751\ \mathrm{cm^3}$. Steel has a density of $7.86\ \mathrm{g/cm^3}$ and a kilogram of it displaces $127\ \mathrm{cm^3}$.

Sea level air has a density of $0.0012\ \mathrm{g/cm^3}$, so the buoyant force on $751\ \mathrm{cm^3}$ of keratin is then $0.92\ \mathrm {gf}$ and the buoyant force on $127\ \mathrm{cm^3}$ of steel is $0.155\ \mathrm {gf}$.

This means that if we weigh the keratin in a vacuum, it will weigh $1\ \mathrm{kgf}$ but in air it will weigh $999.08\ \mathrm {gf}$. If we weigh the steel in a vacuum, it will weigh $1\ \mathrm{kgf}$ but in air it will weigh $999.85\ \mathrm {gf}$.

If we place the two bodies – keratin and steel, one kilogram 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 $(999.85-999.08)\ \mathrm {gf}$ or $0.77\ \mathrm {gf}$.

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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, 2018 at 6:48
  • $\begingroup$ Thank you sir, this is exactly the reply I expected and had hoped to get. $\endgroup$ Dec 20, 2018 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, 2018 at 7:14
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    $\begingroup$ 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, 2018 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. $\endgroup$ Dec 20, 2018 at 16:51

This depends on how you define "weight". If "weight" is defined as the gravitational force from the environment gravitator (here the Earth), then both will, of course, weight the same. If "weight", however, is defined by a scale reading and the scale and objects weighed are immersed in an atmosphere, then you're right: the feathers will "weigh" less as buoyancy helps them to float a little.

How much? Well, we need some figures for the density of all three materials involved. Archimedes' principle tells you that the buoyant force equals the weight (in the gravitational sense) of the displaced air, which, in turn, is determined by the volume of the object. If we denote that by $V_\mathrm{obj}$, we have, then,

$$F_B = \rho_\mathrm{air} V_\mathrm{obj} g$$

In terms of the mass $m_\mathrm{obj}$ and density $\rho_\mathrm{obj}$ of the object, we of course have

$$V_\mathrm{obj} = \frac{m_\mathrm{obj}}{\rho_\mathrm{obj}}$$


$$F_B = \frac{\rho_\mathrm{air}}{\rho_\mathrm{obj}} m_\mathrm{obj}g$$

If we denote the gravitational weight $w$, equal to

$$w = m_\mathrm{obj}g$$

and the scale or effective weight $w_\mathrm{eff}$, we have the very neat expression

$$w_\mathrm{eff} = w - F_B = \left(1 - \frac{\rho_\mathrm{air}}{\rho_\mathrm{obj}}\right) w$$

in other words, all you need is simply the ratio of densities between the object and the immersing medium, plus the actual gravitational weight - for an exact one kilogram, and standard Earth gravity, this is $w = 9.806\ 65\ \mathrm{N}$.

Now iron has a density of 7800 g/L ("steel" will vary depending on the amount of added carbon, but this should be close), equiv. $\mathrm{kg/m^3}$, while air has about 1.2 g/L, hence the iron piece will have

$$\left(1 - \frac{\rho_\mathrm{air}}{\rho_\mathrm{obj}}\right) = \left(1 - \frac{1.2}{7800}\right) \approx 0.99984$$

hence its $w_\mathrm{eff}$ is reduced from its original weight by about 0.016%, or to 9.8051 N.

For the feathers, this:


suggests their density is actually surprisingly higher - at about 1000 g/L, though considerably variable. In this case, the feathers should retain about 99.88% of their "true" weight, meaning a weight reduction of 0.12%, or a $w_\mathrm{eff}$ of 9.7948 N.

Neither of these are very noticeable, hence you would likely not feel anything interesting holding either in your hands.

(As to why the density reported there is more than you might think, it's because this is the density of the actual material feathers are made from. This is, also, the relevant density to use in this calculation because the air interacts, as a gas with no surface tension, directly at the atomic scale of the fibers in the feathers, thus the material, while the "apparent" low density is by considering density to be the sparse appearance of the material, which is how we would hold feathers in our hand, but not how air "holds" them.)


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, 2018 at 0:45
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    $\begingroup$ This would be better as a comment $\endgroup$ Dec 20, 2018 at 1:22
  • $\begingroup$ I have used the above to elaborate on the question. Thanks. $\endgroup$ Dec 20, 2018 at 1:28
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    $\begingroup$ @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, 2018 at 4:03

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. $\endgroup$ Dec 20, 2018 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. $\endgroup$ Dec 20, 2018 at 3:12
  • $\begingroup$ Yes good points. $\endgroup$ Dec 20, 2018 at 5:25

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