Re: [Phys-L] Weight of equal masses

[Zeev Wurman <zeev@ieee.org>]

Such arguments have been made for decades if not centuries, and the
pseudo-precise calculations don’t add much to the discussion.

To convince students a much simpler arrangement is offered, making
essentially the same point: pour a pound of bird-shot into a rubber
balloon, tie it off, and drop it into a water container; do the same to
another balloon, inflate it, tie it off, and drop it into water. Anyone
will accept that the inflated balloon “weighs” less in water. Now just ask
for explanation which should be readily given.

On Sat, Jan 9, 2021 at 2:06 PM Brian Whatcott <betwys1@sbcglobal.net> wrote:

> You may be familiar with the  question: which is heavier,  one tonne of
> feathers, or one tonne of steel?If you take the view that weight  is due
> solely to gravitational  force, then there is no doubt how you would answer.
> Suppose you take the slightly more sophisticated view  that  weight is the
> net force due to gravitational force less centrifugal force in a rotating
> frame, and though the value of g varies from place to place,  on this
> account, you may still hold that their weights are the same.
> But now, suppose you take the slightly more sophisticated view yet, that
> weight is what a scales indicate, where gravitational force is opposed by
> both centrifugal force in this rotating frame and a buoyancy force due to
> the air displaced.
> I will readily concede the inadvisability of mentioning centrifugal force,
> but you soon ascertain that  the  density of this steel is 8000 kg/m^3 ,
> the air density is 1.2 kg/m^3and for convenience we agree a density value
> for the feathers equal to that for  a closed  hollow tube  of keratin,
> which is 2.4 kg/ m^2 and which forms 15% of the feather volume.Let us next
> agree to calculate a cumulative volume for the steel: 1000/8000 m^3or 0.125
> m^3  which displaces 0.15 m^3 of air, which  weighs  1.2 *0.15= 0.18
> kg.Hence in mid latitudes for which the g constant incorporating a
> centripetal component is 9.81 N/kg, the steel weighs (1000kg  - 0.18)*9.81
> newtons  =  9808 N .
> Finally we calculate the volume of the feathers using a density figure
> of 85% of 1.2 kg/m^3 and 15% of 2.4 kg /m^3 = (1.02 kg + 0.36 kg)/m^3 =
> 1.38 kg/m^3We quickly find the volume of 1000  kg feathers is 1000/1.38 m^3
> = 725 m^3discounting the external volume of air trapped   by the
> feathers. This gives a buoyancy of 725 * 1.2 * 9.81 newtons.This leaves us
> with the weight of feathers as measured by a scales:1000 *9.81N -
> 725*1.2*9.81 N = 9810N - 8535N = 1275 N
> This leaves us with the uncomfortable weight comparison of 1 tonne of
> steel weighing  more than seven  times as heavy as 1 tonne of feathers.(An
> argument due to Prof. Cristobal Cortes, U.Zaragoza)
> If I object that the feathers, which are hollow and filled with air  are
> leaky, then I can discount  the internal air volume, and count only the
> volume of the keratin structure which is 15% of the feather's volume, then
> the effective buoyancy is greatly reduced to  15/100 of 8535N = 1280 N
> giving a total feather weight of 8530N In THIS case, the steel weighs only
> 9808/8530  of the feather weight - a mere 15% more.           I welcome
> critiques of Cristobal's and my estimates....
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