Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-L] Weight of equal masses

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 <> 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....
Forum for Physics Educators

Sent from my so-called smartphone, whose spellchecker has a will of its