Do Bathroom Scales Account for Earth's Rotation?

[IWantToLearn]

i know that bathroom scales measures the normal force, hence the equivalence mass

but if we took Earth rotation into account then the normal force will be less by a value mv²/R

according to this relation
N = mg -mv²/R

Example : a 70 kg man standing on a bathroom scale on the equator , where the radius of the Earth at the equator is R = 6378 km , and the time to complete one rotation is T = 24 hours
[assuming that the gravitational acceleration is constant everywhere on the surface of the Earth and equals to g = 9.8 m/s²]
doing the mass : we found that the normal force (the reading of the scale in Newtons)
equals N = 683.64 N
while the gravitational force must be equal to mg = 686 N , this means that the reading is less than the actual value (2.36 N less)
if we did the same calculations near the north pole where the radius is R = 6,356 km
there will be less by a value (2.35 N less)
both readings are not accurate, i know that all scales are calibrated to indicate the mass

now my questions is :
1- These small deviations from the actual value, are they neglected?
2- If they aren't, and the manufacturers take into account the Earth rotation and where in the Earth you are, is that means that if bought a bathroom scale from london it will not be accurate for me to use it in kenya?
3- what about the change of the value of g itself from place to place on the surface of the earth?

 

[HallsofIvy]

For typical bathroom scales, yes, those deviations are neglected. Manufactures of bathroom scales do not do any complicated calculations- they put a known weight on the scale and calibrate the scale to read that known value.

The difference between scale readings in London and Kenya are too small to be of importance.

The change in g is the change in gravitational force.

 

[HomogenousCow]

Huhh thinking about this problem is confusing myself, I'm wondering if the scale would detect the centripetal acceleration at all seeing that it itself (or whatever mechanism responsible) is also accelerating at said rate.

 

[SW VandeCarr]

I'm not sure what you mean by "geocentric" in this context. The physical surface of the Earth for all practical purposes is the surface of maximum gravitational potential. The acceleration at the this surface (g) varies according to latitude and locally for a variety of factors. The Earth bulges slightly at the equator but centrifugal acceleration is maximal here counteracting g for a net g' of 9.780 ms^{-2}. At the poles the Earth's radius is smaller, but the absence of centrifugal acceleration results in a net g of 9.832ms^{-2}.

This is probably very close to the range of g on the Earth's surface and I doubt you would notice it on a typical bathroom scale. However if you were weighing pure gold, the difference between 9.780*M kg and 9.832*M kg might be of interest.

 

[IWantToLearn]

so the answer that no manufacturer make these bathrooms to be fit in one place london for example (geocentric) , all bathroom scales are for global use everywhere in the earth


HallsofIvy : your comment is very true, sorry that i didn't notice that , thanks :)

The change in g is the change in gravitational force.

 

[SW VandeCarr]

so the answer that no manufacturer make these bathrooms to be fit in one place london for example (geocentric) , all bathroom scales are for global use everywhere in the earth

For very accurate scales, they might standardize to 45 N latitude using a standard kilogram mass, however it's the difference that is more important. At the extremes a kilogram mass of pure gold would be about 0.0052 heavier at the poles compared to the equator. At 1200 USD per ounce (28.3g) gold would be worth about 6.24 USD more per ounce at the pole than at the equator if you relied on weight rather than mass. For a kilogram of gold, that's 6.24*35.2=219.65 USD.

 

[IWantToLearn]

For very accurate scales, they might standardize to 45 N latitude using a standard kilogram mass, however it's the difference that is more important. At the extremes a kilogram mass of pure gold would be about 0.0052 heavier at the poles compared to the equator. At 1200 USD per ounce (28.3g) gold would be worth about 6.24 USD more per ounce at the pole than at the equator if you relied on weight rather than mass. For a kilogram of gold, that's 6.24*35.2=219.65 USD.

So i can say that "for scales that is used to measure very valuable materials, or some scientific computing, they must design scales that takes into account Earth rotation and where on Earth we will use these scales, so now i can say that a scale made in london will not be suitable to be used in Kenya"

 

[SW VandeCarr]

So i can say that "for scales that is used to measure very valuable materials, or some scientific computing, they must design scales that takes into account Earth rotation and where on Earth we will use these scales, so now i can say that a scale made in london will not be suitable to be used in Kenya"

Yes. You would need a reference standard for pricing in Newtons, and make the necessary corrections for local deviations since we can't directly measure mass in any practical sense. Frankly, I don't know if this is done. However, I don't think you can take advantage of this to make money because 1) moving enough gold to make it worthwhile would be expensive and 2) the price movements in the markets would generally be greater than the g variations involved.

 

[IWantToLearn]

Yes. You would need a reference standard in Newtons, and make the necessary corrections for local deviations since we can't directly measure mass in any practical sense. Frankly, I don't know if this is done.

i need to be sure of that, for example i want to know if gold scale manufacturers puts a label on their products that they are suitable to be used in london not in Kenya :)

However, I don't think you can take advantage of this to make money because 1) moving enough gold to make it worthwhile would expensive and 2) the price movements in the markets would generally be greater than the g variations involved.

don't worry, i will not try to go to sell gold to people in the north pole :)

 

[Baluncore]

A mass on the Earth's surface is subjected to two forces. Firstly the downward force due to gravity,(spherical), and secondly a very much smaller outward force due to rotation, (cylindrical). The sum of those two forces is what we “feel” gives us weight.

Ideally the mean sea level surface of the Earth must be an equipotential. If it were not, the Earth's mantle and sea water would flow to make it so. The weight you “feel” on an equipotential surface should be the same worldwide. The ideal figure of the Earth will correct for the rotational force.

If scales are calibrated at maybe 250 metres above sea level, the majority of users will be close enough to that height above the sea-level equipotential to not notice the error.

 

[D H]

The change in g is the change in gravitational force.

That's not right, for a couple of reasons. One is that the units are wrong. g has units of acceleration, not force. The other is that g at some spot on the Earth's surface is the magnitude of the resultant vector of the gravitational acceleration toward the Earth and centrifugal acceleration due to the Earth's rotation.

So i can say that "for scales that is used to measure very valuable materials, or some scientific computing, they must design scales that takes into account Earth rotation and where on Earth we will use these scales, so now i can say that a scale made in london will not be suitable to be used in Kenya"

It depends on the purpose of the measurement. If you want apparent weight (aka scale weight), a an ideal spring scale is the ideal device for your measurement. If you want mass, the best apparatus is a balance scale, which measures mass rather than (apparent) weight. An ideal balance scale will say that an ounce of gold "weighs" one ounce at the north pole, atop the highest mountain near the equator, and even on the surface of the Moon.

 

[A.T.]

The weight you “feel” on an equipotential surface should be the same worldwide.

Weight is determined by the gradient of the potential, not the potential itself. Why does equal potential imply equal weight?

 

[D H]

Weight is determined by the gradient of the potential, not the potential itself. Why does equal potential imply equal weight?

It doesn't. It's a trivial matter to construct an equipotential surface for which the gradient of the potential is not constant. An equipotential surface that is an oblate spheroid, for example, does the trick.

 

[A.T.]

It's a trivial matter to construct an equipotential surface for which the gradient of the potential is not constant. An equipotential surface that is an oblate spheroid, for example, does the trick.

So even if the Earth was a perfect equipotential surface, a scale would still show different weights at the pole and equator.