How to calculate the uncertainty for weight?

[jfnn]

Homework Statement

Hello, I am calculating the uncertainty for a course I am in. I am specifically calculating the uncertainty for the weight of coins in my experiment.

Homework Equations

The weight of coins in my experiment is equal to the number of coins I used in the part, multipled by g, multipled by the mass of the coin, which is 0.00627 kg.

w= n * g* m[/B]

The Attempt at a Solution

w (0.00627)(9.8)(1)
w = (0.00627)(9.8)(3) --> I put three there because three coins are used
w = 0.00627*9.8*4 --> I put 4 there because four coins were used[/B]


(Basically, i used coins from 1-10)

I know that the uncertainty for multiplication is the absolute value of the weight, multiplied by the uncertainty in the mass/mass value

For one coin,
Therefore, uncertainty = (0.05643)(0.002/0.00627)

Therefore uncertainty is 0.001 N?

Is this correct? I am so confused.. It changes if I have a different weight.

 

[kuruman]

Therefore, uncertainty = (0.05643)(0.002/0.00627)

What do the numbers 0.05643 and 0.002 represent? How did you get them?
It is best do think of the uncertainty as a percentage. For example, if you have a 10% uncertainty for one coin of weight 1 ounce, then the uncertainty is ±0.1 oz because that's 10% of 1. If you have 10 coins, the uncertainty is 10% of 10 oz. or 1 oz. That's why it changes if you have a different weight.

 

[Ray Vickson]

What do the numbers 0.05643 and 0.002 represent? How did you get them?
It is best do think of the uncertainty as a percentage. For example, if you have a 10% uncertainty for one coin of weight 1 ounce, then the uncertainty is ±0.1 oz because that's 10% of 1. If you have 10 coins, the uncertainty is 10% of 10 oz. or 1 oz. That's why it changes if you have a different weight.

This is an over-simplification.

If different coins can have slightly different weights (due do different wear-and-tear, different dirt deposits, perhaps very slightly different weights right from the mint, etc.) then the "uncertainty" in the weight of 10 coins is NOT 10 times the uncertainty in one coin; rather, it is about $\sqrt{10} \approx 3.16$ times the uncertainty in one coin. That is because uncertainties of (independent---i.e., unrelated) coins are added together to form an overall uncertainty. That total uncertainty would only be 10 times the single-coin uncertainty if every coin was identical, so that either all of them are a bit overweight by the exactly same (unknown) amount, or all are underweight by exactly the same (unknown) amount. That distinction is fundamental to the application of probability and statistics in the real world. It is also the reason why things appear to us to behave smoothly and predictable, even though they are composed of zillions of atoms moving (at least a bit) in random ways. If uncertainty multiplied instead of added, all matter would behave so erratically that life would be impossible, chemistry would not work, physics would fail, etc.

For more material related to this topic, see, eg,
http://web.mit.edu/fluids-modules/www/exper_techniques/2.Propagation_of_Uncertaint.pdf (especially page 5), or
http://news.bbc.co.uk/2/hi/science/nature/2157975.stm
or
http://virgo-physics.sas.upenn.edu/uglabs/lab_manual/Error_Analysis.pdf
among others.

 

[kuruman]

This is an over-simplification.

Indeed it is and you are correct to point this out. Considering OP's confusion regarding why the uncertainty in the total weight increases as the number of coins increases, it occurred to me that a simplified, though unrealistic, example of identical coins each differing from the "true" value the same way, would perhaps be a zeroth order qualitative explanation of the basic principle at work.

 

[jfnn]

Thank you for your help. Problem solved!