# Determining the uncertainty of the coefficient of friction

1. Jan 31, 2016

### Nick tringali

1. The problem statement, all variables and given/known data

So far i know that the uncertainty of the force of friction is +/- 0.1 and the uncertainty of the force normal is +/- 0.1 also.
2. Relevant equations
Force of friction=(myoo)(force normal)
Myoo= force friction / force normal

3. The attempt at a solution
My data is
Force of friction | myoo | fnormal
0.0 | 0.0 | 0.0
1.4 | 0.264 | 5.3
1.9 | 0.260 | 7.3
3.2 | 0.262 | 12.2
4.1 | 0.272 | 15.1
4.6 | 0.269 | 17.1
Thats my calculated data so first I added 0.1 to1.4 and got 1.5 then I added 0.1 to 5.3 and got 5.4 then divided 1.5/5.4 and got 0.27777 (which is my max myoo value) now to get the min value I subtracted 0.1 from 1.4 and 5.3 and divided again 1.3/5.2 and got .25 for the min myoo value then i subtracter 0.27777 from 0.25 and got my uncertanty to be 0.0277, the problem is i tryed the same prossess with another set of values from my data table and got a different uncertainty. For the last set of numbers with frictonal force of 4.6 i got the uncerntanty to be 0.00855 (i did the same process as before) All of my class mates are doing different ways.

2. Jan 31, 2016

### Simon Bridge

This does not make sense to me ...
You need to be clear about what is being attempted ... myoo is the coefficient of friction I take it?

I cannot tell what you have done -

Let f = ffriction is the friction force, N = fnormal is the normal force, and mu or $\mu$ is myoo is the coefficient of friction. (saves typing)
You would normally want to measure f and N to calculate mu.
You say you have calculated all three for the table ... your problem statement gives uncertainties for f and N, which suggests a measurement for f and N.

OK: assuming you have friction $f\pm\sigma_f$ and normal force $N\pm\sigma_N$ then $\mu = f/N$ - f and N are independant and divided so you propagate the relative errors (percentage errors).

The rule is as for $z=xy$, i.e. $$\left(\frac{\sigma_z}{z}\right)^2=\left(\frac{\sigma_x}{x}\right)^2+\left(\frac{\sigma_y}{y}\right)^2$$

... this you'd do for every line, giving you the uncertainty for each value of mu.

Better: plot a graph of f vs N - the slope will be mu, and you can calculate the uncertainty of the slope.
Also better - find the mean and standard deviation of the mu column of your table ... the value of mu is $\mu = \mu_{ave} \pm \sigma /\sqrt{n}$ where n is the number of entries in the columb.

Note: your table mu value for f=0 is wrong. But I don't know what you actually did: I'm just guessing.
It may help you to see:
https://www.physicsforums.com/threads/uncertainty-of-an-average.612633/

Last edited: Jan 31, 2016
3. Jan 31, 2016

### Nick tringali

Thanks simon, but what do those symbols in the numerator represent?

That is uncertainty right?

Would each value of myoo have there own different uncertainty then, I was looking for the uncertainty for the myoo column as a whole, sorry if im not following you 100%. Should i just use that formula you gave?

I also have a graph like that and the slope too of the line of best fit.

Last edited by a moderator: Feb 1, 2016
4. Feb 1, 2016

### Staff: Mentor

You can edit your posts if you want to add something. I merged your posts.

The symbols in the numerators are the uncertainties, right.
Correct. You can calculate an overall uncertainty out of those different values later.
Yes. Or use the fit approach.

Unrelated:
A (positive) fraction gets maximal if the numerator is maximal and the denominator is minimal. 1.5/5.2 is larger than 1.5/5.4. But that approach is not right anyway.

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