# Help with varying coefficients of friction

Hello everyone, my question isn't exactly a "Here's the data how do I do it" but rather why is it giving me "this" answer.

I was doing a lab in class the other week, involving a pulley system which has one mass on a table and the other hanging. We had a balance to measure the masses and a motion sensor(timed for one second 5 times to get an average) to determine the acceleration. The objective to the lab was "to determine how the coefficient of kinetic friction depends upon the speed, surface area, and mass of a block sliding on a horizontal surface." Now correct me if I'm wrong but I thought that the coefficient of friction was independent and didn't vary for a particular surface? The data we collected ended up having different coefficients with the below equation.

Equation

uk(coefficient) = (m1g-(m1+m2)a)/(m2g)

Taken data and plug n chug

One example of data we took was m1(hanging mass)=.05kg, m2(sliding mass)=.05kg and a(acceleration)=.563m/s^2

Now I plug this into the equation:
(.05*9.81-(.05+.1235).563)/(.1235*9.81)

And I get roughly .324 for the coefficient.

Now if I were to add some mass, that shouldn't change the coefficient should it?

Another set of data we took was m1=.1kg, m2=.2235kg, a=.911m/s^2

Plugging into the same equation I get roughly .313 for the coefficient.

Is there something that was done wrong? I would have assumed that the coefficient would stay the same for each run but this equation is giving me different answers. What is going on? Thank you.

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PhanthomJay
Homework Helper
Gold Member
The coefficient is ideally independent of speed, surface area, and mass, and is a function only of the material of the objects in contact. In a lab experiment, you'll never get 'ideal'. Your results are valid to 2 sig figures, you got .31 in one case, and .32 in another. That would make me more than happy. Well that wasn't the most extreme cases. There were 5 sets of data(of which each was done 5 times to get a closer average) which ranged from .276(block on its side for less surface area) and the greatest being .360(.09 hanging mass, .1235 sliding mass and ave accel of 2.09m/s^2) Is that still pretty close? Seems to me its pretty separated.

This is confusing me because there are 3 questions asking if mass, surface area and speed affected the coefficient which it shouldn't be but appears to have. I'm not sure what I would answer with, based off my knowledge or based off the lab data.

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PhanthomJay
Homework Helper
Gold Member
Actually, an object with a mass of .05 kg has only one significant figure (I'm not so good with sig figure rules), so when you got .31 and .32 for a result earlier, that's really a uk of 0.3 for both. That .4 you got later seems to be an outlier. Anyway, the kinetic friction coef must be determined experimentally. Throw out the outliers, and take an average, and write up an error analysis.

ideasrule
Homework Helper
Yes, that's still pretty close for this type of lab. F=uN is an approximation, and a bad one at that.

I didn't think about the sig figs. That would result in the coefficient being .3(well without the .360 one)That gives me something to note in the lab write up. Thanks :D