A way of finding the coefficient of friction
September 9, 2010 1 Comment
It feels really nice when you propose a theory and later find out that it is right. This is what happened with me. Here’s a page from Journal de Brouillons.
“
Another method of finding co-efficient of friction μ
Date: 1/3/2009
I’ve found a new method to find the co-efficient of friction of bodies cylindrical in shape. It requires a thread and something which can measure force like a spring balance and some weights.



The things are arranged as in the diagram.
The weight of the weight when it is free is found. Let it be T1.
Now the maximum and minimum tensions(for which the system is in equilibrium) of the string connected to the spring balance is found. Let them be T2max and T2min.
ΔT = T2max – T1 = T1 – T2min
Consider a small section of the string wrapped around the object. Let it subtend an angle dθ.
Then the normal force due to this segment
= T sin(dθ/2) + (T+dT) sin(dθ/2)
≈ Tdθ
Under limiting conditions;
f= fmax =μsN = μsTdθ
For the string to be equilibrium
T + dT = T + f
dt= f = μsTdθ
T2∫T1 dT/T = 0∫θμsdθ
ln(1+ΔT/T) = μsθ
μs=ln(1+ΔT/T)/θ
if ΔT/T is small enough (if ΔT/T =1/50 the following approximation gives 1% error.)
μs=ΔT/Tθ
However the spring balance may have some friction and it will change the values. So it should be taken into consideration.
I would also like to ….
…
”
The remaining text has some other methods of finding the coefficient friction. And now these are the results I got
| T1 | T2 |
| 50 | 25 |
| 100 | 58 |
| 150 | 85 |
| 200 | 117 |
| 250 | 141 |
| 300 | 175 |
| 350 | 209 |
| 400 | 246 |
| 450 | 284 |
| 500 | 293 |
And this is the table of values which tell about the relationship between T2 and θ. It is exponentially decreasing.
| Angle of Wrapping | T2 |
| π | 175 |
| 3π | 80 |
| 5π | 30 |
| 7π | 13 |
| 9π | 5 |
| 11π | 3 |
And ΔT/T was never small enough.


