- Level: International Baccalaureate
- Subject: Physics
- Word count: 2055
Determination of Coefficient of Friction
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Introduction
DETERMINATION OF THE COEFFICIENT OF FRICTION
Aim:
- To determine µ kinetic between a wooden block and a wooden plane;
- To determine µ static between a wooden block and a wooden plane.
Equipment:
- Wooden block;
- Wooden plane;
- Spring scale;
- Meter rule;
- Weights.
There are two parts of my investigation, so I will precede them separately.
Determination of µ kinetic between a wooden block and a wooden plane
This is the table which I filled during my determination:
RAW DATA | DATA PROCESSING | |||
Weight of the block/N; ±0.05N | Number of weights added | Friction force/N; ±0.1N | Normal force/N; ±0.05N | µ kinetic; ±0.2 |
0.60 | 0 | 0.15 | 0.60 | 0.2 |
1 | 0.30 | 1.60 | ||
2 | 0.50 | 2.60 | ||
3 | 0.65 | 3.60 | ||
4 | 0.75 | 4.60 |
Recording raw data:
First of all, I prepare my working place and start my determination. All my measurements are recorded to the table above. For more accurate results of µ kinetic I recorded data with 5 different weights.
The smallest graduation of the spring scale is 0.1 N. According to this, the absolute uncertainty of weight of the block is ±0.05 N. I do not add additional uncertainty as I did not encounter any further difficulties in weight measurement.
I used weights provided by my teacher. Those weights were precisely 1 N each. In the table I only provide the number of them and therefore I take it without uncertainty.
Once again the smallest graduation of the spring scale is 0.
Middle
Now the biggest deviation from the mean should be taken as the absolute uncertainty of the µ kinetic. The biggest deviation is 0.25 – 0.19468 = 0.05532 ≈ 0.06. I would use this absolute uncertainty if there is no bigger uncertainty when calculated for each situation. Now I will calculate the absolute uncertainty of my µ kinetic measurements for each situation. To do that, I will have to add relative uncertainties by using this formula:
When I added 0 weights, -> ∆µ = 0.1875 ≈ 0.2;
When I added 1 weight, ∆µ = 0.0684 ≈ 0.07;
When I added 2 weights, ∆µ = 0.0421 ≈ 0.04;
When I added 3 weights, ∆µ = 0.0303 ≈ 0.03;
When I added 4 weights, ∆µ = 0.0235 ≈ 0.02;
Obviously, the biggest uncertainty is 0.2. Therefore:
µkinetic ± ∆µkinetic = 0.2 ± 0.2
Now I can compare my result with literature’s. In Giancoli Physics page 97 it is provided that the coefficient of kinetic friction of wood on wood is equal to 0.2. This is exactly the same value as I have counted. The percentage discrepancy is equal to 0%. However, the percentage uncertainty is equal to 100%. I will discuss these finding in conclusion and evaluation part after the determination of static friction coefficient.
Determination of µ static between a wooden block and a wooden plane
This is the table which I filled during this determination:
RAW DATA | DATA PROCESSING | ||||||
Length of the plane/cm; ±0.1 cm | Number of weights attached | Height just before starting to move/cm; ±0.3 cm | Angle α°, ±0.4° | µ static, ±0.06 | |||
1st reading | 2nd reading | 3rd reading | Mean | ||||
51.3 | 0 | 13.9 | 13.4 | 14.0 | 13.8 | 15.6 | 0.22 |
1 | 11.2 | 12.6 | 11.9 | 11.9 | 13.4 | ||
2 | 10.5 | 10.0 | 10.9 | 10.5 | 11.8 | ||
3 | 10.5 | 9.9 | 9.9 | 10.1 | 11.4 | ||
4 | 9.2 | 8.9 | 8.9 | 9.0 | 10.1 |
Recording raw data:
Conclusion
Furthermore, some systematic errors have occurred as I had to do a lot of calculations and roundings during the data processing part. Also, the instruments may have been badly calibrated and this could have affected my determination. However, systematic errors are not so important because even if they even were encountered, they were very small. Another thing is with random errors as they were really significant because the percentage uncertainty shows quite high result.
I could provide several suggestions to improve the determination. First of all, I would rather use bigger and longer plane and bigger block. Then, as I would still use the same equipment with same absolute uncertainties, the percentage uncertainty would be reduced significantly. The uncertainty would be less important and more accurate results would come. Also, human factor uncertainty would be reduced because it would be easier to pull uniformly or to lift the plane. However, my suggestions would only lesser the uncertainties, but they would not totally cancel them.
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