For each situation with different number of weights, the normal force differs as well. I calculated the normal force in this way:
Normal force = weight of the block + n x weight of one weight -> Normal force = 0.60 + 1n, where n is the number of weights added.
When I added 0 weights, normal force = 0.60 N;
When I added 1 weight, normal force = 1.60 N;
When I added 2 weights, normal force = 2.60 N;
When I added 3 weights, normal force = 3.60 N;
When I added 4 weights, normal force = 4.60 N.
To determine the µ kinetic I will use the formula:
Ffr. = µkinetic x FN -> µkinetic = Ffr. / FN
Further I provide the calculations of the µ kinetic for every situation:
When I added 0 weights, µkinetic = 0.15 / 0.60 = 0.25;
When I added 1 weight, µkinetic = 0.30 / 1.60 = 0.1875;
When I added 2 weights, µkinetic = 0.50 / 2.60 = 0.1923;
When I added 3 weights, µkinetic = 0.65 / 3.60 = 0.1806;
When I added 4 weights, µkinetic = 0.75 / 4.60 = 0.1630.
As the coefficient of friction is always the same between two same surfaces, my results should be the same. However, they differ and therefore, I will calculate the arithmetic mean of my results:
µkinetic= (0.25+0.1875+0.1923+0.1806+0.1630)/5 = 0.19468.
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:
Recording raw data:
First of all I measured the length of the wooden plane. I used a meter rule which had a smallest graduation of 0.1 cm. Therefore, I should take the absolute uncertainty of length measurement as ±0.05cm, but I decided to take it as ±0.1cm due to the fact that wooden plane could not have been in a perfect rectangular shape and therefore length at different positions could be different.
I decided to investigate and to check the static coefficient of friction in different situations, so I include the number of weights attached to the wooden block. I provide them as integer numbers.
Further I collected raw data of the height just before which the wooden block started to move. I collected these readings with a meter rule which smallest graduation was 0.1cm and therefore the absolute uncertainty of height measurements should be ±0.05cm but I decided to take the absolute uncertainty as ±0.3cm due to the fact that wooden plane was lifted by a human and therefore it was hard to identify the correct height.
I provide all this information in the table above.
Data processing:
Now I will calculate the mean of height just before starting the block to move.
(13.9+13.4+14.0)/3 = 13.8 cm
(11.2+12.6+11.9)/3 = 11.9 cm
(10.5+10.0+10.9)/3 = 10.5 cm
(10.5+9.9+9.9)/3 = 10.1 cm
(9.2+8.9+8.9)/3 = 9.0 cm
Now I need to find the angle between the wooden plane and the surface. I will find it for every situation in this way: sinα=height/plane length -> α=arcsin(height/plane length):
α=arcsin(13.8/51.3)=15.6°
α=arcsin(11.9/51.3)=13.4°
α=arcsin(10.5/51.3)=11.8°
α=arcsin(10.1/51.3)=11.4°
α=arcsin(9.0/51.3)=10.1°
Now I will determine the absolute uncertainty of angle measurements. To do this I will have to add relative uncertainties for every situation and the biggest found uncertainty will be taken as the absolute uncertainty for the angle measurements.
Δα/α = Δl/l + Δh/h -> Δα = α (Δl/l + Δh/h)
Δα = 15.6 (0.1/51.3 + 0.3/13.8) = 0.4°
Δα = 0.4°
Δα = 0.4°
Δα = 0.4°
Δα = 0.4°
Now I will find the static coefficient of friction. I will derive it from the free body diagram:
mgx=Ffr. and mgy=FN -> sinα*mg = Ffr. and cosα*mg = FN
Also Ffr. = µ * FN -> sinα*mg = µ * cosα*mg -> µ = sinα/cosα
µ = sin15.6/cos15.6 = 0.28
µ = sin13.4/cos13.4 = 0.24
µ = sin11.8/cos11.8 = 0.21
µ = sin11.4/cos11.4 = 0.20
µ = sin10.1/cos10.1 = 0.18
As the coefficient of friction between two surfaces is always permanent, so I have to find only one value for it. To do that I will calculate the mean:
µ = (0.28+0.24+0.21+0.20+0.18)/5 = 0.22
Now I will determine the absolute uncertainty for each situation will. To do this I will have to add relative uncertainties:
Δ µ / µ = Δα/α + Δα/α -> Δ µ = µ*2Δα/α
Δ µ = 0.28*2*0.4/15.6 = 0.01
Δ µ = 0.01
Δ µ = 0.01
Δ µ = 0.01
Δ µ = 0.01
However the biggest deviation from the mean is 0.06, so this number will be taken as the absolute uncertainty.
Now I can compare my result with literature’s. In Giancoli Physics page 97 it is provided that the coefficient of static friction of wood on wood is equal to 0.4. The percentage discrepancy is equal to 45%. Also, the percentage uncertainty is equal to 27%. I will discuss these finding in conclusion and evaluation part.
Conclusion and evaluation:
The first part of my determination, where I had to find the kinetic coefficient was done more precisely, but not more accurately. The percentage discrepancy compared with the literature’s was 0% what is a great achievement, but however the percentage uncertainty was 100%. In the second part discrepancy was 45% and percentage uncertainty was 27%. The percentage uncertainty of the static friction coefficient is quite good, but still the result and determined coefficient is quite far from given literature’s value. Therefore, I have to state the weaknesses and limitations of my determination.
Some errors were encountered and the percentage uncertainties are quite big despite the fact that I tried to be as accurate as possible. Next time I may try to improve my determination with suggestions provided further and then more accurate results could come. First of all, the main weakness of the determination was human factor as it was needed either to pull uniformly or to lift the plane very slowly and uniformly as well. It was clearly the weakest part of all determination as uniformity was very difficult to achieve. Of course, calculating means helped me to determine the coefficients more precisely, but huge uncertainties were still left. Moreover I could blame the instruments as I had to use quite many of them, but the uncertainty they provided was relatively small.
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.