Group number: 7
Section SLN: 7242S
TA’s name:
Part 3: Wooden block with one weight.
Mass of the wooden block, mb = 0.122 kg
Mass of the heavy weight 1, mw1 = 0.494 kg
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Part 4: Statistical distribution of data.
The attached histogram has been created for the
static / kinetic (select one)frictional force data.
III. Data Analysis.
Part I: Determination of the coefficient of static friction
- Based on the data collected for the wooden block carrying two weights and sliding along the lab table the coefficient of static friction was determined as follows:
(Show equation and sample calculation)
fs = μs N
where fs =Average frictional force = 7.889 N
N = Total normal load = (0.122 + 0.494 + 0.497) x 9.8 = 10.9074 N
So, coefficient of static friction (μs) = fs / N = 7.889/10.9074 = 0.723
- This coefficient has an error associated with it that propagates from the uncertainty in the force of static friction as follows:
(Show equation and sample calculation)
Δ μs = Δ fs/ N
where, Δ fs is uncertainty = ±0.15845 N
Δ μs = Δ fs/ N = ±0.15845/10.9074 = ±0.0145
- Based on the data collected for the wooden block carrying one weight and sliding along the lab table the coefficient of static friction was determined as follows:
(Show equations and sample calculation)
fs = μs N
where fs =Average frictional force = 4.247 N
N = Total normal load = (0.122 + 0.497) x 9.8 = 6.0662 N
So, coefficient of static friction (μs) = fs / N = 4.247 /6.0662 = 0.7
- This coefficient has an error associated with it that propagates from the uncertainty in the force of static friction as follows:
(Show equation and sample calculation)
Δ μs = Δ fs/ N
where, Δ fs is uncertainty = ±0.1157 N
Δ μs = Δ fs/ N = ±0.1157 /6.0662 = ±0.01
Part I: Determination of the coefficient of kinetic friction
- Based on the data collected for the wooden block carrying two weights and sliding along the lab table the coefficient of kinetic friction was determined as follows:
(Show equation and sample calculation)
fs = μk. N
where fs =Average frictional force = 5.666N
N = Total normal load = (0.122 + 0.494 + 0.497) x 9.8 = 10.9074 N
So, coefficient of kinetic friction (μk) = fs / N = 5.666/10.9074 = 0.52
- This coefficient has an error associated with it that propagates from the uncertainty in the force of kinetic friction as follows:
(Show equation and sample calculation)
Δ μk = Δ fs/ N
where, Δ fs is uncertainty = ±0.05101 N
Δ μs = Δ fs/ N = ±0.05101 /10.9074 = ±0.005
- Based on the data collected for the wooden block carrying one weight and sliding along the lab table the coefficient of kinetic friction was determined as follows:
(Show equations and sample calculation)
fs = μs N
where fs =Average frictional force = 2.767 N
N = Total normal load = (0.122 + 0.497) x 9.8 = 6.0662 N
So, coefficient of static friction (μs) = fs / N = 2.767 /6.0662 = 0.456
- This coefficient has an error associated with it that propagates from the uncertainty in the force of kinetic friction as follows:
(Show equation and sample calculation)
Δ μs = Δ fs/ N
where, Δ fs is uncertainty = ±0.0682547 N
Δ μs = Δ fs/ N = ±0.0682547 /6.0662 = ±0.011
Part II: Evaluating the distribution of selected frictional force data.
For the static / kinetic (choose one) force data set the percentage of measurements that falls within one standard deviation of the mean was calculated to be:
(Show calculation)
For this reason, percentage of measurements that falls within one standard deviation of the mean was calculated to be
(11+11)/(5+11+11+7+1) = 63% ,
i.e. there is a 63% chance that the mean value is within one stand. dev. of the mean value (±σn-1) which was obtained by 35 measurements.
For the same selected experimental set of data as above, the percentage of measurements that falls within two standard deviations of the mean is:
(Show calculation)
For this reason, percentage of measurements that falls within two standard deviation of the mean was calculated to be
(5+11+11+7)/(5+11+11+7+1) = 97% ,
i.e. 97% of the data points within 2σn-1’s of measurements.
IV. Results.
- Coefficients of static and kinetic friction – class data.
(Enter your group data in different color. All values should be presented with
uncertainties)
- Statistical distribution of measurements for the force of static / kinetic (select one)
friction – comparison with normal distribution.
V. Discussion & Conclusion.
The experiment was carried out successfully and safely. By carrying out this experiment knowledge of determining coefficient of static and kinetic friction was gained. Moreover, it has been found out that the errors were truly random and normally distributed. So, objective of this experiment was met.
- Data Table with columns showing the maximum force of static friction for part 1, the average kinetic force of friction for part 1, the maximum force of static friction for part 3, the average force of kinetic friction for part 3.
- Graphs displaying statistics (mean value and uncertainty in the mean value) for all four columns of your data.
- Histogram created for one set of data from part 1 (maximum force of static friction or the average force of kinetic friction) with clearly marked regions: mean +/- std. deviation, mean +/- 2std. deviations.
