Kinetic Friction Experiment #13 Joe Solution E01234567 Partner- Jane Answers PHY 221 Lab Instructor- Nathaniel Franklin Wednesday, 11 AM-1 PM Lecture Instructor Dr. Jacobs Abstract The purpose of this experiment was to examine kinetic friction and what factors affect it. We pulled a wood block across a surface to determine whether the surface area of the block or the type of surface affects friction. The surface area did not appear to affect friction, since there was a small percent difference of 6.16% between different surface areas. The type of surface did as the difference in values was large, at 72.2%. (75 words)
Results Object Mass (kg) Wood Block 0.0641 Trombone Paperclip 0.001 Butterfly Paperclip 0.003 Hanger 0.005 Data Table 1- Masses of Objects M+m (kg) F N (N) Tension (N) Coefficient of Kinetic Friction A B C A B C 0.0641 0.628 0.137 0.169 0.0821 0.218 0.268 0.131 0.114 1.12 0.229 0.258 0.109 0.205 0.231 0.0974 0.164 1.61 0.318 0.348 0.155 0.198 0.217 0.0966 0.214 2.10 0.408 0.408 0.189 0.194 0.194 0.0903 0.264 2.59 0.498 0.498 0.219 0.192 0.192 0.0847 0.314 3.08 0.587 0.528 0.249 0.191 0.171 0.0810 Average 0.200 0.212 0.0968 Data Table 2 Finding the Coefficient of Kinetic Friction
Figure 1- Graph of F T vs F N to find values of µ k Comparison % Difference Part A to Part B 6.16 Avg of A&B to C 72.2 Part A 3.09 Part B 11.0 Part C 10.9 Data Table 3- Percent Differences. Error Analysis One source of error was the assumption that the block was not accelerating in the horizontal direction. We made this assumption by setting all of the accelerations besides gravity to 0 in the equations found in the lab manual. I noticed that the block did appear to accelerate once enough mass was added to the hanger to cause the block to move. If the block accelerated, we would need to alter our equations used throughout out the experiment. Looking back over the equations, our values of F T and F k would change. F T and F k would become slightly smaller due to the acceleration. This would mean the coefficients for each trial would also decrease. The calculations below show how the formulas for the tension force and friction would change.
(9) (10) Since the hanging mass and block are connected by the string, they should accelerate at the same rate, so a=a x. Next, we can plug in the value of F T from equation 9 into equation 10, noting that F k should still be equal to µ k F N Solving this equation for µ k, we find a new formula for our value of µ k. If we use this new formula to calculate our coefficients assuming that there was a small acceleration of 0.02m/s, we would see that each of the coefficients found in parts A, B and C would decrease. Below is a sample calculation as well as a recreation of Data Table 1 using the new calculation to find µ k F N (N) Hanging Mass (kg) Tension (N) Coefficient of Kinetic Friction A B C A B C A B C 0.628 0.014 0.017 0.008 0.137 0.169 0.0821 0.216 0.266 0.128 1.12 0.023 0.026 0.011 0.229 0.258 0.109 0.203 0.229 0.0951 1.61 0.032 0.036 0.016 0.318 0.348 0.155 0.195 0.214 0.0944 2.10 0.042 0.042 0.019 0.408 0.408 0.189 0.192 0.192 0.0880 2.59 0.051 0.051 0.022 0.498 0.498 0.219 0.190 0.190 0.0825 3.08 0.060 0.054 0.025 0.587 0.528 0.249 0.188 0.169 0.0787 Average 0.197 0.210 0.0944 Data Table 4- Factoring in Acceleration into the Coefficient of Kinetic Friction Using this new data, we can also recreate Figure 1, the graph of F T vs. F N.
Figure 2- Recreation of Figure 1 Accounting for Acceleration The slopes have slightly changed compared to the slopes found in the experiment. If we then compare the new averages and slopes that account for acceleration just as we did before using a percent difference, we find a percent difference of 1.80% for part A, 9.8% for part B, and 8.4% for part C. With a reduction of percent difference, this source of error can be held accountable for some of the error in the experiment. Another source of error is that there was friction in the pulley that was not accounted for. The string and pulley were in contact. Two surfaces in contact will generate friction as one surface moves past the other. This would mean that the friction between the pulley and the string was not accounted for. We can alter our equations to include this term. F S represents the force of friction due to the string sliding along the pulley. As we can see by the final equation, this factor would reduce the values we calculated for the coefficient of friction. For example, if we estimate the force of friction between the string and pulley to be 0.002N, we would find the following new results given the data we collected in the experiment. A sample calculation is shown below as well as a new data table that used the new calculation for the coefficient.
M+m (kg) F N (N) Hanging Mass (kg) Tension (N) Coefficient of Kinetic Friction A B C A B C A B C 0.0641 0.628 0.014 0.017 0.008 0.137 0.169 0.0821 0.214 0.265 0.128 0.114 1.12 0.023 0.026 0.011 0.229 0.258 0.109 0.203 0.230 0.0956 0.164 1.61 0.032 0.036 0.016 0.318 0.348 0.155 0.197 0.215 0.0954 0.214 2.10 0.042 0.042 0.019 0.408 0.408 0.189 0.194 0.194 0.0893 0.264 2.59 0.051 0.051 0.022 0.498 0.498 0.219 0.192 0.192 0.0840 0.314 3.08 0.060 0.054 0.025 0.587 0.528 0.249 0.190 0.171 0.0803 Average 0.198 0.211 0.0953 Data Table 5- Factoring in the Friction Between the Pulley and String Again, we can see that the averages are lower for each part of the experiment when compared to previous values. We can then compare these values to the graph of F T vs. F N in a percent difference as was done previously. When compared, we find a percent difference of 2.24% from part A, 10.3% from part B, and 9.3% for part C all of which are reduced when compared to the percent differences found during the experiment. Questions for Thought 1. The value of friction does depend on the normal force between the two objects. We saw this in equation 7,. We can see that the frictional force is proportional to the normal force through the coefficient of friction. As the normal force increases, the frictional force increases. 2. The percent difference between my graphs and the corresponding averages appear to be close to one another. The percent differences were 3.09%, 11.0%, and 10.9% for parts A, B, and C respectively. These percent differences are under 15% and are reasonable when comparing averages to best fit lines. 3. Looking at the graphs and averages, we have found a few different values of the coefficient. Of our values found for µ k, I think that the slopes found for our best fit lines give us the most accurate measurement. Excel uses the method of least-squares when calculating the best-fit line for a given data set. This is a much more accurate approach for finding a best fit value than one found just taking an average. 4. If we wanted to reduce error in the experiment, we may want to use a material with a higher value of µ k, like rubber. By using a higher coefficient, any small changes in the value of the coefficient will be harder to detect as the actual value will be much higher. If we can keep the uncertainty at about the same value, but increase the value it applies to, like the coefficient, we can reduce the error.