This sample report is meant to be a guideline; please be aware that no report is perfect.
Consult the lab manual and online resources, and review previous comments from your
GA, when you write your report.
This sample report is intended for PHY 223/224 students only. If you are a PHY 221/222
student you need to reference the sample long report on the course website and use the
sections called for in the weekly rubric.
Kinetic Friction
Experiment # 13
Jane Answers
E012345678
Partner – Joe Solution
PHY 221
Lab Instructor – Mr. Smith
Wednesday, 11:00 AM – 12:50 PM
Lecture Instructor – Dr. Winters
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 or the type of surface
affects friction. We compared the coefficients of kinetic friction for different surface areas and
types and found the percent difference was 6.35% between different surface areas and 72.6%
between different surface types. The surface area did not affect friction, whereas the type of
surface did.
Results
Object Mass (kg)
Mblock
0.0641
mTrombone
0.001
mButterfly
0.003
mhanger
0.005
Table 1: Mass Values. These are the masses of the different objects that we use in the experiment.
M+m (kg) FN (N)
0.0641
0.1141
0.1641
0.2141
0.2641
0.3141
0.629
1.119
1.610
2.100
2.591
3.081
mClips (kg)
A
B
C
0.014 0.017 0.008
0.023 0.026 0.011
0.032 0.036 0.016
0.042 0.042 0.019
0.051 0.051 0.022
0.060 0.054 0.025
A
0.137
0.226
0.314
0.412
0.500
0.589
FT (N)
B
C
0.167 0.078
0.255 0.112
0.353 0.157
0.412 0.190
0.500 0.216
0.530 0.249
Average
A
0.218
0.202
0.195
0.196
0.193
0.191
0.199
µk
B
0.265
0.228
0.219
0.196
0.193
0.172
0.212
C
0.125
0.100
0.098
0.091
0.083
0.081
0.096
Table 2: Experimental Data. This is the data we obtained to determine the coefficient of kinetic
friction for the block.
𝐹! = 𝑚! 𝑔 = 0.0050𝑘𝑔 + 6 ∗ 0.0030𝑘𝑔 ∗ 9.81
𝑚
= 0.226 𝑁
𝑠!
𝐹! = 𝑀 + 𝑚 𝑔 = 0.0641𝑘𝑔 + 0.0500𝑘𝑔 ∗ 9.81
𝜇! =
𝐹!
𝐹!
𝜇! =
0.226 𝑁
1.119 𝑁
𝑚
= 1.119 𝑁
𝑠!
𝜇! = 0.202
We note that the coefficient of friction is a unitless quantity as expected.
Figure 1: Tension vs. Normal Force. This is a plot of the tension force versus the normal force
on the block for parts A, B, and C. Note that the y-intercept for each line has been set to zero as
when the normal force is zero the friction will be zero.
Comparison
Part A & Part B
Avg of Parts A & B to Part C
µk, avg, A to slope from Part A
µk, avg, B to slope from Part B
µk, avg, C to slope from Part C
% Difference
6.35
72.2
3.09
11.0
10.9
Table 3: Comparison of µk values. This shows that there is very little difference in the
coefficient of friction for different surface areas but a large difference for various types of
surfaces. This also indicates that the average coefficient of friction from each of the parts is
close to the value from the slopes of the graph.
% 𝐷𝑖𝑓𝑓 =
|𝜇!,!"#,! − 𝜇!,!"#,! |
∗ 100%
1
𝜇
+
𝜇
!,!"#,!
2 !,!"#,!
% 𝐷𝑖𝑓𝑓 =
|0.199 − 0.212|
∗ 100%
1
2 0.199 + 0.212
% 𝐷𝑖𝑓𝑓 = 6.35%
Error Analysis
One source of error is the assumption that the block was not accelerating in the x direction. 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 FT and fk would
change. FT and fk 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 formula for
the tension force would change after applying Newton’s Second Law to the hanging mass.
Σ𝐹! = 𝐹! − 𝑚! 𝑔 = −𝑚! 𝑎!
(1)
F! = m! 𝑔 − 𝑚! 𝑎!
(2)
Also, the next calculation shows how the formula for the force of friction would change after
applying Newton’s Second Law to the wooden block.
Σ𝐹! = 𝐹! − 𝑓! = 𝑀 + 𝑚 𝑎!
(3)
𝑓! = 𝐹! − 𝑀 + 𝑚 𝑎!
(4)
Since the hanging mass and block are connected by the string, they should accelerate at the same
rate, so 𝑎! = 𝑎! . Next, we can plug in the value of FT from Equation (2) into Equation (4),
noting that fk should still be equal to µkFN
𝑓! = 𝑚! 𝑔 − 𝑚! 𝑎 − 𝑀 + 𝑚 𝑎 = 𝜇′! 𝐹!
Solving this equation for µ’k, we find a new formula for our value of µ’k.
𝜇′! =
𝑚! 𝑔 − 𝑚! 𝑎 − 𝑀 + 𝑚 𝑎
𝐹!
(5)
If we use this new formula to calculate our coefficients, assuming that there was a small
!
acceleration of 0.02 ! , 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 Table 2 using the new calculation to find
µ’k.
𝜇′! =
𝜇′! =
𝑚! 𝑔 − 𝑚! 𝑎 − 𝑀 + 𝑚 𝑎
𝐹!
0.0139𝑘𝑔 ∗ 9.8
𝑚
𝑚
𝑚
− 0.0139𝑘𝑔 ∗ 0.02 ! − 0.641𝑘𝑔 + 0.00𝑘𝑔 0.02 !
!
𝑠
𝑠
𝑠
0.628𝑁
𝜇′! = 0.214
M+m (kg) FN (N)
0.0641
0.1141
0.1641
0.2141
0.2641
0.3141
0.629
1.119
1.610
2.100
2.591
3.081
mClips (kg)
A
B
C
0.014 0.017 0.008
0.023 0.026 0.011
0.032 0.036 0.016
0.042 0.042 0.019
0.051 0.051 0.022
0.060 0.054 0.025
A
0.137
0.225
0.313
0.411
0.499
0.587
FT (N)
B
C
0.166 0.078
0.255 0.112
0.352 0.157
0.411 0.190
0.499 0.215
0.529 0.249
Average
A
0.216
0.199
0.193
0.194
0.191
0.189
0.197
µ'k
B
0.245
0.207
0.199
0.176
0.173
0.152
0.192
C
0.124
0.099
0.097
0.090
0.083
0.080
0.096
Table 4: Experimental Data with Acceleration. This is the new data that we obtained after
!
accounting for the acceleration of the block of 0.02 !! when calculating the coefficient of kinetic
friction.
Using this new data, we can also recreate Figure 1, the graph of FT vs. FN.
Figure 2: Tension with Acceleration vs. Normal Force. This is a new plot of the tension versus
!
the normal force on the block after accounting for the acceleration of the block of 0.02 !! .
The slopes have slightly changed compared to the slopes found in the experiment. If we then
calculate new percent differences between the slopes and the µavg values, we find a percent
difference of 1.80% for part A, 9.8% for part B, and 8.4% for part C. Since this is a reduction of
percent difference, this source of error is valid.
Another source of error is that there was friction in the pulley. The pulley is not entirely
frictionless at the point of rotation. Thus, we need to account for the friction between the pulley
and the axis of rotation. The tension in the string from the hanging mass will have to overcome
this friction before acting on the block. So the tension that acts on the block is less than the
tension in the string at the hanging mass. If we assume the friction in the pulley acts as some
force, FP, against the tension, we can alter our equations to include this term.
Σ𝐹! = 𝐹! − 𝑓! − 𝐹! = 𝑀 + 𝑚 𝑎!,!"#$% = 0
f! = F! − F! = µμ"! 𝐹!
𝜇"! =
𝑚! 𝑔 − 𝐹!
𝐹!
We can see in the above equation that the coefficient of friction decreases. For example, if we
estimate 𝐹! = 0.005 𝑁, we find the following new result given the data we collected in the
experiment. A sample calculation is shown below for the new calculation of the coefficient, µ”k.
𝜇"! =
𝜇"! =
𝑚! 𝑔 − 𝐹!
𝐹!
𝑚
− 0.005𝑁
𝑠!
0.628𝑁
0.014𝑘𝑔 ∗ 9.81
𝜇"! = 0.209
If we apply this calculation to the rest of the data from Part A, we obtain a new average value for
µ”k of 0.196. Again, we can see that the average is lower when compared to the original value of
0.199. If we compare this value with the value from the graph of FT vs. FN, Figure 1, we find a
new percent difference of 2.24% for part A, which is reduced from 3.09%. Thus we can say that
this is a valid source of error in our experiment.
Questions For Thought
1. The magnitude of the force of friction does depend on the normal force between the two
objects. We saw this in the following equation, 𝑓! = 𝜇! 𝐹! . We can see that the frictional
force is proportional to the normal force through the coefficient of friction. In addition,
we saw, in Table 2, that when the normal force was 1.12 N the tension required to
overcome friction was 0.226 N, whereas when the normal force was 2.59 N the tension
was 0.500 N. Since the tension required to overcome the friction increases as the normal
force increases and from Newton’s Second Law applied to the wooden block we know
𝐹! = 𝑓! , we can see that as the normal force increases, the force of friction increases.
2. The percent difference between my graphs and the corresponding averages from Table 2
are small, and the values are 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. Therefore the slopes
do represent the coefficients of friction.
3. We can find, when looking at the graphs and averages, a few different values for the
coefficient of kinetic friction. I think that the slopes found for our best fit lines give us
the most accurate measurement for µk. 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 value for the coefficient of kinetic friction than one found by just taking an
average, since the best-fit line helps to account for error in the data set.
4. If we want to reduce error in the experiment, we may want to use a material with a higher
value of µk, such as rubber. If we use a higher coefficient, any small changes in the value
of the coefficient will make less impact, as the actual value will be much higher.