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.

If you are a PHY 221/222 student you need determine which sections are called for in the

weekly rubric before you start to write your report.

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.

Introduction

The purpose of this experiment is to explore the properties of kinetic friction. Specifically, we

will examine how the surface area of the object and the type of surfaces in contact with each

other affect the kinetic friction.

Friction is a force that always opposes the motion of an object. There are two types of friction –

static and kinetic. Kinetic friction occurs when two objects are moving against one another with

some part of their surfaces in contact. Kinetic friction opposes the motion of the object and is

proportional to the normal force acting on an object. The proportionality constant is the

coefficient of friction, µk. The equation for this relationship is:

𝑓! = 𝜇!𝐹! (1)

In order to calculate the coefficient of kinetic friction, consider a simple setup with a block of

mass, M, on a horizontal surface with a tension force, FT, being applied via a string attached to a

mass, m, hanging over a massless, frictionless pulley (refer to Figure 1). We can determine the

friction acting on the object by applying Newton’s Second Law on the block and the hanging

mass. We begin by examining the forces acting on the hanging mass; they are

Σ𝐹! = 𝐹! −𝑚𝑔 = −𝑚𝑎! (2)

where FT is the force of tension in the string, m is the mass of the hanging object, g is the

acceleration due to gravity, and ay is the acceleration of the hanging object. If we assume that the

hanging mass is not accelerating, and thus 𝑎! = 0 !!!

, we can solve the above equation for FT and

find the following expression.

𝐹! = 𝑚𝑔 (3)

In addition, we can look at the forces acting on the block resting on the table. We must consider

the forces separately for the x and y directions. Let’s begin with the forces acting in the y

direction on the block.

Σ𝐹! = 𝐹! −𝑀𝑔 = 𝑀𝑎!   (4)

where FN is the normal force acting on the block, M is the mass of the block, and ay is the

acceleration of the block in the y direction. Since the block isn’t accelerating in the y direction,

we can set 𝑎! = 0 !!!

and solve the equation for FN.

𝐹! = 𝑀𝑔 (5)

Now, we need to examine the forces acting on the block in the x direction by taking the sum of

the forces.

Σ𝐹! = 𝐹! − 𝑓! = 𝑀𝑎! (6)

fk is the force due to kinetic friction. If we assume that the block only just starts to move and is

not accelerating, we can set 𝑎! = 0 !!!

. Since the string attached to the block is also attached to

the hanging mass, we can assume that FT is the same for both the block and the hanging mass.

Solving the above equation for FT, we find the following.

𝐹! = 𝑚𝑔 = 𝑓! (7)

If we solve Equation (1) for µk and substitute Equation (7) for fk, we obtain the following

equation for the coefficient of kinetic friction.

𝜇! =𝐹𝑇𝐹𝑁

= 𝑚𝑔𝑀𝑔 =

𝑚𝑀 (8)

In this experiment we use a small, rectangular wooden block with Teflon tape on one side to

examine the coefficient of kinetic friction. The Teflon tape allows us to see the effect of different

surface types on the coefficient of friction. We use trombone and butterfly paperclips on a hanger

to apply a force of tension to the block. We vary the mass (M) of the block by adding small

masses to the top of it, this will increase the normal force.

Results

We measured the mass of the wooden block, one trombone paperclip, one butterfly paperclip,

and the hanger with the digital scale and recorded the data in Table 1 below.

Table 1: Mass Values. These are the masses of the different objects that we use in the experiment.

We placed a small, rectangular wooden block on the surface of a table and attached a hanging

mass to the block via a string and a pulley, as shown in Figure 1 below.

Figure 1: Experimental Setup. This is a drawing of the setup that we used in this experiment.

We determined the amount of force necessary to overcome the static friction by adding

paperclips to the hanger until the block just started to move. We recorded the mass of the clips

and the hanger in Table 2. In the first line in Table 2, the added mass, m, is equal to zero. In each

successive trial we added 0.050kg to the block up to a maximum amount 0.250kg.

In Part B, we turned the block so that a different side, with a different surface area, was in

contact with the table. The procedure was then repeated with this new surface to find the

coefficient of friction. In Part C, the block was flipped again, this time placing the side with the

Teflon tape in contact with the table. The same trials and calculations were repeated in order to

obtain a value for the coefficient of kinetic friction. Table 2 shows all of the data collected from

each part. Part A represents the first set of trials. Part B represents the second set where the

Object Mass (kg)Mblock 0.0641mTrombone 0.001mButterfly 0.003mhanger 0.005

y

x

surface area was changed. Part C represents the trials run with the Teflon tape side in contact

with the table.

Table 2: Experimental Data. This is the data we obtained to determine the coefficient of kinetic friction for the block.

We calculated the tension in the string using Equation (3). A sample calculation is shown below

using the data from Part A trial 2, where 6 butterfly paperclips are needed to move the block.

𝐹! = 𝑚!𝑔   = 0.0050𝑘𝑔 + 6 ∗ 0.0030𝑘𝑔 ∗ 9.81𝑚𝑠!  = 0.226  𝑁

We calculated the normal force acting on the block using Equation (5). A sample calculation is

shown below for this trial.

𝐹! = 𝑀 +𝑚 𝑔   = 0.0641𝑘𝑔 + 0.0500𝑘𝑔 ∗ 9.81𝑚𝑠!  1.119  𝑁

We used these two values in Equation (8) to find a value for the coefficient of kinetic friction for

this trial.

𝜇! =𝐹!𝐹!

𝜇! =0.226  𝑁1.119  𝑁

𝜇! = 0.202

It is important to note that the coefficient of kinetic friction is a unitless quantity.

A B C A B C A B C0.0641 0.629 0.014 0.017 0.008 0.137 0.167 0.078 0.218 0.265 0.1250.1141 1.119 0.023 0.026 0.011 0.226 0.255 0.112 0.202 0.228 0.1000.1641 1.610 0.032 0.036 0.016 0.314 0.353 0.157 0.195 0.219 0.0980.2141 2.100 0.042 0.042 0.019 0.412 0.412 0.190 0.196 0.196 0.0910.2641 2.591 0.051 0.051 0.022 0.500 0.500 0.216 0.193 0.193 0.0830.3141 3.081 0.060 0.054 0.025 0.589 0.530 0.249 0.191 0.172 0.081

0.199 0.212 0.096

µk

Average

M+m (kg) FN (N)mClips (kg) FT (N)

We notice a few trends in the data shown in Table 2. We can see that as the mass of the block

increases, the total force of tension also increases. We can also see that, despite the change in

surface area, the coefficients of friction do not vary by much between parts A and B. If we take a

percent difference between these two values, we find the following result.

%  𝐷𝑖𝑓𝑓 =|𝜇!,!"#,! − 𝜇!,!"#,!|12 𝜇!,!"#,! + 𝜇!,!"#,!

∗ 100%

%  𝐷𝑖𝑓𝑓 =|0.199− 0.212|12 0.199+ 0.212

∗ 100%

%  𝐷𝑖𝑓𝑓 = 6.35%

This percent difference is relatively small and leads us to conclude that sides A and B have the

same value for the coefficient.

We can also see in the table that side C’s coefficient of friction is very different from that of

sides A and B. If we take an average of the coefficients in side A and B, 0.206, and compare it to

side C, the percent difference, using the same calculation as above, is 72.6%. So we can

conclude that the coefficient of kinetic friction does depend on the type of surface. One last trend

we note is that the first trial of each part is slightly higher than the rest and the value of the

coefficient seems to decrease slightly as we increase the mass of the block.

We used the data in Table 2 above to plot FT vs. FN for each part so we could see trends in the

data more clearly. According to Equation (8), the slope of these lines should be equal to the

coefficient of friction for each set of trials. The percent differences are shown in Table 3.

Comparison % Difference µk, avg, A to slope from Part A 3.09 µk, avg, B to slope from Part B 11.0 µk, avg, C to slope from Part C 10.9

Table 3: Comparison of µk values. This indicates that the average coefficient of friction from each of the parts is close to the value from the slopes of the graph.

Figure 2: 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 and the tension required to move the block will be zero.

We can see that the best-fit lines for data sets A and B are very close to one another and have

very similar values for the slope, which is the coefficient of kinetic friction. So we say that the

surface area has no significant influence on friction. The slope from Part C is very different from

A and B so we can say that the type of surface has a significant influence on friction.

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! (5)

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!

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′! =𝑚!𝑔 −𝑚!𝑎 − 𝑀 +𝑚 𝑎

𝐹!

𝜇′! =0.0139𝑘𝑔 ∗ 9.8𝑚𝑠! − 0.0139𝑘𝑔 ∗ 0.02

𝑚𝑠! − 0.641𝑘𝑔 + 0.00𝑘𝑔 0.02𝑚𝑠!

0.628𝑁

𝜇′! = 0.214

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 2, the graph of FT vs. FN.

Figure 3: 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 µk values, we find a percent difference

A B C A B C A B C0.0641 0.629 0.014 0.017 0.008 0.137 0.166 0.078 0.216 0.245 0.1240.1141 1.119 0.023 0.026 0.011 0.225 0.255 0.112 0.199 0.207 0.0990.1641 1.610 0.032 0.036 0.016 0.313 0.352 0.157 0.193 0.199 0.0970.2141 2.100 0.042 0.042 0.019 0.411 0.411 0.190 0.194 0.176 0.0900.2641 2.591 0.051 0.051 0.022 0.499 0.499 0.215 0.191 0.173 0.0830.3141 3.081 0.060 0.054 0.025 0.587 0.529 0.249 0.189 0.152 0.080

0.197 0.192 0.096Average

M+m (kg) FN (N)mClips (kg) FT (N) µ'k

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"! =𝑚!𝑔 − 𝐹!

𝐹!

𝜇"! =0.014𝑘𝑔 ∗ 9.81𝑚𝑠! − 0.005𝑁

0.628𝑁

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.

Conclusion

In this experiment, we examined kinetic friction and looked at the factors that affect the value of

the coefficient of kinetic friction. We examined whether or not the surface area of the objects

affects the coefficient. We used different surface areas in Part A and Part B and found the

coefficients of kinetic friction we very similar with a percent difference of only 6.35%. Thus, we

can say that the surface area of an object does not affect the coefficient of kinetic friction or the

force of friction. We tested whether or not the type of surface has an effect on the coefficient of

kinetic friction by changing the type of surface used in Part C from what was used in Parts A and

B. We found the coefficient of kinetic friction in Part C and the average coefficient of kinetic

friction from Parts A and B were very different with a large percent difference of 72.6%. Thus

we can say that the type of surface significantly affects the coefficient of kinetic friction and the

force of friction.

The magnitude of the force of friction does depend on the normal force between the two objects.

We saw this in Equation (1), 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 Equation (7) states 𝐹! = 𝑓!, we can see that

as the normal force increases, the force of friction increases.

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.

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.

Finally, if we wanted to reduce error in the experiment, we may want to use a material with a

higher value for the coefficient of kinetic friction, such as rubber. If we use a higher coefficient,

any small changes in the measured value of the coefficient will make less impact as the actual

value will be much higher.