Experiment 5: Coefficients of Friction
Purpose
(1) To become familiar with the concepts of static and kinetic friction.
(2) To measure the coefficients of static and kinetic friction for a plane.
Theory
I. Static Friction on a Horizontal Board
If an object, resting on a horizontal surface is pulled by a horizontal force F (Fig 1), the
surface will exert a force of friction fs (force of static friction) which exactly balances F until F
reaches a critical value Fcrit. Fcrit is the maximum value of the static force of friction fmax.
Until this point, the net force is zero and the object remains at rest. Above this point the net force
is not zero and the object will move. Experiments show that fmax is proportional to the normal
force N:
○,1
fmax = Fcrit = s N
where s is the coefficient of static friction.
22
Experiment 5
In your experiment, an object of mass M will be sitting on an inclined plane in a
horizontal position. A force F will be applied by hanging a mass m from a string attached to the
mass. (Fig. 2) You will gradually increase the hanging mass until the force from its weight mg
equals the maximum force of static friction fmax. Beyond that point the object would slide since
the net force is no longer zero .
II. Static Friction on an Inclined Board
On an inclined plane the force of gravity Mg is downward. There are components of the force of
gravity parallel and perpendicular to the surface of the plane. Figure 3 shows the components
parallel and perpendicular to the surface of the inclined plane. The normal force N on the block is
equal to the component of the force of gravity that is perpendicular to the surface
Mblock g coss. The gravitational force parallel to the plane which can cause the block to slide
down is Mblock g sins. If the block doesn’t move, the force of friction is equal in magnitude
23
Experiment 5
and opposite in direction to the force of gravity down the slope (parallel to the surface of the
plane). Equation 1 still applies: fmax = s N = Mblock g coss.
III. Kinetic Friction
Once the object is moving, the frictional force is called the force of kinetic friction fk:
2 fk = k N
where k is now the coefficient of kinetic friction. It is constant regardless of the speed of the
object. If the frictional force fk is less than fmax, the motion can be maintained by applying a
force F that is less than Fcrit. When the force F exactly equals the force of kinetic friction fk,
the net force will be zero and the moving object will move at a constant speed. As in the static
case, the applied force will be due to either hanging weights or the force of gravity on the block
itself.
Apparatus
(1) Friction board
(2) Slotted weights
(3) Equal arm balance
Procedure  Part I: Determination of s with horizontal board
(1) Measure and record the mass Mblock of your block.
(2) Place the board in horizontal position. Put the block on the board and adjust the pulley so
that the string is horizontal.
(3) Hold the block above the board and load the hanger with a few weights. Lower the block
and weights and release everything very gently. The block should remain at rest. If it moves no
matter how gently you release it, try again with fewer weights. You may have exceeded the force
of static friction.
(4) If the block remains at rest increase the load further. Try releasing the block at different
spots on the board to determine its general behavior. Find the largest load for which the block
will stay at rest. Record the value of the weights mweights. The total load mload is the value of
mweights plus the mass of the hanger mhanger.
Part II: Determination of s with inclined board
(5) Remove the block and disengage the hanger. Raise the board to about 8 degrees from the
horizontal. Place the block on the board and release it very gently. It should remain at rest.
24
Experiment 5
(6) Gradually increase the angle and determine the largest angle s for which the block will stay
at rest. Record s with an accuracy of 0.1 degrees.
Part III: Determination of k with horizontal board
(7) Place the board in horizontal position. Put the block on the board and adjust the pulley so
that the string is horizontal.
(8) Put some weights on the hanger and tap the board with your finger so that the block slides
slowly at a constant speed. Your goal is the make the force from the weights equal the force of
kinetic friction. If you do not have enough weight, the block will stop. If you have too much
weight, the block will accelerate. Determine the smallest load mweightswhich will make the
block slide very slowly without stopping.
(9) Repeat this procedure with extra masses Mextra of 1, 2, 3 and 4 kg put on top of the block.
Record in a table the smallest loads for each which will result in a slow uniform motion.
Remember that we must add the mass of the hanger to the mass of the weights to get the total
load.
Mextra
(g)
mweights
(g)
mload = mweights
(g)
+ mhanger
0
1000
2000
3000
4000
Part IV: Determination of k with inclined board
(10) Remove the block and disengage the hanger. Raise the board to about 5 degrees from the
horizontal. Place the block on the board. Tap the board slightly. If the block moves and then
stops the force of gravity from the block’s mass is too small to overcome the force of friction.
The angle is too small.
(11) Gradually increase the angle and determine the angle k for which the block will
consistently creep down (after tapping the board) without acceleration. Record k with an
accuracy of 0.1 degrees.
Before you leave the lab
Make sure you know how to draw a graph of your results in Part III and how to analyze it.
25
Experiment 5
Lab Report
Calculation of s
1. Calculate s from the equation
s Mblock g = mload g using your data.
2. Calculate sfrom your data for
Part II using Equation 3 from
right.
From Equation 1 , we know that fmax = s N. We
also know fmax = mload g. Therefore,
fmax = s N = s Mblock g
fmax = mload g
s Mblock g = mload g
In Part II, the maximum force of friction is equal to the maximum
force of gravity down the slope (ie parallel to the surface of the
plane) for which the block doesn’t slide.
fmax = Fgravity
s Mblock g cos s = Mblock g sin s
scos s = sin s
s= sin s = tan s
Equation:
3
cos s
3. Display the accuracy of your work by using the following formula:
s(Part I) - s(Part II)
.5 [s(Part I) + s(Part II)]
Notice that the denominator is the average of the Part I and Part II values for s.
Calculation of k
4. Make a plot of mload vs Mextra from your data table for Part III and draw a line of best fit.
Don’t forget to include the value for Mextra= 0. Calculate the slope and find the y intercept of
your best fit line.
26
Experiment 5
5. Write down the slope “m” of your
best fit line. “m” = k
6. (a) Calculate Mblock from the y
intercept. Use your experimental
value of k from above in (5):
y intercept = k Mblock
Mblock = y intercept
k
We can calculate k from the equation for this plot:
Since the block is moving at a constant speed, the force of
kinetic friction fk equals the applied force Fapplied.
fk = Fapplied
fk = k N = k (Mextra + Mblock)g.
Fapplied=mload g
k (Mextra + Mblock) g = mload g
Therefore, canceling g:
mload = k Mextra + k Mblock
This equation is in the form of a linear plot y = mx + b. In our
equation “y” is mload, “x” is Mextra, the slope “m” is k and the y
intercept “b” is k Mblock.
(b) Find the percent discrepancy between the value of Mblock found from the y intercept and the
value found from the equal arm balance. Use the equal arm value as the accepted value.
7. Show the relationship between
k and k for Part IV. Calculate k
from your data.
In Part IV, since the block is not accelerating, we know that the
force of kinetic friction equals the force of gravity down the
slope.
fk = Fgravity
k N = k (Mblock) g cos k
Fgravity = (Mblock) g sin k
k (Mblock) g cos k = (Mblock) g sin k
8. Display the accuracy of your work by using the following formula:
k(Part III) - k(Part IV)
.5 [k(Part III) + k(Part IV)]
Here the denominator is the average of the Part III and Part IV values for k.
27