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  Static Friction
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.
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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 mass until the force from its weight equals the
maximum force of static friction fmax. Beyond that point the object will slide since the net force
is no longer zero .
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 the coefficient of kinetic friction. It is constant regardless of the speed of the object.
The frictional force fk is less than fmax so motion can be maintained (but not started) by a force
F which is less than Fcrit .
If the force F exactly equals the force of kinetic friction fk then the net force will be zero
and the moving object will move at a constant speed (there is no acceleration).
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Experiment 5
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
(1) 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.
(2) 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
(1) Place the board in horizontal position. Put the block on the board and adjust the pulley so
that the string is horizontal.
(2) 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.
(3) 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.
24
Experiment 5
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
mhanger
(g)
+
0
1000
2000
3000
4000
Part IV: Determination of k with inclined board
(1) 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.
(2) 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.
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Experiment 5
Lab Report
Calculation of s
1. From Equation 1 , we know that fmax = s N and we also know fmax = mload g.
Therefore,
s N = s Mblock g = mload g
Calculate s from this equation, using your data.
2. 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 force of gravity 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. (See
Figure 3.)
In Part II, the maximum force of friction is equal to the maximum force of gravity down
the slope (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
3
cos s
Calculate sfrom your data for Part II using Equation 3 .
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)]
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Experiment 5
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.
Calculate the slope of the plot and find its y intercept.
5. We can calculate k by finding the equation for this plot: Since the block is moving at
constant speed, the force of kinetic friction equals the applied force. The applied force F = mload
g. From Equation 2 , we know fk = k N = k (Mextra + Mblock)g. Therefore
fk = Fapplied
k N = k (Mextra + Mblock) g = mload g
mload = k Mextra + k Mblock
This is in the form of a linear plot with mload as “y”, Mextra as “x”, a slope = k and a y
intercept of k Mblock. Write down k from the slope.
6. Calculate Mblock from the y intercept using your experimental value of k found in (2)
above as follows:
y intercept = k Mblock
Mblock = y intercept
k
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. 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
Show the relationship between k and k for Part IV. Calculate k from your data.
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)]
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