November 17, 2002 

The Coefficient of Friction (µ)

Abstract:

To determine the coefficient of friction, µ, by find the force necessary to overcome friction.

The coefficient of static friction was verified by using inclines.

  *        *        *

                The first part of this lab was to determine the force necessary to overcome friction. To do this, a block with weights in it was pushed by the force probe, which was attached to the GAX Grapher computer program. We graphed force as a function of time.  When the object started to move, the escalating line on the graph peaked and went down, then leveled. The peak point on this graph was used to determine static friction, the leveled points to determine kinetic friction.

                To determine the force of the weights for each run, we first had to find the force exerted by each individual weight:

The first surface we used was a block of wood. For each surface, we did three tests: one with all the weights in the block, one with all but the 1000g, and one with all but the 500g and the 200gA.

                This is the graph created for the first run, all the weights in the block on the wood:

Run 1 (Force versus Time)

The peak point here was 10.8N. This value was used to determine static friction. To determine µ_s, we used the equation

F_f = µmg

The force was plugged in as the peak point, the mass was the total of all the masses, and g remains the same in all equations. Thus, we were finding µ. Here is an example of how we did this first calculation:

F_f = µmg

10.8 = µ(2.338)(9.8)

µ = .471

To verify this value of µ, we used inclines. When the force of friction and the force down an incline are equal, an object will stay in place. When the object has begun to slide down the incline, however, the force of friction has been overcome. By catching the angle at which the object on the incline begins to slide, the coefficient can be determined:

µmg (cos theta) _{(force down the incline)} = mg (sin theta) _{(force up the incline)}

µ = tan (theta)

The angle at which the block with the weights in it began to slide on the wood was 34˚. The tangent of this angle is .675. This value SHOULD be the same as the first calculated value of mu, and should serve as a verification. Ours, however is not. To calculate percent error, the difference between the two values was taken, and divided by what we should have gotten, which in this case was the tangent of theta. For this run, the percent error was 30.2%.

                The second type of friction present is kinetic friction. The formula for kinetic friction is the same,

F_f = µmg

Here, however, the force is not the peak of the graph, but the average point at which the graph levels off. For this run, this was 7.0.

F_f = µmg

7 = µ(2.338)(9.8)

µ = .3055

This mu is the coefficient of kinetic friction. It could not be verified in the same manner as which the other was verified.

                This same process was repeated five more times, twice on the wood, and then the same three trials on a Physics book that was present in the lab room. All the calculations were carried out in the same manner. Here are the results:

Run 2 (Force versus Time)

Run 3 (Force versus Time)

Run 4 (Force versus Time)

Run 5

no graph – see error analysis

Run 6 (Force versus Time)

To recap:

Error Analysis

** In our work, one of our biggest mistakes was in the saving of the graphs we created graphing force versus time. The fourth and fifth graphs (all items on the book, and all minus the 1 kg weight on the book) both are the same picture as saved, but they were not really the same result; thus, there was a saving mistake.

However, we could still do part of the calculations for this run because we knew the angle at which static friction was overcome, and could thus work backwards to find what the peak of the graph would have been. To find mu, we took the tangent of this theta, 25˚. Then, we plugged µ, mass, and gravity back into the first equation, F  =  µmg to find F.

F_f = µmg

F = .466(1.338)(9.8)

F = 10.7 N

The biggest percentage of error between two values of  µ was on this run, which was undoubtedly due to this fudging of calculations in order to find part of the process.

                The more standard error we had in this graph was with µ itself. While the two values of the coefficient of static friction as found from the graph and from the incline exercise should have been the same, ours were greatly different, resulting in error. As shown by the table, our percentages of error were most commonly in the 30-percentile, with two much higher, and one at a mere 1.5%.

In this lab, there was a lot of room for error. Potential places for error include: random scratches in the wood board which would increase the friction for only a moment, the uneven force applied by us to the object, the inconsistency of the program (for instance, the forces applied by the two different 200 g weight were 1.92 N and 1.95 N, but since both weights were supposed to be the same mass, they should have exerted the same force) and perhaps the inaccuracy of the mass of the objects, and misreading of the force graphs.