To determine the coefficient of kinetic friction between two selected surfaces by two different methods
An incline, a wooden box, a weight hanger, a set of slotted weights, a cord and pulley system, a sheet of graph paper, and a scientific calculator
The Coefficient of friction is defined as the ratio of friction force to the normal force, μ = F /N . Consider the following two cases: One measures the force of static friction and the other the force of kinetic friction.
At constant velocity to the Right, we must have F = Fk . This means that measuring the pulling force F is like measuring the force of static friction Fk .
On the verge of motion to the Right, F = Fs . This means that measuring the pulling force F is like measuring the force of static friction Fs .
μ k = Fk /N . Here, F = Fk .
Fk = Force of kinetic friction, μk = coefficient of kinetic friction, and N = normal to the contacting surfaces.
μ s = Fs/N . Here, F = Fs .
Fs = Force of static friction, μs = coefficient of static friction, and N = normal to the contacting surfaces.
The force of friction always acts against the direction of motion. Note that Fk < Fs and consequently, μk < μs .
If the externally applied force F just equals the force of static friction, Fs, then the object is on the verge of slipping, and the coefficient of friction involved is called the coefficient of static friction, μs.
If the externally applied force F just equals the force of kinetic friction, Fk, then the object slides at a constant velocity, and the coefficient of friction involved is called the coefficient of kinetic friction, μk .
In order to find the coefficient of kinetic friction between two surfaces, the following two experimental procedures are to be followed:
I) The Inclined surface Method:
In this case the tangent of the angle at which a block slides down on an incline at a constant velocity gives the coefficient of kinetic friction between the bottom of the block and the top of the incline. μk = tanθk
The proof is as follows:
| First, it is easy to prove that when the sides of
two angles are perpendicular to each other, the two angles are
equal, as shown Fig. 3 below.
In each of the two right triangles the sum of angles must be 180o.
Each has a 90o-angle. Also, Angles 1 and 2 are opposite
and therefore equal. Their third
angles must therefore
be equal. If one is θ ,
the other has to be equal to θ as
well. This argument is used in Fig.
Note: N is the same thing as FN, the normal force.
In Fig. 4, the weight force W is the force of gravity acting on the block causing a normal force to develop by the incline. This normal force N < W.
W can be replaced with F┴ = Wcosθ & F|| = Wsinθ. Verify this by paying more attention to the right triangles in Fig. 4.
Also, note that for equilibrium, we must have N = F┴ or N = Wcosθ. Why?
At a certain special angle of inclination, θk, at which the block slides down at constant velocity, Fk and F|| become equal:
Fk = F|| or Fk = Wsinθk. Why? Since, μk = Fk /N , we may write:
μ k = Fk /N = Wsinθk/(Wcosθk), (verify) or
μk = tanθk
II) Horizontal surface Method:
Referring to Fig. 5, it is clear that M1 moves to the right because of F = M2g. F causes a tension in the cord that transmits to block M1 that pulls it. M1 is resisted by the friction force Fk . If F is much greater than Fk, motion definitely occurs and the system accelerates. We want motion to occur but at zero acceleration that means at a constant velocity. For that to happen the pulling force F must exactly equal the friction force Fk . If you keep increasing M2 until F = M2g equals Fk , with a little tapping on the surface motion will occur at constant velocity. At constant velocity of M1 to the right, we may write:
μ k= Fk /N = F /W or, μ k = M2g /M1g , or,
μ k = M2/M1
In this experiment, since the coefficient of kinetic friction is to be measured, all cases of constant speed motion of the block must be accompanied by tapping the incline. This causes vibration in the system and avoids the measurement of the coefficient of static friction instead. The goal is to measure the coefficient of kinetic friction.
I) Inclined Plane Method:
Measure the mass of the wooden box and record its value. Place the wooden box on the incline and add a mass of 200 grams to it. Gradually increase the inclination angle until the box slides down at a constant velocity (See Fig. 4). Read the value of the angle and name it θk. Use the following formula to calculate μk.
μk = tanθk .
Repeat this procedure 4 more times, with 300, 400, 500, and 600 grams added to the box in different cases according to Table 1 under Data. Measure the angle at which the block slides down at a constant velocity each time. Make sure that the added weight in each case is evenly distributed in the box. Using μk = tanθk, find μk in each case and then find the average of those five values. Name it (μk)I that means the μk found by Method I.
II) Horizontal Plane Method:
Set the plane in the horizontal position and put M1 = 200 grams inside the box. Attach a cord and a weight hanger to the box as shown in Fig. 5. Gradually increase the load on the weight hanger until the box slides at a constant velocity. Record load M2 that includes the mass of the weight hanger. Repeat this procedure 4 more times for different masses of 300, 400, 500, and 600 grams added to the box subsequently, and find the necessary hanging weight in each case that pulls the box at constant velocity. In each of the five cases, calculate μk by using the following formula:
μk = M2 / M1
Finally, find the average of the five values you obtained for μk and name it (μk)II that means the μk found by Method II.
Calculate the percent difference between the mean values of the coefficient of kinetic friction found by Method I and the value found by Method II ;in other words, compare the mean value of Method I to the mean value of Method II.
Graph the values of Fk versus N found in Method II.
This is the same thing as graphing the values of M2g versus M1g. Note that M2g versus M1g is the same as M2 versus M1.
(μk)I = tanθk
|box+200 =||box+200 =|
|box+300 =||box+300 =|
|box+400 =||box+400 =|
|box+500 =||box+500 =|
|box+600 =||box+600 =|
Comparison of the results:
Write the percent difference formula down and calculate the percent difference.
State your conclusions of the experiment.
Provide a discussion if necessary.
1) Is the coefficient of kinetic friction the same for two surfaces regardless of the normal force? Why? Is your answer the same for the coefficient of static friction?
2) How does the value of the coefficient of static friction compare with the value of the coefficient of kinetic friction for the same two surfaces?
3) How does the slope of the graph (in Case 4) compare with the coefficient of friction?