To determine the coefficient of kinetic friction between two selected surfaces by applying two different methods
An incline, a wooden block, 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 force of friction to the normal force, μ = F / N . Consider the following two cases (Figures). One is for static friction and the other for kinetic friction
At constant velocity to the Right, F = F k.
On the verge of motion to the Right, F = Fs
μ k = Fk /
Here, F = Fk
Fk = Force of kinetic friction, μk = coefficient of kinetic friction, N = Normal force or the force perpendicular to the contacting surfaces.
μ s = Fs /
Here, F = Fs
Fs = Force of static friction, μs = coefficient of static friction, N = Normal force or the force perpendicular to the contacting surfaces.
The force of friction acts against the direction of motion. Note that Fk < Fs and consequently, μk < μs .
If the externally applied force (F) is just equal to 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) is equal to the force of kinetic friction, Fk, then the object slides at constant speed, 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) Inclined Plane Method:
In this case the tangent of the angle at which a block slides down on an incline at constant velocity gives the coefficient of kinetic friction between the bottom of the block and the top of the incline. [ μ k = tan θk ]
The reason 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 right triangles the sum of angles must be
180°. Each has a 90°-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. 4
Note : N is the same thing as FN, the normal force.
Note that if weight (W) is the only force acting on the block causing a normal force to be developed, on an incline, N < W, but on a horizontal surface N = W.
In Fig. 4, W can be replaced with (F┴ = W cos θ) and (F|| = W sin θ). Verify this by paying more attention to the figure.
Also, note that for equilibrium, we must have N = F┴ or N = W cos θ. Why?
At a certain angle of inclination, θk, at which the block slides down at constant velocity, Fk and F|| become equal:
Fk = F||, or Fk = W sin θk. Why?
Since, by definition μk = Fk / N, we may write:
μ k = ( Fk / N ) = (W sin θk) / (W cos θk), (verify) or μk = tan θk.
II) Horizontal Plane 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 and pulls it. M1 is resisted by the friction force Fk. With tapping the horizontal surface, F = Fk only when M1 slides to the right at constant velocity. At a 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. You are trying 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.g to it. Gradually elevate the incline until the box slides down at constant velocity (See Fig. 4). Clamp the plane in this position and read the value of the angle (θk). Use the following formula to calculate μk.
μk = tan θk
Repeat this procedure four more times, once with 300.gr, once with 400.gr, ... and once with 600.gr of mass added to the box, and measure the angle at which the block slides down at constant velocity each time. Each time, make sure that the added weight is evenly distributed in the box. Using (μk = tan θk), find μk in each case and then average 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.gr 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 constant velocity. Record the load (M2) that includes the mass of the weight hanger. Repeat this procedure four more times, once with 300.gr, once with 400.gr, ..., and once with 600gr added to the box, 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 = (Fk/N) = (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 value 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.)
|Method I||Method II|
|(M1)1 = box + 200g||(θk)1 =||(M1)1 = box + 200 g||(M2)1 =|
|(M1)2 = box + 300g||(θk)2 =||(M1)2 = box + 300 g||(M2)2 =|
|(M1)3 = box + 400g||(θk)3 =||(M1)3 = box + 400 g||(M2)3 =|
|(M1)4 = box + 500g||(θk)4 =||(M1)4 = box + 500 g||(M2)4 =|
|(M1)5 = box + 600g||(θk)5 =||(M1)5 = box + 600 g||(M2)5 =|
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.
Questions for Discussion: