Experiment 5

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 = ms N

where ms is the coefficient of static friction.


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 = mk N

where mk 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).


Apparatus

(1) Friction board

(2) Slotted weights

(3) Equal arm balance

Procedure - Part I: Determination of ms 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 ms 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 qs for which the block will stay at rest. Record qs with an accuracy of 0.1 degrees.

Part III: Determination of mk 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. 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 mk 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 qk for which the block will consistently creep down (after tapping the board) without acceleration. Record qk 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.


Lab Report

Calculation of ms

1. From Equation 1 , we know that fmax = ms N and we also know fmax = mload g. Therefore,

ms N = ms Mblock g = mload g

Calculate ms 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 cosqs. The gravitational force parallel to the plane which can cause the block to slide down is Mblock g sinqs. (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

ms Mblock g cos qs = Mblock g sin qs

mscos qs = sin qs

3 ms= sin qs = tan qs

cos qs

Calculate msfrom your data for Part II using Equation 3 .

3. Display the accuracy of your work by using the following formula:

ms(Part I) - ms(Part II)

.5 [ms(Part I) + ms(Part II)]

Calculation of mk

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 mk 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 = mk N = mk (Mextra + Mblock)g. Therefore

fk = Fapplied

mk N = mk (Mextra + Mblock) g = mload g

mload = mk Mextra + mk Mblock

This is in the form of a linear plot with mload as “y”, Mextra as “x”, a slope = mk and a y intercept of mk Mblock. Write down mk from the slope.

6. Calculate Mblock from the y intercept using your experimental value of mk found in (2) above as follows:

y intercept = mk Mblock

Mblock = y intercept

mk

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

mk N = mk (Mblock) g cos qk

Fgravity = (Mblock) g sin qk

mk (Mblock) g cos qk = (Mblock) g sin qk

Show the relationship between qk and mk for Part IV. Calculate mk from your data.

8. Display the accuracy of your work by using the following formula:

mk(Part III) - mk(Part IV)

.5 [mk(Part III) + mk(Part IV)]

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