Coefficient of Friction
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1 HOUSTON COMMUNITY COLLEGE SYSTEMS SOUTHWEST COLLEGE COLLEGE PHYSICS I PHYS 1401 PRE LAB QUESTIONS Due before lab begins. Coefficient of Friction 1) Explain briefly the different types of frictional forces. 2) How does the coefficient of friction depend upon the area of contact? Explain. 3) Give some examples in which the force of friction causes the object to accelerate. 4) If you push on a heavy box that is at rest, you must exert some force to start its motion. However, once the box is in motion you need a smaller force to maintain that motion. Why? 1
2 5) Find the force needed to pull a mass of 30 kg at constant velocity on a rough incline making an angle of 40 0 with horizontal and with µ k = 0.3 OBJECTIVE The nature of the frictional force between two wooden surfaces will be investigated. This will lead to the observation of the properties of friction. The following objectives will be accomplished: 1. the determination of the coefficients of static friction, µ s, and kinetic friction, µ k, between two wood surfaces by sliding the block down the board. 2. Determination of the coefficient of friction by using the board as an inclined plane. 3. Comparison of the two values of µ s from the two different methods. 4. Demonstration that the coefficients of friction are independent of the surface area of contact. 5. Verification that the coefficient of static friction, µ s, is in general slightly larger than the coefficient of kinetic friction, µ k, for any same type of surfaces. MATERIALS 1. Wooden Block 2. Weight hanger 3. Wooden Board 4. Slotted weights 5. Triple-beam balance 6. Pulley 7. Protractor 8. String 9. Paper towel 10. Two sheets of Cartesian graph paper 2
3 INTRODUCTION Friction is a resistive force that occurs due to motion between two contacting surfaces. The two surfaces must either be sliding or attempting to slide over another resulting in a force that is tangent to the two surfaces and in a direction that tends to oppose the motion of each surface. In general, frictional force depends on the nature and physical properties of the surfaces in contact. It slows down the motion of an object and is directed along the two surfaces in contact in the opposite direction of the motion. Friction is always called a non-conservative force because some energy is needed to overcome it. The following conclusions can be made of friction, Friction between two surfaces is: (i) (ii) (iii) (iv) Independent of the surface area of contact Independent of the sliding speed of the surfaces Proportional to the normal force acting on the surfaces Acts in a direction opposite to the direction of motion of each surface. On the other hand the coefficients of friction µ s and µ k, which are constants, depend on the nature of the surfaces in contact and are independent of the normal force. THEORY: There are two types of frictional forces: a) Static friction ( f s ) This exists when the object is at rest relative to the surface. This force must be overcome in order to make the object start moving. It is given by f s µ s n. b) Kinetic friction ( f k ) This exists when the object is in motion and is given by f k = µ k n (where µ s and µ k are the coefficient of static and kinetic friction respectively, and n is the normal force which presses the two surfaces together). In general, f s > f k and µ s > µ k because it takes a larger force to start an object sliding (static friction) than to keep it sliding (kinetic friction). As mentioned above the coefficient of friction, µ, (pronounced mu) is depends on the nature of the surfaces and is independent of the area of contact. In this experiment, using the board in a horizontal position, 3
4 we measure the frictional force f k and f s as they vary with respect to the normal force n. Using the second law of motion, we can calculate the coefficients of friction between block and board. Another way to find µ s is to set up the board as an incline plane. The coefficient of friction µ s is related to the maximum angle θ m to which the board can be elevated before the block starts its motion is: In the case of an inclined plane, θ m can be adjusted. When the block is placed on the plane, the angle is slowly increased till a point is reached at which the block begins to slip. Accordingly, the forces on block include the normal force, N, which acts perpendicularly to the plane, and the components of the weight of the block, mgcosθ m, which acts in the opposite direction. Since the plane is in equilibrium as far as perpendicular motion is involved, these two forces are equal, thus N = mgcosθ m (1) Where is the angle at which the block just begins to slip down the inclined plane. There are also two forces parallel to the plane. There are the component of the weight of the block, mgsinθ m, which acts down the plane, and the frictional force f s which acts up the plane. At the point that the blockb just slips, the maximum frictional force is exerted, and these two forces are equal, that is f s = µsn = mgsinθ m (2) but N == mgcosθ m in equation (1) above. When the two equations are combined, the result is Hence µ s = tanθ m in this case. µ s = mgsinθ m = mgsinθ m = tanθ m (3) N mgcosθ m EXPERIMENTAL PROCEDURE 1. Weigh the wooden block on the triple beam balance and record its weight 2. Set up the block on the board with the largest surface in contact with the board surface and string attached to the block run over a pulley. Place some weights on the weight hanger. Slightly increase the load on the weight hanger until the block begins to move slowly with a constant speed after it has been started with a very small 4
5 push. Make sure you wipe away any dust from the surfaces. Record the weight placed on the weight hanger including the mass of the weight hanger. 200 g on the surface of the block. Slowly increase the load on the hanger until the block starts slowly moving with constant speed after given a small push. Don t forget to include the mass of the hanger. 3. Repeat Procedure 2 above placing masses of 200, 400, 600, 800 and 1000 g successively on top of the wood block. Record the total weights placed on the weight hanger including the weight hanger. The coefficient of kinetic friction can be obtained from these data. 4. Turn and place the smallest side of the wood block on the board and repeat Procedure 2 above. 5. Set up the block again as in procedure 2 and place a mass of 600 g on it. Place weights gently on the hanger and increase them slowly until the block just starts its motion without any push. Repeat a total of 3 independent trials and record your data. With this data you can find the coefficient of static friction. 6. Remove the pulley and set up the board as an inclined plane. Place the wood block on it with its largest surface being in contact. Gently and smoothly tip the board until block starts to slide down the plane. Measure the angle θ m with a protractor and record it in the table. Repeat this procedure 3 independent times. REPORT FORM Weight of block Procedure 2 Normal and Friction force for kinetic friction Mass on block 200 g 400 g 600 g 800 g 1000 g Total Normal Force, N Friction Force, N µ k from graph Procedure 3 Normal and Friction force for static friction Mass on block 600 g 600 g 600 g Averages Total Normal Force, N Friction Force, N µ s Deviation 5
6 Procedure 4 Inclined plane for static friction Measurement Angle θ m µ s Deviation Averages Percent difference between the two values for µ s CALCULATIONS 1) From procedure 2, draw a graph using the data of the force of friction as the ordinate and those of the normal force as the abscissas. Draw the best straight line joining most of the points. Obtain the slope of this graph. The slope so obtained is the coefficient of kinetic friction, µ k, between the wooden block and board. 2) From procedure 3, calculate the value for µ s between block and board. Obtain their average value and average deviation. 3) From procedure 4, calculate the value for µ s between block and board. Find their average value and average deviation. Compare the two values for µ s for wood on wood by calculating their percent difference which is given by: Percent difference = [difference of the two values/ average] x 100 % 6
7 POST LAB QUESTIONS 1. Explain why the block should move at constant speed in procedure Derive equation 1, µ s = tan θ m ( apply 2 nd law of motion for constant velocity) 3. In procedure 4, find the acceleration of the block as it breaks away if µ k = 0.2 and θ m = Compare your experimental value with the book value for the coefficient of kinetic friction for wood( µ k = 0.2). Find the percent error and give reasons for it. 5. Is the frictional force in this experiment only due to the surface of contact between block and board? Explain. 7
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