The Effect of Mass and Contact Surface Area of Block on Frictional Force and Coefficient of Friction - Lab Report Example

?Summary Sheet Objective: To be able to determine the limiting angle of repose and find out the effects of mass and contact surface area of block on frictional force and coefficient of friction Method: Weigh the wood block and record the weight in N. Set the block on the board with its largest surface in contact with the board’s surface. Through a pulley, attach the block to the weight hanger by a cord and place some weights on the hanger increasing the load to keep the block sliding slowly with constant speed. Record the load and the hanger’s weight and repeat this procedure placing masses of 200, 400, 600, 800, and 1000 g successively on top of the wood block. Turn the wood block on its side and repeat former task with a mass of 400 g on top of the block then turn the same block with the largest contact surface with the plane and place 400 g on top of the block, gradually increasing the load on the hanger until the block just starts to move. Set up the board as an inclined plane and place the wood block on the plane with its largest surface in contact, and gradually tip the plane up until the block just starts to slide down. Results: Coefficient of Kinetic Friction (based on the graph) = 0.323 ?k = 0.307 (procedure 4) ; ?s = 0.3710 (procedure 6) ?s = 0.4073 (procedure 5) ; ?k = 0.3575 (procedure 7) Applications: The concept may be applied by an engineer or specialist who needs to know the type/ property, mass of material, and angles of impending motion suitable for use in problems where friction is a significant factor in design and construction. Calculations (1) Using MS Excel, Based on this, ?k = 0.323 (2) Using the data of Procedure 4: total normal force = 6.388 N and friction force = 1.962 N and since fk = ?k * FN then, ?k = 1.962 N / 6.388 N ---? ?k = 0.307 (3) From the data of Procedure 5: ?s = force to start moving the block / total normal force Trial 1: ?s = 2.7468 N / 6.389 N ---? ?s = 0.4299 Trial 2: ?s = 2.354 N / 6.389 N ---? ?s = 0.3684 Trial 3: ?s = 2.707 N / 6.389 N ---? ?s = 0.4237 Average value (?s) = (0.4299 + 0.3684 + 0.4237) / 3 = 0.4073 Deviation (trial 1) = 0.4299 - 0.4073 = 0.0226 , |0.0226| = 0.226 Deviation (trial 2) = 0.3684 - 0.4073 = -0.0389 , |-0.0389| = 0.0389 Deviation (trial 3) = 0.4237 - 0.4073 = 0.0164 , |0.0164| = 0.0164 (4) From the data of Procedure 6: ?s = tan (?max) Trial 1: ?s = tan (19°) ---? ?s = 0.3443 Trial 2: ?s = tan (23°) ---? ?s = 0.4245 Trial 3: ?s = tan (19°) ---? ?s = 0.3443 Averaage value (?s) = (0.3443 + 0.4245 + 0.3443) / 3 = 0.3710 Deviation (trial 1) = 0.3443 - 0.3710 = -0.0267 , |-0.0267| = 0.0267 Deviation (trial 2) = 0.4245 - 0.3710 = 0.0535 , |0.0535| = 0.0535 Deviation (trial 3) = 0.3443 - 0.3710 = -0.0267 , |-0.0267| = 0.0267 Difference between two values of ?s = 0.4073 - 0.3710 = 0.0363 (5) From the data of Procedure 7: ?s = tan (?max) Trial 1: ?s = tan (21°) ---? ?s = 0.3839 Trial 2: ?s = tan (19°) ---? ?s = 0.3443 Trial 3: ?s = tan (19°) ---? ?s = 0.3443 Averaage value (?s) = (0.3839 + 0.3443 + 0.3443) / 3 = 0.3575 Deviation (trial 1) = 0.3839 - 0.3575 = 0.0264 , |0.0264| = 0.0264 Deviation (trial 2) = 0.3443 - 0.3575 = -0.0132 , |-0.0132| = 0.0132 Deviation (trial 3) = 0.3443 - 0.3575 = -0.0132 , |-0.0132| = 0.0132 Questions & Answers (1) Explain in your own words why it is necessary that the block move at constant velocity in Procedures 2 – 4. The block must move in constant velocity so that no acceleration occurs which would create a net force that would affect determination of normal force and friction. (2) (a) How does the coefficient of friction depend upon the normal force between the surfaces in contact? (b) How does it depend upon the area of the surfaces in contact? The coefficient of friction decreases with increasing normal force between the surfaces of contact. However, coefficient of friction does not depend upon the contact surface area because in the experiment, values of normal forces stay the same while the force to keep the block moving uniformly does not differ much between that of the flat position and the friction produced when block is positioned on side. (3) How does the coefficient of static friction compare with the coefficient of kinetic friction for the same surfaces, areas, and normal masses? In this connection explain what happened to Procedure 5. Basically, by common knowledge, coefficient of static friction works for frictional force that acts upon an object at rest while coefficient of kinetic friction is associated with frictional force of an object in motion. In the experiment, as the mass of the block increases, force due to weight increases as well and the normal force correspondingly so that a larger force due to friction must be overcome to maintain the constant motion of the block. (4) Calculate the force needed to pull a mass of 20 kg at a uniform slow speed up a plane inclined at an angle of 30° with the horizontal if the coefficient of kinetic friction is 0.20. Solution: Let P = unknown force Summation of forces along rotated y-axis: FN - 20(9.81)*cos(30°) = 0 ---? FN = 169.914 N f = ?k*FN = (0.20)*(169.914) = 33.983 N then, summation of forces along the rotated x-axis: P - f - 20*(9.81)*sin(30°) ---? P = 33.983 N + 20(9.81)(sin 30°) N P = 132.08 N (5) Suppose the force in Question 5 is a push applied horizontally on the mass instead of a pull directed parallel to the plane. (a) Calculate the value this force must have to keep the mass moving up the plane at constant speed. (b) Is there an angle for the plane at which no horizontal force can keep the mass moving at constant speed? If so, what is it? Solution: Again, let P = unknown force By summing forces along the vertical axis: FN - W = 0 FN = W = 20*(9.81) = 196.2 N Then, summation of forces along the horizontal axis: P - f = 0 f = ?k*FN = (0.20)*(196.2) = 39.24 N Thus, P = 39.24 N Read More