Friction. Lab report Example | Topics and Well Written Essays - 750 words

Friction. - Lab Report Example 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 ÃŽ ¼k = 0.307 (procedure 4) ; ÃŽ ¼s = 0.3710 (procedure 6) ÃŽ ¼s = 0.4073 (procedure 5) ; ÃŽ ¼k = 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 (19Â °) ---? ?s = 0.3443 Trial 2: ?s = tan (23Â °) ---? ?s = 0.4245 Trial 3: ?s = tan (19Â °) ---? ?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 (21Â °) ---? ?s = 0.3839 Trial 2: ?s = tan (19Â °) ---? ?s = 0.3443 Trial 3: ?s = tan (19Â °) ---? ?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