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Unformatted text preview: homework 29 BAUTISTA, ALDO Due: Apr 12 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. A sphere of weight W is held against a wall by a string being pulled at an angle as shown. R O F The torque equation about point O , the center of the sphere, leads to : ( f is the magnitude of the frictional force, and weight W = Mg ) 1. F sin cos = f 2. F = f correct 3. W = f 4. F cos 2 = f 5. F + W = f 6. F sin = f Explanation: For static Equilibrium, X i F i = 0 , X i ~ i = 0 . From X i ~ i = 0, we get c = cc about O , the center of the sphere. That is, F R = f R, where c = F R and cc = f R . Or F = f. Question 2 Part 2 of 2. 10 points. The vertical component of the force equa tion leads to : 1. F cos + W = f 2. F sin = W 3. F sin = f + W 4. F sin = f 5. F sin + f = W correct Explanation: From X i ( F i ) y = 0, we get F sin + f = W. Question 3 Part 1 of 1. 10 points. In the figure the radius of the rod is . 221 cm, the length of aluminum part is 1 . 3 m, the copper part is 2 . 79 m. For alu minum, Youngs modulus is 70000000000 Pa, for copper, 110000000000 Pa. Aluminum Copper r L a L b Determine the elongation of the rod if it is under a tension of 7790 N Correct answer: 2 . 23056 cm (tolerance 1 %). Explanation: The crosssectional area of the wire A = (0 . 221 cm) 2 = 1 . 53439 10 5 m 2 m 2 , and the tension = 7790 N N throughout both homework 29 BAUTISTA, ALDO Due: Apr 12 2006, 4:00 am 2 pieces. Let us compute the elongation of each part separately: For aluminum: L al = L F Y A = (1 . 3 m)(7790 N) (70000000000 Pa)(1 . 53439 10 5 m 2 ) = 0 . 00942862 m . Similarly, for the copper part: L cu = L F Y A = (2 . 79 m)(7790 N) (110000000000 Pa)(1 . 53439 10 5 m 2 ) = 0 . 012877 m . So, the total elongation = 0 . 0223056 m = 2 . 23056 cm . Question 4 Part 1 of 1. 10 points. A stainlesssteel orthodontic wire is applied to a tooth that is out of line by 40 . The wire has an unstretched length of 4 . 5 cm and a diameter of 0 . 22 mm. 40 40 If the wire is stretched 0 . 11 mm, find the magnitude of the force on the tooth. (Disre gard the width of the tooth). Youngs modu lus for stainless steel is 180000000000 Pa. Correct answer: 21 . 5023 N (tolerance 1 %). Explanation: Given : L = 4 . 5 cm = 0 . 045 m , = 40 , r = 0 . 11 cm = 0 . 00011 m , L = 0 . 11 mm = 0 . 00011 m , and Y = 180000000000 Pa . 40 40 F F We need the tension required to stretch the wire by 0 . 11 mm: Y = F L A L F = Y A L L = Y ( r 2 ) L L The resultant force exerted on the tooth has components X F x = F cos  F cos = 0 and X F y = F sin  F sin so F net = 2 F sin = 2 Y r 2 L sin L = 2 (180000000000 Pa) (0 . 00011 m) 2 (0 . 00011 m) sin 40 . 045 m = 21 . 5023 N ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
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