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Unformatted text preview: homework 08 GUPTA, SHIVANI Due: Feb 17 2008, 11:00 pm 1 Question 1, chap 5, sect 5. part 1 of 1 10 points The pulley system is in equilibrium and the pulleys are weightless and frictionless. The weights are 164 N, 25 N, and 31 N. The acceleration of gravity is 9 . 8 m / s 2 . 164 N 25 N 31 N T Find the tension T . Correct answer: 369 . 5 N (tolerance 1 %). Explanation: Let : W 1 = 164 N , W 2 = 25 N , W 3 = 31 N , and g = 9 . 8 m / s 2 . W 1 W 2 W 3 4 7 5 3 6 1 2 8 T 7 T 1 T 7 T 7 T 8 T 1 T 1 T 6 T 6 T 2 T 5 T 6 T 5 T 5 T 4 T 3 T T T We will start at the loose end with ten sion T and work toward the weight W 1 by first moving to the left and down, then by moving up. The equation will come from the equilibrium at the suspended weight W 1 . 1) At the weight W 2 , T 1 = W 2 + T . At pulley 1, T 2 = 2 T 1 = 2 W 2 + 2 T . 2 ) Starting up from pulley 2, T 3 = 2 T . At the weight W 3 , T 4 = W 3 + T 3 = W 3 + 2 T . At pulley 3, 2 T 5 = T 4 T 5 = T 4 2 = W 3 2 + T . At pulley 4, T 5 = 2 T 6 homework 08 GUPTA, SHIVANI Due: Feb 17 2008, 11:00 pm 2 T 6 = T 5 2 = W 3 4 + T 2 . At pulley 5, T 7 = 2 T 6 = W 3 2 + T . At pulley 6, T 8 = 2 T 7 = W 3 + 2 T . 3) Finally, for equilibrium at W 1 , T 7 + T 8 + T 6 = W 1 + T 1 + T 2 7 4 W 3 + 7 2 T = W 1 + 3 W 2 + 3 T 7 W 3 + 14 T = 4 W 1 + 12 W 2 + 12 T T = 2 W 1 + 6 W 2 7 2 W 3 = 2 (164 N) + 6 (25 N) 7 2 (31 N) = 369 . 5 N . Question 2, chap 5, sect 5. part 1 of 1 10 points Three strings (labeled A , B , and C ) at tached to the sides of a square box are tied together by a knot as shown in the figure. The tension in the string labeled C is 52 N. 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 y Distance(m) x Distance (m) Three Strings Knotted Together A T A B T B C 52 N Figure: Drawn to scale. Calculate the magnitude of the tension in the string marked A . Correct answer: 49 . 5827 N (tolerance 3 %). Explanation: Basic Concepts: summationdisplay F x = ma x , (1) summationdisplay F y = ma y , (2) summationdisplay F z = ma z , and summationdisplay = I . There is no acceleration, so a x = a y = 0 . Solution: Using the knot (3 m , 5 . 5 m) as the coordinate origin (see figure below), we have x a = 0 m 3 m = 3 m , y a = 2 . 5 m 5 . 5 m = 3 m , r a = radicalBig x 2 a + y 2 a = 4 . 24264 m , x b = 2 . 5 m 3 m = . 5 m , y b = 10 m 5 . 5 m = 4 . 5 m , r b = radicalBig x 2 b + y 2 b = 4 . 52769 m , x c = 10 m 3 m = 7 m , y c = 0 . 5 m 5 . 5 m = 5 m , r c = radicalBig x 2 c + y 2 c = 8 . 60233 m ,321 1 2 3 4 5 6 754321 1 2 3 4 y Distance(m) x Distance (m) Three Strings Knotted Together A T A B T B C 52 N homework 08 GUPTA, SHIVANI Due: Feb 17 2008, 11:00 pm 3 Figure: Drawn to scale....
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This note was uploaded on 10/13/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Acceleration, Friction, Work

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