hw23 - alexander (jra2623) homework 23 Turner (92510) 1...

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Unformatted text preview: alexander (jra2623) homework 23 Turner (92510) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A mobile consists of four weights hanging on three horizontal rods of negligible mass. 20 N W 1 W 2 W 3 4 m 9 m 6 m 5 m 5 m 4 m Find the unknown weight on the lower rod. Correct answer: 8 . 88889 N. Explanation: Let : W = 20 N , r 1 = 9 m , and l 1 = 4 m . Applying rotational equilibrium about an axis through the point of suspension of the lowest rod in the mobile, summationdisplay vector = l 1 W - r 1 W 1 = 0 W 1 = l 1 W r 1 = (4 m) (20 N) 9 m = 8 . 88889 N . 002 (part 2 of 3) 10.0 points Find the unknown weight on the middle rod. Correct answer: 24 . 0741 N. Explanation: Let : r 2 = 5 m and l 2 = 6 m . Applying rotational equilibrium about an axis through the point of suspension of the middle rod in the mobile, summationdisplay vector = l w W 2- r 2 ( W + W 1 ) = 0 W 2 = r 2 ( W + W 1 ) l w = (5 m) (20 N + 8 . 88889 N) 6 m = 24 . 0741 N . 003 (part 3 of 3) 10.0 points Find the unknown weight on the upper rod. Correct answer: 66 . 2037 N. Explanation: Let : r 3 = 4 m and l 3 = 5 m . Applying rotational equilibrium about an axis through the point of suspension of the top rod in the mobile, summationdisplay vector = l e ( W + W 1 + W 2 )- r 3 W 3 = 0 W 3 = l 3 ( W + W 1 + W 2 ) r 3 = 5 m 4 m (20 N + 8 . 88889 N +24 . 0741 N) = 66 . 2037 N . 004 10.0 points The angle of the inclination is 36 . 6 , the outer part of the large pulley has a radius of 3 r and alexander (jra2623) homework 23 Turner (92510) 2 the inner part of the large pulley has a radius...
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas at Austin.

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hw23 - alexander (jra2623) homework 23 Turner (92510) 1...

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