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Unformatted text preview: sousse (res2468) HW 10 Kleinman (58225) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This HW covers Ch. 14. 001 (part 1 of 2) 10.0 points A 0 . 106 kg meterstick is supported at its 31 . 1 cm mark by a string attached to the ceil ing. A 0 . 602 kg mass hangs vertically from the 4 . 71 cm mark. A mass is attached somewhere on the meterstick to keep it horizontal and in both rotational and translational equilibrium. The force applied by the string attaching the meterstick to the ceiling is 23 . 9 N. The acceleration of gravity is 9 . 81 m / s 2 . a) Find the value of the unknown mass. Correct answer: 1 . 72829 kg. Explanation: Basic Concept: The first (translational) condition of equi librium: F s m ms g m 1 g m 2 g = 0 . Let : m ms = 0 . 106 kg , m 1 = 0 . 602 kg , F s = 23 . 9 N , and g = 9 . 81 m / s 2 . Solution: m 2 g = F s ( m ms + m 1 ) g m 2 = F s g ( m ms + m 1 ) = 23 . 9 N 9 . 81 m / s 2 (0 . 106 kg + 0 . 602 kg) = 1 . 72829 kg 002 (part 2 of 2) 10.0 points b) Find the point where the mass attaches to the stick. Correct answer: 0 . 39133 m. Explanation: Basic Concept: The second (rotational) condition of equi librium (axis of rotation at the zero mark) m 1 g d 1 + m 2 g d 2 + m ms g d ms F s d s = 0 . Let : d ms = 50 cm , d 1 = 4 . 71 cm , and d s = 31 . 1 cm . Solution: m 2 gd 2 = F s d s ( m 1 d 1 + m ms d ms ) g d 2 = F s d s m 2 g m 1 d 1 + m ms d ms m 2 = (23 . 9 N)(0 . 311 m) (1 . 72829 kg)(9 . 81 m / s 2 ) (0 . 602 kg)(0 . 0471 m) 1 . 72829 kg (0 . 106 kg)(0 . 5 m) 1 . 72829 kg = . 39133 cm 003 (part 1 of 3) 10.0 points A uniform brick of length 31 m is placed over the edge of a horizontal surface with the maximum overhang x possible without falling. g x 31 m Find x for a single block. Correct answer: 15 . 5 m. Explanation: Let : L = 31 m . Basic Concepts: The definition of the center of mass ( n bricks): x m n summationdisplay i =1 x i m i n summationdisplay i =1 m i = 1 n n summationdisplay i =1 x i , sousse (res2468) HW 10 Kleinman (58225) 2 where x i is the center of mass position of the i th brick and m i is the mass of the i th brick. Solution: The center of mass of a single brick is in its middle or 1 2 of a bricks length from its maximum overhang. Since x 1 L = 1 2 , as measured from the maximum overhang, x m L vextendsingle vextendsingle vextendsingle n =1 = 1 2 1 = 1 2 . x m = 1 2 L = 1 2 (31 m) = 15 . 5 m . g x 31 m 004 (part 2 of 3) 10.0 points Two identical uniform bricks of length 31 m are stacked over the edge of a horizontal sur face with the maximum overhang x possible without falling....
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This note was uploaded on 04/08/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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