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Unformatted text preview: Gozick, Brandon – Homework 17 – Due: Nov 10 2006, 5:00 pm – Inst: D Weathers 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Two weights attached to a uniform beam of mass 15 kg are supported in a horizontal po sition by a pin and cable as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 43 kg 19 kg 2 . 5 m 7 . 6 m 42 ◦ 15 kg What is the tension in the cable which sup ports the beam? Correct answer: 0 . 595278 kN. Explanation: Let : m = 15 kg , M 1 = 43 kg , M 2 = 19 kg , L 1 = 2 . 5 m , L 2 = 7 . 6 m , and θ = 42 ◦ . Basic Concepts: X ~ F = 0 X ~ τ = 0 . Solution: The sum of the torques about the pivot is T L 2 sin θ mg L 2 2 M 1 gL 1 M 2 gL 2 = 0 Solving for the tension T T = m g L 2 2 + M 1 g L 1 + M 2 g L 2 L 2 sin θ = (15 kg)(9 . 8 m / s 2 ) µ 7 . 6 m 2 ¶ (7 . 6 m)sin θ + (43 kg)(9 . 8 m / s 2 )(2 . 5 m) (7 . 6 m)sin θ + (19 kg)(9 . 8 m / s 2 )(7 . 6 m) (7 . 6 m)sin θ = 0 . 595278 kN . keywords: 002 (part 1 of 1) 10 points A uniform plank of length L = 7 . 7 m and mass 37 kg rests horizontally on a scaffold, with ‘ = 1 . 3 m of the plank hanging over one end of the scaffold. The acceleration of gravity is 9 . 8 m / s 2 . L l x How far can a painter of mass 210 kg walk on the overhanging part of the plank x before it tips? Correct answer: 0 . 449286 m. Explanation: Basic Concepts: In equilibrium, X ~ F = 0 X ~ τ = 0 Solution: Considering the edge of the scaf fold to be the pivot point, then X τ = 0 m g • L 2 ‘ ‚ M g x = 0 Gozick, Brandon – Homework 17 – Due: Nov 10 2006, 5:00 pm – Inst: D Weathers 2 Therefore, x = m M • L 2 ‘ ‚ = (37 kg) (210 kg) × • (7 . 7 m) 2 (1 . 3 m) ‚ = 0 . 449286 m ....
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This note was uploaded on 09/12/2008 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.
 Fall '08
 Weathers
 mechanics, Work

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