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Online Homework Answers 5 (Oct. 12, 2006)

Online Homework Answers 5 (Oct. 12, 2006) - Hood Charles...

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Hood, Charles – Homework 5 – Due: Oct 12 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Consider four vectors ~ F 1 , ~ F 2 , ~ F 3 , and ~ F 4 , where their magnitudes are F 1 = 58 N, F 2 = 38 N, F 3 = 17 N, and F 4 = 54 N. Let θ 1 = 140 , θ 2 = - 150 , θ 3 = 27 , and θ 4 = - 51 , measured from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vec- tor ~ F , where ~ F = ~ F 1 + ~ F 2 + ~ F 3 + ~ F 4 ? Correct answer: 32 . 4143 N. Explanation: Basic Concepts: Vector components fig- ure1 Solution: The x components of the forces ~ F 1 , ~ F 2 , and ~ F 3 are F 1 x = F 1 cos(140 ) = - 44 . 4306 N F 2 x = F 2 cos( - 150 ) = - 32 . 909 N F 3 x = F 3 cos(27 ) = 15 . 1471 N F 4 x = F 4 cos( - 51 ) = 33 . 9833 N . and the y components are F 1 y = F 1 sin(140 ) = 37 . 2816 N F 2 y = F 2 sin( - 150 ) = - 19 N F 3 y = F 3 sin(27 ) = 7 . 71784 N F 4 y = F 4 sin( - 51 ) = - 41 . 9659 N . The x and y components of the resultant vec- tor ~ F are F x = F 1 x + F 2 x + F 3 x + F 4 x = ( - 44 . 4306 N) + ( - 32 . 909 N) + (15 . 1471 N) + (33 . 9833 N) = - 28 . 2092 N F y = F 1 y + F 2 y + F 3 y + F 4 y = (37 . 2816 N) + ( - 19 N) + (7 . 71784 N) + ( - 41 . 9659 N) = - 15 . 9664 N Hence the magnitude of the resultant vector k ~ F k is k ~ F k = q F 2 x + F 2 y = q ( - 28 . 2092 N) 2 + ( - 15 . 9664 N) 2 = 32 . 4143 N 002
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