Online Homework Answers 3 (Sept. 28, 2006)

Online Homework Answers 3 (Sept. 28, 2006) - Hood Charles...

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Unformatted text preview: Hood, Charles – Homework 3 – Due: Sep 28 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 9 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 Given two vectors ~ F 1 , and ~ F 2 . Where the magnitude of these vectors are F 1 = 89 N , and F 2 = 57 N . And where θ 1 = 53 ◦ , and θ 2 = 10 ◦ . The angles are measure from the positive x axis with the counter-clockwise angular direc- tion as positive. What is the magnitude of the resultant vec- tor k ~ F k , where ~ F = ~ F 1 + ~ F 2 ? Correct answer: 136 . 346 N. Explanation: Drawing a diagram to scale of the vectors, we have θ F 1 F 2 F Scale: 10 N where : F 1 = 89 N , θ 1 = 53 ◦ , F 1 x = 53 . 5615 N , F 1 y = 71 . 0786 N , F 2 = 57 N , θ 2 = 10 ◦ , F 2 x = 56 . 134 N , F 2 y = 9 . 89795 N , and F = 136 . 346 N , θ = 36 . 4345 ◦ , F x = 109 . 696 N , F y = 80 . 9765 N . The x components of the forces ~ F 1 and ~ F 2 are F 1 x = F 1 cos(53 ◦ ) = 53 . 5615 N F 2 x = F 2 cos(10 ◦ ) = 56 . 134 N . And the y components are F 1 y = F 1 sin(53 ◦ ) = 71 . 0786 N F 2 y = F 2 sin(10 ◦ ) = 9 . 89795 N . The x and y components of the resultant vec- tor ~ F are F x = F 1 x + F 2 x = (53 . 5615 N) + (56 . 134 N) = 109 . 696 N F y = F 1 y + F 2 y = (71 . 0786 N) + (9 . 89795 N) = 80 . 9765 N . Hence the magnitude of the resultant vector k ~ F k is k ~ F k = q F 2 x + F 2 y = q (109 . 696 N) 2 + (80 . 9765 N) 2 = 136 . 346 N ....
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Online Homework Answers 3 (Sept. 28, 2006) - Hood Charles...

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