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# hw6 - This print-out should have 11 questions...

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Unformatted text preview: 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 2) 10.0 points Consider four vectors vector F 1 , vector F 2 , vector F 3 , and vector F 4 , where their magnitudes are F 1 = 43 N, F 2 = 36 N, F 3 = 19 N, and F 4 = 54 N. Let θ 1 = 120 ◦ , θ 2 = − 130 ◦ , θ 3 = 20 ◦ , and θ 4 = − 67 ◦ , measured from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vec- tor vector F , where vector F = vector F 1 + vector F 2 + vector F 3 + vector F 4 ? Correct answer: 34 . 026 N. Explanation: Basic Concepts: Vector components fig- ure1 Solution: The x components of the forces vector F 1 , vector F 2 , and vector F 3 are F 1 x = F 1 cos(120 ◦ ) = − 21 . 5 N F 2 x = F 2 cos( − 130 ◦ ) = − 23 . 1404 N F 3 x = F 3 cos(20 ◦ ) = 17 . 8542 N F 4 x = F 4 cos( − 67 ◦ ) = 21 . 0995 N . and the y components are F 1 y = F 1 sin(120 ◦ ) = 37 . 2391 N F 2 y = F 2 sin( − 130 ◦ ) = − 27 . 5776 N F 3 y = F 3 sin(20 ◦ ) = 6 . 49839 N F 4 y = F 4 sin( − 67 ◦ ) = − 49 . 7073 N . The x and y components of the resultant vec- tor vector F are F x = F 1 x + F 2 x + F 3 x + F 4 x = ( − 21 . 5 N) + ( − 23 . 1404 N) + (17 . 8542 N) + (21 . 0995 N) = − 5 . 6868 N F y = F 1 y + F 2 y + F 3 y + F 4 y = (37 . 2391 N) + ( − 27 . 5776 N) + (6 . 49839 N) + ( − 49 . 7073 N) = − 33 . 5474 N Hence the magnitude of the resultant vector bardbl vector F bardbl is bardbl vector F bardbl = radicalBig F 2 x + F 2 y = radicalBig ( − 5 . 6868 N) 2 + ( − 33 . 5474 N) 2 = 34 . 026 N 002 (part 2 of 2) 10.0 points What is the direction of this resultant vector vector F ?...
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hw6 - This print-out should have 11 questions...

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