hw06 - alexander(jra2623 – homework 06 – Turner...

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Unformatted text preview: alexander (jra2623) – homework 06 – Turner – (92510) 1 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 with magnitudes F 1 = 55 N, F 2 = 37 N, F 3 = 29 N, and F 4 = 46 N, θ 1 = 130 ◦ , θ 2 = − 150 ◦ , θ 3 = 31 ◦ , and θ 4 = − 68 ◦ , measured from the positive x axis with counterclockwise positive. What is the magnitude of the resultant vec- tor vector F = vector F 1 + vector F 2 + vector F 3 + vector F 4 ? Correct answer: 25 . 6337 N. Explanation: The components of a vector are F x = F cos θ and F y = F sin θ , so F 1 x = (55 N) cos(130 ◦ ) = − 35 . 3534 N F 2 x = (37 N) cos( − 150 ◦ ) = − 32 . 043 N F 3 x = (29 N) cos(31 ◦ ) = 24 . 8578 N F 4 x = (46 N) cos( − 68 ◦ ) = 17 . 2319 N F 1 y = (55 N) sin(130 ◦ ) = 42 . 1324 N F 2 y = (37 N) sin( − 150 ◦ ) = − 18 . 5 N F 3 y = (29 N) sin(31 ◦ ) = 14 . 9361 N F 4 y = (46 N) sin( − 68 ◦ ) = − 42 . 6505 N . The components of the resultant vector are F x = F 1 x + F 2 x + F 3 x + F 4 x = ( − 35 . 3534 N) + ( − 32 . 043 N) + (24 . 8578 N) + (17 . 2319 N) = − 25 . 3066 N and F y = F 1 y + F 2 y + F 3 y + F 4 y = (42 . 1324 N) + ( − 18 . 5 N) + (14 . 9361 N) + ( − 42 . 6505 N) = − 4 . 08191 N , so the magnitude of the resultant vector is bardbl vector F bardbl = radicalBig F 2 x + F 2 y = radicalBig ( − 25 . 3066 N) 2 + ( − 4 . 08191 N) 2 = 25 . 6337 N ....
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hw06 - alexander(jra2623 – homework 06 – Turner...

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