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Unformatted text preview: hiarker (srh959) – homework 06 – Turner – (58220) 1 This printout should have 11 questions. Multiplechoice 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 = 48 N, F 2 = 33 N, F 3 = 29 N, and F 4 = 62 N. Let θ 1 = 130 ◦ , θ 2 = − 140 ◦ , θ 3 = 22 ◦ , and θ 4 = − 69 ◦ , measured from the positive x axis with the counterclockwise 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: 32 . 2353 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(130 ◦ ) = − 30 . 8538 N F 2 x = F 2 cos( − 140 ◦ ) = − 25 . 2795 N F 3 x = F 3 cos(22 ◦ ) = 26 . 8883 N F 4 x = F 4 cos( − 69 ◦ ) = 22 . 2188 N . and the y components are F 1 y = F 1 sin(130 ◦ ) = 36 . 7701 N F 2 y = F 2 sin( − 140 ◦ ) = − 21 . 212 N F 3 y = F 3 sin(22 ◦ ) = 10 . 8636 N F 4 y = F 4 sin( − 69 ◦ ) = − 57 . 882 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 = ( − 30 . 8538 N) + ( − 25 . 2795 N) + (26 . 8883 N) + (22 . 2188 N) = − 7 . 02621 N F y = F 1 y + F 2 y + F 3 y + F 4 y = (36 . 7701 N) + ( − 21 . 212 N) + (10 . 8636 N) + ( − 57 . 882 N) = − 31 . 4603 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 ( − 7 . 02621 N) 2 + ( − 31 . 4603 N) 2 = 32 . 2353 N 002 (part 2 of 2) 10.0 points What is the direction of this resultant vector vector F ?...
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This note was uploaded on 04/23/2010 for the course PHY 303K taught by Professor Nui during the Spring '09 term at University of TexasTyler.
 Spring '09
 NUI
 Physics, Work

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