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Unformatted text preview: hernandez (ejh742) – homework 06 – Turner – (58120) 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 = 35 N, F 2 = 30 N, F 3 = 13 N, and F 4 = 68 N. Let θ 1 = 130 ◦ , θ 2 = − 140 ◦ , θ 3 = 20 ◦ , and θ 4 = − 51 ◦ , 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: 41 . 9683 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 ◦ ) = − 22 . 4976 N F 2 x = F 2 cos( − 140 ◦ ) = − 22 . 9814 N F 3 x = F 3 cos(20 ◦ ) = 12 . 216 N F 4 x = F 4 cos( − 51 ◦ ) = 42 . 7938 N . and the y components are F 1 y = F 1 sin(130 ◦ ) = 26 . 8115 N F 2 y = F 2 sin( − 140 ◦ ) = − 19 . 2836 N F 3 y = F 3 sin(20 ◦ ) = 4 . 44626 N F 4 y = F 4 sin( − 51 ◦ ) = − 52 . 8459 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 = ( − 22 . 4976 N) + ( − 22 . 9814 N) + (12 . 216 N) + (42 . 7938 N) = 9 . 53083 N F y = F 1 y + F 2 y + F 3 y + F 4 y = (26 . 8115 N) + ( − 19 . 2836 N) + (4 . 44626 N) + ( − 52 . 8459 N) = − 40 . 8718 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 (9 . 53083 N) 2 + ( − 40 . 8718 N) 2 = 41 . 9683 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 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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