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Unformatted text preview: gonzalez (ng4897) – 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 = 46 N, F 2 = 24 N, F 3 = 29 N, and F 4 = 67 N. Let θ 1 = 140 ◦ , θ 2 = − 150 ◦ , θ 3 = 22 ◦ , and θ 4 = − 63 ◦ , 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: 31 . 2919 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(140 ◦ ) = − 35 . 2381 N F 2 x = F 2 cos( − 150 ◦ ) = − 20 . 7846 N F 3 x = F 3 cos(22 ◦ ) = 26 . 8883 N F 4 x = F 4 cos( − 63 ◦ ) = 30 . 4173 N . and the y components are F 1 y = F 1 sin(140 ◦ ) = 29 . 5682 N F 2 y = F 2 sin( − 150 ◦ ) = − 12 N F 3 y = F 3 sin(22 ◦ ) = 10 . 8636 N F 4 y = F 4 sin( − 63 ◦ ) = − 59 . 6975 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 = ( − 35 . 2381 N) + ( − 20 . 7846 N) + (26 . 8883 N) + (30 . 4173 N) = 1 . 28297 N F y = F 1 y + F 2 y + F 3 y + F 4 y = (29 . 5682 N) + ( − 12 N) + (10 . 8636 N) + ( − 59 . 6975 N) = − 31 . 2656 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 (1 . 28297 N) 2 + ( − 31 . 2656 N) 2 = 31 . 2919 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 03/10/2010 for the course EE PHY 303K taught by Professor Gordito during the Spring '10 term at École Normale Supérieure.
 Spring '10
 GORDITO

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