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Unformatted text preview: kaplan (hmk378) – Homework 4 – Weathers – (17104) 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 = 50 N, F 2 = 28 N, F 3 = 26 N, and F 4 = 54 N. Let θ 1 = 120 ◦ , θ 2 = − 150 ◦ , θ 3 = 23 ◦ , and θ 4 = − 58 ◦ , 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: 7 . 14237 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 ◦ ) = − 25 N F 2 x = F 2 cos( − 150 ◦ ) = − 24 . 2487 N F 3 x = F 3 cos(23 ◦ ) = 23 . 9331 N F 4 x = F 4 cos( − 58 ◦ ) = 28 . 6156 N . and the y components are F 1 y = F 1 sin(120 ◦ ) = 43 . 3013 N F 2 y = F 2 sin( − 150 ◦ ) = − 14 N F 3 y = F 3 sin(23 ◦ ) = 10 . 159 N F 4 y = F 4 sin( − 58 ◦ ) = − 45 . 7946 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 = ( − 25 N) + ( − 24 . 2487 N) + (23 . 9331 N) + (28 . 6156 N) = 3 . 29998 N F y = F 1 y + F 2 y + F 3 y + F 4 y = (43 . 3013 N) + ( − 14 N) + (10 . 159 N) + ( − 45 . 7946 N) = − 6 . 33432 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 (3 . 29998 N) 2 + ( − 6 . 33432 N) 2 = 7 . 14237 N 002 (part 2 of 2) 10.0 points What is the direction of this resultant vector vector F ? Note: Give the angle in degrees, use coun terclockwise as the positive angular direction, between the limits of − 180 ◦ and +180 ◦ from the positive x axis....
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This document was uploaded on 10/26/2011 for the course PHYS 1710 at North Texas.
 Fall '08
 Weathers
 mechanics, Work

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