Homework 4-solutions

Homework 4-solutions - kaplan(hmk378 Homework 4...

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kaplan (hmk378) – Homework 4 – Weathers – (17104) 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 , 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 counter-clockwise 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. Correct answer: 62 . 4819 . Explanation: ( 4 . 6, 8) ( 4 . 5, 2 . 6) (4 . 4, 1 . 9) (5 . 3, 8 . 5) 62 .
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