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Homework 3

# Homework 3 - Gozick Brandon Homework 3 Due 5:00 pm Inst D...

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Gozick, Brandon – Homework 3 – Due: Sep 15 2006, 5:00 pm – Inst: D Weathers 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Consider four vectors ~ F 1 , ~ F 2 , ~ F 3 , and ~ F 4 , where their magnitudes are F 1 = 56 N, F 2 = 27 N, F 3 = 12 N, and F 4 = 60 N. Let θ 1 = 140 , θ 2 = - 140 , θ 3 = 18 , and θ 4 = - 54 , measured from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vec- tor ~ F , where ~ F = ~ F 1 + ~ F 2 + ~ F 3 + ~ F 4 ? Correct answer: 31 . 1721 N. Explanation: Basic Concepts: Vector components fig- ure1 Solution: The x components of the forces ~ F 1 , ~ F 2 , and ~ F 3 are F 1 x = F 1 cos(140 ) = - 42 . 8985 N F 2 x = F 2 cos( - 140 ) = - 20 . 6832 N F 3 x = F 3 cos(18 ) = 11 . 4127 N F 4 x = F 4 cos( - 54 ) = 35 . 2671 N . and the y components are F 1 y = F 1 sin(140 ) = 35 . 9961 N F 2 y = F 2 sin( - 140 ) = - 17 . 3552 N F 3 y = F 3 sin(18 ) = 3 . 70821 N F 4 y = F 4 sin( - 54 ) = - 48 . 541 N . The x and y components of the resultant vec- tor ~ F are F x = F 1 x + F 2 x + F 3 x + F 4 x = ( - 42 . 8985 N) + ( - 20 . 6832 N) + (11 . 4127 N) + (35 . 2671 N) = - 16 . 902 N F y = F 1 y + F 2 y + F 3 y + F 4 y = (35 . 9961 N) + ( - 17 . 3552 N) + (3 . 70821 N) + ( - 48 . 541 N) = - 26 . 192 N Hence the magnitude of the resultant vector k ~ F k is k ~ F k = q F 2 x + F 2 y = q ( - 16 . 902 N) 2 + ( - 26 . 192 N) 2 = 31 . 1721 N 002 (part 2 of 2) 10 points What is the direction of this resultant 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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Homework 3 - Gozick Brandon Homework 3 Due 5:00 pm Inst D...

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