HW 06 solution

# HW 06 solution - gutierrez(ig3472 – HW6 – ditmire...

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Unformatted text preview: gutierrez (ig3472) – HW6 – ditmire – (58216) 1 This print-out should have 13 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 Three masses are arranged in the ( x,y ) plane as shown. y [m]- 5- 3- 1 1 3 5 x [m]- 5- 3- 1 1 3 5 9 kg 8 kg 5 kg What is the magnitude of the result- ing force on the 9 kg mass at the ori- gin? The universal gravitational constant is 6 . 6726 × 10 − 11 N · m 2 / kg 2 . Correct answer: 3 . 84529 × 10 − 10 N. Explanation: Let: m o = 9 kg , ( x o ,y o ) = (0 m , 0 m) , m a = 8 kg , ( x a ,y a ) = (- 3 m , 1 m) , and m b = 5 kg , ( x b ,y b ) = (1 m ,- 3 m) . Applying Newton’s universal gravitational law for m o and, F ao = G m o m a ( x a- x o ) 2 + ( y a- y o ) 2 = (6 . 6726 × 10 − 11 N · m 2 / kg 2 ) × (9 kg) (8 kg) (- 3 m) 2 + (1 m) 2 = 4 . 80427 × 10 − 10 N , where tan θ a = y a x a θ a = arctan parenleftbigg y a x a parenrightbigg = arctan parenleftbigg 1 m- 3 m parenrightbigg = 161 . 565 ◦ , so the components of this force are F a x = F a cos θ a = ( 4 . 80427 × 10 − 10 N ) cos 161 . 565 ◦ =- 4 . 55773 × 10 − 10 N and F a y = F a sin θ a = ( 4 . 80427 × 10 − 10 N ) sin 161 . 565 ◦ = 1 . 51924 × 10 − 10 N . Applying Newton’s law for m o and m b , F bo = G m o m b ( x b- x o ) 2 + ( y b- y o ) 2 = (6 . 6726 × 10 − 11 N · m 2 / kg 2 ) × (9 kg) (5 kg) (1 m) 2 + (- 3 m) 2 = 3 . 00267 × 10 − 10 N , where θ b = arctan parenleftbigg y b x b parenrightbigg = arctan parenleftbigg- 3 m 1 m parenrightbigg = 288 . 435 ◦ , so the components of this force are F b x = F b cos θ b = ( 3 . 00267 × 10 − 10 N ) cos 288 . 435 ◦ = 9 . 49528 × 10 − 11 N and F b y = F b sin θ b = ( 3 . 00267 × 10 − 10 N ) sin 288 . 435 ◦ =- 2 . 84858 × 10 − 10 N . gutierrez (ig3472) – HW6 – ditmire – (58216) 2 The magnitude of the resultant force is F = radicalBig F 2 x + F 2 y = radicalBig ( F a x + F b x ) 2 + ( F a y + F b y ) 2 = bracketleftbig(- 4 . 55773 × 10 − 10 N +9 . 49528 × 10 − 11 N ) 2 + ( 1 . 51924 × 10 − 10 N- 2 . 84858 × 10 − 10 N ) 2 bracketrightBig 1 / 2 = 3 . 84529 × 10 − 10 N . 002 (part 2 of 2) 10.0 points Select the figure showing the direction of the resultant force on the 9 kg mass at the origin....
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## This note was uploaded on 04/13/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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HW 06 solution - gutierrez(ig3472 – HW6 – ditmire...

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