This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hiarker (srh959) homework 06 Turner (58220) 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 = 48 N, F 2 = 33 N, F 3 = 29 N, and F 4 = 62 N. Let 1 = 130 , 2 = 140 , 3 = 22 , and 4 = 69 , 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: 32 . 2353 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(130 ) = 30 . 8538 N F 2 x = F 2 cos( 140 ) = 25 . 2795 N F 3 x = F 3 cos(22 ) = 26 . 8883 N F 4 x = F 4 cos( 69 ) = 22 . 2188 N . and the y components are F 1 y = F 1 sin(130 ) = 36 . 7701 N F 2 y = F 2 sin( 140 ) = 21 . 212 N F 3 y = F 3 sin(22 ) = 10 . 8636 N F 4 y = F 4 sin( 69 ) = 57 . 882 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 = ( 30 . 8538 N) + ( 25 . 2795 N) + (26 . 8883 N) + (22 . 2188 N) = 7 . 02621 N F y = F 1 y + F 2 y + F 3 y + F 4 y = (36 . 7701 N) + ( 21 . 212 N) + (10 . 8636 N) + ( 57 . 882 N) = 31 . 4603 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 ( 7 . 02621 N) 2 + ( 31 . 4603 N) 2 = 32 . 2353 N 002 (part 2 of 2) 10.0 points What is the direction of this resultant vector vector F ?...
View Full
Document
 Spring '09
 NUI
 Physics, Work

Click to edit the document details