Hw7

Download Document
Showing pages : 1 - 2 of 4
This preview has blurred sections. Sign up to view the full version! View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: schneiter (tcs639) HW07 Ross (89152) 1 This print-out should have 9 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 the 693 N weight held by two cables shown below. The left-hand cable had tension T 2 and makes an angle of θ 2 with the ceiling. The right-hand cable had tension 540 N and makes an angle of 42 with the ceiling. The right-hand cable makes an angle of 42 with the ceiling and has a tension of 540 N . 693 N T 2 5 4 N 4 2 θ 2 a) What is the tension T 2 in the left-hand cable slanted at an angle of θ 2 with respect to the wall? Correct answer: 520 . 62 N. Explanation: Observe the free-body diagram below. F 2 F 1 θ 1 θ 2 W g Note: The sum of the x- and y-components of F 1 , F 2 , and W g are equal to zero. Given : W g = 693 N , F 1 = 540 N , θ 1 = 42 , and θ 2 = 90 - θ . Basic Concept: Vertically and Horizontally, we have F x net = F x 1- F x 2 = 0 = F 1 cos θ 1- F 2 cos θ 2 = 0 (1) F y net = F y 1 + F y 2- W g = 0 = F 1 sin θ 1 + F 2 sin θ 2- W g = 0 (2) Solution: Using Eqs. 1 and 2, we have F x 2 = F 1 cos θ 1 (1) = (540 N) cos42 = 401 . 298 N , and F y 2 = F 3- F 1 sin θ 1 (2) = 693 N- (540 N) sin42 = 693 N- 361 . 331 N = 331 . 669 N , so F 2 = radicalBig ( F x 2 ) 2 + ( F y 2 ) 2 = radicalBig (401 . 298 N) 2 + (331 . 669 N) 2 = 520 . 62 N ....
View Full Document