# HW 6 - pampalone(tp8327 – HW#6 – swaminathan –(10500...

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Unformatted text preview: pampalone (tp8327) – HW#6 – swaminathan – (10500) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 2 s. A passenger in the elevator is holding a 5 . 5 kg bundle at the end of a vertical cord. The acceleration of gravity is 9 . 8 m / s 2 . What is the tension in the cord as the ele- vator accelerates? Correct answer: 56 . 65 N. Explanation: T mg a elevator g Let h be the distance traveled and a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression and solve for acceleration of the elevator: h = v t + 1 2 a t 2 = 1 2 a t 2 = ⇒ a = 2 h t 2 . The equation describing the forces acting on the bundle is F net = ma = T − mg T = m ( g + a ) = m parenleftbigg g + 2 h t 2 parenrightbigg = (5 . 5 kg) bracketleftbigg 9 . 8 m / s 2 + 2 (1 m) (2 s) 2 bracketrightbigg = 56 . 65 N . 002 (part 1 of 2) 10.0 points Hint: sin 2 θ + cos 2 θ = 1 . Consider the 672 N weight held by two cables shown below. The left-hand cable had tension 490 N and makes an angle of θ 2 with the ceiling. The right-hand cable had tension 510 N and makes an angle of θ 1 with the ceiling. 672 N 4 9 N 5 1 N θ 1 θ 2 a) What is the angle θ 1 which the right- hand cable makes with respect to the ceiling? Correct answer: 43 . 4721 ◦ . 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 = 672 N , F 1 = 510 N , and F 2 = 490 N . Basic Concepts: summationdisplay F x = 0 F x 1 = F x 2 F 1 cos θ 1 = F 2 cos θ 2 (1) pampalone (tp8327) – HW#6 – swaminathan – (10500) 2 F 2 1 cos 2 θ 1 = F 2 2 cos 2 θ 2 (2) and summationdisplay F y = 0 F y 1 + F y 2 + F y 3 = 0 F 1 sin θ 1 + F 2 sin θ 2 − F 3 = 0 F 1 sin θ 1 = − F 2 sin θ 2 + F 3 F 2 1 sin 2 θ 1 = F 2 2 sin 2 θ 2 − 2 F 2 F 3 sin θ 2 + F 2 3 , since (3) F 3 sin θ 3 = F 3 sin 270 ◦ = − F 3 , and F 3 cos θ 3 = F 3 cos 270 ◦ = 0 . Solution: Since sin 2 θ + cos 2 θ = 1 and adding Eqs. 2 and 3, we have F 2 2 = F 2 1 − 2 F 1 F 3 sin θ 1 + F 2 3 sin θ 1 = F 2 1 + F 2 3 − F 2 2 2 F 1 F 3 θ 1 = arcsin parenleftbigg F 2 3 + F 2 1 − F 2 2 2 F 1 F 3 parenrightbigg = arcsin bracketleftbigg (672 N) 2 + (510 N) 2 2 (510 N) (672 N) − (490 N) 2 2 (510 N) (672 N) bracketrightbigg = 43 . 4721 ◦ . 003 (part 2 of 2) 10.0 points b) What is the angle θ 2 which the left-hand cable makes with respect to the ceiling? Correct answer: 40 . 9952 ◦ . Explanation: Using Eq. 1, we have cos θ 2 = F 1 F 2 cos θ 1 θ 2 = arccos parenleftbigg F 1 F 2 cos θ 1 parenrightbigg = arccos parenleftbigg 510 N 490 N cos 43 . 4721 ◦ parenrightbigg = 40 . 9952 ◦ ....
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## This note was uploaded on 04/28/2011 for the course PHYS 105 taught by Professor Ken during the Spring '08 term at NJIT.

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HW 6 - pampalone(tp8327 – HW#6 – swaminathan –(10500...

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