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Unformatted text preview: pampalone (tp8327) – HW#4 – swaminathan – (10500) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 7 . 5 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar. When the car accelerates to the right, the rope makes an angle of 46 ◦ with the vertical The acceleration of gravity is 9 . 8 m / s 2 . a 7 . 5 kg 4 6 ◦ Find the acceleration of the car. ( Hint: vectora object = vectora car ) Correct answer: 10 . 1482 m / s 2 . Explanation: Given : m = 7 . 5 kg , θ = 46 ◦ , and g = 9 . 8 m / s 2 . T T sin θ T cos θ θ m g Vertically summationdisplay F y = T cos θ − m g = 0 T cos θ = m g . (1) Horizontally, summationdisplay F x = T sin θ = m a . (2) Dividing Eqs 1 and 2, we have T sin θ T cos θ = a g tan θ = a g a = g tan θ = ( 9 . 8 m / s 2 ) tan46 ◦ = 10 . 1482 m / s 2 . 002 (part 1 of 2) 10.0 points Hint: sin 2 θ + cos 2 θ = 1 . Consider the 691 N weight held by two cables shown below. The lefthand cable had tension 570 N and makes an angle of θ 2 with the ceiling. The righthand cable had tension 490 N and makes an angle of θ 1 with the ceiling. 691 N 5 7 N 4 9 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: 35 . 4419 ◦ . Explanation: Observe the freebody diagram below. F 2 F 1 θ 1 θ 2 W g Note: The sum of the x and ycomponents of F 1 , F 2 , and W g are equal to zero. Given : W g = 691 N , pampalone (tp8327) – HW#4 – swaminathan – (10500) 2 F 1 = 490 N , and F 2 = 570 N . Basic Concepts: summationdisplay F x = 0 F x 1 = F x 2 F 1 cos θ 1 = F 2 cos θ 2 (1) 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 sin270 ◦ = − F 3 , and F 3 cos θ 3 = F 3 cos270 ◦ = 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 (691 N) 2 + (490 N) 2 2 (490 N) (691 N) − (570 N) 2 2 (490 N) (691 N) bracketrightbigg = 35 . 4419 ◦ . 003 (part 2 of 2) 10.0 points b) What is the angle θ 2 which the lefthand cable makes with respect to the ceiling? Correct answer: 45 . 4707 ◦ . 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 490 N 570 N cos35 . 4419 ◦ parenrightbigg = 45 . 4707 ◦ ....
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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.
 Spring '08
 KEN

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