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**Unformatted text preview: **miller (zdm77) – Homework04 – Chiu – (60180) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An object placed on an equal arm balance requires 12 kg to balance it. When placed on a spring scale, the scale reads 12 kg. Now everything (balance, set of masses, scale, and object) is transported to the moon, where the force of gravity is 1 6 that on earth. What are the new readings on the balance and spring scale, respectively? 1. 12 kg, 2 kg correct 2. 12 kg, 72 kg 3. 12 kg, 12 kg 4. 2 kg, 2 kg 5. 2 kg, 12 kg Explanation: The equal arm balance measures mass , whereas the spring scale, properly speaking, measures weight . Since an object’s mass is an inherent property, it will not change when we go to the moon. The apparent weight, how- ever, will be less, since the gravitational force exerted by the moon on the object is smaller. 002 (part 1 of 2) 10.0 points Consider the 694 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 410 N and makes an angle of 38 ◦ with the ceiling. The right-hand cable makes an angle of 38 ◦ with the ceiling and has a tension of 410 N . 694 N T 2 4 1 N 3 8 ◦ 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: 547 . 152 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 = 694 N , F 1 = 410 N , θ 1 = 38 ◦ , 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) = (410 N) cos38 ◦ = 323 . 084 N , and miller (zdm77) – Homework04 – Chiu – (60180) 2 F y 2 = F 3- F 1 sin θ 1 (2) = 694 N- (410 N) sin38 ◦ = 694 N- 252 . 421 N = 441 . 579 N , so F 2 = radicalBig ( F x 2 ) 2 + ( F y 2 ) 2 = radicalBig (323 . 084 N) 2 + (441 . 579 N) 2 = 547 . 152 N . 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: 53 . 8087 ◦ . Explanation: Using Eq. 2, we have θ 2 = arctan parenleftbigg F y 1 F x 1 parenrightbigg = arctan parenleftbigg 252 . 421 N 323 . 084 N parenrightbigg = 53 . 8087 ◦ . 004 (part 1 of 2) 10.0 points Two masses m 1 and m 2 are connected in the manner shown. m 2 m 1 T 1 T 2 a = g 2 The system is accelerating downward with acceleration of magnitude g 2 ....

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