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Unformatted text preview: madding (cmm3632) – HW04 – markert – (56475) 1 This print-out should have 15 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 A light, inextensible cord passes over a light, frictionless pulley with a radius of 7 . 3 cm. It has a(n) 16 kg mass on the left and a(n) 9 . 5 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2 . 6 m apart. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 6 m 7 . 3 cm ω 16 kg 9 . 5 kg At what rate are the two masses accelerat- ing when they pass each other? Correct answer: 2 . 49804 m / s 2 . Explanation: Let : R = 7 . 3 cm , m 1 = 9 . 5 kg , m 2 = 16 kg , h = 2 . 6 m , and v = ω R. Consider the free body diagrams 16 kg 9 . 5 kg T T m 2 g m 1 g a a Since the larger mass will move down and the smaller mass up, we can take motion downward as positive for m 2 and motion up- ward as positive for m 1 . Apply Newton’s second law to m 1 and m 2 respectively and then combine the results: For mass 1: summationdisplay F 1 : T- m 1 g = m 1 a (1) For mass 2: summationdisplay F 2 : m 2 g- T = m 2 a (2) We can add Eqs. (1) and (2) above and obtain: m 2 g- m 1 g = m 1 a + m 2 a a = m 2- m 1 m 1 + m 2 g = 16 kg- 9 . 5 kg 16 kg + 9 . 5 kg (9 . 8 m / s 2 ) = 2 . 49804 m / s 2 . 002 (part 2 of 2) 10.0 points What is the tension in the cord when they pass each other? Correct answer: 116 . 831 N. Explanation: T = m 1 ( g + a ) = (9 . 5 kg) (9 . 8 m / s 2 + 2 . 49804 m / s 2 ) = 116 . 831 N . 003 10.0 points madding (cmm3632) – HW04 – markert – (56475) 2 A force F = 81 . 2 N acts at an angle α = 32 ◦ with respect to the horizontal on a block of mass m = 25 kg, which is at rest on a horizontal plane. The acceleration of gravity is 9 . 81 m / s 2 . If the static frictional coefficient is μ s = . 84, what is the force of static friction? 25 kg μ s = 0 . 84 8 1 . 2 N 32 ◦ 1. 148 . 166 N 2. 206 . 01 N 3. 0 N 4. 57 . 8437 N 5. 242 . 155 N 6. 43 . 0294 N 7. 169 . 865 N 8. 263 . 854 N 9. 36 . 1447 N 10. 68 . 8615 N correct Explanation: Let : m = 25 kg , F applied = 81 . 2 N , α = 32 ◦ , μ s = 0 . 84 , and g = 9 . 81 m / s 2 m F a p p l i e d α mg N F s Basic Concepts: F applied,x = F applied cos α F applied,y = F applied sin α F y,net = N - F applied,y- mg = 0 F x,net = ma x = F applied,x- F s = 0 when at rest....
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