OLDHW25 - oldhomewk 25 PAPAGEORGE MATT Due Apr 1 2008 4:00...

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oldhomewk 25 – PAPAGEORGE, MATT – Due: Apr 1 2008, 4:00 am 1 Question 1, chap 13, sect 1. part 1 of 1 10 points A rod can pivot at one end and is free to rotate without friction about a vertical axis, as shown. A force vector F is applied at the other end, at an angle θ to the rod. L m F θ If vector F were to be applied perpendicular to the rod, at what distance d from the axis of rotation should it be applied in order to produce the same torque vector τ ? 1. d = L 2. d = L tan θ 3. d = 2 L 4. d = L sin θ correct 5. d = L cos θ Explanation: The torque the force generates is τ = F L sin θ . Thus the distance in question should be L sin θ . Question 2, chap 13, sect 1. part 1 of 4 10 points A 1800 kg block is lifted at a constant speed of 4 cm / s by a steel cable that passes over a massless pulley to a motor-driven winch. The radius of the winch drum is 25 cm and the winch drum has a mass of 1100 kg. The acceleration due to gravity is 9 . 81 m / s 2 . 1800 kg 25 cm 4 cm / s 1100 kg What force must be exerted by the cable? Correct answer: 17 . 658 kN (tolerance ± 1 %). Explanation: Let : m = 1800 kg , v = 4 cm / s = 0 . 04 m / s , and r = 25 cm = 0 . 25 m . Note: There is no acceleration in this sys- tem and the mass of the winch drum is not required. Because the block is lifted at constant speed, the cable tension is T = m g = (1800 kg) (9 . 81 m / s 2 ) · kN N = 17 . 658 kN . Question 3, chap 13, sect 1. part 2 of 4 10 points What torque does the cable exert on the winch drum? Correct answer: 4 . 4145 kN · m (tolerance ± 1 %). Explanation: The torque is τ = T r = (17 . 658 kN) (0 . 25 m) = 4 . 4145 kN · m . Question 4, chap 13, sect 1.
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