# a5_001 - Vertically T-W 1 = 0 ⇒ T = W 1 if static...

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PHY 2053, Section 3794, Fall 2009, Quiz 5 — Whiting 1. An object with mass 3.00 kg rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass 4.00 kg. a) Find the acceleration of each object. For the mass on the horizontal surface, we have T = m H × a. For the mass hanging vertically, we have T - m V × g = - m V × a. By subtracting the second equation from the ±rst, we can eliminate T , and obtain m V × g =( m H + m V ) × a. Solving for a we ±nd: a = m V × g m H + m V = 4 . 00 × 9 . 8 3 . 00 + 4 . 00 =5 . 60 m / s 2 . b) Find the tension in the cable. We can substitute this result into the ±rst equation to solve for T . We get: T = m H × a =3 . 00 × 5 . 60 = 16 . 8N . We can check this in the second equation. On the left we get: T - m V × g = 16 . 8 - 4 . 00 × 9 . 8 = 16 . 8 - 39 . 2= - 22 . 4N , while the right side also gives us: - m V × a = - 4 . 00 × 5 . 60 = - 22 . 4N . 2. a) What is the minimum force of friction required to hold the system shown in equilibrium?
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Unformatted text preview: Vertically: T-W 1 = 0 ⇒ T = W 1 if static. Horizontally: T-F R = 0 ⇒ F R = T. So F R = W 1 = 50 . 0 N . b) What coe²cient of static friction between the 100.-N block and the table ensures equilibrium? Vertically: N-W 2 = 0 ⇒ N = W 2 . Now, F R = μ s × N ⇒ μ s = F R /N = W 1 /W 2 = 0 . 500 . c) If the coe²cient of kinetic friction between the 100.-N block and the table is 0.250, what hanging weight should replace the 50.0-N weight to allow the system to move at constant speed once it is set in motion? As a = 0, we still have F R = W 1 from a) but now F R = μ k × N = μ k × W 2 from b). So W 1 = μ k × W 2 = 25 . 0 N ....
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