PHY302K-Sai-HW8-solutions

# PHY302K-Sai-HW8-solutions - Oslund (hno56) Homework 8 sai...

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Oslund (hno56) – Homework 8 – sai – (58015) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Two objects, oF masses 10 kg and 18 kg, are hung From the ends oF a stick that is 70 cm long and has marks every 10 cm, as shown. 10 kg 18 kg A B C D E F G 10 20 30 40 50 60 IF the mass oF the stick is negligible, at which oF the points indicated should a cord be attached iF the stick is to remain horizontal when suspended From the cord? 1. E 2. B 3. A 4. F correct 5. C 6. D 7. G Explanation: Let : # = 70 cm , m 1 = 10 kg , and m 2 = 18 kg . ±or static equilibrium, τ net = 0. Denote x the distance From the leFt end point oF the stick to the point where the cord is attached. m 1 gx - m 2 g [ # - x ]= τ =0 ( m 1 - m 2 ) x = m 2 # x = m 2 # m 1 - m 2 = (18 kg) (70 cm) 10 kg - 18 kg = 45 cm . ThereFore the point should be point F . 002 10.0 points A wheel oF radius R and negligible mass is mounted on a horizontal Frictionless axle so that the wheel is in a vertical plane. Three small objects having masses m , M , and 2 M , respectively, are mounted on the rim oF the wheel, as shown. 60 R M 2 M m IF the system is in static equilibrium, what is the value oF m in terms oF M ? 1. m = 5 M 2 2. m =2 M 3. m = 3 M 2 correct 4. m = M 2 5. m = M Explanation: Since the system is in static equilibrium, the net torque is zero: τ = mg R + M g R cos 60 - (2 M ) gR = + 1 2 M g R - 2 M g R = - 3 2 M g R , so m = 3 M 2 .

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Oslund (hno56) – Homework 8 – sai – (58015) 2 003 (part 1 of 2) 10.0 points A solid bar of length L has a mass m 1 . The bar is fastened by a pivot at one end to a wall which is at an angle θ with respect to the horizontal. The bar is held horizontal by a vertical cord that is fastened to the bar at a distance a distance x cord from the wall. A mass m 2 is suspended from the free end of the bar. T m 2 m 1 θ x cord L Find the tension T in the cord. 1. T = ± 1 2 m 1 + m 2 ²± L x cord ² g cos θ 2. T = ± m 1 + 1 2 m 2 L x cord ² g 3. T = ± m 1 + 1 2 m 2 L x cord ² g sin θ 4. T =( m 1 + m 2 ) g cos θ 5. T =0 6. T = ± m 1 + 1 2 m 2 L x cord ² g cos θ 7. T m 1 + m 2 ) ± L x cord ² ³ g 2 ´ 8. T m 1 + m 2 ) g sin θ 9. T = ± 1 2 m 1 + m 2 L x cord ² g sin θ 10. T = ± 1 2 m 1 + m 2 L x cord ² g correct Explanation: Basic Concepts: Static Equilibrium: µ F . µ τ . Torque: τ = r F = rF We can begin by either summing the torques or the forces and setting them equal to zero. In this case, since we know little about the reaction force between the bar and the wall, it is easiest to begin by examining the torques around the connection between the bar and the wall, because the reaction forces will produce no torques around that point. µ τ = - Lm 2 g + x cord T - L 2 m 1 g . Solving this for T gives T = ± 1 2 m 1 + m 2 L x cord ² g . 004 (part 2 of 2) 10.0 points Find the the horizontal component of the force exerted on the bar by the wall. (Take the positive direction to be right.) 1. F x = ± 1 2 m 1 + m 2 L x cord ² g cos θ 2. F x correct 3. F x m 1 + m 2 ) g cos θ 4. F x = ± 1 2 m 1 + m 2 L x cord ² g sin θ 5. F x = ± m 1 + 1 2 m 2 L x cord ² g sin θ 6. F x m 1 + m 2 ) g sin θ 7. F x m 1 + m 2 ) ± L x cord
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## This note was uploaded on 02/22/2012 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.

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PHY302K-Sai-HW8-solutions - Oslund (hno56) Homework 8 sai...

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