HW 12 - johnson(mjj622 HW 12 Coker(56625 This print-out...

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johnson (mjj622) – HW 12 – Coker – (56625) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment deals with concepts from Chapter 13, the next to last chapter to be covered on Quiz 3, coming up November 10. 001 10.0 points A force of F = F 1 ˆ i + F 2 ˆ j + F 3 ˆ k acts at the point r = r 0 ˆ i . What is the total torque about the origin of the coordinate system? 1. r 0 parenleftBig F 2 ˆ k - F 3 ˆ j parenrightBig correct 2. r 0 parenleftBig F 3 ˆ k + F 2 ˆ j + F 1 ˆ i parenrightBig 3. r 0 parenleftBig F 2 ˆ k - F 1 ˆ j parenrightBig 4. r 0 parenleftBig F 2 ˆ k + F 3 ˆ j parenrightBig 5. r 0 radicalBig F 2 1 + F 2 2 + F 2 3 6. r 0 parenleftBig F 1 ˆ k + F 3 ˆ i parenrightBig 7. r 0 parenleftBig F 1 ˆ i + F 2 ˆ j parenrightBig Explanation: From the definition of the cross product, ˆ i × ˆ j = + ˆ k and ˆ i × ˆ k = - ˆ j , and ˆ i × ˆ i = 0. Applying the definition of torque, vector τ = r × F = parenleftBig r 0 ˆ i parenrightBig × parenleftBig F 1 ˆ i + F 2 ˆ j + F 3 ˆ k parenrightBig = r 0 bracketleftBig F 1 parenleftBig ˆ i × ˆ i parenrightBig + F 2 parenleftBig ˆ i × ˆ j parenrightBig + F 3 parenleftBig ˆ i × ˆ k parenrightBigbracketrightBig = r 0 parenleftBig F 2 ˆ k - F 3 ˆ j parenrightBig 002 10.0 points A horizontal bar with a mass m suspended from one end is held by a cord with tension T , fastened at a distance D from the end of the bar that is attached to a wall with a frictionless pin. m D θ Tension T Pin What is the torque about the pin exerted on the bar by the cord? 1. D T (1 - sin θ ) 2. D T (1 - cos θ ) 3. D T sin θ 4. D T cos θ correct 5. D T tan θ 6. D T (1 - tan θ ) Explanation: The distance from the pin to the cord is D cos θ , so the torque is T × D cos θ = D T cos θ . 003 10.0 points A uniform, rigid rod of mass M and length L is pivoted frictionlessly at its upper end. If the rod is dropped from an initially horizontal position, it swings freely through a vertical position, as shown. A B C D In which of the four positions illustrated is the net torque about the pivot on the rod
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johnson (mjj622) – HW 12 – Coker – (56625) 2 greatest, and in which of the four positions is it the least? 1. Not enough information is provided. 2. Greatest at A ; least (zero) at D correct 3. Greatest at D ; least (zero) at A 4. The same in all four positions, since the force M g acts with lever arm L 2 in all four positions. Explanation: The only forces acting on the rod are the force of the pivot keeping the rod in place, and the gravitational force. The pivot force has a zero lever arm and as such exerts no net torque. At A, the lever arm is L 2 and is perpendicular to the gravitational force at A , yielding the maximum torque M g L 2 . As the rod falls, the lever arm shortens until at position D it is 0 since the line of force passes through the pivot point.
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