# solution_7 - ouseph (jao888) – HW07 – tsoi – (92940)...

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Unformatted text preview: ouseph (jao888) – HW07 – tsoi – (92940) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 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 sin θ correct 3. d = √ 2 L 4. d = L cos θ 5. d = L tan θ Explanation: The force generates a torque of τ = F L sin θ , so the distance is L sin θ . 002 10.0 points A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and per- pendicular to the page. Four forces are ex- erted tangentially to the rims of the wheels, as shown below. F 2 F F F 2 R 3 R What is the magnitude of the net torque on the system about the axis? 1. τ = 5 F R 2. τ = F R 3. τ = 0 4. τ = 14 F R 5. τ = 2 F R correct Explanation: The three forces F apply counter-clockwise torques while the other force 2 F applies a clockwise torque, so τ = summationdisplay F i R i = (- 2 F ) (3 R ) + F (3 R ) + F (3 R ) + F (2 R ) = 2 F R . 003 10.0 points Two identical massless springs are hung from a horizontal support. A block of mass 3 . 9 kg is suspended from the pair of springs, as shown. The acceleration of gravity is 9 . 8 m / s 2 . k k 3 . 9 kg ouseph (jao888) – HW07 – tsoi – (92940) 2 When the block is in equilibrium, each spring is stretched an additional 0 . 26 m. The force constant of each spring is most nearly 1. k ≈ 24 N / m 2. k ≈ 220 N / m 3. k ≈ 42 N / m 4. k ≈ 56 N / m 5. k ≈ 49 N / m 6. k ≈ 12 N / m 7. k ≈ 74 N / m correct 8. k ≈ 36 N / m 9. k ≈ 90 N / m 10. k ≈ 50 N / m Explanation: Let : m = 3 . 9 kg , x = 0 . 26 m , and g = 9 . 8 m / s 2 . k x k x mg m Due to equilibrium of the forces, 2 k x = mg k = mg 2 x = (3 . 9 kg) (9 . 8 m / s 2 ) 2 (0 . 26 m) = 73 . 5 N / m ≈ 74 N / m . keywords: 004 (part 1 of 3) 10.0 points A floodlight with a mass of 24 kg is used to illuminate the parking lot in front of a library. The floodlight is supported at the end of a horizontal beam that is hinged to a vertical pole, as shown. A cable that makes an angle of 22 ◦ with the beam is attached to the pole to help support the floodlight. Assume the mass of the beam is negligible when compared with the mass of the floodlight. 24 kg 22 ◦ Find the force F T provided by the cable. The acceleration of gravity is 9 . 81 m / s 2 ....
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## This note was uploaded on 07/09/2011 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas at Austin.

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solution_7 - ouseph (jao888) – HW07 – tsoi – (92940)...

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