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CHP11_Practice - Here’s the tube(2.4 meters long with its...

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CHP11 Practice Sheet 1. F 2. F 3. T 4. T 5. T 6. C 7. D 8. B (I have some issues with the wording of this problem) 9. D 10. C 11. B 12. A 13. D 14. We need to balance the torques exerted on opposite sides of the fulcrum: Torque of girl = Torque of boy (30 kg) (10 m/s 2 ) (4.0 m) = (40 kg) (10 m/s 2 ) (d) (where d = distance from fulcrum where boy should sit) Solving for d: d = 3 m 15. ( You are not responsible to know how to do this problem for the exam. ) The 40 kg. tube alone can be considered as a point mass concentrated half way along its length. (That is, the C.G. of the tube by itself is in the center of the tube, 1.2 m from either end.) To find the CG of the combination of tube and 60 kg. mass at one end of the tube, think of the combination as being a 40 kg mass up half the length of the tube and the 60 kg. mass at the far end. The CG of the combined object is a weighted average of the mass’s distance from a reference point in this way:
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Unformatted text preview: Here’s the tube (2.4 meters long) with its load: 40 kg 60 kg __________X__________M 0 m 1.2 m 2.4 m (40 kg) (1.2 m) + (60 kg) (2.4 m) Dist of CG from 0 meter end is = ______________________________ = 1.92 m (40 kg + 60 kg) Again, I wouldn’t ask you to calculate this, but I’d like you to have a general feel for where the CG of such a contraption lies. 16. The torque about some axis due to an applied force exerted at a distance from the axis is equal to the component of that force perpendicular to the lever arm connecting the force’s point of contact to the axis times the length of the lever arm. (continued….) For two people balanced on a seesaw, one exerts a clockwise torque about the fulcrum which is balanced by the other person’s equal and opposite (counterclockwise) torque, producing a net torque of zero about the fulcrum....
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