PHYS18Fall2009MT2solns

PHYS18Fall2009MT2solns - Physics 18 Fall 2009 Midterm 2...

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Unformatted text preview: Physics 18 Fall 2009 Midterm 2 Solutions For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back. Please sit every other seat, and please dont cheat ! If something isnt clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the class. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck ! 1. A 5.0 kg cat and a 2.0 kg bowl of tuna fish are at opposite ends of a 4.0 m long seesaw. How far to the left of the pivot must a 4.0 kg cat stand to keep the seesaw balanced? Solution This is just a statics problem since none of the components of the system is moving. The bowl and the smaller cat will tend to twist the seesaw one way, while the heavier cat will twist it the other way. Thus, the seesaw is subject to three torques. If the system is to balance, then the sum of the torques must be zero, i = net = 0. So, we just need to add up the different torques, noting that the bowl and smaller cat will lead to a positive torque, while the bigger cat will give a negative torque. So, net = bowl + small cat- big cat = 0 . Now, in general, = rF sin . The force each component is just its mass times gravity. Since gravity is always pointing straight down, and since the seesaw is horizontal, the angle of each torque is 90 . Thus, we have net = bowl + small cat- big cat = 0 ....
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PHYS18Fall2009MT2solns - Physics 18 Fall 2009 Midterm 2...

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