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Unformatted text preview: pulling on M to the right is the same as the tension T pulling on m up and (II) that they have the same acceleration or they move together, or they move at the same rate, etc. T T mg a a Using the above implications we can not write the two equations for M and m that we need. We only need the equation for M in the xdirection and the equation for m in the ydirection. The other equations give us no information. We chose right and down as the positive directions. We then have: F net on M = Ma T = Ma F net on m = ma mg − T = ma mg − Ma = ma ma + Ma = mg a ( m + M ) = mg a = m m + M g = 0.5 1 + 0.5 g = g 3 = 3.27 m s 2 T = Ma = (1)(3.27) = 3.27N Substituting T=Ma into equation 2, we get: There’s a shortcut to skip straight to this step! Can you think of it? It will save us all the work from parts b and c. Part a) 33% Q W E R T Part b) 33% Q W E R T Part c) 33% Q W E R T...
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This note was uploaded on 09/08/2010 for the course PHY 7B PHY7B taught by Professor Cebra during the Winter '10 term at UC Davis.
 Winter '10
 Cebra

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