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Unformatted text preview: 3.29) Atwoods Three a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a38a37 a39a36 a38a37 a39a36 a38a37 a39a36 a117 a117 a117 2 m 3 m m a54 x + Define the positions of the three masses as x 1 , x 2 , and x 3 , with the positive x direction defined as up in the figure. Summing the forces on each mass reveals m x 1 =  mg 2 m x 2 = 2  2 mg 3 m x 3 =  3 mg with the additional constraint x 1 + 2 x 2 + x 3 = 0 Substitute this into our equation for x 2 and see m x 1 m x 3 = 2  2 mg Now add this to the equation for x 1 and obtain m x 3 = 3  3 mg And add three times this equation to the equation for x 3 to find the tension 0 = 10  12 mg = 6 5 mg Plug this back into our equations of motion and get m x 1 = 6 mg/ 5 mg = mg/ 5 2 m x 2 = 12 mg/ 5 2 mg = 2 mg/ 5 3 m x 3 = 6 mg/ 5 3 mg = 9 mg/ 5 and finally we compute accelerations... x 1 = g/ 5 x 2 = g/ 5 x 3 = 3 g/ 5 1 3.30) Atwoods Four a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a30a29 a31a28 a30a29 a31a28 a30a29 a31a28 a117 a30a29 a31a28 a117 x x 3 x 2 x 1 Label the position of the mass in the upper right as x . The positions of the pulleys with string passing under them will be labelled as...
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 Spring '10
 deHeer
 mechanics, Force, Mass

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