Analytical Mech Homework Solutions 142

Analytical Mech Homework Solutions 142 - L = m1 ( x1 + x2 )...

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() ( ) ( 11 2 2 1 2 3 2 3 4 2 3 2 L mx x m x x m m x =+ + + + ±± ± ) 1234 2 L gm m m m x ( ) ( ) 12 1 1234 2 43 3 mmx mmmmx mmx gmmmm + +++ + − = +−− ( 323 423 3 L mxxmxx x =− ± ) 34 3 L x ( ) 2 3 mmx mmx gmm −+ +=− For , mm , , and 1 = 2 4 = 3 2 m = m 4 = : , 53 3 mx mx mg −= 3 5 x xg = 3 38 2 mx mx mx mg −+ −= , 23 3 mx mx mg −+ = 32 1 3 x = + Substituting into the second equation: 22 2 99 82 55 33 x g xxg −++−−= g 2 88 8 15 15 x g = , 2 11 g x = 1 31 0 6 51 1 1 1 x gg  =   3 2 4 1 1 1 x == Accelerations: : 1 m 5 11 x += : 2 m 7 11 x : 3 m 3 11 x : 4 m 5 11 x −− = 10.6 See figure 10.5.3, replacing the block with a ball. The square of the speed of the ball is calculated in the same way as for the block in Example 10.5.6.
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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