07_P54InstructorSolution

07_P54InstructorSolution - 7.54. Model: The hanging masses...

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7.54. Model: The hanging masses 12 3 , , and mm m are modeled as particles. Pulleys A and B are massless and frictionless. The strings are massless. Visualize: Solve: (a) The length of the string over pulley B is constant. Therefore, () ( ) B3 BA B A B3B 2 yy yy L y yyL −+ − =⇒= The length of the string over pulley A is constant. Thus, ( ) A2 A1 A A12 2 yy yyL yyy −+ −== B3B 12 A 22 yyL yyL ⇒− = 321 2 constant yyy ⇒+ + = This constraint implies that 20 m / s dy dy dy dt dt dt ++= 2 y yy vvv = ++ Also by differentiation, 2 m / s . aaa (b) Newton’s second law for the masses ,,, mmm and pulley A is 3 3 2 2 A1 1 1 y Tm gm aTm a −= BA N TT
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The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint are five equations for the five unknowns (two tensions and three accelerations). To solve for A , T multiply the 3 m equation by 2, substitute BA 24 , TT = then divide each of the mass equations by the mass. This gives the three
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at University of Georgia Athens.

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07_P54InstructorSolution - 7.54. Model: The hanging masses...

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