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An Atwood's machine consists of masses of 2 and 2.5 kg suspended from a massless, frictionless pulley. When the system is released from rest, how long

does it take for the larger mass to descend 1.6 m? What is the tension in the cord connecting the two masses during the descent?



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Tension T on the string attached to 2 kg is same as the tension in string attached to 2.5 kg.  Acceleration of the two weight system be a.

T - 2 g = 2 a
2.5 g - T  = 2.5 a

adding : 0.5 g = 4.5 a  => a = g/9
T = 2 g + 2 a = 20 g /9

Time t ,              s = 1/2 a t²    =>  t = 2 * 1.6 * 9/ 9.81  seconds

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Solution: We note that since an answer for the tension in the connecting rope is requested, the 'systems' approach is insufficient to solve the complete problem. However, we will take this approach to solve the first part of the problem.   Fig. 1 illustrates the 'system' which has a total mass of 4.5 kg. As a system the only 'external' forces present are the weight of A, and the weight of B, (as shown). The system accelerates in a 'counterclockwise sense', and the 2nd law gives us:  


         S Fccw  =   WA ‑  WB  =  M a x .


Hence, the acceleration of the system is:            a x =     {(2.5)(9.8) - (2)(9.8)}/(2 + 2.5) =  1.09 m/sec2 .


The equation of motion for 'A' would be:   x(t) = (1/2) a x t2 . Thus, if it descends 1.6 m in a time t', we have:


                  x(t') =  1.6  = (1/2)(1.09) t'2  à   t'  =  1.71 seconds.


As noted above, we cannot find the tension since the tension is 'internal' to the system, and hence doesn't appear in the 2nd law. To have the tension appear we must consider the two masses individually.


The force diagram for A & B are drawn & coordinate systems are chosen. (Note: 'x' is chosen upward for B so that the acceleration of B will be the same as that for A.) Applying Newton's Law to each object, we have:   For A:      S Fx =   WA  ‑ T  =  mA  a x  (1)   For B:      S Fx =   T  ‑   WB   =  mB  a x  (2)  

We have a set of 2 equations in 2 unknowns (a & T). Solving, we find:


     a x =  1.09 m/sec2 ;   T  =  21.8 N .


Note that the equation for the system is obtained if equations (1) & (2) above are added together.


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