Posted by nikhil on Tuesday, February 1, 2011 at 1:49am.
If the pulley is accelerated at rate a, it is the same as if the weights became m(g +a) and 2m(g+a), in a coordinate system moving with the pulley, with gravitational accleration g + a.
There would be additional acceleration of the weights at a rate
a' = (g+a)(m/3m) = (g+a)/3
(2m would accelerate at rate a' down and m would accelerate up)
For the cable tension, let us apply Newton's second law to 2m.
2m*(g+a)-T = 2m*a' = 2m*(g+a)/3
T = 2m*(g+a)(2/3)= (4m/3)(g+a)
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