1. Initially the system of objects shown in the following figures is held motionless. All surfaces, pulley, and wheels are frictionless. At the instant after the system of objects is released, m2 moves vertically downward. Caculate the acceleration of m1, m2 and M, and the tension T in the string. I attached a photo of the figure, including my professor's solution as well. So you can actually see how my professor approached this problem: Assume: acceleration of m2: a acceleration of M: A Then the acceleration of m1: a-A So we have: m2g-T=m2a T=MA T=m1(a-A) What confused me is the expression of accelerations: As for why acceleratoin of m1 is equal to a-A, I can fully understand. Because: a(m1->ground)=a(m1->A)+a(A->ground) But now, lets look at m2 a(m2->ground)=a(m2->A) + a(A->ground) As we all know that, a(m2->A) is pointing downwards. However, there should be an acceleration horizontally put on m2 for the second item, a(A->ground). Thats according to the equation, right? But now, I find that, there is no horizontal force exerted on m2. That is what is confusing me. And I dont know whether my professor's solution is correct or not. I think he should be correct but I really need further explaination on the solution. Can anyone help me out? Thank you! https://www.physicsforums.com/showthread.php?t=687489 Figure is here.