Why Does a2 Equal m1g/m2 in an Accelerating Pulley System?

[CausticPhantom]

a1 = a2
a1 = F/(M+m1+m2) | Force/(Mass of entire system) = acceleration of entire system
*a2 = (m1g)/(m2+m1) | Force = m1g; acceleration of m2 and m1 = m1g/(m1+m2)

My answer: F = (m1g)(M+m1+m2)/(m2+m1)
The book's answer: F = (m1g)(M+m1+m2)/(m2)
*This step is what leads me to a slight variation of the book's provided answer.

I've looked through the forums, and have done a lot of thinking myself, and I believe what it comes down to is a false assumption. I understand that the force pulling m1 and m2 is m1g, and that the tension that then pulls m2 is equal to m1g, leading to an acceleration of m1g/m2 and subsequently the correct answer, but I do not understand what is incorrect about concluding that the system (looking at m2 and m1) as a whole accelerates at m1g/(m1+m2). Gut feeling tells me that the conclusion I've made is incorrect because the acceleration would be m1g/(m1+m2) in an inertial frame, but not in this case because it is part of an accelerating system.

Why does a2 = m1g/m2 rather than a2 = m1g/(m1+m2)?

 

[Orodruin]

but I do not understand what is incorrect about concluding that the system (looking at m2 and m1) as a whole accelerates at m1g/(m1+m2).

You have here considered m1 and m2 as a system, but the force you quote is not the force on that system, it is the force on m2 only.

 

[CausticPhantom]

You have here considered m1 and m2 as a system, but the force you quote is not the force on that system, it is the force on m2 only.

Is there another force that I would need to account for, such as the applied force on M?

 

[Orodruin]

Is there another force that I would need to account for, such as the applied force on M?

There are forces acting from the wall on m1 and from the pulley on the rope. Computing these forces is not necessary if you chose a better system to consider.