# Homework Help: Accelerating Pulley System

1. Sep 24, 2016

### CausticPhantom

• Member advised to use the homework template for posts in the homework sections of PF.

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)?

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Last edited: Sep 24, 2016
2. Sep 24, 2016

### Orodruin

Staff Emeritus
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.

3. Sep 24, 2016

### CausticPhantom

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

4. Sep 24, 2016

### Orodruin

Staff Emeritus
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.