Does a weight and force of same magnitude on a pulley make any difference? I am confronting a confusion here.Here it goes.
Imagine a single fixed pulley system with one end of the rope attached to a mass m.
If you pull the other end with a force 2mg  ,the mass will have a resultant force of mg   and will accelerate up with acceleration a.  
Now I connect a mass $2m$ to the side of the rope instead of the 2mg force.
Note that the other end still has the mass m.
Calculating net acceleration,
At mass m
$$T-mg=ma$$
At mass 2m
$$2mg-T=2ma$$
Adding both equations,
$$mg=3ma$$
or
$$a=g/3$$
I'm confused. How does that make sense ?
In both cases force acting is same,but acceleration produced is different.I suspect some foolish conceptual error. Please help.
 A: In the first case the force is pulling only a $1m$ mass.  In the second case the force is pulling a total of $3m$ mass ($2m$ on the one side, $1m$ on the other).  The mass is 3 times greater, so the acceleration is one-third the original.
A: The first case is understandable in which you pull the rope with $2mg$, which becomes the tension in the rope. Since the rope is massless and inextensible the block of mass $m$ (say A) feels a force of $2mg$ upwards due to the rope.
In the second case, we have a block of mass $2m$ (say B).
Here's the free body digram of B :
 
Tension is an action-reaction force. The block pulls the rope with $T$, and the rope pulls the block with $T$. If you look at at it the tension is pulling the block A upwards, not the weight of block B.
Let's make an assumption. If $T$ were equal to $2mg$ then block $B$ would be at rest and block A will move up with an acceleration $g$. This scenario is obviously not possible because the string is inextensible and the acceleration of the system must be the same, whatever it may be. 
So we need that value of tension for which the whole system moves with same acceleration. But at least you know it won't be equal to $2mg$. 
To get the value of tension you would have to form the equations of motion just like you have in your question. 
Another way is to look at the system as a whole. There is a total of $2mg$ driving force and $mg$ of opposing force and the mass to carry is $3m$. This will give you the same value of acceleration. This can also be understood by looking at the equations of motion. Tension being an internal force has no effect on the acceleration of centre of mass and on adding the equations of motion it gets cancelled out. All you are left with is the weight pulling the system downwards, which is an external force for the block-pulley system.
$$a=\frac{\text{Total driving force} - \text{Total opposing force}}{\text{Total mass}}=\frac{2mg-mg}{3m}=\frac{g}{3}$$
