Well, for one thing your question is not very clear! What are you trying to do?For another, this is the homework forum and you haven't shown that you have tried anything yourself!
of cours i tried but it is a exercaise from book and i have an answer at the end. But my solution is incorrect. So i ask here. And i tried to count forces work on thread on my picture.
Uhmm, let a_{0} be the acceleration of the movable pulley with respect to the ground.
Positive direction downward.
Let a_{1} be the acceleration of the m1 with respect to the ground.
Let a'_{2} be the acceleration of the m2 with respect to the movable pulley (that's how the m2 moves in the movable pulley's view).
Let a'_{3} be the acceleration of the m3 with respect to the movable pulley (that's how the m3 moves in the movable pulley's view.).
So the acceleration of m2 with respect to the ground is a_{2} = a'_{2} + a_{0}, the acceleration of m3 with respect to the ground is a_{3} = a'_{3} + a_{0}.
And you have a_{0} = -a_{1} (The acceleration of the movable pulley and the m1 is the same in magnitude but in opposite direction).
Since the mass of the movable pulley is negligible, the resultant force acts on it must be [itex]\vec{0}[/itex].
You can use Newton's 2nd law to solve this problem.
Viet Dao,