Let's try to visualise the system at initial setup at a position when the m1 is placed on the table and the string is pulled over the pully with a bit of slack hanging down in an arc between the pully and m1, and connected to m2, after wounding over the pully with a vertical drop of x.
After we let go, the m2 and the weight of vertical drop of the string will pull the horizontal part of the string over the table to an equilibrium, assuming the shape of a catenary suspended between the pully and m1.
This will be the initial equilibrium and it is guaranteed if m3 is lighter than m1 and m2.
At the moment the system starts to accelerate, the friction of m1 and the vertical component of the string tension at m1 (half its weight) will be smaller than the m2 plus vertical length of the string. This vertical tension is turned into the horizontal projection of the string tension over tha catenary arc, however this tension is continuously increased by the weight of the string along the length of it.
$$m_{2} + m_{3}(x/l) \geq\ \mu (m_{1} + m_{3}\frac{(1-x)}{2l} )$$
Now we need to calculate the sag of the catenary part of the string to calculate its length and mass, we already know its horizontal and vertical reactions.
I let you figure that out.
