Painter on a Scaffold

1. The problem statement, all variables and given/known data

A painter of mass M stands on a scaffold of mass m and pulls himself up by two ropes which hang over pulleys, as shown. He pulls each rope with a force F and accelerates upwards with a uniform acceleration a. Find a – neglecting the fact that no one could do this for long.

2. Relevant equations

[tex]F=M\ddot{y}[/tex]
[tex]\zeta=(M+m)[/tex]
[tex]F=\zeta\ddot{y}[/tex]
For the painter:
[tex]2T-N_1-Mg=M\ddot{y}[/tex]
For the scaffold alone:
[tex]2T-mg-N_2=m\ddot{y}[/tex]
For the entire system:
[tex]T_\Sigma -\zeta g=\zeta\ddot{y}[/tex]

3. The attempt at a solution
I have assumed that:
1 – [itex]N_2=Mg[/itex] because of Newton’s 3rd Law.
2 – The acceleration of the entire system is [itex]\ddot{y}[/itex].

After solving for numerous equations, I checked the solutions section and got a hint: if [itex]M=m[/itex] then [itex]a=g[/itex]. After plugging in values, I did not get [itex]g[/itex] but instead [itex]2g[/itex] and many other values. My first approach:

Entire system: [itex]T_\Sigma -\zeta g=\zeta\ddot{y}[/itex]
So [tex]T_\Sigma=\zeta(g+\ddot{y})[/tex] and [tex]\ddot{y}=T_\Sigma -g=2F-g[/tex]
For the painter: [itex]2F-N_1-Mg=M\ddot{y}[/itex]. Since [itex]N_1=Mg[/itex], that means that:
[tex]\zeta (g+\ddot{y})=M\ddot{y}[/tex] which leads to: [tex]\ddot{y}(M-\zeta)=\zeta g[/tex]. Solving for the acceleration, we get: [tex]\ddot{y}=\frac{\zeta g}{(M-\zeta)}[/tex].
When I let M and m equal 1, I got twice the acceleration. Where did I go wrong? I had many more attempts but this one seems like the clearest one to me.

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