**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.

http://ift.tt/1g13SVZ