# 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

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

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

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

Entire system: $T_\Sigma -\zeta g=\zeta\ddot{y}$
So $$T_\Sigma=\zeta(g+\ddot{y})$$ and $$\ddot{y}=T_\Sigma -g=2F-g$$
For the painter: $2F-N_1-Mg=M\ddot{y}$. Since $N_1=Mg$, that means that:
$$\zeta (g+\ddot{y})=M\ddot{y}$$ which leads to: $$\ddot{y}(M-\zeta)=\zeta g$$. Solving for the acceleration, we get: $$\ddot{y}=\frac{\zeta g}{(M-\zeta)}$$.
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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