Solving for Acceleration in a Pulley System

[Radarithm]

Homework Statement

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.

Homework 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}

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.

 

[dauto]

1st, there is a sign mistake in the equation for the painter

2nd, your application of the 3rd law makes no sense. Those forces are not a pair of action and reaction. One is a normal force and the other is a gravitational force. A pair of action and reaction will always have the same nature. Pair normal with normal and gravity with gravity. Also the pair of forces never-ever-ever act on the same object. It's always Object A acts on object B and object B reacts on object A.

 

[jbunniii]

For the painter:
2T-N_1-Mg=M\ddot{y}
For the scaffold alone:
2T-mg-N_2=m\ddot{y}

Isn't $N_1$ the normal force of the scaffold acting on the painter? It should be upward. Also, if $N_2$ is the normal force of the painter on the scaffold, then $N_1 = N_2$.

I have assumed that:
1 - N_2=Mg because of Newton's 3rd Law.

As noted above, $N_1 = N_2$, but I don't see why these would be equal to $Mg$ (the weight of the painter). Indeed, if $M < m$ then there won't be any contact force at all.

 

[Radarithm]

[STRIKE][/STRIKE]I was able to derive a second equation, but I get 6.2 m/s^2:
Painter: 2F+N_1-Mg=M\ddot{y}
Scaffold alone: 2F-N_2-mg=m\ddot{y}

N_2=N_1
N_2=2F-m(g-\ddot{y})

System itself: F-\zeta g=\zeta\ddot{y}

T-\zeta g=\zeta\ddot{y}
2F=\zeta (g+\ddot{y})

This means that if I plug this into the equation for the painter:
2\zeta(g+\ddot{y})-g(m-M)=\ddot{y}(M+m)
M\ddot{y}+m\ddot{y}+2\zeta\ddot{y}=g(2\zeta -m-M)
\ddot{y}(M+m+2\zeta)=g(2\zeta -m-M)
g\frac{(2\zeta -m-M)}{(M+m+2\zeta)}

The mistake is staring me right in the face.

 

[jbunniii]

[STRIKE][/STRIKE]I was able to derive a second equation, but I get 6.2 m/s^2:
Painter: 2F+N_1-Mg=M\ddot{y}
Scaffold alone: 2F-N_2-mg=m\ddot{y}

N_2=N_1
N_2=2F-m(g-\ddot{y})

Already you've made a sign error in that last equation, assuming it is supposed to be a rearrangement of the "scaffold alone" equation above.

 

[dauto]

Also, the equation 2F=ζ(g+y¨) for the whole system is wrong. The left side is off by a factor of 2.

 

[Born]

So I think this post is dead but I'll give it a shot and try to revive it. I'm sorry if it seems inappropriate but I must also check my answer to this problem

Here is my my analysis:

$2T+N_{scaffold}-M_{parinter}g=M_{painter}a_{painter}$

$2T-N_{painter}-m_{scaffold}g=m_{scaffold}a_{scaffold}$

since $a_{painter}=a_{scaffold}$ and $N_{painter}=N_{scaffold}$ and we can solve for simply $a$ (acceleration of the painter and scaffold) by adding the equations and diving by $M+m$

$a=\frac{4T-(M+m)g}{M+m}$

Now it turns weird. I can assume by Newton's Third Law that $F$ of the painter should create an equal and opposite tension on the rope, yielding

$a=\frac{4F-(M+m)g}{M+m}$

which agrees with the clue given in the book, but I'm a little hesitant with this since I feel the scaffold (and the painter just by standing on it) must somehow also be affecting the tension.

Any help?