# Painter on a Scaffold

1. Mar 26, 2014

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.

2. Mar 26, 2014

### 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.

3. Mar 26, 2014

### jbunniii

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 $N_1 = N_2$.
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.

4. Mar 26, 2014

[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.

5. Mar 26, 2014

### jbunniii

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.

6. Mar 26, 2014

### dauto

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

7. Aug 26, 2014

### 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?

Last edited: Aug 27, 2014