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Painter on a Scaffold

  1. Mar 26, 2014 #1

    Radarithm

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    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.
     
  2. jcsd
  3. Mar 26, 2014 #2
    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.
     
  4. Mar 26, 2014 #3

    jbunniii

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    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.
     
  5. Mar 26, 2014 #4

    Radarithm

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

    [tex]N_2=N_1[/tex]
    [tex]N_2=2F-m(g-\ddot{y})[/tex]

    System itself: [itex]F-\zeta g=\zeta\ddot{y}[/itex]

    [tex]T-\zeta g=\zeta\ddot{y}[/tex]
    [tex]2F=\zeta (g+\ddot{y})[/tex]

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

    The mistake is staring me right in the face. :confused:
     
  6. Mar 26, 2014 #5

    jbunniii

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    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.
     
  7. Mar 26, 2014 #6
    Also, the equation 2F=ζ(g+y¨) for the whole system is wrong. The left side is off by a factor of 2.
     
  8. Aug 26, 2014 #7
    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
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