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SIN contest Tension and Acceleration question

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem and solution are attached, but I don't think the solution is correct.


    2. Relevant equations



    3. The attempt at a solution
    The only thing I don't understand is why Joe needs to supply the force to move both people.
    It seems to me that gravity is providing the force to move John, and Joe is simply applying just enough force to give Brian the same acceleration as Joe. So P = (60)(0.75g)

    Even if Joe doesn't apply any force at all, Brian could continuously pull down on the rope with F=60g and keep himself from falling. And John would accelerate without Joe applying any force at all.

    Am I correct?
    Thank you for reading.
     

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  2. jcsd
  3. Feb 8, 2013 #2

    TSny

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    But wouldn't that cause John to slip on the surface of the block which violates the conditions of the problem?
     
  4. Feb 8, 2013 #3
    Ok, but I don't understand why the solution has p = (60 + 80)a when gravity is responsible for the movement of John and the force is only applied to Brian. Shouldn't it be p = 60a?
     
  5. Feb 8, 2013 #4

    TSny

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    Since John and Brian remain at rest with respect to the block, the system acts as one object with total mass equal to the sum of the masses of John and Brian. So, the "inertia" that must be overcome by Joe's force is the total inertia in the system (i.e., the total mass of the system).
     
  6. Feb 8, 2013 #5

    TSny

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    You are correct that the tension in the string provides the necessary force to accelerate John. But the string goes around the pulley and the string exerts both a vertical and horizontal component of force on the pulley-block system. These tension forces are indicated in blue in the attachment. Also, there is a horizontal normal force acting on Brian and an equal reaction force acting back on the block (shown in green). So, the the block experiences two horizontal forces to the left which are indicated by the brown arrows. Since the block is massless, the net force on the block must be zero (otherwise it would have infinite acceleration!). So, Joe must provide a horizontal force to the right equal to the sum of the tension force and the normal force on Brian. Can you see that these two force have magnitudes of mjohna and mbriana, respectively?
     

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  7. Feb 8, 2013 #6
    Sorry for being slow but I don't quite understand.
    Why is the left pointing blue force acting on the massless block?
    It seems to me that the left pointing green reaction force is the only force acting directly on the block, and the horizontal force acting on Brian F=60A equals the reaction force, so shouldn't the reaction force also be 60A?
     
  8. Feb 8, 2013 #7

    TSny

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    Consider the section of the string that at some moment is in contact with the pulley as shown. We have one of those "massless" strings found only in physics problems! So, the net force on the section of string must be zero. The forces on the segment are the tension forces at each end and the normal force on the rope segment coming from the pulley. This force is labeled ##F_{Nrope}##. The rope exerts an equal but opposite force back on the pulley: ##F_{Npulley}##

    So we must have

    ##\vec{T}_1 + \vec{T}_2 + \vec{F}_{Nrope} = 0##

    Hence ##\vec{T}_1 + \vec{T}_2 - \vec{F}_{Npulley} = 0##

    So, ## \vec{F}_{Npulley} = \vec{T}_1 + \vec{T}_2 ##

    This shows that the rope exerts a force on the pulley-block system that is equivalent to the two tension forces. So Joe will need to exert a force that will balance out both ##\vec{T}_1## and the normal force from Brian.
     

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  9. Feb 8, 2013 #8
    Thank you so much TSny, I understand now!
     
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