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Tension of the rope with 2 objects

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Referring to
    https://www.physicsforums.com/showthread.php?t=561140

    The figure shows two 2.5 kg blocks connected by a rope that has a mass of 450g. The entire assembly is accelerated upwards at 7.4 m/s^


    What is the tension at the at the top end of the rope, near block A




    My question:

    To find the tension near block A, why don't you need to include the mass in the bottom block but only the mass of the rope and the mass of the first block?
     
  2. jcsd
  3. Mar 3, 2013 #2

    haruspex

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    You do indeed need to include the mass of the lower block, not the mass of the upper block. Is that what you meant to ask?
     
  4. Mar 3, 2013 #3

    ohh, that is not what i meant but you are right. I was confused about it. But actually why don't u need the mass of the upper block? I thought u are suppose to include everything
     
  5. Mar 3, 2013 #4

    haruspex

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    What needs to be included depends on what info you have. If you knew the force F and the acceleration and the mass of the upper block, you could compute the tension just underneath the upper block from that alone. Here, you know the acceleration and the mass of everything below the upper block, so you can compute the tension from that.
     
  6. Mar 3, 2013 #5
    do u mean the m1 and F = mass rope + m2?
     
  7. Mar 3, 2013 #6
    Because you only want the forces acting on that individual object, so you would exclude the mass of the lower block, I think? Its not directly acting on the first block (no contact)

    when you draw the free body diagram you would only draw all the forces in respect to one object, so one block.
     
  8. Mar 3, 2013 #7

    haruspex

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    You're mixing masses and forces there, and the diagram has A, B, not m1, m2.
    I'll add two more symbols, Tu for the tension in the rope immediately below A and Td for the tension in the rope immediately above B.
    Suppose you knew F (in the diagram, that's the tension above A), the acceleration, a, and the mass of A, mA. From analysing forces on A you could deduce that Tu = F-(a+g) mA.
    But instead you know a, mA, mB and mr (the mass of the rope). Now Tu = F-(a+g) mA is still true, but it doesn't help because you don't know F. So you analyse the forces on the rope and the lower block instead. For the lower block, Td = (a+g) mB; for the rope Tu - Td = (a+g) mr. From those, you can eliminate Td to find Tu.
     
  9. Mar 3, 2013 #8
    umm sorry, i still have trouble drawing the FBD
    Do you set the point on mass A or Tu?
    if I set it on A, then Tu is pulling down so it is a downward force, and I got T = F+ m(g-a)
     
  10. Mar 3, 2013 #9

    haruspex

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    You're right, I was being inconsistent about the directions in which I was measuring a and g. I was measuring a upwards and g downwards, since those are the directions in which they would be positive. But if we fix on up as positive (so g goes negative) then Tu = F-(a-g) mA etc. Just change g to -g in each of my equations.
     
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