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Pulley and spring displacement

  1. Mar 12, 2017 #1
    • Homework posted in wrong forum, so no template
    Hey guys, I'm wondering why the displacement of the mass mounted in this setup is 2(x1+x2)?. I'm assuming that when the pulley rotates(when the mass is pulled downwards) the distance between the initial point and the end point of its center and the distance traveled by the mass are equal, meaning that both the mass and the pulley moved the same distance. However, I don't know what other phenomenon is adding displacement to the mass To make it double. I would be really grateful if you clarify on this matter.
    IMG_1126.JPG
     
  2. jcsd
  3. Mar 12, 2017 #2
    In pulley problems, before doing any force balances, the first step must be to consider the kinematics of the motion. In particular, you need to determine the displacements, under the constraint that the total length of rope is constant. If pulley 1 moves up x1 and pulley 2 moves down x2, what is the change in length of the section of rope between pulley 2 and mass m? How much does m2 move downward?
     
  4. Mar 12, 2017 #3
    My guess is that the change in length is x1+x2 and to me that's the same distance that m2 has moved downwards.
     
  5. Mar 12, 2017 #4
    Thank you for your answer Mr.Chestermiller
     
  6. Mar 12, 2017 #5
    I'm still stuck in this, I do understand that the rope's length remains constant but I'm still not getting why the distance travelled by m2 is the double do the distance travelled by the pulleys. I'm begging to think that both the rotation and translation of each pulley adds the same amount of displacement to m2 which would explain why the displacement is doubled but I may be getting more confused.
     
  7. Mar 12, 2017 #6
    That's not correct. If pulley 1 moves up x1, then the part of the rope on the left decreases in length by x1. If pulley 1 moves up x1 and pulley 2 moves down x2, then the part of the rope in the middle decreases in length by (x1+x2). So, in order for the total length of rope to remain unchanged, the part of the rope on the right must increase in length by 2x2+x1. This is how much the mass moves down relative to pulley 2. But pulley 2 has moved down by x2. So the total amount that the mass m moves down is x2+2x1+x2=2(x1+2x2)
     
  8. Mar 12, 2017 #7
    Ok, so pulley 1 goes up and makes the left side of the rope decrease X1 units but it also makes the middle part of the rope decrease x1 units and pulley 2 makes the middle part of the rope decrease in x2 units but it still adds x2 units to the position of the mass? Why is that?
     
  9. Mar 12, 2017 #8

    haruspex

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    Imagine pulley 2 moving down by some distance x2, but the other pulley and the mass remaining fixed. How much slack is introduced on each side of the pulley? How far down will the mass have to move to take up all that slack?
     
  10. Mar 13, 2017 #9
    Read the last two lines of my post #6.
     
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