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Block and tackle and tension forces

  1. Apr 10, 2012 #1
    Please, I need your help with this... it's starting to annoy me.

    index.php?action=dlattach;topic=32002.0;attach=6425.jpg

    I just don't know where are the tension forces... I get really confused doing this kind of things with pulleys and ropes.
    Look at pulley 4, I've seen that the tension force in the right is upward but why? Isn't that the pulley just changes the direction of the force? So then it would be downward.. I just cannot understand why is it upward.

    Thanks!
     
  2. jcsd
  3. Apr 10, 2012 #2

    tiny-tim

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    Hi Hernaner281 :smile:

    Tension is always "away" …

    if a rope is pulling a body, the force of tension is always away from the body

    if a rope is pulling two bodies (with the rope in between) the force of tension is always away from both bodies

    every tiny little bit of the rope is in tension, every tiny little bit is being stretched by equal and opposite forces at its ends.
     
  4. Apr 10, 2012 #3
    Hmmm, okay.. but in this irritating case with so many PULLEYS (the ones below with mass), are the forces of tension equal? Yes, they are, aren't they? Thanks for your reply!
     
  5. Apr 10, 2012 #4

    tiny-tim

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    It looks like one continuous rope, so yes, the tension is the same all the way along. :smile:

    (that's assuming that each pulley is frictionless

    if there's friction on a pulley, you'd need to work out the normal force, find the friction force, and then the difference in tension on either side of that pulley would be the friction force)

    (it's also assuming that the rope is light, ie weightless … if the weight can't be ignored, then the tension at the top of any vertical rope will be greater than at the bottom)
     
  6. Apr 10, 2012 #5
    The problem says that pulley 3 + 4 have a weight Pp and the block has a weight of PL. Apart from that it does say that there's no friction and it tells me to find the minimum external force which can lift L a height of h. I analised and concluded that when the minimum force is applied the velocity is constant so the total force has to be zero.
    But to do that I have to analise the forces in Pp and that's where I fail!

    Thank you!!
     
    Last edited: Apr 10, 2012
  7. Apr 10, 2012 #6
    Don't leave me tiny-tim!! :D hehe
     
  8. Apr 11, 2012 #7

    tiny-tim

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    Hi Hernaner28! :smile:

    (just got up :zzz:)
    That's correct :smile: … the upward force from the rope must equal the downward weight.

    Since pulleys 3 and 4 are rigidly connected by that bar, do a free body diagram for pulleys 3 and 4 as a single body

    what do you get? :smile:
     
  9. Apr 11, 2012 #8

    that pulley system has a Mechnical Advantage (MA) of 5.

    with that said F = ( Pp + Pp + PL ) / 5

    if you actually have that pulley system in your class then you can run a little physics learning experiment.

    Lift an object 10 cm off of the ground and measure how much of the rope you pulled towards you. MA = amount of rope you pulled ( in cm ) divided by 10 cm. the centimeters cancel out and your left with a unitless number the MA.
     
  10. Apr 11, 2012 #9
    Oh yeah! There are 4 equal tension forces in block 3 and 4, so they 4 have to be equal to the weight of the block. Now that I know that the force is three quarters less it asks me to say how much rope I have to pull, I know that I have to reel four times more but I cannot realize that looking at the diagram!
    Thanks for your help1
     
  11. Apr 11, 2012 #10

    tiny-tim

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    technically, it's slightly less than 4, since the diagram shows that one of the sections of rope is at an angle to vertical :wink:
     
  12. Apr 11, 2012 #11
    Ahh yes you're right! And how could you calculate the distance of pulling the rope? I cannot realize that is 4 more times...
     
  13. Apr 11, 2012 #12

    tiny-tim

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    just move the bottom block up 1 cm …

    how much shorter do you have to make each section of rope?
     
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