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Please help me tackle this problem

  1. Apr 1, 2006 #1
    Please help me tackle this problem!!

    I really need some help. I seem to be having lots of PROBLEMS with PHYSICS.

    Our lab group project (due Monday!) is to design and test four pulley systems to lift our prof who will be sitting in a chair in the front of the class. We decided to split up the systems we came up with. So, this is the one my lab partners gave me to analyze.

    I have to come up with the mechanical advantage of this system.
    I tried to draw a force diagram. But I don't know what to do next. I don't think I can get it to work!!!!

    I need to get this done because I will be demonstrating this pulley system on Monday morning in front of the class!!! I don't want to look foolish up there. I really need some help. Thanks!

    Attached Files:

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  3. Apr 1, 2006 #2
    How do attachments get approved?
  4. Apr 1, 2006 #3


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    You need to wait till someone looks at it and decide it does not contain any offensive material. This takes quite a long time (1 day?) normally (maybe the new owners will be quicker). What some poeple tend to do is post their pictures on another site that do not monitor the contents and then they insert a link to the picture in their post. This makes the picture immediately avaible.
  5. Apr 1, 2006 #4
    My attachment now shows up. Yay!
    Can anyone help?
  6. Apr 1, 2006 #5


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    My suggestion would be to draw a free body diagram for each pully separately.

    Take the pully at the bottom, for example. It will have a force equal to L acting down through the center, and two forces "T1" acting up at the edges.

    Repeat the same for the other two pullies and see if you can solve for whatever you need to solve for.

    Keep in mind that if you are ignoring friction then the tension throughout a single strand of rope is constant.
  7. Apr 2, 2006 #6


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    The question you need to answer is how are the tension in the cables on opposite sides of a pulley related?
  8. Apr 2, 2006 #7
    I think they are the same in the ideal case.
  9. Apr 2, 2006 #8


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    That is correct, now you need to think about using Newton's second law.

  10. Apr 2, 2006 #9


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    Yes it can be explained as follows: If there is no friction in a pulley (it turns easily, that is it is not resisting the turning motion) it will apply a force normal to the string only. This means it cannot disturb the tension in the string as it goes around it - it only changes the direction of the tension not its magnitude, just like with centripetal motion.

    You drew the forces that the cable are exerting on the pulleys, load and roof at the ends of all the cables. These are the same as the tensions in the cables at these point in the diagram (not neccesarily the same for all though).

    The next point that needs clarification is how are the tensions at the opposite ends of a cable related when it is straight?
  11. Apr 2, 2006 #10
    The tensions are opposite in direction. In the ideal case, they have the same magnitude.
  12. Apr 2, 2006 #11


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    Correct. If your friend is pulling on the end of a cable that you are holding on to you will feel the pull on it on the other side. If you were to hold it at the halfway point you will feel the same pull. The cable serves to transmits the pull from one point to another without changing it (if it had a lot of mass the weight of it would also need to be added to the tension).

    With these two bits of information you should now be able to solve the tensions in the system.

    Starting from the middle pulley on the right the tension on its right hand side will be f - the applied force to keep the system of pulleys and the load, L in equilibrium - the same tension appears on its left hand side. Continuing to the pulley at the bottom the same tension is transmitted by the cable to its right hand side, which is also the case for the tension on its lefthand side. These two tensions need to keep the load in equilibrium. At the pulley at the bottom we have that
    so the applied force needs to be half of the load in order to keep the system in equilibrium.
  13. Apr 3, 2006 #12
    Thanks for everyone's help. But something doesn't look right with the design.
  14. Apr 3, 2006 #13
  15. Apr 3, 2006 #14


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    So that's what was going on! :eek:
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