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Tricky pulley problem

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    "What are the input forces needed to lift the 200 N load for the arrangement shown?"

    Notice how there are two movable pulleys that move in opposite directions as the weight is lifted. That is the confusing part to me.

    2. Relevant equations

    Ma=Fo/Fi

    3. The attempt at a solution

    Attempt: Ma=Fo/Fi

    Fi=Fo/Ma

    Fi=200 N/3

    Fi= 66.6 N
     

    Attached Files:

  2. jcsd
  3. Nov 10, 2012 #2

    Doc Al

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    Staff: Mentor

    Set up equations for equilibrium. Note that there are two strings involved and thus two tensions.
     
  4. Nov 10, 2012 #3
    Yes, but wouldn't the tension be equally distributed among the 3 lengths of cable supporting the weight bearing pulley? That would give it an ideal mechanical advantage of 3, would it not?

    Or are you saying that for mechanical advantage, I only have to take the tension of the two lengths of cable to which the force is being applied?
     
    Last edited: Nov 10, 2012
  5. Nov 10, 2012 #4

    Doc Al

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    Staff: Mentor

    No reason to think so. Set up the equations and find out.

    Assuming the usual--that the pulleys are massless and frictionless--the tension is uniform along each string. But the two strings will have different tensions.
     
  6. Nov 10, 2012 #5
    I'm just a bit lost on the setting up the equations. I know, for example, that if you just had 1 stationary pulley attached to the ceiling, you are just changing the direction of the force, so the ideal mechanical advantage is 1, giving the equation:

    Ideal Mechanical Advantage = Force out / Force in = 1

    And if you had a movable pulley attached to the weight, the tension would be shared by the two parts of the cable attached to the movable pulley, giving the equation:

    Ideal Mechanical Advantage = 2 * Force out / force in = 2

    I have a pretty good understanding of how to set up an equation for a system with just one movable pulley, or a set of pulleys attached to eachother.

    But I'm a bit lost when it comes to having 2 different cables and two different pulleys moving in different directions. It's not as simple as making a free body diagram as it is with just one movable pulley. How do I set up an equation for two different pulleys?
     
    Last edited: Nov 10, 2012
  7. Nov 10, 2012 #6

    Doc Al

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    Staff: Mentor

    Start by labeling the tension in the two strings. Let T1 be the tension in the string that is being pulled and T2 be the tension in the other string.

    Now set up an equation for each of the two movable pulleys. What forces act? What must the sum of the forces equal?
     
  8. Nov 10, 2012 #7
    Ok I think I got it now. I just set up equations for both of the pulleys and substituted the T2 from the second equation for the 2T1 in the first equation. That way I could solve for T1, which also equals the force required for equilibrium.

    Am I on track here?
     

    Attached Files:

  9. Nov 10, 2012 #8

    Doc Al

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    Staff: Mentor

    Exactly correct. (Except that you solved for T2 instead of T1. Fix that.)
     
  10. Nov 10, 2012 #9
    Correction: substituted wrong tension. T1 = Fi, so solving for T1, Fi = 50 N
     

    Attached Files:

  11. Nov 10, 2012 #10

    Doc Al

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    You got it. :approve:
     
  12. Nov 10, 2012 #11
    Thanks a lot! I would not have been able to figure that out without your help!
     
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