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Static equilibrium pulley problem

  1. Jun 18, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img849.imageshack.us/img849/3716/staticeqpulley.jpg [Broken]

    Hopefully the image showed up. If not, here's the link: http://img707.imageshack.us/img707/3716/staticeqpulley.jpg" [Broken]

    Sorry for the low quality pictures. Hopefully it will do though. The question is: Relate the traction force F3 to the weight mg and the angle theta.

    2. Relevant equations

    They give us that the tension T is uniform throughout the string (i.e. F1, F2 = T). F3 acts along the horizontal.


    3. The attempt at a solution

    I resolved the diagonal forces into horizontal and vertical components. Since θ1 and θ2 are equal I just call them both θ. The system is in static equilibrium so the total force is zero. So I expand the equilibrium equations out as follows:

    -F1cosθ - F2cosθ + F3 = 0

    and

    -F1sinθ + F2sinθ = 0

    Also, T = mg

    Do I just sum the two equations above and set them equal to mg? That doesn't make sense since the sum is 0, and mg is certainly not 0. Or should I sum them and subtract mg from the sum?

    Any help would be appreciated! Thanks.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 18, 2011 #2

    PhanthomJay

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    Hi, virtuoso735, welcome to PF! :smile:

    F1 = F2, but F3 is not part of the string. F3 does not = F1 or F2.
    yes.
    correct.
    yes, but since F1 = F2, this just tells you that 0 = 0, a well known fact.:wink: . But it also tells you that θ1 = θ2, something you assumed or realized earlier.
    yes, and T = F1 = F2
    Try it again, and see what you get.
     
  4. Jun 18, 2011 #3
    Is this right?

    I replace the F1 and F2 with T. So I get -Tcosθ - Tcosθ + F3 - mg = 0.

    -2Tcosθ + F3 = mg

    Then I replace the T with mg, so -2mgcosθ + F3 = mg

    Is this the relationship?
     
  5. Jun 18, 2011 #4

    PhanthomJay

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    No. You are looking at forces in the x direction. You had the right equation originally, but now you have added an mg term. Substitute 'T' with 'mg'.
     
  6. Jun 18, 2011 #5
    Sorry, I'm not understanding what you mean. Which was the right equation I had originally? Should I not have added the two cosine terms together? So is -mgcosθ - mgcosθ + F3 - mg = 0 any good?
     
  7. Jun 18, 2011 #6

    ehild

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    Your original equation for the horizontal force components was correct. "-F1cosθ - F2cosθ + F3 = 0"

    You know that F1=F2=T and T=mg. And you need F3 in terms of mg and theta.

    ehild
     
  8. Jun 18, 2011 #7
    Does that work out to F3 = 2mgcosθ ?

    Here is another similar problem, which I used the same process on:

    [PLAIN]http://img708.imageshack.us/img708/5553/pulley2q.jpg [Broken]

    There are horizontal and vertical components to this problem as well.

    For the horizontal component I get: -F1cosθ + F2 = 0.
    For the vertical component I get: F1sinθ = 0 [Is this right? This component is positive since it is going up, right? Should it be equal to 0]

    Summing them up gives: -F1cosθ + F2 + F1sinθ = 0.
    Simplify to: -mgcosθ + F2 + mgsinθ = 0
    mg(sinθ - cosθ) = F2

    Does this make sense?

    Also, how would the mass of the pulley affect the angle?
     
    Last edited by a moderator: May 5, 2017
  9. Jun 18, 2011 #8

    PhanthomJay

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    Yes!
    You have given us the sketch, but not the question.
    That should be sin, not cos, otherwise this is correct.
    It should not be equal to 0, and it's cos, not sin. You have F1 cos theta acting up, but what force acts down? Their sum is equal to 0.
    No.
    It would be less.
     
    Last edited by a moderator: May 5, 2017
  10. Jun 19, 2011 #9
    Sorry, the question is: relate the traction force F2 to the weight mg and the angle theta.

    I tried it again, hopefully this is closer.

    Horizontal component: -F1sinθ + F2 = 0
    Vertical component: F1cosθ - mg = 0

    Substitute F1 and F2 with mg and sum them, which leaves:
    -mgsinθ + F2 + mgcosθ - mg = 0

    Solving for F2:
    F2 = mg(sinθ - cosθ + 1)

    Is this right?
     
  11. Jun 19, 2011 #10
    Sorry, the question is: relate the traction force F2 to the weight mg and the angle theta.

    I tried it again, hopefully this is closer.

    Horizontal component: -F1sinθ + F2 = 0
    Vertical component: F1cosθ - mg = 0

    Substitute F1 and F2 with mg and sum them, which leaves:
    -mgsinθ + F2 + mgcosθ - mg = 0

    Solving for F2:
    F2 = mg(sinθ - cosθ + 1)

    Is this right?
     
  12. Jun 19, 2011 #11

    PhanthomJay

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    No.
    If the pulley is frictionless, the tension in the string wrapped around it will be equal...this should have been stated in the problem. So F2 is equal to mg, but F1 is not (it's a different string). You should start to get used to drawing free body diagrams.

    And the answer is....???
     
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