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Broken leg & pulleys

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data
    assume that the total downward force exerted at A & B by the leg is 38N. find the weight W, attached on the lower puller

    2. Relevant equations
    please see attached diagram

    3. The attempt at a solution

    i have written roman numerals to help determine which equation refers to which point

    at point I
    [tex] W = T [/tex]


    at point II
    in the X direction
    [tex] T \cos 25 + F_{1} = F_{A} cos \theta [/tex]
    in the y direction

    [tex] T \sin 25 = F_{A} \sin \theta [/tex]

    at point III
    in the X direction
    [tex]0=T \cos 20 +T \cos 25 [/tex]

    in the Y direction

    [tex] T \sin 20 + T \sin 25 [/tex]

    at point IV

    in the X direction

    [tex] T \cos 20 + T \cos 60 = 0 [/tex]


    in the Y direction

    [tex] -T \sin 20 - T\sin 60 = 0 [/tex]

    And that the end of this, can we say that T = 38N?

    Please let me know if all the equations are correct?
    Thank you for your input and assistance!
     

    Attached Files:

  2. jcsd
  3. Sep 23, 2015 #2

    PhanthomJay

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    Gold Member

    Well I'm not sure of this problem but if the lower leg is horizontal and takes only an axial pulling force F1 , then draw the FBD of the pulley at the foot and sum vertical forces to solve for T and thence mg. F-a is given I think as a vert force of 38 N at A so I don't know why you are breaking it into components. You are also missing a tension force in your equation at that pulley.
     
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