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Statics, Distributed load

  1. Apr 17, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-4-17_22-32-56.png
    find internal normal, shear, and moment forces at point C, P=8kn

    3. The attempt at a solution
    for the most part I can solve this entire question but looking at the solution for it one of my equations is different and I cannot figure out why.
    my equation for finding T(tension in cable)
    ΣMa=0: 8(2.25)-T(.75/.96)=0
    gives T=28.84kN

    equation 2
    ΣMa=0: -T(0.6)+8(2.25)=0 ←????????
    gives T=30.00kN
     
  2. jcsd
  3. Apr 17, 2015 #2
    I don't see where your 1st equation comes from.

    2nd equation looks right.
     
  4. Apr 17, 2015 #3
    but why exactly the T(0.6) in 2nd equation
     
  5. Apr 17, 2015 #4
    Isn't it the lever arm 0.5m+0.1m?
     
  6. Apr 17, 2015 #5
    how does all of the tension in the cable get transmitted to the lever arm over the pulley?
     
  7. Apr 17, 2015 #6
    Just as anything else. If the pulley is not rotating then the tension in the cable is constant for its entire length.
     
  8. Apr 18, 2015 #7

    PhanthomJay

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    You didn't respond to paisiello2 comment in post 2, but it appears you tried to treat this as a truss at the joint where load P is applied by isolating the joint and assuming all the vert applied load is taken by the cable. But some of it is taken in shear by the lower member. Trust your equilibrium equations when determining support reactions from externally applied loads.
     
  9. Apr 18, 2015 #8
    Ahhhhhhh OK because the cable is only in the x direction after it goes over the pulley all of its y components are transferred to the beam. So if I finish my first equation I end up with.

    ΣM=0: 8(2.25)-T(cos33.7)(2.25)+T(cos33.7)(1.5)=0
    And I get T=30
     
  10. Apr 19, 2015 #9
    Not sure what you did there but if you draw a proper free body diagram you will see directly that the 2nd equation in your OP is the most straight forward.
     
  11. Apr 19, 2015 #10
    It is clearly the most straiforward. But how is the component of the tension at the end of the beam not included in equation?
     
  12. Apr 19, 2015 #11
    Draw a free body diagram. The rope is only cut at one location. The rest is internal.
     
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