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Finding Tension of the boom

  1. May 31, 2009 #1
    1. The problem statement, all variables and given/known data
    A 1820N uniform boom is suppoted by a cable. The boom is pivoted at the bottom, and a 1100N object hangs from its end.
    The boom has a length L=13 m and is at an angle of 48 degree above the horizontal. At support cable is attached to the boom at a distance of .77L from the foot of the boom and its tension is perpendicular to the boom.
    Find the tension in the cable holding up the boom.


    2. Relevant equations



    3. The attempt at a solution


    1100N * L * cos(48)
    + 1820N* 0.5L * cos(48)
    - T * 0.77L
    =0

    Since L is not relevant in this equation you can just cancel it out.
    I got answer to be
    1746.691583N
    which of course is wrong.
    Could you please them me what I am doing wrong.
     
  2. jcsd
  3. May 31, 2009 #2

    Doc Al

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    Staff: Mentor

    Looks right to me.
     
  4. May 31, 2009 #3
    ok well...similar to his question

    Find the vertical components of the reaction
    force on the boom by the floor.

    i found that the horizontal component of the reaction
    force on the boom by the floor is just Tsin(theta).

    so how do i find teh vertical components?
     
  5. May 31, 2009 #4

    Doc Al

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    Staff: Mentor

    Hint: What does the net vertical force on the boom equal?
     
  6. Jun 3, 2009 #5
    is it the
    Normal of y= Fg(boom)+Fg(block)-Tcos(theata)???
     
  7. Jun 3, 2009 #6

    nvn

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    Science Advisor
    Homework Helper

    Yes, that's right.
     
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