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Moments - Crane

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A crane with jib AB, 7m long. Centre of gravity 3m from A, inclined at 30 degrees to the vertical.
    The horizontal tie is attached at C, 5m from A.
    If the mass of the jib is 200kg and the load at B is 600kg, what is the tension in the tie.

    [​IMG]

    2. Relevant equations
    [tex]
    \sum {M_{CW} } = \sum {M_{ACW} }
    [/tex]

    3. The attempt at a solution
    OK, well if we take the pivot about the point A, we have;
    [tex]
    \begin{array}{l}
    \sum {M_{CW} } = \sum {M_{ACW} } \\
    \left( 3 \right)\left( {200 \times 9.8} \right) + \left( 7 \right)\left( {600 \times 9.8} \right) = \left( 5 \right)\left( {T\sin 30} \right) \\
    \Rightarrow T = \frac{{5880 + 41160}}{{2.5}} = 18816N \\
    \end{array}
    [/tex]

    This tension seems very high for the system.

    Could someone please explain the principle behind the Tension and angle components of the crane, because I don't think I fully understand it.

    Thanks in advance
    Steven.
     
  2. jcsd
  3. Nov 10, 2007 #2

    PhanthomJay

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    You are not evaluating the moments correctly. The moment of a force about a point is equal to the force times the perpendicular distance from the line of action of the force to the point.
     
  4. Nov 10, 2007 #3
    Draw a free body diagram. That should help you see what you need to do to solve this. Its actually just a one step problem.
     
  5. Nov 10, 2007 #4
    thanks for your reply Jay

    "The moment of a force about a point is equal to the force times the perpendicular distance from the line of action of the force to the point."

    So for taking the moments about the point A:
    The centre of mass for the jib has moment; (3sin30)(200x9.8)

    Following this procedure;
    (3sin30)(200x9.8) + (7sin30)(600x9.8) = 5T
    therefore T = 4704N
     
  6. Nov 10, 2007 #5

    PhanthomJay

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    Almost there. But the perpendicular distance from the tension force tothe point is (5)sin60, right?
     
  7. Nov 10, 2007 #6
    ya but thats not perpendicular.. thats horizontal, its COS not sin
     
  8. Nov 10, 2007 #7
    ya exactly cos30 or sin60.. either works
     
  9. Nov 10, 2007 #8
    OK so this perpencidular distance we are looking for the is the distance 'x' in the diagram yea?

    [​IMG]

    cos30 = x / 5
    the distance x is the 5cos30

    Using this;
    (3sin30)(200x9.8) + (7sin30)(600x9.8) = 5cos30T
    therefore T = 5.43x10^3 N
     
  10. Nov 10, 2007 #9
    [​IMG]

    The perpendicular distance, for each of the forces, will be the shortest distance from the green dot to each dotted line.

    Its the distance to the line of action of the force, which is infinite.
     
  11. Nov 10, 2007 #10

    PhanthomJay

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    Correct.
     
  12. Nov 10, 2007 #11
    excellent :)
    thanks for your help Vidatu and PhanthomJay
     
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