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Math Help - Vectors - Tension of cable and thrust along boom on a crane.

  1. #1
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    Vectors - Tension of cable and thrust along boom on a crane.

    Hey, I am having trouble coming up with a solution to a problem. I do not know how to set it up, nor do I understand what one of the terms means.

    Question
    A 2000 kg steel girder is raised by a crane and held in position as shown in the diagram below. Find the tension in the cable OA and the thrust along the boom OB. (Round to the nearest whole number.)

    I attached the diagram to this post. http://www.mathhelpforum.com/math-he...1&d=1273077422 this is link incase it doesn't work.

    I do not understand what "the thrust along the boom" is. I found a definition saying thrust is "pushing (or compressive) force exerted on the structure". What direction should the vector be in? I do not know how to set up the triangle to find either of these values. I have had similar problems related to something hanging from a ceiling from two cables (I was given the angle of each cable to the ceiling) and I was to find the tension in each cable. All I had to do was make a triangle with the N pushing down on one side, and I knew all the angles so I use the sine law to find each tension.
    Can anyone help me in solving this problem? Any help is greatly appreciated!
    Attached Thumbnails Attached Thumbnails Vectors - Tension of cable and thrust along boom on a crane.-vector-crane-diagram.jpg  
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  2. #2
    MHF Contributor ebaines's Avatar
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    To picture what's happening in the boom, imagine that it was too weak to support the load - how would it fail? It's pretty obvious that it would collapse due to compressive forces (as opposed to being pulled apart by tension). So theh "thrust along the boom" means the compressive forces in the boom. These compressive forces are generated by (a) the tension in the cable pressing down on the boom at the top, and (b) the supporting force from the ground holding the bottom of the boom up. Thee two forces must be equal and opposite.

    To solve this type of problem you need to set up a free body diagram that shows the horizontal and vertical components of force at the top of the boom and at the bottom. Then solve such that the sum of forces in the horizontal direction is zero, and the sum of forces in the vertical direction is zero.
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  3. #3
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    Quote Originally Posted by ebaines View Post
    To picture what's happening in the boom, imagine that it was too weak to support the load - how would it fail? It's pretty obvious that it would collapse due to compressive forces (as opposed to being pulled apart by tension). So theh "thrust along the boom" means the compressive forces in the boom. These compressive forces are generated by (a) the tension in the cable pressing down on the boom at the top, and (b) the supporting force from the ground holding the bottom of the boom up. Thee two forces must be equal and opposite.

    To solve this type of problem you need to set up a free body diagram that shows the horizontal and vertical components of force at the top of the boom and at the bottom. Then solve such that the sum of forces in the horizontal direction is zero, and the sum of forces in the vertical direction is zero.
    How would I go about drawing that diagram? So there would be 19600N going down and 19600 going up? What about the horizontal components?
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  4. #4
    MHF Contributor ebaines's Avatar
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    Call the tension in the rope T and the compression in the boon B. The vertical forces at the top of the boom are:

    W: the weight of the object, pulling down
    T sin(40): the tension in the rope acting at the top of the boom pressig down.
    B sin(60): the compressive force in the boom, upward.

    So:

    (eq 1) Sum of forces in the vertical direction = 0 = B sin(60) - W - Tsin(40)

    Now in the horizontal direction - you have:
    T cos(40) pulling to the left
    B cos(60) pushing to the right

    So:
    (eq 2) Sum of forces in horizontal direction = 0 = Bcos(60) - Tcos(40)

    You now have two equation in 2 unknowns. Solve!
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