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Problem in Rigid Bodies

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data

    The upper portion of the crane boom consists of the jib AB,
    which is supported by the pin at A, the guy line BC, and the
    backstay CD, each cable being separately attached to the
    mast at C. If the 5-kN load is supported by the hoist line,
    which passes over the pulley at B, determine the magnitude
    of the resultant force the pin exerts on the jib at A for
    equilibrium, the tension in the guy line BC, and the tension
    T in the hoist line. Neglect the weight of the jib. The pulley
    at B has a radius of 0.1 m.

    2. Relevant equations

    Rigid Bodies Equillibrium

    3. The attempt at a solution

    First of all, the problem seems to be BIG and contains lots of 'unnecessary' details. I am confused here in WHAT to take as my body to draw my FBD. Is it the jib AB or DAB... I followed my common sense and took AB.. Not only that but the JIB AB attached to the pulley as one system. Now, we ignore CD. I solved it and I got the correct answer by choosing moment about A by 3 different forces. Tension along BC along y axis and two tension around the pulley. (Even the horizontal tension has a distance from A)... NOW MY QUESTION STARTS!. The book gives one surprising FBD. It is included in the word document. One for the JIB and the other for the pulley. My question is WHY is there a 5 kN Horizontal and Vertical AT THE CENTER of the pulley. The forces are tangential.. How is it drawn inside? I suppose this is the reaction of the jib on the pulley. But HOW??... I am sorry for elongating the post. Thanks for bearing with me and being patient.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

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  3. Apr 3, 2014 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    The book is showing you in isolation the forces acting on the sheave at B around which the 5 kN load is supported.

    Since a wire rope or other type line is being used to support the load, this line acts in tension. By isolating the sheave with its own FBD, the reaction forces on the pin at B are determined, and these forces can be applied to the FBD of the jib, eliminating the need to show the sheave, the load, and the line supporting the load.

    You have a vertical load on pin B due to supporting the 5 kN load from the sheave, and because the load wire wraps around the sheave and runs horizontally along the top of the jib, there is a 5 kN load acting horizontally as well.

    Yes, the forces act tangentially to the sheave, but the forces acting on pin B are what are applied to the jib.

    I think one reason such an elaborate problem set up was given was to show the student how to analyze such a situation, isolate the key components for developing a FBD, and to show how loads from other parts of the mechanism not included in the critical FBD are applied.
     
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