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Math Help - Stuck on a statics problem..

  1. #1
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    Stuck on a statics problem..

    Hello I am stuck on this mechanics problem. Any help would be hugely appreciated, because I am just starting with statics. Thank you in advance .

    A diagram of the problem can be found here :http:////img97.imageshack.us/i/79504950.png

    A small crane is used in demolition and swings a large rigid ball of mass M1, suspended on the lifting cable over a pulley of radius R, by rotating around the vertical axis zz with a steady velocity ω. The crane consists of a supporting uniform boom structure of mass M2 and length L inclined at an angle of θ to the horizontal plane. The boom is raised or lowered by the support cable that pulls horizontally and is attached at a distance of b from the pivot supporting the boom.

    Note that the lifting cable is parallel to the boom.
    The crane is shown in the static state and the ball is supported on the cable of length shown. For the dynamic conditions described, and for values:
    h = 2 m; L = 2.5 m; R = 0.2 m; M1 = 250 kg; M2 = 200 kg;
    ω = 1.0rad/s; θ = 60; b =1.5 m; r = 0.15 m;

    Determine the following:

    a) The radius R2 about which the ball moves about the zz axis.
    b) The tension in the lifting cable.
    c) The tension in the support cable.
    d) The reactive pivot force at the base of the boom.
    e) The axial force distribution along the boom.
    f) The torque exerted by the lifting cable drum that has radius r.
    You may assume that the mass of the cables and pulley are negligible and that the pulley is frictionless.



    This problem is related to the following topics: Method of Joints,method of sections,equilibrium equations, distributed loads along a line(F=integral_L w dx)

    I did a free body diagram,including the boom, the lifting cable and the base. I examine it as a truss.
    F1=250kg*9.8=2450N( is this the tension in the lifting cable-Q.b?) , F2=200*cos60*9.8=980N ( ans. to Q.c ?)
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  2. #2
    Senior Member
    Joined
    Nov 2010
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    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by frisky View Post
    ...

    ...
    I did a free body diagram,including the boom, the lifting cable and the base. I examine it as a truss.
    F1=250kg*9.8=2450N( is this the tension in the lifting cable-Q.b?) , F2=200*cos60*9.8=980N ( ans. to Q.c ?)
    WOW!

    Nobody answered this??? Well, this problem IS rather involved.

    Generally speaking, a Free-Body diagram should only contain ONE simple object. You should have one for the ball , one (a different one) for the boom, one for the pulley, one for the base, one for the lifting cable, etc.

    When the the crane is rotating about a vertical axis, the portion of the lifting cable between the ball and the pulley is not vertical. The vertical component of the tension in this cable is equal to (250kg)·(9.8m/sē)=2450N. The horizontal component of the tension is equal to the centripetal force required to make the ball follow a circular path.

    There are several forces on the boom. The supporting cable does not support all of the boom's weight. However, it does support some other forces which are transmitted through the boom.
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