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Static equilibrium diagram help

  1. Dec 11, 2005 #1
    hi ive been looking at this problem for hours and im not sure if im approaching it right.

    Q: The boom in the figure [check attachment] is uniform and weighs 2.20kN. The magnitude of the force exerted on the boom by the hinge is...

    i knowthe boom has a mass of 224.261 kg, and summation tau = TLsin theta - Mg 2/3 L - mg 1/2 L if my free body diagram is right, not sure though the wire attaching at 2/3 L is really throwing me off hopefully some one can shed some insight on this.

    i apologize ahead for my crapy diagrams but its the best i can do on ms paint.

    Attached Files:

  2. jcsd
  3. Dec 11, 2005 #2


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    Staff: Mentor

    For static equilibrium, the sum of the moment forces = 0.

    Apply the weight (mg) of the boom at its center of mass, L/2.

    The tension T in the wire is applied at 2/3 L.

    The forces, boom weight (mg) and tension (T), must be resolved into components normal to the boom for the sum of the moments.
  4. Dec 11, 2005 #3

    ok so i have:

    summation tau = 0 = T 2/3 L sin 60 - mg 1/2 L
    T = 1.91 kN

    and 0 = Tension + Weight + Force

    so F = -4.11 kN

    is that right?
  5. Dec 11, 2005 #4


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    Staff: Mentor

    the weight of the boom acts downward (vertical), so one must use the Tension component normal to the boom.

    In the second case, the reaction at the base of the boom must push (balance) the Tension and Weight components which are parallel to the boom.

    -- Clarification

    Basically, in this type of statics problem, the sum of the forces in horizontal (x), [itex]\Sigma{F_x}[/itex]=0, sum of forces in vertical (y), [itex]\Sigma{F_y}[/itex]=0, and sum of moments [itex]\Sigma{M}[/itex]=0. A positive moment is counterclockwise, negative moment is clockwise. The moment is the product of the normal force (perpendicular) at point of application and the moment arm (distance between point of application and pivot point).
    Last edited: Dec 11, 2005
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