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Static equilibrium of a crane

  1. Mar 10, 2009 #1
    Static equilibrium!!!

    1. The problem statement, all variables and given/known data

    A 25-ft crane supported at its lower end by a pin is elevated by a horizontal cable as shown in the figure. A 250-pound load is suspended from the outer end of the crane. The center of gravity of the crane is 10 feet from the pin, and the crane weighs 200 pounds. What is the tension in the horizontal cable?

    l12g28.gif

    2. Relevant equations


    3. The attempt at a solution

    I would just like to know if my equations are missing anything. I generally can solve problems of the nature, but the configuration is different than anything I have done thus far and I am not completely clear on the forces that interact with the crane, especially at the pin/wall.

    My equations are

    ∑Fx=Rcos30-T=0
    ∑Fy=Rsin30-200-250
    ∑tau=T(25)cos30-(200)(10)sin30-(250)(25)sin30

    Are those correct or is my analysis wrong?
     
  2. jcsd
  3. Mar 10, 2009 #2
    Re: Static equilibrium!!!

    If it helps, imagine rotating the picture so the crane is laying horizontally, then draw the components of the forces that act perpendicular to the crane at their respective locations (see picture attached). That way it is easier to picture the angles of each force acting on the crane. Unless I'm mistaken, the only mistake you made was your calculation for the angles. I wrote my solution below if you want to look at it.

    My solution:

    There are three forces on the crane, whose magnitudes are:

    1. The load’s weight w_l=250lb
    2. The crane’s weight w_c=200lb
    3. The wire’s tension T =??? (the target variable)

    Each force creates a torque on the system, whose axis of rotation is on the pin.
    The torques are:

    1. Load torque: Tau_1=(250lb)(25ft)sin60
    2. Crane torque: Tau_2=(200lb)(10ft)sin60
    3. Tension torque: Tau_3=T(25ft)sin30

    In order for the system to be in rotational equilibrium, the net torque must be zero, so

    Tau_3 = Tau_1 + Tau_2
    T(25ft)sin30=(250lb)(25ft)sin60+(200lb)(10ft)sin60
    T=[(250lb)(25ft)sin60+(200lb)(10ft)sin60]/ (25ft)sin30≈570lb

    Hopefully my algebra didn't fail me.
     

    Attached Files:

  4. Mar 11, 2009 #3
    Re: Static equilibrium!!!

    Isnt there a force being exerted by the wall/pin? If the wall wasnt there, the beam would fall down and to the left?
    Is the wall's force not in the equation because the components of that force act at a 90 degree angle to the origin and therefore have no torque? Wouldnt the component for the x axis be in the Etau equation?
     
  5. Mar 11, 2009 #4
    Re: Static equilibrium!!!

    bump.
     
  6. Mar 12, 2009 #5
    Re: Static equilibrium!!!

    I'm sorry, I admit I completely forgot about the normal force from the wall. The wall does in fact exert a force on the beam. If it didn't, then the tension and weight would slingshot the beam down and to the left like you said.

    However in this case, the point of rotation for the beam is on the wall where this normal force acts, thus the normal force vector points THROUGH the rotation axis and does not contribute any torque (thus leaving my solution coincidentally unchanged).

    In short, your analysis looks right, only you seem to have the wrong angles in your sines. Just redraw a diagram and keep an eye on the angles you are using for the sines and cosines.
     
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