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Physics Homework: Moments and Tension

  1. Nov 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A flagpole of mass 25kg and 2.5m in length is held in a horizontal position by a cable shown in the diagram.
    The centre of gravity of the flagpole is at a distance of 1.5m from the fixed end. Determine:
    a the tension T in the cable;
    b the vertical component of the force at the fixed end of the pole.
    [​IMG]

    2. Relevant equations

    moment = Fx
    x = d sinθ
    ac/w moment = c/w moment
    Fy = F sinθ

    3. The attempt at a solution

    I attempted part a and think I have got the answer, however I would be grateful for somebody to verify this for me.

    If we call the distance between the fixed end of the flagpole to its centre of gravity x1 and its weight F1 then moment1 = F1x1 = 25 x 9.81 x 1.5 = 367.875 Nm.

    If we then call the length of the flagpole d and the tension T F2, and we know that x2 = d sinθ, then moment2 = F2 x 2.5 sin30 = F2 x 1.25 and therefore F2 = moment2 / 1.25.

    The principle of moments states that when an object is in equilibrium , the anticlockwise moments equal the clockwise moments and therefore moment1 = moment2.

    Therefore the tension T = 367.875 / 1.25 = 294.3 N.

    Where I am really stumped is part b. I know how to work out the vertical component of a force - Fy = F sinθ - that's pretty easy. However when the question asks for the force at the fixed end of the pole i'm not quite sure what that is and thus I can't work out the vertical component.

    If anybody could enlighten me of what that force is I'd be very pleased. :smile:
     
  2. jcsd
  3. Nov 4, 2008 #2
    Your solution to part a looks good. What you did was select your pivot at the wall, then set the clockwise torques (what you call "F1x1") equal to the counterclockwise torques ( which I like to call "T sin theta")... to "balance the torques.

    By selecting your pivot at the wall, you made sure that any moments created by forces from the wall are zero because the lever arm is zero. There are actually forces from the wall on the beam -- a normal force Fx (that keeps the beam from breaking through the wall and a friction force Fy that keeps it from sliding down.

    For part b, what you look to be doing is now looking at balancing the ALL the forces.
    You started to do this with the vertical components, but I think you made an error. List ALL the forces that are vertical (include the component of the tension, the weight and an unknown Fy at the wall pivot point). Note: the friction keeps the beam from sliding down based on its weight, but the tension helps this friction force.
    Then, do the same for the horizontal part. The normal force will work against the tension pulling the beam toward the wall.
     
  4. Nov 4, 2008 #3
    Thank you - I understand it now.
     
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