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Simple torque problem

  1. Nov 10, 2007 #1
    This is a pretty simple problem, I am just a bit confused.

    Consider a 10m-long homogeneous ladder of mass M that leans in equilibrium against a vertical frictionless wall. Identify the forces acting on the ladder and evaluate their magnitude (relative to the weight W=Mg).

    I can't draw the diagram.

    However, the length of the base subtended by the ladder to the wall is 6m, and so the height is 8m.

    The equation given for sum of torques is 3Mg - F8, where F is the force of the normal at the wall. Where abouts would the fulcrum be? Maybe then i can understand this equation.
  2. jcsd
  3. Nov 10, 2007 #2


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    since the ladder is in equilibrium, a force equal to F must be acting parallel to the ground at the bottom of the ladder towards the wall. Wieght of the ladder acts through center of mass and vertically downward. A normal reaction equal to weight of the ladder acts at the end of the ladder in the vertical direction. Now consider clockwise moment and anticlockwise moment and find the sum
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