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Static equilibrium problem:A-shaped ladder

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    A ladder is made in the shape of the letter A. Treat the two sides of the ladder as identical uniform rods, each weighing 455 N, with a length of 3.60 m. A frictionless hinge connects the two ends at the top, and a horizontal wire, 1.20 m long, connects them at a distance 1.40 m from the hinge, as measured along the sides. The ladder rests on a frictionless floor. What is the tension in the wire?

    2. Relevant equations

    Sum of the forces=0

    Sum of the torques=0

    3. The attempt at a solution

    I understand that the ladder can be treated as two separate poles since they are symmetrical. I am just confused as to how to solve it without knowing what the angle is. I tried taking the weight times length (lever arm?) and having that be equal to the tension and the answer I got was 1365N which was incorrect. Any suggestions? Thanks!
     
  2. jcsd
  3. Oct 9, 2011 #2

    PhanthomJay

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    You can calculate the angles from geometry. The floor is frictionless. The weight of each diagonal rod acts down at its midpoint measured along the diagonal. If you take a free body diagram of one leg of the ladder and sum torques about the top, you must consider the sum of torques of all forces about that point to solve for the tension force. You can use the cross product rule to calculate torques (Torque = r X F). You might first want to caculate the force reactions from the floor.
     
  4. Oct 11, 2011 #3
    Oh ok that makes sense. Thanks!
     
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