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Equilibrium of a rigid body under coplanar forces

  1. Apr 7, 2012 #1
    1. The problem statement, all variables and given/known data

    The foot of a ladder rests against a wall,and its top is held by a horizontal tie rope. The ladder weighs 100 N and its center of gravity is 0.4 of its length from the foot. A 150 N child hangs from a rung that is 0.2 of the length from the top. Determine the tension in the tie rope and the components of the force on the foot of the ladder. The angle between the wall and the ladder is 37°

    2. Relevant equations

    Sum of torques = 0
    Sum of forces = 0

    3. The attempt at a solution
    Sum of torques = (60.2N)(0.4L) + (90.3N)(0.8L) + (x)(L)
    x= -96.32 N
    The tension of the rope only has an x component so I don't think this calculation helps. I'm comfortable doing calculations when the beam is horizontal but I can't figure this one out.
  2. jcsd
  3. Apr 7, 2012 #2


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    Staff: Mentor

    You've used the wrong trig function to determine the portion of the weights acting perpendicularly to the ladder. Draw a diagram.

    The tension in the rope is going to have components that run along the ladder and perpendicular to the ladder. The component that's perpendicular to the ladder is what is keeping the ladder from rotation about the foot pivot point.
  4. Apr 7, 2012 #3
    Thanks for the reply. I did draw a diagram. Is the perpendicular force at the end of the beam the hypotenuse of the tension force?
    Last edited: Apr 7, 2012
  5. Apr 8, 2012 #4


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    Staff: Mentor


    The vector representing the tension in the rope will be the hypotenuse of the force triangle. Components will be perpendicular to and along the ladder.

    Attached Files:

  6. Apr 8, 2012 #5
    Thank you. I understand now. I really appreciate the help.
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