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Ladder on a wall held in place by a peg

  1. Jan 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Figure attached
    A uniform ladder is 10 m long and weighs 400 N. It rests with its upper end against a frictionless vertical wall. Its lower end rests on the ground and is prevented from slipping by a peg driven into the ground. The ladder makes a 30 degree angle with the horizontal. The magnitude of the force exerted on the peg by the ladder is:
    W = 400N
    L = 10m
    Answer: 470N
    2. Relevant equations
    Torque = r x F

    3. The attempt at a solution
    Calculating torque about the lower end, I was able to find force exerted by the wall
    Tweight = Twall
    400N(5cos30) = Fwall(10sin30)
    Fwall = 346.4N

    the forces I have so far are: normal force, weight and force from the wall.
    Are there two other forces from the peg(perpendicular to the ladder going down) and the force of the ladder on the peg(perpendicular to ladder up)?
     

    Attached Files:

  2. jcsd
  3. Jan 8, 2017 #2

    haruspex

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    You can thonk of the force from the peg as a single force or as composed of two forces at right angles to each other; in the latter choice, you could use vertical and horizontal or perpendicular and parallel to the ladder.
    I suggest using vertical and horizontal.
     
  4. Jan 8, 2017 #3

    ehild

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    In your figure, the peg looks as part of the ladder and perpendicular to the leg of the ladder. But the peg should stand out from the ground, probably vertical. Was the figure attached to the problem, or have you drawn it yourself?
    You have to collect the forces acting on the ladder, and their sum should result zero. Determine the force of the peg on the ladder. The force of the ladder on the peg is equal in magnitude and of opposite direction according to Newton's third law.
     
  5. Jan 9, 2017 #4

    CWatters

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    The usual form of this problem has friction with the ground rather than a peg preventing it slipping. Unless there is additional info in the original problem statement I would solve the peg version the same way. eg assume the peg can only provide a horizontal force.
     
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