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Ladder against wall - what is friction needed to prevent ladder from slipping

  1. Jul 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A uniform 15 kg ladder whose length is 5.0 m stands on the floor and leans against a vertical wall, making an angle of 25 degrees with the vertical. Assuming that the friction between the ladder and the wall is negligible, what is the minimum amount of friction between the ladder and the floor which will keep the ladder from slipping?

    2. Relevant equations

    Fg = mg
    Torque = distance x force

    3. The attempt at a solution

    I really am not sure about my approach. I figure I should try and find the force on the horizontal on the ground and that should be the friction needed.

    Fg = 15 x 9.8 at teh center of the ladder =147 N
    Torque also at the center = 147 x 2.5m = 367.5
    The gravity should be going straight down and the torque perpendicular to the ladder. This gievs me the hypotenuse and one side of a triangle, left with the ground as the horizontal which solves as 337 N but the correct answer is 34 N.
    I'm guessing I would need to use the angle in there somewhere.

    Thanks for any help.
  2. jcsd
  3. Jul 8, 2007 #2

    Can anybody give me a tip in what we to start/approach this question?

  4. Jul 8, 2007 #3


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    Staff Emeritus
    Science Advisor

    Sum of the moments is zero.

    Determine the forces and their corresponding moment arms.
  5. Jul 9, 2007 #4
    Sorry, still dont get it. I tried using a gravity arm downwards, a torque perpendicular and then solving the horizontal as the friction.

    Any other suggestions? What am I missing?
  6. Jul 9, 2007 #5


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    Science Advisor
    Homework Helper

    Draw the three forces acting on the ladder. i) gravity (which acts at the center of mass, so produces no torque, ii) normal+friction force at the ground and iii) normal force from the house (which must be horizontal since there is no friction). Set the sum of all the forces equal to zero and the sum of all the produced torques equal to zero. That's what Astronuc said in a very concise manner.
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