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Equillibrium Ladder

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform ladder of mass m and length L leans at an angle θ against a frictionless wall, Fig. 9-61. If the coefficient of static friction between the ladder and the ground is 0.43, what is the minimum angle at which the ladder will not slip?

    Pic:
    http://www.webassign.net/giancoli5/9_61.gif


    2. Relevant equations

    ?? tan(theta) = .43?

    3. The attempt at a solution

    23.26
     
  2. jcsd
  3. Jan 16, 2010 #2

    tiny-tim

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    Hi tigerwoods99! :smile:

    (have a theta: θ :wink:)
    I don't think so.

    What equations did you use?
     
  4. Jan 16, 2010 #3
    Hi! I am really not sure its more or less a wild guess using trigonometric functions.
     
  5. Jan 17, 2010 #4

    tiny-tim

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    (just got up :zzz: …)
    thought so! :rolleyes:

    ok … call the normal reaction force (from the ground) N, and the horizontal reaction force H …

    how much is N? :smile:
     
  6. Jan 17, 2010 #5
    I think this is right.


    τorque = 1/2mgx - FNx + Ffy

    at max angle
    μmgy = 1/2mgx

    the maximum angle is θ = tan−1(1/(2μ))
     
  7. Jan 17, 2010 #6

    tiny-tim

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    Yup! :biggrin:

    Easy, isn't it? :wink:
     
  8. Jan 17, 2010 #7
    haha yea i guess.. though i haven't learned this yet so i had to figure it out.
     
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