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Ladder on rough ground against rough wall

  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A ladder on the rough floor is leaning against a vertical rough wall. The ladder has length l and mass m. The coefficients of friction are μ for both contact surfaces. What is the smallest angle between the ladder and the floor?

    2. Relevant equations
    ∑F=ma
    ∑τ = F⊥ r
    fs ≤ μsN

    3. The attempt at a solution

    Snapshot.jpg
    ∑Fy = 0
    N2 + f1 - W = 0

    ∑Fx = 0
    f2 - N1 = 0
    f2 = N1

    ∑τ about the ground = 0
    N1lsinθ + f1lcosθ - W(l/2)(cosθ) = 0

    For θ to be the smallest angle, what is the condition? Is it f1 = μN1 or f2 = μN2 or both of them need to happen at the same time?
     
    Last edited: Apr 15, 2015
  2. jcsd
  3. Apr 15, 2015 #2

    ehild

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    Can the ladder move if the friction is static at some of the contact surfaces at the ends of the ladder?
     
  4. Apr 15, 2015 #3
    I think the ladder can't move if one the friction is static. So both must happen at the same time.
     
    Last edited: Apr 15, 2015
  5. Apr 15, 2015 #4
    Okay, I finally managed to solve it. θ = tan-1[1-μ2/(2u)]. Thanks ehild!
     
  6. Apr 15, 2015 #5

    ehild

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    Are you sure that ##\theta= \tan^{-1}\left(1 - \frac{ \mu^2 }{2 \mu }\right)## ? Why you do not simplify with μ? Or some parentheses missing? :devil:
     
  7. Apr 15, 2015 #6
    Oh it is a typo! Should be θ = tan-1[(1-μ2)/(2μ)]. Thanks anyway!
     
  8. Apr 15, 2015 #7

    ehild

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    It is correct now. Nice work!
     
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