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Torque / Static Equilibrium Question

  1. Oct 15, 2006 #1
    I need help with the following question.

    1. A uniform ladder of mass M and length L leans at an angle A against a frictionless wall. If the coefficient of static friction between the ladder and the ground is Q, what is the minimum angle at which the ladder will not slip?

    Answer so far:
    let T stand for torque

    Tladder + TNwall = 0
    rFcosA + rFsinA = 0
    LFcosA = -LFsinA
    mgcosA / sinA = Ffriction
    mg = FfrictiontanA
    (1 / QcosA) = tanA
    A = tan^-1(1 / QcosA)

    *The real answer is supposed to be A = tan^-1(1 / 2Q)

    I was wondering if you could tell me what I did wrong, and what I should do to get the right answer. Thanks!
     
    Last edited: Oct 15, 2006
  2. jcsd
  3. Oct 15, 2006 #2
    Can cosA equal Q, which is actually mu?
     
  4. Oct 15, 2006 #3

    OlderDan

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    Maybe its just the notation I can't follow, but I'm not seeing some things I think I should see. What is the normal force of the floor on the ladder? What is the friction force in terms of the normal force? What is the normal force of the wall on the ladder. Is that your r? I can't see you having a net zero torque equation with less than three torques.

    Please define your variables.
     
  5. Oct 15, 2006 #4
    I put my pivot point at the base of the ladder so the normal force of the floor on the ladder would be 0. The friction force would be equal in magnitude to the normal force the wall exerts on the top of the ladder. r is the length of the ladder (L) because it represents the length of the lever arm a.k.a. ladder.

    T represents torque
    Q represents mu, the coefficient of friction
    A represents theta, the angle
    M represents the mass of the ladder
    F represents force

    Does this help?
     
  6. Oct 15, 2006 #5
    Got It!

    I was able to get the answer, so you do not have to reply. I found my mistake! :)
     
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