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Minimum u to keep ladder from slipping

  1. Aug 26, 2007 #1
    I've read similar posts and have tried the problem several times but don't get the right answer.

    1. The problem statement, all variables and given/known data
    A uniform ladder with a mass of 15 kg leans against a frictionless wall at a 65 degree angle. Find the required friction coefficient (u) at the floor that will allow a 100kg person to stand 2/3 of the way up the ladder without slipping.

    The answer is .301 but I get u is 1.38.

    2. Relevant equations
    t = r x f fg = 9.8 m

    3. The attempt at a solution
    Force of friction = force of wall

    Normal force = fg of the ladder + fg of the person = 15*9.8 + 100*9.8 = 1127N

    torque = 0 = torque of ladder + torque of person - torque of wall = 15*9.8*cos(25)*(1/2) + 100*9.8*cos(25)*(2/3) - (Force of wall)*cos(65)
    I solved for Force of Wall to be 1558.7 N (I know that's wrong because it's supposed to be less than the normal force)

    Force of friction = force of wall = 1558.7 = u*Fn = u*1127 u = 1.38

    I think I messed up in the torque equation but I don't know exactly where. Help?
  2. jcsd
  3. Aug 26, 2007 #2

    Doc Al

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    Staff: Mentor

    Check your trig: Why are you using cos(25) to find the torque due to the weights?
  4. Aug 26, 2007 #3
    Switching the cos and the sin gives the right answer. I guess i misunderstood the placement of the 65 degree angle
  5. Aug 26, 2007 #4

    Doc Al

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    Staff: Mentor

    I assume that the ladder is 65 degrees with respect to the floor--but it's not clearly specified in the problem statement.
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