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Ladder - Equilibrium

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img206.imageshack.us/img206/8178/ladderjt1.png [Broken]

    A uniform 5.0-kg ladder is leaning against a frictionless vertical wall, with which it makes a 15* angle. The coefficient of friction between ladder and ground is 0.26. Can a 65-kg person climb to the top of the ladder without it slipping? If not, how high can the person climb? If so, how massive a person would make the ladder slip?

    2. Relevant equations
    \sum {\vec \tau } = \vec 0 \hfill \\
    \sum {\vec F} = \vec 0 \hfill \\

    3. The attempt at a solution

    So, I choose the end of ladder that is touching the floor to be the axis. Now I want to find the sum of the torques, then set them equal to zero. What I don't quite get in the equation is the sin(90-15) in the first term in the following equation my professor gave me:

    F_W L\sin (90^ \circ - 15^ \circ ) - M_p gL\sin (15^ \circ ) - M_L g\frac{L}
    {2}\sin (15^ \circ ) = 0

    sin(90-15)? Technically, it should be the angle between the r and F....but this doesn't make sense to me....or perhaps 15* is meant to be the other angle in the triangle?

    If anyone can help, it would be greatly appreciated. Thank you!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 3, 2008 #2


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    Homework Helper

    Hi RedBarchetta,

    Yes, it's the other angle; the ladder makes a 15 degree angle with the wall, so the angle with the floor is 75 degrees. Once you correct your diagram, do you see why the angles your professor chose are correct?
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