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Ladder against a wall

  1. Sep 16, 2013 #1
    Hi all, this is a problem with which I led myself in circles with for hours, because I kept forgetting to factor in a different thing each time. Long story short, I thought I'd cracked it, but my answer and the given answer are out by a square. Please see my efforts below!

    1. The problem statement, all variables and given/known data
    A uniform ladder of mass M and length L stands on rough horizontal ground and leans against a vertical wall with which it makes an angle θ. A man of mass m stands on the ladder a distance x from the bottom.

    a) obtain expressions for the normal reaction and the friction force acting at the point of contact between the ladder and the ground.


    2. Relevant equations

    The normal is obviously (m+M)g.

    Other than that, it's static equilibrium, and the equation I used for balancing was the moments, where τNW is the moment due to the normal FNW to the wall at the top of the ladder, and τMg and τmg are the two moments of the ladder and man:

    1. τNW = τMg + τmg

    3. The attempt at a solution

    Pretty simple. I have the ladder to be length L, so I set the origin of turning at the base of the ladder. I want to find FNW because it's equal to the frictional force FF at the base of the ladder.

    The radius component for the moment τNW is this Lcosθ. The moment is thus: FNWLcosθ.
    The radius component for τMg is (L/2)sinθ, and for τmg is (x/L)sinθ.
    So the two moments are: [(MgL)/2]sinθ and [(mgx)/L]sinθ

    Equation 1. then becomes:

    FNWLcosθ = [(MgL)/2]sinθ + [(mgx)/L]sinθ

    Which evidently becomes, solving for FNW:

    FNW = (M/2 + mx/L2)gtanθ

    In the given answer, the mx/L2 term is simply mx/L. But the position of the man along the ladder was given as x from the bottom, so this should be written x/L, right? Which means dividing out the L on the FNW side will always create a squared term.

    Any insight appreciated.

    Jack
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2013 #2
    The moment arm of mg should be x sinθ ,not (x/L)sinθ .The distance of the man is given from the bottom of the ladder .This distance is x not x/L .Just as the moment arm of Mg is (L/2)sinθ,likewise that of mg will be x sinθ .The torque due to mg ,τmg=mgxsinθ.

    This will give you the correct answer.
     
  4. Sep 16, 2013 #3
    Why would there be an extra L? Your "man" term in the equation is not dimensionally consistent with the other two terms. Your term would be correct if he were at a fractional location x/L of the total ladder length. But that's not the case. He is at an absolute distance x up the ladder.
     
  5. Sep 16, 2013 #4
    This formula is obviously wrong:

    FNW = (M/2 + mx/L2)gtanθ

    The second term in the parenthesis has units of mass/length so you cannot add it to the M/2.

    Now why did you get this?
    You misinterpret the meaning of x.
    They say "distance x from the bottom". So x is clearly a distance, in meters or whatever unit you use.

    The moment of the force should be a force times a distance. For the weight of the man, the "radius component" is x*sinθ so the moment will be mgx*sinθ.
    I don't really see why would you think to divide by L.

    Edit. Sorry, I was writing (slowly :)) while Chestermiller posted already.
     
  6. Sep 16, 2013 #5
    Yup, you're all clearly correct. Thanks so much for pointing it out. I really don't know why I was trying to divide by the L. I had established it way early on in my workings and didn't think to question it later. Losing the forest for the trees!

    Cheers everyone :)
     
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