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Homework Help: Static Equilibrium - When will this guy bail off his ladder?

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A window cleaner of mass 95 kg places a 22kg ladder against a frictionless wall at a angle 65 degrees with the horizontal. The ladder is 10 m long and rests on a wet floor with a coefficient of static friction equal to .40. What is the max lenght that the window cleaner can climb before the ladder slips?

    2. Relevant equations

    sum F(x)=F(fr)-F(w)=0
    sum F(y)=F(n)-mg=0
    sum torque=F(w) * l (sin theta) - mg * 1/2 cos theta=0

    (Using the point of the ladder on the ground as the pivot point)

    3. The attempt at a solution

    Using this equation: FW (10) sin(65) - 22 (9.8) (5) cos(65) - 95 (9.8)(x)cos(65) = 0

    I then used Fr-Fw=0
    Fn(Us) = FW
    (1146.6*.40) = 458.64

    I then input this into the equation

    FW=22 [(9.8) (5) cos(65) - 95 (9.8)(x)cos(65)]/ (10) sin(65)

    x = 9.41 m for the distance of the guy up the ladder. I've checked and re-checked and keep getting this answer, but intuitively it seems like that's too high since it's a wet floor that the ladder's on. Is the coefficient of static friction of 0.40 high enough to make this reasonable? Thanks for anyone who is willing to take a look!

  2. jcsd
  3. Apr 8, 2009 #2


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    Is the coefficient of static friction of 0.40 high enough to make this reasonable?
    Not necessarily. It is the angle of ladder that prevents the slipping. Walking on the slippery floor we take smaller steps.
  4. Apr 10, 2009 #3
    Hmm..ok so being one who doesn't often hang out on ladders, I'm trying to figure out whether 65 degrees is enough to prevent slipping. Is it possible that 9.41 m would be a reasonable answer? I came out with the same answer after trying it multiple times, so I can't see where I could be going wrong. Thanks for anyone's opinion!
    Last edited: Apr 10, 2009
  5. Apr 10, 2009 #4


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    What is the answer?
  6. Apr 10, 2009 #5
    Hi rl.bhat, the answer I came up with x=9.41 m on the 10 m ladder. What do you think?
  7. Apr 10, 2009 #6


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    I too got the same answer. Is it wrong?
  8. Apr 10, 2009 #7
    yay! That's good news. I'm submitting it for an assignment so I'm glad to hear it!
    I'll let you know whether it's right when I get it back :) Thanks for helping!
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