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Homework Help: Static Friction -- A ladder leaning against a wall starts to slip

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    A ladder,10 m long and having a mass of 25 kg, rests on a rough horizontal floor and is supported by a rough vertical wall. The ladder is inclined at 60 degrees as shown and starts to slip when a person having a mass of 75 kg has climbed halfway up it. The coefficient of static friction at the wall is 0.2. Assume the weight of the ladder to be concentrated at its midpoint. Calculate the coefficient of static friction at the floor.

    2. Relevant equations

    3. The attempt at a solution

    I've been stuck at this question for 3 days and I am still unable to find the coefficient of static friction at the floor. I have drawn the free body diagram and tried to solve it, getting a static friction of 0.3. Please tell me what is wrong with my workings, I don't want to ponder over such this question any longer.(sorry for my terrible handwriting) 20141105_225852.jpg 20141105_225841.jpg
  2. jcsd
  3. Nov 5, 2014 #2
    Please explain your 4th line.
  4. Nov 5, 2014 #3
    Notice that the ladders just started to slip, so: Is it ok to say that the net force on the Y axis equals zero?

    Also think about this: Isn't it necessary to add a rotation equation to solve this problem?
  5. Nov 5, 2014 #4
    I was trying to use the principle of moments, taking the ground contact as the point of reference
  6. Nov 5, 2014 #5
    Uhm...so I can only equate Y forces as 0 when I only take the weight of the ladder into account ? So do I equate X when I take both the ladder and the person's weight into account ? Also, what do you mean by rotational equation ? Is it the principle of moments ? If it is, Then I have used it in my fourth line.
  7. Nov 5, 2014 #6
    Your handwriting is essentially unreadable. Please type out your equations, and define your variables. You should have a force balance in the x direction, a force balance in the y direction, and a moment balance around some convenient center of rotation.

  8. Nov 6, 2014 #7
    Sorry, maybe my comment just confused you. What I wanted to say is that you can only make an equation equal to ZERO, being the sum of forces in X or in Y, or the sum of moments around a point, if there is no acceleration. However, in the statement it is mentioned that the system just started to slip. So it kind of looks like there is some acceleration going on here.
    My question: is it OK in this case to equal all the equations to ZERO? Why?

    Once you understand this, with these three equations (Sum of FX, Sum of FY and Sum of Moments around one point) you can solve the problem.
  9. Nov 6, 2014 #8
    A clearer picture of my free body diagram.

    F(n)y+F(r)y=981 N (weight of the ladder and the person combined)
    Using principle of moments, take the contact between the ladder and the floor as the pivot.
    981 N* 5cos(60°)m=F(n)x*10sin(60°)m [10m being the ladder's length]

    so...where have I gone wrong ?
  10. Nov 6, 2014 #9


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    Moment balance misses term for Fny
  11. Nov 6, 2014 #10
    Thank you for your help, I've already found the answer
  12. Nov 6, 2014 #11
    Thank you so much for pointing that out ! I've managed to get my answer
  13. Nov 6, 2014 #12
    No problem. I didn't help much, I think.

    It's still interesting to understand why you can equal the balances to ZERO, though..
  14. Nov 6, 2014 #13
    The moment balance is incorrect because you left out the moment from the friction force F(n)y.

    They meant to say that the ladder is on the verge of slipping, but static friction has not quite released yet. They weren't very precise about how they worded it.

  15. Nov 6, 2014 #14
    Yes, that's what I had figured out but as you said, they weren’t very precise on the explanation, and a misunderstanding of this detail changes a lot in the output. I just wanted to make sure the OP understands what that means and to think about what the implications were in the case of the ladder being actually falling, having a system out of equilibrium where the balances are not longer equal to ZERO. Just to add a little extra kick to the problem, that’s all..
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