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Stability: Ladder Problem

  1. Dec 16, 2014 #1
    1. The problem statement, all variables and given/known data
    image.jpg

    Where it says ''from the bottom'' I assumed it's referring to a distance along the ladder. So:


    image.jpg

    Data:

    ##w_{ladder} = 98.0\ N##

    ##w_{person} = 686\ N##

    ##d_1 = 4\sqrt(2)\ m##

    ##d_2 = 1\ m##

    ##d_3 = 2/3\ m##


    2. Relevant equations

    ##\sum \tau = 0##

    ##\sum F = 0##

    3. The attempt at a solution

    Diagram:

    image.jpg image.jpg

    I chose my pivot point to be the bottom of the ladder. Hence, torques due to normal force ##N## and friction ##f## are zero. ##ccw## is positive and ##cw## is negative.

    ##0 = w_pd_2 + w_ld_3 - F_wd_1##

    ##F_wd_1 = w_pd_2 + w_ld_3##

    ##F_w = \frac{w_pd_2 + w_ld_3}{d_1} = \frac{686\ N \cdot 1.00\ m + 98.0\ N \cdot 2/3\ m}{4\sqrt{2}\ m} = 133\ N##

    ##F_w = f = 133\ N##

    Then:

    ##N = w_p + w_l = 686\ N + 98.0\ N = 784\ N##

    The force at the top of the ladder is just ##133\ N##.
    The force at the bottom is ##\small{\sqrt{N^2+ f^2} = \sqrt{ (784\ N)^2+ (133\ N)^2} = 795\ N}##. Doubt: this magnitude, whether it's correct or not, acts along the ladder, right? (as a tension would?).

    The textbook's solutions are ##126\ N## and ##751\ N##, respectively. It seems I haven't missed anything, so I don't know where the mistake is.
     
    Last edited: Dec 16, 2014
  2. jcsd
  3. Dec 16, 2014 #2

    BvU

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    Which direction does Fw point ?
     
  4. Dec 16, 2014 #3
    To the right.

    And in a cw direction about the pivot point.
     
  5. Dec 16, 2014 #4

    BvU

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    No vertical component ?
     
  6. Dec 16, 2014 #5
    Components of ##F_w##? Mmm, It didn't occur to me that there could be vertical components there. Has that of a ''frictionless rain gutter'' something to do with it?
     
  7. Dec 16, 2014 #6
    It seems that I also forgot to take ##F_w## into account when finding the force at the bottom.
     
  8. Dec 16, 2014 #7

    haruspex

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    Frictionless says there will only be a normal force. Normal means perpendicular to the surface of contact. What direction will that be? You should assume the ladder projects slightly beyond the point of contact with the gutter.
     
  9. Dec 17, 2014 #8

    BvU

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    Consider a more extreme case where the ladder is at 45 degrees (or even less) wrt the ground. Would it lean ON the gutter, as well as against the gutter ?
     
  10. Dec 17, 2014 #9
    I think I'm getting the idea. It would be something like this:

    gutter.jpg


    However I'm not sure what is the correct direction that ##F_w## takes.

    I think the sharp corner of the gutter can be viewed as a fulcrum. Should ##F_w## act perpendicular to the edge of the gutter, as bisecting it? If that were the case, my drawing wouldn't be totally accurate. Would it be in that direction, independent of the angle that the ladder makes with the ground?.
     
  11. Dec 17, 2014 #10

    haruspex

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    No. It might help to rotate the problem around to a more familiar set-up. Imagine a table on flat level ground, but the legs of the table are sharp wedges, meeting the ground at some odd angle. Which way would the normal force act?
    More generally, as I wrote, normal means perpendicular to the contact plane. I'm not sure how contact planes are defined generally, but if either surface is differentiable then it will be the tangent plane which passes through the point of contact. Taking the ladder uprights to be rectangular in section, they have a differentiable surface. Or if you take them to be arbitrarily thin rods then it must be a plane containing that rod and containing the edge line of the gutter - there is only one such. Sounds complicated, but the underlying idea is fairly intuitive.
     
  12. Dec 18, 2014 #11

    BvU

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    Another approach: ladder pushes against gutter. In which direction ? Gutter pushes back, opposite direction (no friction).

    If there were a rubber hose between the two, how would it compress ? To a vertical ellipse ? Or to an ellipse with the long axis parallel to the ladder ?
     
  13. Dec 18, 2014 #12
    I would say to an ellipse with the long axis parallel to the ladder, which would mean that the force exerted by the gutter on the ladder is perpendicular to it.

    In the other case (leaning against a wall), the wall becomes the differentiable surface and the rubber hose would compress parallel to it. It makes sense.

    If the gutter weren't frictionless, friction would act parallel to the ladder's structure, right?
     
  14. Dec 18, 2014 #13

    haruspex

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    Yes, I agree with all of that.
     
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