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Ladder and circle question

  1. Nov 27, 2006 #1
    Basically wondering what shortest possible length of this ladder is, one that reaches 2 meters high. The ground its on is quarter of a circle with radius 1 with an extra length of 1 on top. Weight is 40 (so 40*gravity for force) and µ (coefficient of friction) between ground and ladder is 0.8


    I sort of worked out an answer, but its two complex for me to work out so i thought i'd post here

    i started off by modeling the curve bit as y = -(1 - (x^2))^0.5 ; 0<x<1

    w = 1 - x

    h = 1 - y (y is negative)

    The length can therefore be written as

    L = (h^2 + w^2)^0.5

    L = ((1-y)^2 + (1-x)^2)^0.5

    Next i worked out angles:

    tan(theta1) = dy/dx

    dy/dx = x/(1 - (x^2))^0.5

    tan(theta1) = x/(1 - (x^2))^0.5

    tan(theta2) = (1 - y)/(1 - x)

    Forces must be the same in all directions:

    S + Rsin(theta1) = Fcos(theta1) (x axis)

    40g = Rcos(theta1) + Fsin(theta1) (y axis)

    g = 9.8 (gravity)

    Magnitude of rotation around base of ladder must be zero:

    (1/2)(40g)(L)cos(theta2) = (L)(S)sin(theta2)

    Friction = µR

    F = 0.8R

    Cancelling down i got two equations with L and x... unfortunatly one is so complicated that i have no idea how to put them together. I'm also worried that i made a mistake somewhere, since it got so complicated, is there possibly an easier/less complicated way of working it out than i did?

    First equation, pretty easy to see where i got this (replaced y with x):

    L = ((1 + (1 - (x^2))^0.5)^2 + (1 - x)^2)^0.5

    Second equation (possibly incorrect): :surprised

    0 = ((196(1-x))/(1+(1-(x^2))^0.5))/(0.8-(x/(1-(x^2))^0.5)) - ((392)/(1+((0.8x)/((1-x)^2))))

    or written in paint:


    Which i cancelled down to (possibly even more incorrect)

    0.8x^2 + 2.8x -1.6 + x(1 - x^2)^0.5 - 0.6(1 - x^2)^0.5 = 0

    So i have no idea how to put those together, or work out when x is zero. Putting them into a graphing program i got:


    (blue is first red is second)

    this means there would be two possible places at which the ladder would be at limiting equilibrium... unsure whether this should happen.

    So can anyone help?
    Last edited: Nov 27, 2006
  2. jcsd
  3. Nov 27, 2006 #2


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    If you calculate torque about the base of the ladder, and the net force in the horizontal axis, you should get two equations, which, with the information you already have, can be used to answer this question.
  4. Nov 27, 2006 #3
    I already have both of those?


    (1/2)(40g)(L)cos(theta2) = (L)(S)sin(theta2)


    S + Rsin(theta1) = Fcos(theta1) (x axis)

    I just tried cancelling everything down again and got the two equations again:



    L = ((1 + (1 - (x^2))^0.5)^2 + (1 - x)^2)^0.5

    I don't know how to put them together... or how to solve the first one.
    Last edited: Nov 27, 2006
  5. Nov 27, 2006 #4


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    The maximum force of friction on the bottom of the ladder is:
    [tex]40 \mu g[/tex]

    The only other horizontal fore acting on the ladder is the normal force on the wall. So, if I read your notation you should have:
    [tex]F_h=40 \mu g+S=0[/tex]
    [tex]S=40 \mu g[/tex]
  6. Nov 27, 2006 #5
    Doesnt the normal force come out at a 90degre angle to whatever an object is resting on? Then R would change as you moved the ladder up the slope. I am confused whether you ment maximum force of friction as in when the ladder is at the bottom of the slope or whether you are saying that R is always 40g.
  7. Nov 28, 2006 #6


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    The normal force exerted by the floor on the ladder, and gravity are the only two forces acting on the lader in the vertical direction. Since this is a statics problem, they must cancel - so they're of equal magnitude in opposite directions.
  8. Nov 28, 2006 #7
    Ok, got an answer of 1.81 thanks for the help
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