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Homework Help: Equillibrium ladder on a wall problem

  1. Dec 4, 2007 #1
    1. The problem statement, all variables and given/known data
    http://img90.imageshack.us/img90/4989/scanuy8.png [Broken]
    http://img125.imageshack.us/img125/8163/fbdzh7.png [Broken]

    3. The attempt at a solution
    Been working on this one for at least an hour now =/

    [tex]n_{1}[/tex] is the normal force of the wall on the ladder
    [tex]n_{2}[/tex] is the normal force of the ground on the ladder

    [tex]\sum F_{x} = f_{s} - n_{1} = 0[/tex]
    [tex]f_{s} = n_{1}[/tex]

    [tex]\sum F_{y} = n_{2} - m_{ladder}g - m_{weight}g = 0[/tex]
    [tex]n_{2} = 150(9.8) = 1470N[/tex]

    [tex]\tau = Fdcos\theta[/tex]
    [tex]\theta = sin^{-1}(4.5/5) = 64[/tex]
    [tex]\sum\tau = (n_{1})(5)cos(64) - (m_{ladder}g)(2.5)(cos(64)) - (m_{weight}g)(5)(cos(64)) = 0[/tex]
    [tex]n_{1} = 1131N[/tex]

    And since f_{s} = n_{1}, the answer I get is 1131N, which is wrong. If anyone can see where I messed up, I'd greatly appreciate any help.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 4, 2007 #2
    It just occured to me that it is asking for the entire force from the ground, not just the frictional force.

    [tex]F_{tot} = \sqrt{n_{2}^2 + f_{s}^2} = \sqrt{1131^2 + 1470^2} = 1852 N[/tex]

    Still off =/

    *EDIT* using the answers as I guide, I figured out a way to make my calculations match an answer, but I have a feeling it is wrong...

    [tex]f_{s-max} = n_{2} * \mu_{s} = 1470(0.40) = 588N[/tex]
    [tex]F_{tot} = \sqrt{n_{2}^2 + f_{s-max}^2} = \sqrt{1470^2 + 588^2} = 1583 N[/tex]

    Which matches an answer, but I think the ladder should be sliding down with all the weight...
    Last edited: Dec 4, 2007
  4. Dec 4, 2007 #3


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    well i'm not sure what you got anymore, but i notice that in your original post you incorrectly calculated the torque of the normal force at the wall you used cos64 when you should have used sin 64......that should give you the correct answer..the friction force at the base is not necessarily (mu)n2, it is equal or less than that value depending on your corrected torque equation.
  5. Dec 4, 2007 #4
    [tex]n_{1}(5)sin64 = (M_{ladder}g)(2.5)cos64 + (M_{weight}g)(5)cos64[/tex]
    [tex]n_{1} = 550[/tex]
    [tex]F_{tot} = \sqrt{550^2 + 1470^2} = 1570[/tex]

    And 550 < [tex]f_{s-max}[/tex] of 588 :D

    Thanks for the help!
  6. Dec 4, 2007 #5


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    You're welcome, you did a good job with this problem, just a small trig error threw it off. It's sometimes easier to use torque as the product of the force times the perpendicular distance rather than torque = Fdsintheta, where theta is the angle between the force and displacement vectors, because it's easy to mess up on the correct theta value.
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