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Frictional force exerted

  1. Apr 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A 5.00-m long uniform ladder leans against a smooth wall and its base rests on a rough floor. The ladder has a mass of 18.0 kg and its base is a distance of 2.30 m from the wall. A person of mass 70.0 kg climbs 2.50 m up the ladder. If the ladder is to remain in place, what frictional force must be exerted by the floor on the ladder?

    2. Relevant equations
    t = FR
    t =0
    N=mgcosθ
    ΣF=0
    Σt = 0
    -mLgbg-mPgbP+Fn,w*a = 0
    Fn,w= (mLgbg + mPgbP ) / a

    ΣFx = 0
    Fs - Fn,w = 0
    Fs = Fn,w = (mLgbg + mPgbP ) / a

    3. The attempt at a solution
    wall (a)= sqrt(30.29)= 5.50m, bL=2.30m, bP=cos(67.3)*cP
    c= 5.00m, cP=2.50m
    θL= tan-1(5.50/2.30)= 67.3°

    Fs = ((18.0)(9.81)(2.30)+(70.0)(9.81)(.965)) / 2.5 = 194 N

    I'm not completely sure if I found all forces for this problem, did I find everything for the frictional force?
     
  2. jcsd
  3. Apr 13, 2016 #2
    angle of the ladder with the ground = arc cos(2.3/5) = 62.61°. I do not know which angle you find out to be 30.5 and how? You have to label all the possible forces acting on the ladder and take moments about a convenient point. I think it is center of mass of teh ladder. You think why and what is the convenience. modify your reply accordingly then I would comment.
     
  4. Apr 13, 2016 #3

    haruspex

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    It is rarely necessary to calculate the actual angles from the sides of the right-angled triangle. Just work in terms of the trig functions of the angles.
     
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