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Homework Help: Equilibrium Ladder Question, can get 1st part but not 2nd

  1. Apr 1, 2012 #1
    Equilibrium Ladder Question, can get 1st part but not 2nd!!

    A symmetrical ladder of mass M = 20.8kg leans against a smooth frictionless wall so the top of the ladder is height h = 6.58m above the floor and the bottom of the ladder is distance d = 2.19 m from base of wall. The floor is also frictionless so a horizontal wire connects bottom of ladder to wall so ladder does not slip.

    With nobody on ladder, here is what I did to get T, magnitude of tension in wire:

    ƩF = 0

    F wire = F wall

    L of ladder = sq. root of 6.58^2 + 2.19^2 = sq. root of 48.0925 = 6.9348m

    Angle θ b/t wire and ladder --> tan θ = 6.58 / 2.19; so θ = 71.59°

    Center of mass for ladder = 6.9348 / 2 = 3.46743m

    mg down = (20.8)(9.81) = 204.048 N

    (6.93487 x sin 71.59°)(F wall) - (3.46743 cos 71.59°)(204.048) = 0

    F wall = 33.9 N, so F wire = 33.9 N

    Now, suppose wire will snap when the magnitude of the tension is T = 197 N. What is x, the distance a man of mass m = 78.3 kg can climb up along ladder before the wire snaps?

    mg down = (78.3 + 20.8) x 9.81 = 972.171

    (6.93487 times sin 71.59)(197) = (x cos 71.59)(972.171)

    and I get x = 4.22m but this is wrong...

    What am I doing wrong here? I know I did first part right (without man on ladder) but what I am doing wrong for this part??
  2. jcsd
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