1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Equilibrium Ladder Question, can get 1st part but not 2nd
  1. Equilibrium Question (Replies: 0)

Loading...