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Classic Ladder Problem

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A ladder that is 10ft long, weighs 30lbs, and has a 180 lb man standing 5ft up it, resting at 60 degrees on a wall

    2. Relevant equations
    equilibrium, and torque

    3. The attempt at a solution
    180lbs cos60° * 2.5ft + 30lbs cos60° * 10ft = Fwall sin60° * 10ft (all trig functions used to find component of force that is perpendicular to the ladder)
    and when I solve for Fwall i get ≈43.3lbs

    then I would simply set the force of friction (or as he has in his problem, the person standing at the bottom of the ladder on a frictionless floor, god know how that works, lol) to that as they are the only horizontal forces, then the normal force from the ground would be the two weights added together. So Ffriction is also 43.3lbs and Fnormal is 210lbs.
  2. jcsd
  3. Oct 30, 2013 #2
    What is the problem??
  4. Oct 30, 2013 #3


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    Staff: Mentor

    Let's start all over....

    The self-weight of a uniform ladder acts through the ladder's midpoint.

    The question says the man is half way up the ladder, your working implies he is one quarter of the way up the ladder. Which shall we use?

    What are you solving for? I'm guessing the reaction at the floor?
  5. Oct 30, 2013 #4


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    Staff: Mentor

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