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The old ladder-sliding-the-wall problem!

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    PQ represents a 40 ft ladder with end P against a wall and end Q on level ground. R represents the angle where the wall meets the ground. If the ladder is slipping down the wall, what is the distance RQ at the instant when Q is moving along the ground 3/4 as fast as P is moving down the wall?

    2. Relevant equations

    a^2 + b^2 = c^2

    3. The attempt at a solution

    PR^2 + RQ^2 = PQ^2

    2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
    = 2(40)0
    = 0
    x(dx/dt) + y(dy/dt) = 0

    What do I do now?!
  2. jcsd
  3. Mar 6, 2009 #2
    Now find a relation between dx/dt and dy/dt from the question, if you assume PR=x and QR=y what relation do you have in the rate of change of these distances with time?

    You already have a relation between x and y.
    Last edited: Mar 6, 2009
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