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The old ladder-sliding-the-wall problem!

  • Thread starter matalo
  • Start date
1. Homework Statement

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. Homework 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?!
 
193
1
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:

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