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Rates of Change Sliding Ladder Application

  1. Aug 11, 2005 #1
    "a 5m ladder is placed against a wall. the top of the ladder is sliding down the wall at 0.5m per second. at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 3m from the wall. ? "

    Dirac.
     
  2. jcsd
  3. Aug 11, 2005 #2

    lurflurf

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    let x(t), y(t) be the distance respectively from the wall and floor.
    1) relate x and y
    2) differentiate the relationship to relate x' and y'
    3) find write y(t) in terms of t then find x(t),x'(t),y'(t)
    4) use x(ti)=3 to find x'(ti)
     
  4. Aug 11, 2005 #3
    Could you please do a step-by-step solution

    Dirac.
     
  5. Aug 11, 2005 #4

    lurflurf

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    No.
    so the ladder, a piece of floor and a piece of wall form a right triangle.
    where the legs are x,y and the hypotenus is 5 can you write an equation relating these quantities?
     
  6. Aug 11, 2005 #5
    Yes, ok but what is x(t)
     
  7. Aug 11, 2005 #6

    HallsofIvy

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    Dirac: TRY!! If you can't get it show us what you have done and we'll give suggestions. The important thing is that you try yourself.
     
  8. Aug 11, 2005 #7
    In a simpler way, if the ladder is moving down at a rate of 0.5m/s which is 5m above the ground, the ladder is slidding at the same time, so what you need to do is find the time it takes for the ladder to fall using v=dx/dt or v=x/t and then using the same equation use that elapsed time with the bottom of the ladder that is 3m away from the wall. does this make sense?
     
  9. Aug 11, 2005 #8
    Yes, I can do it, but I get two solutions for t from

    (x^2)=5t-0.25(t^2)

    Dirac.
     
    Last edited: Aug 11, 2005
  10. Aug 11, 2005 #9

    lurflurf

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    x is the distance from the bottom of the ladder to the wall measured along a line along the floor that is perpendicular to the wall.
    y is the distance from the top of the ladder to the floor measured along a line along the wall that is perpendicular to the floor
    also I am assuming the wall is perpendicular to the floor
    thus we have a right triangle with legs x,y and hypotenus 5
     
  11. Aug 11, 2005 #10
    I dont understand how you have a quadratic, this is a linear problem

    I get from your values that the velocity of the sliding is 0.333m/s. But dont quote me!!!

    I f you show me your working perhaps i can help

    hhh79bigo
     
  12. Aug 11, 2005 #11
    It is not linear

    By Pythagoras'

    (x^2)+(y^2)=(r^2)

    y=5-0.5t

    r=5

    =>(x^2)=25-(25-5t+0.25(t^2))

    =>(x^2)=5t-0.25(t^2)
     
  13. Aug 11, 2005 #12
    Oh ok, done it now.

    Dirac.
     
  14. Aug 11, 2005 #13

    lurflurf

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    x and y should be positive
    then there is only one solution
    There is also an easier way
    we know
    x*x'+y*y'=0
    so
    x'=y*y'/x
    we know y'=-.5 x=3 y=4 so x' is easy to find
     
  15. Aug 11, 2005 #14

    HallsofIvy

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    And you STILL haven't shown us what you have done! Don't just give us your (wrong) answer. Show us how you got it.

    In fact, your answer makes no sense. The problem does not even ASK you to find t! If you let x be the distance from the wall to the foot of the ladder, the problem asks you to find dx/dt.

    Now do what lurflurf suggested to begin with: Draw a picture, look at the right triangle in the picture and write an equation relating the parts of the picture. That will be a "static" equation- your picture is kind of like a snapshot of the sliding ladder. To get a "dynamic" equation (a moving picture) differentiate the entire equation with respect to t (even though there is no t in the equation!)

    For the same reason, the answer to hhh79bigo's question is "No, that makes no sense at all- you were not asked to find the time the ladder takes to fall to the floor."
     
  16. Aug 11, 2005 #15
    For the same reason, the answer to hhh79bigo's question is "No, that makes no sense at all- you were not asked to find the time the ladder takes to fall to the floor."[/QUOTE]

    Forgive me for being ignorant

    I apologise I thought that the top of the ladder was 5m of the ground

    hhh79bigo
     
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