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Ladder problem - is there a mistake here ?

  1. Oct 27, 2006 #1

    Fermat

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    I have a ladder which is propped up against a wall. The ladder slides down the wall and is constrained to move such that the speeeds of the ends of the ladder against the wall (and floor) are a constant value. i.e. [tex]\dot x = \dot y = constant[/tex]

    The problem question is: What is the angular velocity of the ladder about its centre when the angle, of the ladder against the wall, is @=15 degrees ?

    BUT, if the ladder length is L, then x² + y² = L² and differentiating this expression wrt time gives,[tex]2x\dot x + 2y\dot y = 2L \dot L = 0[/tex]
    Therefore, [tex]x\cdot\dot x = -y\cdot\dot y[/tex]
    However, if we are given [tex]\dot x =\dot y[/tex] (numerically), then this means that x = y!
    But if x = y, then the ladder must be at an angle of 45 degrees, yes?
    So, this means that the ladder has only one orientation, @ = 45 degrees, such that the speeds of the ends of the ladder are the same.
    Is this right? Or have I made a mistake in my earlier working ?
     
  2. jcsd
  3. Oct 28, 2006 #2

    OlderDan

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    Your conclusion is correct. The ladder could slide with one end at constant speed, but not both. I don't know how you could do the problem anyway with just the information given, unless you are supposed to find the answer in terms of the constant velocity.
     
  4. Oct 28, 2006 #3

    Andrew Mason

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    Perhaps you could give us the exact wording of the question. If the speed of the end against the wall is constant, the speed of the ends against the floor cannot be constant let alone the same speed.

    As a function of angle [itex]\theta[/itex] of the ladder to the floor:

    [tex]\dot y = \frac{d}{dt}L\sin\theta = -L\cos\theta\dot\theta = K[/tex]

    means that:

    [tex]\dot\theta = -K/L\cos\theta[/tex]

    [tex]\dot x = \frac{d}{dt}L\cos\theta = -L\sin\theta\dot\theta = L\sin\theta(K/L\cos\theta) = K\tan\theta[/tex]

    So if the speed of the end down the wall is constant, the speed along the floor is proportional to [itex]\tan\theta[/itex]

    AM
     
    Last edited: Oct 28, 2006
  5. Oct 28, 2006 #4

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    The other point was:if both ends had different tangential velocities, which they would have if both ends had the same speed ( a given of the problem) but a non-45 degree inclination, then could they have an angular velocity about the ladders centre? I would have said they would only have an angular velocity about some point, the distance of which from and end would be proportional to the relative end velocites? Is that right ?
     
  6. Oct 28, 2006 #5

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    Here is the wording of the question as it was given to me:

    What is the angular velocity of a 12-ft ladder about its center as it begins to slide off a perpendicular wall at an angle of 15 degrees to the wall, with each end having a relative speed of 8in/sec to their respective surfaces?
     
  7. Oct 28, 2006 #6

    Fermat

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    I wish I could stay for some answers, but gotta go to work :(
     
  8. Oct 28, 2006 #7

    Andrew Mason

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    The short answer is that if the wall is perpendicular to the floor and the other end of the ladder is on the floor, the question poses an impossible set of facts.

    AM
     
  9. Oct 28, 2006 #8

    Fermat

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    Many thanks.
     
  10. Oct 28, 2006 #9

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    Update

    Just had my answer accepted :biggrin:

    I told the questioner that I thought his question was wrong and that the angle should be 45 degrees, and worked out an answer based on that assumption.
    He insisted that it should be 15 degrees. But he's just come back and accepted my original answer. So looks like he checked up and found out it was a typo after all!!
     
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