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Plank leaning against a wall

  1. Dec 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A plank of length 2L leans against a wall.It starts to slip downward without friction.At what fraction of the initial height does the top of the plank lose contact with the wall .

    2. Relevant equations



    3. The attempt at a solution

    Let the normal force from the vertical wall on top of the plank be N1.
    Let the normal force from the floor on bottom of the plank be N2.
    Let x and y represent the coordinate of the CM with origin at the bottommost point of the wall.
    Let θ be angle which plank makes with the floor at any instant.

    The top of the plank loses contact when N1 = 0.

    Now,we have

    N2-Mg = Md2y/dt2 (1)

    N1 = Md2x/dt2 (2)

    N1Lsinθ-N2LCosθ = [M(2L)2/12]d2θ/dt2

    On simplifying , N1Lsinθ-N2LCosθ = (ML2/3)d2θ/dt2 (3)


    Another relation we can obtain is x=Lcosθ

    Differentiating,dx/dt = -Lsinθ(dθ/dt)

    Again differentiating , d2x/dt2 = -L[cosθ(dθ/dt)2+sinθd2θ/dt2] (4)

    How should I proceed further ?

    I would be grateful if some member could help me with the problem .


    Edit :Fixed errors .Erroneously typed d2x/dt2 in place of d2y/dt2 and vica versa .
     

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    Last edited: Dec 27, 2013
  2. jcsd
  3. Dec 27, 2013 #2
    Tanya,

    Good start. You need to express both x and y in terms of θ. (Incidentally, in the coordinate system you are using, shouldn't x = Lsinθ)? Once you have the second derivatives of x and y with respect to t expressed in terms of the time derivatives of θ, you can substitute these into eqns 1 and 2, and solve for N1 and N2. You then substitute these results into the moment balance. This gives an equation entirely in terms of θ.

    Chet
     
  4. Dec 27, 2013 #3

    TSny

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    I believe x = Lcosθ is correct for the x coordinate of the cm. [Edit: Oh, I now see the confusion. Equations (1) and (2) imply x is vertical and y is horizontal, while x = Lcosθ implies x is horizontal.]

    I think you can avoid the force and torque equations (1) and (3) by just using energy along with your equation (4).
     
    Last edited: Dec 27, 2013
  5. Dec 27, 2013 #4
    Sorry for the needless confusion in OP .I have fixed the errors .
     
  6. Dec 27, 2013 #5
    For using energy should I use : Loss in PE = Gain in Kinetic energy

    But then , I am not sure what is the initial position of the plank .Is it vertical or is it at an arbitrary angle, say α
     
  7. Dec 27, 2013 #6

    TSny

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    Yes.

    Assume it's an arbitrary angle. It will still work out nicely.
     
  8. Dec 27, 2013 #7
    Let the initial angle be α then,using COE

    mgL(sinα-sinθ) = (2/3)ML2(dθ/dt)2

    Is it correct ? If yes ,how to proceed further ?
     
  9. Dec 27, 2013 #8
    Did the steps I outlined in my post #2 not work?
     
  10. Dec 27, 2013 #9

    TSny

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    That looks good.

    What do your equations (2) and (4) tell you at the instant the plank loses contact with the vertical wall?
     
    Last edited: Dec 27, 2013
  11. Dec 27, 2013 #10

    TSny

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    I apologize for redirecting Tanya towards energy concepts. I happened to find the energy approach to be a nice way to do it. But, of course, the torque and force equations should work, too.
     
  12. Dec 27, 2013 #11
    Since I had erroneously interchanged d2x/dt2 with d2y/dt2 in OP ,when I first tried the approach outlined by you ,ended up with a complicated DE .But that is mistake on my part :smile:.

    I will try again and post the working
     
  13. Dec 27, 2013 #12
    This is what i am getting

    d2θ/dt2 + (3g/2L)cotθ(sinα-sinθ) = 0

    Is it correct ?
     
  14. Dec 27, 2013 #13

    TSny

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    Yes.
     
  15. Dec 27, 2013 #14
    Now comes the most difficult part :frown:,solving the DE .

    This looks to be a second order DE .But the second term is confusing me.

    I will have to look at some reference .Could you please help me identify the DE and the method I have to use .
     
  16. Dec 27, 2013 #15

    TSny

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    You can avoid solving a DE. Can you see a way to use your energy equation to obtain an expression for ##\ddot{θ}##?
     
  17. Dec 27, 2013 #16
    This is simply outstanding :biggrin:.

    Excellent thinking TSny !!!

    Nevertheless ,how can we solve the DE :frown:?
     
  18. Dec 27, 2013 #17

    TSny

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    I don't know. I don't think ##\theta## as a function of time would be very simple.
     

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  19. Dec 27, 2013 #18
    :rofl:

    terrific sense of humour...
     
  20. Dec 27, 2013 #19
    One thing I need to clarify is that θ is measured from the horizontal with clockwise being the direction of increasing θ .

    Does that mean direction of increasing θ has to be considered positive ?
    Does this reflect on the direction of torque as well with clockwise torque being considered positive and ccw negative ?

    Even though I have used this I am having a little doubt .
     
  21. Dec 27, 2013 #20

    haruspex

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    I got that time is proportional to ∫cosec(θ/2).dθ, where θ is the angle to the vertical. This is with being vertical initially. That makes sense in that if it really started from vertical with no initial speed then it would take an infinite time.
     
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