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Leaning pole with friction problem

  1. May 28, 2015 #1
    This is a problem from Mechanics by Kleppner and Kolenkow


    1. The problem statement, all variables and given/known data

    Two identical masses M are pivoted at each end of a massless pole of length L. The pole is held leaning against frictionless surfaces at angle θ, as shown, and then released. Find the initial acceleration of each mass.

    Code (Text):
     Note:
    |
    o
    |  \
    |    \    
    |      \      
    |        \        
    |_____o____

    2. Relevant equations
    y^2+x^2=L^2
    yy''+(y')^2+xx''+(x')^2=0, if the masses are at initially at rest then yy''+xx''=0, x'' = (-y/x)y'' = -tan(θ) y''.

    The forces that act on the upper mass is the weight (Mg), normal force (Ny) , and force of the pole (F) so the equations are:
    My'' = Mg-Fsin(θ) and Ny = Fcos(θ) (No use, can be omitted)

    The forces that act on the lower mass is the weight (Mg), normal force (Nx), force of the pole (F) and friction (f) so the equations are:
    Mx'' = Fcos(θ)-f and Nx-Fsin(θ)-Mg=0
    3. The attempt at a solution
    By manipulation, I ended up getting y'' = (-g)/(1+µtan(θ)+(tan(θ))^2) and so x'' can easily be derived.

    My question here is that K&K didn't include Fsin(θ) in the second equation of the lower mass. This will result to Nx=Mg which then leads to a different answer. Why did K&K exclude Fsin(θ) knowing that it is a force from the pole that contributes to the downward force for the lower mass?




     
    Last edited by a moderator: May 28, 2015
  2. jcsd
  3. May 28, 2015 #2

    haruspex

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    The answer looks right (after setting mu to 0), but there's an easier way.
    Let the accelerations be ax and ay. Resolving along the ladder they must have the same component: ##a_x\sin(\theta)=a_y\cos(\theta)##.
    Take moments about the point where the two normals intersect (so they don't figure in the equations). ##MgL\sin(\theta)=Ma_xL\cos(\theta)+Ma_yL\sin(\theta)##.
     
  4. Jul 16, 2015 #3
    I'm working on this problem now in my book, and I'm not really sure if y^2+x^2=L^2 is really the way to go on this problem. I'd just like to know why it isn't (b-y)^2 + x^2 = L^2 since y is really the distance the top of the pole moves. If that's the case, then the answer the first guy who posted got might not be right. Without making sure of this, I'm a bit stuck.
    Can someone please explain / clear this up? Thanks very much!
     
  5. Jul 16, 2015 #4

    haruspex

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    The OP did not define y explicitly, but appears to have taken it as the height of the top of the pole. If you define it differently you will get a different equation.
     
  6. Jul 16, 2015 #5
    ok. just wondering, aren't we supposed to use b since it was given? and isn't this problem exactly the same as 2.7 in kleppner and kolenkow (the other 'leaning pole' one, I know this one came from 3.1) or am I just missing something here? Is it supposed to be done in a different way in this problem?

    well since I defined it differently I got (y')^2 + y(y'') = (x')^2 + x(x''). Then I evaluated the FBDs and ended up simplifying everything down to 2W = M(y'' + x''). (or basically 2g = y'' + x'') So now I have a constraint equation and another from my FBDs, but I can't think how to solve for y'' and x'' from these two equations and what I have. Is there a way to do this that I'm not thinking of?

    I'd like to be able to work my approach out to the answer. (I have no clue what that answer might look like). and sorry I'm don't exactly follow the 'easier way' you mentioned in the 2nd post as you explained there. I'd like to know the exact steps to that, it's probably a better way than I'm pursuing. I've been trying this problem in whatever way I know for a while and I think I need to stop. Thank you!
     
  7. Jul 16, 2015 #6

    haruspex

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    b? You mean L? By dimensional analysis, it cannot feature in the answer. Only g and theta can be involved.
    No idea, never seen that book.
    I may have taken theta to be the complement of that intended. Looks like I took it as the angle to the vertical, and y measured down from the top, as you have.
    What is the component of ay along the pole? What is the component of ax along the pole? How must they be related?
    That looks wrong. If you want me to check your method you will need to post all your working.
     
  8. Jul 16, 2015 #7
    lol, I do mean b. It's given in this problem's diagram, where the setup is just like another problem in the book shown here (though I don't know if I can assume that motion exerted by the rod is in line with it as was stated there): https://www.physicsforums.com/attachments/kk-2-7-jpg.72173/

    and here are my steps:
    (here F is the force the rope exerts)
    from FBD for top M
    my" = W - Fsin(theta) = 0
    N - Fcos(theta) = 0 --> N = Fcos(theta) (1)

    from FBD for bottom M
    mx" = Fcos(theta)
    N - W - Fsin(theta) = 0 --> N = W + Fsin(theta) (2)

    substituting 1 into 2: Fcos(theta) = W + Fsin(theta) => F(cos(theta) - sin(theta)) = W
    so my" = Fcos(theta) - 2Fsin(theta)
    and mx" = Fcos(theta)
    so: my" = mx" - 2(W-my") = mx" - 2W + my" ==> 2W = m(y" + x") ==> 2g = y" + x"

    maybe there's something here I'm not catching, it's not a great way, but it's the only one I could think of.
    though I thought I might have come close, I'm not sure how to proceed from here to get y" and x",
     
  9. Jul 16, 2015 #8

    haruspex

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    Ok, that was not shown in this thread. Anyway, same story - it cannot feature in the answer, other than replacing functions of theta by functions of b/L.
    You mean force, yes? I don't understand the book's remark about the force only being in line with the rod at the start. If the rod has no mass, the forces it exerts must always be along it. Consider moments about one end of the rod.
    I assume the "=0" there is a typo.
    Eh? You seem to be assuming the two normal forces are equal. Always try to avoid using the same symbol for two different quantities.
     
  10. Jul 17, 2015 #9
    oh. thanks for catching that mistake! :)
     
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