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Stability of hanging picture

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A (rectangular homogeneous) picture is hung with a nail on the wall from a twine. Neglecting the friction of the twine with the nail and that of the wall with the picture,
    a.-Could there exist equilibrium in other positions than the one where the picture is hanging horizontally? If so, are those other positions in stable equilibrium?
    b.-Is the picture in stable equilibrium when hanging horizontally?

    Thanks for any help
    I´m attaching a drawing picture.jpg

    2. Relevant equations
    Torques about a point.


    3. The attempt at a solution

    I chose the nail as reference. Tensions are going at an angle but their line of action intersect at the nail so only the weight could make any torque about that point. Horizontally there{s no torque at all about that point. If I slightly tilt the picture then a torque appears about that point trying to restore it to its original horizontal position. sO I think QUESTION b is: YES.

    Part a. I think it´s more tricky... I{m not sure about the geometry. I´m trying to see if having different lengths for the two sides of the twine make it impossible to vanish the torque generated by the weight.
     
    Last edited: Jan 18, 2013
  2. jcsd
  3. Jan 18, 2013 #2

    SammyS

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    That link doesn't work for me.

    Please check it.
     
  4. Jan 18, 2013 #3

    Simon Bridge

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    What happens when the nail is at the join between the string and the picture?

    In equilibrium, the com will be directly below the nail.
    Stability questions concern what happens when a small push is given to the system.
    Are there positions where the string slides over the nail?
     
  5. Jan 18, 2013 #4
    Oh, ok, I tried to edit it. However, the drawing is not that essential, It{s just a picture hanging on a wall but not on a horizontal position. Both ends of a cord are attached to the frame so the cord lets the picture hang from nail on the wall but a bit skewed.
     
  6. Jan 18, 2013 #5

    haruspex

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    Where do you think the centre of mass of the picture is in relation to the line joining the endpoints of the string? Can you say anything about the two angles the string sections make to the vertical? Consider horizontal forces as well as torque.
     
  7. Jan 19, 2013 #6
    Hi,
    well, for me it´s clear that the center of gravity of the picture is just below the nail when it's hanging horizontally (not slanted). Therefore, any small push would make the line of action of the weight create a torque about the nail. This torque would make it tend to rotate back to its centered position.

    I think the horizontal position is a stable equilibrium one.

    I´m trying to see if the picture can be in equilibrium (even if it´s unstable) in any other positions. I think it is possible for the picture to be slanted and in equilibrium, maybe unstable?
    I'm not sure because I don't know if it's possible that the cord can be slippery and change the length on one side because of no friction


    Considering the horizontal components of the tensions on both sides, they would have to be equal in magnitude. The vertical component of the tensions, if slanted, would be different and would counteract the weight. I'mhaving problems to see if, given a slanted position, the picture can be in any kind of equilibrium
     
  8. Jan 19, 2013 #7
    Equilbrium is when there is no motion right? So torques must cancel out and all forces aswell, this leaves two vectorial equations:
    [itex] \sum\mathbf{\tau} = \vec{0}[/itex]
    [itex] \sum\vec{F} = \vec{0}[/itex]
    Now the picture has two wires holding it up. The weight of the picture is w = mg (in y dir.).
    [itex] \vec{w}-\vec{T_1}+\vec{T_2} = 0 [/itex]
    Let the nail define our rotation axis then the torque eq. becomes(the picture is slightly tilted I imagine by an angle theta and the wires makes an angle phi with horizontal)
    Edit:
    The tensions and the weight's tourqe must cancel out , when the right length is smaller than the left of the wire
    [itex] \sum\vec{\tau} = \vec{H}\times \vec{w} + \vec{\ell}_1\times \vec{T_1} - \vec{\ell_2}\times \vec{T_2} = 0[/itex]
    Use then Newton's 1st law.
     
    Last edited: Jan 19, 2013
  9. Jan 19, 2013 #8
    What does your intuition tell you? Suppose the center of the twine slides sideways by even the slightest amount relative to the frictionless nail? Can this represent a stable configuration for any possible rotational orientation of the picture? If the sliding continues, where will picture end up?
     
  10. Jan 19, 2013 #9
    Yes, my intuition tells me it will continue slipping to one side if not adjusted horizontally. But when I make a Diagram it seems to depend on the length of the picture frame. :s
    For a very rectangular frame it seems that the picture will have a net torque about the nail the will keep making it rotate to the same side, I drew something... picture_1.jpg
    Can somebody confirm my diagram is right in telling me that this will be unstable because of the torque made bythe weight?
     
  11. Jan 19, 2013 #10

    TSny

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    arestes,

    Is the cord attached to the upper edge of the picture (as in your last figure) or is the cord attached along the vertical sides of the picture as shown in your original drawing?
     
  12. Jan 19, 2013 #11

    CWatters

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    If there is no friction between cord and nail then the tension in the cord either side must be the same or it will move over the nail. Perhaps look at what positions give the required symmetry?
     
  13. Jan 19, 2013 #12
    Hi, there's no friction between the nail and wall. Also, the cord should attached to the vertical sides of the frame. I forgot that in my last picture, but I think it still doesn't matter much because the tensions point to the nail and I'm taking moments about the nail. In any case, the first picture is the one that I was given.

    I don't understand why the tension on either side must be the same. I already established that the symmetrical position would be an equilibrium one. Also, I think the horizontal (no t slanted) position would be a stable equilibrium.

    I'm still not sure how to justify that there can be any other equilibrium positions. (even if it's unstable or neutral)


    My last picture suggests that if it's a little bit slanted, there would be a torque that would make it rotate even more* clockwise. But This depends on where I put the center of gravity... If this rectangle were a bit wider in its vertical frame the CG would be on the other side... I'm confused
     
  14. Jan 19, 2013 #13

    CWatters

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    Forget the picture for a moment. Imagine you just have a rope draped over a frictionless pulley. If the tension on one side is greater than the other the rope will move in that direction.
     
  15. Jan 19, 2013 #14

    CWatters

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    If the rope is attached to the sides of the frame what about this case. It might hang there but not be stable. Any slight disturbance and it may not return to this position.
     

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    Last edited: Jan 19, 2013
  16. Jan 19, 2013 #15
    Oh wow, that case looks bizarre but I haven't considered it... However, I see it can only happen if the cord is *long enough*... I guess that never happens with average lengths of the cord. I think the question is asking if it could be in equilibrium for positions near the symmetrical horizontal one.

    I just found out that if the picture was a horizontal rod, it would be unstable but if the picture was a vertical rod it would be stable always... this seems to imply that the dimensions of the picture frame are important?
     
  17. Jan 19, 2013 #16

    CWatters

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  18. Jan 19, 2013 #17

    TSny

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    The string has a fixed length and the ends are attached to fixed points of the picture. So, an ellipse might be relevant. See attachment.

    The points of attachment are the focal points of the ellipse (fig. 2). Here I picked the attachment points at the top corners, but that isn't necessary. If the picture is tilted as shown in fig. 1 then that corresponds to moving the nail from point N to N' on the ellipse and then rotating the entire ellipse clockwise by θ until the line from the COM to N' is vertical and shifting up or down until the nail is in its original place. If the distance N' to COM is greater than the distance from N to COM, then the COM shifts downward (as in fig. 1).

    Whether or not the horizontal position of the picture is stable equilibrium depends on the relative sizes of the quantities shown in fig. 3. If the horizontal orientation is not stable, then you can find an angle of tilt where the picture is in stable equilibrium.
     

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  19. Jan 19, 2013 #18
    Oh thanks! I think I did enough research for this problem! I'll try to access a full version of the paper. I'll contact my teacher about this because it seems it's not a straightforward problem and it involves many more details. Thanks
     
  20. Jan 19, 2013 #19

    haruspex

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    It must always be directly below the nail, if at rest. But I asked where it is in relation to the line joining the endpoints of the string. Let's say it's distance h from there, with those endpoints being in the upper half of the picture.
    Yes.
    No. If the horizontal components are the same but the vertical components different then the tension would be different on one side of the nail from the other. But we're assuming no friction at the nail, so the tensions must be the same.
    What does that tell you about the angles the string makes to the vertical each side of the nail?
    From these considerations, you should be able to obtain an expression for the distance the CoM is below the nail, d = d(w, h, l, tan(theta)), where l is the length of the string, w is the width of the frame, and theta is the angle the frame makes to the horizontal. (Maybe not all of those variables show up.) Can you think of an inequality involving d as a function of theta that represents stability?
     
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