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Uniform right circular cone hanging in equilibrium

  1. Jul 21, 2016 #1
    1. A uniform right circular solid cone of weight W is suspended by two vertical strings attached to the ends A and B of a diameter of its base. If the cone hangs in equilibrium with its vertex vertically below A, find the tension in the strings.

    2. Centre of mass of cone from the vertex is 3/4 h
    Moments formula


    3. I understand we have to take moments to find the tension but 1) I'm not sure if the tension in both strings is the same 2) since the vertex is vertically below A, I'm not sure what the horizontal distance from the weight to A and B is to be able to use moments. I know that the c.o.m of the cone lies at 3/4 h from the vertex but not sure what the coordinates will be in this case and I assume you require the x-coordinate for the horizontal distance .
     
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  3. Jul 21, 2016 #2

    BvU

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    1) definitely not
    2) make a drawing
    You want two tensions, so you'll need two equations. One from moments. Is it clear to you what the other is ?
     
  4. Jul 21, 2016 #3
    I'm not really sure what the other is, could you guide me a little please?
     
  5. Jul 21, 2016 #4
    Is it the resultant force? Sum of upward forces= sum of downward forces?
     
  6. Jul 21, 2016 #5
    This is the diagram I drew.
     

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  7. Jul 21, 2016 #6

    BvU

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    Yes. equilibrium means no net force and no net torque.
    In the diagram you want to avoid the suggestion that A and B are at the same height. In the cone, the center of mass is on the axis. Now you need the horizontal coordinate for the moments.
     
  8. Jul 22, 2016 #7
    That's what I'm not sure of. How do we find the horizontal coordinates? All we know is that the C.O.M lies at 3/4 h from the vertex and the question doesn't give any other details.

    So if I tilt the axis on my diagram so that it's symmetrical then G will lie on the axis-is that what you mean? I don't understand the part about the height being different. If A and B are not at the same height then the cone would no longer be symmetrical?
     
  9. Jul 22, 2016 #8

    BvU

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    upload_2016-7-22_9-29-41.png
    If this is upside down with the vertex below A then B isn't at the same height as A.
    I see what you mean. In such a case, use symbols for unknowns (h, r) and maybe :smile: (*) they will cancel in the final result.

    (*) Didn't work out the whole thing, so I have to trust the exercise composer in this...
     
  10. Jul 22, 2016 #9
    I think the distance from the c.o.m of the cone to points A and B has to be the radius of the cone. If the height is the same then the coordinates from the c.o.m would be according to me (r, h/4)
    and (-r,h/4). If the height is different, we don't even know how different it is so how will we integrate this difference within our equation?
     
  11. Jul 22, 2016 #10
    So I actually made a cone out of paper and hung it in this orientation then traced it onto my notebook. I think even though B is lower than A, the height appears to be the same.
     

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  12. Jul 22, 2016 #11

    BvU

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    2) from post #2

    upload_2016-7-22_10-43-26.png

    Now it's time to choose a coordinate system and write down some equations.
     

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  13. Jul 22, 2016 #12
    Can't we work in Cartesian? Since they don't do polar and cylindrical in A'Levels.
    So Ta+Tb=w
    and we take the moment around w?
     
  14. Jul 22, 2016 #13

    BvU

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    We certainly can work in Cartesian. And the axis can be chosen freely.
     
  15. Jul 22, 2016 #14
    So the axis doesn't need to lie on the line of symmetry of the cone?
     
  16. Jul 22, 2016 #15

    BvU

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    No. You could pick the moments around A (or around the vertex -- same moments) just as well. Choose what you think is easiest.
     
  17. Jul 22, 2016 #16
    I'm afraid I'm still not sure what the coordinates are going to be. I can just think of the radius.
     
  18. Jul 22, 2016 #17

    BvU

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    Why not start with writing out the two equations for the equilibrium relationships, using symbols for whatever is unknown. Then we can look at those unknowns and try to deal with them....

    You already have Ta+Tb=w, so what about these moments ?
     
  19. Jul 22, 2016 #18
    Ok please correct me if I'm wrong.

    Ta+Tb=w
    and
    moment around A (or the vertex) is
    w (r)= Tb (2r)
    I have a strong feeling it can't be this simple.
     
  20. Jul 22, 2016 #19

    BvU

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    is (r) the perpendicular distance between W and A ?
    (2r) between B and A ?
     
  21. Jul 22, 2016 #20
    No it isn't. So I drew in the perpendicular distance, we can use the Pythogoras theorem but one of the heights is missing. Or can we use similar triangles?
     
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