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    (Original post by HashBrowns)
    I used the sine rule for the triangle with two radii as sides two show the length of the longer side is 1.6 cos a. From there I used another triangle to show x = 0.8. Is this wrong?
    'Fraid so. I can't see how a triangle with two radii as sides would help.

    See attached: x=0.8 tan a.
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    Could I see your solution?
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    (Original post by HashBrowns)
    Could I see your solution?
    OK, as this has gone on for a while.

    Spoiler:
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    Let the weight be W, the angle of the rod to the horizonal be \alpha, and the distance of the pin to the point of contact with the cylinder "x", as we had before.


    We know there is no vertical component at the pin, so resolving vertically for one of the rods we have:

    R\cos\alpha=W

    Then taking moments about the pin for one of the rods we have:

    0.5W\cos\alpha=xR

    Refering to the diagram, and considering the triangle BOM, we have

    x=0.8\tan\alpha

    So, substituting into our 2nd equation we have:

    0.5W\cos\alpha=0.8R\tan\alpha

    And substituting for W from your first equation we get:

    0.5R\cos^2\alpha=0.8R\tan\alpha

    R cancels and multiplying through by 10 we have:

    5\cos^2\alpha=8\tan\alpha

    If we now convert the cos^2 into 1/sec^2 = 1/(1+tan^2) and multiply up and re-arrange we have:

    8\tan^3\alpha+8\tan\alpha-5=0

    Which on the face of it looks a bit of a pig!

    If we divide through by 8, we can perhaps see that one root is tan a = 1/2.

    \tan^3\alpha+\tan\alpha-\frac{5}{8}=0

    giving us

    (\tan\alpha-\frac{1}{2})(\tan^2\alpha+  \frac{1}{2}\tan\alpha+\frac{5}{4  })=0

    Now the quadratic has no real roots, so we are left with \tan\alpha=\frac{1}{2}

    Hence we find alpha, and hence the half angle at the pin, and hence the whole angle.

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