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    what do you equate your moments equation to when it 'just about rotates at' a support? when a rod is at equilibrium, you equate it to 0, but i dont know what to do if it isnt. here's an example:

    4) A non uniform rod AB of length 5m and mass 15kg rests horizontally suspended from the ceiling by two vertical strings attached to C and D, where AC = 1m and AD = 3.5m.

    a) Given that the centre of mass is at E where AE = 3m, find the magnitudes of the tensions in the strings.

    I have calculated that the reactions are 117.6N at D and 29.4N at C.

    When a particle of mass 10kg is attached to the rod at F, the rod is just about to rotate about D.

    b) Find the distance AF.

    Now when I take moments, what should the general equation equate to?


    If i read that correctly, if its about to tip, there should be no reaction at C, asign x to AF then resolve about D.

    The way I seem to remember that you solve these questions (counterintuitively imo) is that since it is ABOUT to rotate about D, there must be no reaction at C.

    So if you negate the reaction at C:

    Am not sue on how you use a spoiler thing. So don't read below if you want to suss it out yourself.

    Since it's going to rotate about D, AF>3.5m

    Let x = distance between D and F

    M(D) 0.5*15g - 10gx = 0

    x = 7.5/10

    x = 0.75

    So AF = 3.5+0.75 = 4.25m
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Updated: January 14, 2010
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