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    I've told you twice why the answer is mg sin a. Your answer is wrong, because it does not account for the fact that the weight is attached to the system. (And if it wasn't attached, it would fall off).

    (Original post by Whispering Eye)
    why is the force not mg/sina ?

    sina = mg/F no?
    I think you're constructing your right-angled triangle wrong. :sadnod:
    Think about where the angle a actually is. Remember the whole system is tilted at an angle; perhaps consider rotating your coordinate system to suit, as you would for motion on a slope.

    Either way, you get the gravitational force, mg, along the hypotenuse, and the force F along the opposite side to the angle. Then basic trigonometry gives you:

    \displaystyle \sin(a) = \frac{F}{mg} \implies F = mg\sin(a) :yep:

    You can apply a rudimentary check to this: does it make sense?
    Well, as a\to 0,\ F\to 0;\ \mathrm{and\ as\ }a\to 90^{\circ},\ F\to F_{max}. This is what you would intuitively expect, isn't it? When you apply a force perpendicular to the beam, it'll take the full 'push' from said force, but applying the same force along the length of the beam will have no net effect.
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Updated: February 9, 2010
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