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    Im super terrible at mechanics and im stuck on this one question. Any mechanics nerds free to give me a hand? (Q3)

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    (Original post by zoolita)
    Im super terrible at mechanics and im stuck on this one question. Any mechanics nerds free to give me a hand? (Q3)

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    So just do the normal resolve vertically, then horizontally first.
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    (Original post by zoolita)
    Im super terrible at mechanics and im stuck on this one question. Any mechanics nerds free to give me a hand? (Q3)

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    Have you made any start? Please post what you've tried / thought about so far.

    For this question you can either resolve vertically/horizontally or you can create a triangle of forces.

    You will need to tell us which method you have been using.
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    (Original post by zoolita)
    Im super terrible at mechanics and im stuck on this one question. Any mechanics nerds free to give me a hand? (Q3)
    Probably best not to call us nerds if you want our help...

    In any case, resolve vertically an horizontally and use the fact that the particle is in equilibrium so the horizontal and vertical forces need to equate to zero.

    Let me do it horizontally for you: P \cos 30^{\circ} = 10 + Q\cos 60^{\circ} (this is me saying the force acting to the left is the same as the force acting to the right so that they cancel and the particle is in equilibrium)

    You know what \cos 30^{\circ} = \frac{1}{\sqrt{2}} and \cos 60^{\circ} = \frac{1}{2} is, so you can simply the above equation a bit. But that's not enough, obviously. You have two variables, you'll need two equations.

    What's the second equation you can get in P and Q? Resolve vertically, what do you get?

    Edit: ninja-ed twice over. I'll leave.
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