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Forces in Equilibrium

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data

    A set of forces are in equilibrium. Find "F" and angle "x"

    The following are forces actually on a particle in the center (N being newtons):

    4N to the right along the horizontal
    (3 x root3)N downwards down the vertical
    6N diagonally up right, 60deg from the horizontal
    2N diagonally up left, 60deg from the horizontal
    "F"N diagonally down left, angle "x" is the actual between force "F" and the force pointing directly down.

    2. Relevant equations

    3. The attempt at a solution

    Resolve going right:


    Resolve going down:


    Sub one into the other


    this is where I get stuck lol.

    not sure If Im doing it right but Im stuck ahhhh

    hope you can help
  2. jcsd
  3. Nov 22, 2007 #2


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    Staff Emeritus
    Science Advisor


    so 6*cos(x)= sin(x) and then sin(x)/cos(x)= tan(x)= 6.
  4. Nov 22, 2007 #3
    o rite, i didnt realise sin(x)/cos(x) = tan(x) :D

    thought I was missing a trig identity or something lol

    thanks :D
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