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Thread: finding theta

  1. #1
    Junior Member
    Aug 2010

    finding theta

    Can someone point out where I am going wrong in this problem?

    Three forces are applied to an object, as indicated in the drawing. Force 1 has a magnitude of 33.0 newtons (33.0 N) and is directed 30.0 to the left of the +y axis. Force 2 has a magnitude of 16.0 N and points along the +x axis. What must be the magnitude and direction (specified by the angle θ in the drawing) of the third force 3 such that the vector sum of the three forces is 0 N? magnitude28.6 Nθ2

    The only thing I need to do is find theta. I thought it would work by doing tan^-1(28.6) but that gave me 88 degrees and it didn't work. Suggestions?

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  2. #2
    MHF Contributor
    skeeter's Avatar
    Jun 2008
    North Texas
    the given sketch (for some reason, either an error or an intentional deception) is incorrectly depicted.

    $\displaystyle F_2$ is a 16 N force in the positive x-direction.

    $\displaystyle F_1$ has an x-component = 33cos(60) = 16.5 N in the negative x-direction.

    for there to be equilibrium, F_3 should be in quad IV with an x-component to the right of 0.5 N.

    if the problem is legit, then $\displaystyle \theta$ (as depicted) should be greater than 90 degrees.
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  3. #3
    Senior Member Educated's Avatar
    Aug 2010
    New Zealand
    To solve $\displaystyle \theta$, use the right angle tangent rule:

    $\displaystyle tan(\theta) = \frac{opposite}{adjacent}$

    As skeeter said, there must be a 0.5N force to the right and a 28.6N force downwards to balance everything out.

    For me, I get $\displaystyle 88.998^{\circ}$ in quadrant IV, therefore:
    $\displaystyle \theta = 91^{\circ}$
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