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Directed resultant force

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    http://img6.imageshack.us/img6/4041/meprob210.jpg [Broken]


    This was a plate with two bars attached / coming out of it (A and B, represented by Fa and Fb, respectively). The question says:

    Determine the angle of θ for connecting member A to the plate so that the resultant force of Fa and Fb is directed horizontally to the right. Also, what is the magnitude of the resultant force?

    2. Relevant equations

    law of sines, trig components,

    3. The attempt at a solution

    So going by that, then θ should be 0 since it's horizontally to the right.

    So solving for the red θ above:

    0 = tan-1(Fy/Fx), 0 because of the resultant being directed along the right horizontal axis.

    Fx = (8kcos θ) + (6kcos (-50))

    Fy = (8ksin θ) + (6ksin (-50))


    0 = tan-1( [(8ksin θ) + (6ksin (-50))]/[ (8kcos θ) + (6kcos (-50))] )

    tan-1(0) = 0 so

    0 = (8ksin θ) + (6ksin (-50))/(8kcos θ) + (6kcos (-50))

    and then 0 = (8ksin θ) + (6ksin (-50))

    then θ = sin-1(4596/8000) and I get θ = 35 degrees but this isn't the given answer.

    Did I do something wrong?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 6, 2013 #2

    SteamKing

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    Your calculations are wrong because you forgot that theta is referenced to the vertical axis instead of the horizontal axis. You were very careful to adjust the angle for Fb, and you should have done the same for Fa.
     
  4. Feb 6, 2013 #3
    Ah thanks.


    It's fine then because 35 degrees was the angleabove the horizontal, so then

    theta = 90 - 35 = 55 degrees.
     
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