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Homework Help: Direction of the components of a vector using trigonometry

  1. Dec 17, 2015 #1
    1. The problem statement, all variables and given/known data
    The truss supports a sign board as shown in the figure. Calculate the reactions at A and the force in the member FG produced by the horizontal wind load of 60lb per vertical foot of a sign.

    2. Relevant equations

    3. The attempt at a solution
    Everything is right except for Ax which should equal 257lb in the left direction. Now my problem is when I set up my equations, they always fix my directions (negative answer means wrong assumption). However in this case, I had to use trigonometry to solve for Gx and FG. When I plugged in the numbers, I get the wrong direction. How do I find the proper direction of the forces when using trigonometry?
  2. jcsd
  3. Dec 17, 2015 #2


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    For these sign problems, it is easiest to refer back to your diagram.
    600 is left to right.
    Ax is right to left.
    Gx is left to right.
    So |600| - |Ax| = |Gx|.
  4. Dec 17, 2015 #3
    Well that does indeed give the right answer, but I'm confused about how that equation is set up. According to my diagram and my assumed directions, the equation is:
    ΣFx = Gx-Ax+600 = 0

    So if I rearrange that, it would be :
    ΣFx = 600-Ax = -Gx

    Why is the Gx is your equation positive on the right side?
  5. Dec 17, 2015 #4


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    From the diagram, I see 600lb being applied in the +x direction, which is distributed into components Gx and Ax. So just looking at positive values, |Gx| + |Ax| = 600.

    Perhaps, your application of the angle for Gx was incorrect, and you should have gotten Gx = -343 lb, implying it is applying the force from right to left, and Ax is + 257 lb implying it is applying the force from left to right. Then your equations balance out.
    Are Gx and Ax the resisting forces which are keeping the structure in one place?
  6. Dec 17, 2015 #5
    Well for some reason my book uses negative answers as meaning left and down forces. So it is saying that Ax is indeed right to left. It also implies that FG is going towards the left. The book unfortunately does not provide the answer to Gx, because it is not asked for. However, I do see how 600-343 would equal 257, and although it is positive, that is okay because my Ax direction was assumed properly (right to left). So how would I know the direction of Gx in that case? If Gx is indeed right to left, that would solve my problem.

    I was also never taught to just use positive values as you did, so I am a little confused as to where that comes from, however it does give the right answer. It's just throwing me off because in my diagram the Ax is right to left, so it would be -Ax in the equation, and 600 is positive, so it would be negative once moved to the right side of the equation.

    EDIT: My main question is how do I find the direction of Gx if it was found using only trig? It is clear to me that it must be negative, but how would I figure that out mathematically?
    Last edited: Dec 17, 2015
  7. Dec 18, 2015 #6

    Mister T

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    This is what I get using your drawing:
    ΣFx = Gx+Ax+600 = 0
    ΣFx =+343+Ax+600=0
    Ax = -943 lb

    [My main question is how do I find the direction of Gx if it was found using only trig? It is clear to me that it must be negative, but how would I figure that out mathematically?[/QUOTE]

    Based on your drawing Gx is positive because it points towards the right.
  8. Dec 18, 2015 #7


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    From what Mister T has pointed out, if you want the sum of your forces to balance, you have to change the directions of your arrows. Ax and Gx are pushing back against the structure to hold it in place, bracing it against the 600lb force the wind is applying.
    As for the trig, I would apply the same sign rules you have in place for the drawing, i.e. theta = 0 refers to pointing directly to the right. In your drawing, from the angle theta, the adjacent leg goes toward the left, so make that a -2, and the opposite leg goes up from the base, so make that a +5, then ## \tan^{-1} \theta = -5/2 \implies \theta = -68.2 ° \text{ or } 111.8 °##. These angles also agree with the convention of right, up = positive and down, left = negative.
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