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Finding an unknown force at an unknown angle

  1. Feb 4, 2015 #1
    1. The problem statement, all variables and given/known data

    So we have a force of unknown magnitude acting on these struts at an angle θ measured from strut AB.
    The component of the force acting along AB is 600lb, and the magnitude of the force acting along BC is 500lb.
    If Φ = 60°, what is the magnitude of F and the angle θ?

    Hibbler.ch2.p13.jpg

    2. Relevant equations

    Fcos(θ) = 600lb

    3. The attempt at a solution

    Ok. So, it'll probably help if I knew the third angle of the triangle formed.
    180° = ɣ + (60° + 45°)
    180° - 105° = ɣ
    75° = ɣ

    Great. So, I know that Fcos(θ) = 600lb, and Fcos(75° - θ) = 500lb
    hm. Fcos(θ)/600 = 1 = Fcos(75 - θ)/500
    500Fcos(θ) = 600Fcos(75° - θ)
    5Fcos(θ) = 6Fcos(75° - θ)
    5cos(θ) = 6cos(75° - θ)
    0 = 5cos(θ) - 6cos(75° - θ)

    Originally I tried finding where z = 5cos(θ) - 6cos(75° - θ) intersected with z = θ + η where η = 75° + θ, but I couldn't get Wolfram Alpha to understand what I was talking about. Here, I see I should have just left η as
    75° - θ, but even still, I have to *ask* Wolfram Alpha what θ works for 0 = 5cos(θ) - 6cos(75° - θ) when
    0<=θ<=75° (it gives me an angle of ~30.7°).

    Worse, since I couldn't figure it out, I figures if I gave in on the magnitude of F, I could still find the angle. Mastering Engineering told me F = 870lb. So, if Fcos(θ) = 600, θ=arccos(600/F) and arccos(600/870) ≈ 46.4°.
    Which was wrong. The θ it wanted was ~34°

    Which means everything I did was wrong.
    So what triangle magic do I do to get from the initial problem, to the final F=870 θ=34°, without doing something so complicated I need Wolfram Alpha to crunch it out?
     
  2. jcsd
  3. Feb 4, 2015 #2

    Svein

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    Science Advisor

    A hint: cos(x-y) = cos(x)cos(y) + sin(x)sin(y). Apply that to cos(75° - θ).
     
  4. Feb 4, 2015 #3

    Doc Al

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    Staff: Mentor

    Good. I suggest using a trig identity to simplify the expression Fcos(75° - θ).
     
  5. Feb 4, 2015 #4
    While that does get everything in simpler terms of θ, that doesn't address that 870cos(34) = 721.3 and not 600.
    My error is a lot farther up.
     
  6. Feb 4, 2015 #5

    Doc Al

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    Staff: Mentor

    Start over with the two simpler equations. You'll get a different value for θ and F.
     
  7. Feb 4, 2015 #6
    what do you mean by that? Because F=870 and θ≈34° are set in stone Those are the answers for this problem.
     
  8. Feb 4, 2015 #7

    Doc Al

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    Staff: Mentor

    And yet you know that:

    Says who?

    Why not just solve it using that trig substitution. You'll solve it easily without needing Wolfram.
     
  9. Feb 4, 2015 #8
    Says The homework. Because I've already gotten this question wrong, and the answers were displayed. I'm asking about this question here so I know what to do when this sort of problem comes up again.
     
  10. Feb 4, 2015 #9

    Doc Al

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    Staff: Mentor

    Do you not agree that the given answers do not work? That they contradict the problem statement?

    I would solve it in the manner described above, so you can get a straightforward answer.
     
  11. Feb 4, 2015 #10
    The given answers don't work for Fcos(θ)=600, no. But that could mean either Mastering Engineering flubbed or, more likely, I flubbed the starting equations.
     
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