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Challenging Free Body Diagram Problem (Statics)

  1. Jan 29, 2016 #1
    Screen Shot 2016-01-29 at 8.30.10 AM.png 1. The problem statement, all variables and given/known data
    Given the picture, find theta, and the magnitude of the force going from point A to B.

    I have asked multiple people how to solve this problem and they have no idea, I end up getting 2 equations with 2 unknowns but then when I try to solve for theta I get 70cos(theta) - 80sin(theta) = 30 or something like that and idk what to do with that.

    I'm a tutor at my college so I get how someone might just want to half way give me the answer and let me solve the rest of it because most of the time you learn more that way, but with this problem, I think it would be best is someone could just walk me through it, I've already stared at it for hours with multiple people/teachers and we can't get it.

    Thanks to anyone that attempts this! I really appreciate it!
     

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  2. jcsd
  3. Jan 29, 2016 #2

    BvU

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    Re idk: But you do know that ##\cos\alpha \cos \beta - \sin\alpha \sin\beta = \cos(\alpha+\beta)## so dividing left and right by ##\sqrt{70^2+80^2}## should help you solve your goniometric equation....
     
  4. Jan 29, 2016 #3
    hmm, i'll try that and see if it works. I just realized that it's actually 70.6sin(theta) - 30cos(theta) = 30, so how would that trig identity change?

    Thanks
     
  5. Jan 29, 2016 #4

    BvU

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    Use symbols, not numbers.

    ##a \cos \beta - b\sin\alpha = c ## becomes ##\cos\alpha \cos \beta - \sin\alpha \sin\beta = \cos(\alpha+\beta)## if

    ##\cos\beta = {\displaystyle a\over\displaystyle \sqrt{a^2+b^2}}##,

    ##\sin\beta = {\displaystyle b\over\displaystyle \sqrt{a^2+b^2}}##,

    ##\cos(\alpha+\beta) = {\displaystyle c\over \displaystyle \sqrt{a^2+b^2}}##
     
  6. Jan 29, 2016 #5
    what I'm saying is I accidentally wrote it backwards, I wrote cos(theta) - sin(theta) = constant, but it should've been sin(theta) - cos(theta) = constant

    So would it just be sin(alpha)sin(beta) - cos(alpha) cos(beta) = sin( alpha+beta) ?

    and there's only one angle so why are you writing alpha and beta?
     
  7. Jan 29, 2016 #6

    BvU

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    Oh, sorry, typo:
    $$a \cos \beta - b\sin\beta = c$$
    simple goniometric equation. Introduce ##\alpha## and ##\gamma =\alpha + \beta## as shown and you get ##\cos(\alpha+\beta) = \cos \gamma## which you should be able to solve.
    And you of course can deal wiith the sign change straightforwardly by keeping the signs, or with ##- \cos\phi = \cos(\pi-\phi)##
     
  8. Jan 29, 2016 #7

    gneill

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    Sorry, but as a matter of forum policy helpers will not give complete solutions to homework problems. They can give hints and suggestions or point out errors, suggest strategy or topics to research, but they won't do the work for you. This is spelled out in the forum rules.
     
  9. Jan 29, 2016 #8

    BvU

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    I find that hard to believe. What kind of help do you need ?
     
  10. Jan 29, 2016 #9
    I just don't get how to do this problem lol like I don't even know if the approach to finding theta that I was asking you about is the right way to do it.

    And I have worked with 3 students for about an hour each, someone I work with who is also a teacher and has a bachelors in math and a masters in systems engineering, and my statics professor. My statics professor just suggested that I do certain things but some of the things he suggested wouldn't help me with the problem.
     
  11. Jan 29, 2016 #10

    BvU

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    Not making much headway here, are we. How about following the suggestion to use symbols instead of numbers ? From the picture attached I decipher and re-write:
    $$ \eqalign { F_{AB}\;\cos\gamma &= mg\;(1+\cos\theta) \\
    F_{AB}\;\sin\gamma &= mg \; \sin\theta }$$ so what's more straightforward now than to eliminate ##F_{AB}## and ##mg## by dividing the second one by the first one to get the goniometric equation $$ \tan\gamma = {\sin\theta \over 1+\cos\theta}$$Then amongst all these learned persons there will be one who says: hey, why don't we express the righthand side in ##\theta\over 2## ? No need to use the brute force approach (that works for numbers) I described earlier.

    You've got all weekend to work this out !

    And working with symbols instead of (rounded) numbers means you get the exact answer instead of some lukewarm approximation :smile:

    --
     
  12. Jan 29, 2016 #11
    thanks man I appreciate you helping me out with all of this.

    I found out that theta is actually twice the given angle of 23 though lol..so theta is 46.
     
  13. Jan 30, 2016 #12

    BvU

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    Eureka ! And it fits with the extremes ##\theta = 0## or ##\theta = \pi/2## :smile:
     
  14. Jan 30, 2016 #13

    Nidum

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    Tensions in DA and CA are the same . Line of action of resultant of the two tensions bisects the angle theta .

    Continuation of line of action of resultant must run along axis of link AB .

    By inspection : gamma = half theta .
     
    Last edited: Jan 30, 2016
  15. Jan 30, 2016 #14
    I feel like it's almost cheating to do that problem that way lol. Would the only other way be to do all of that trig?
     
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