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Pulling a rope on a wall, finding gamma

  1. Aug 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine coordinate direction angle γ of the resultant force acting at B.
    Hibbler.ch2.p97.jpg


    2. Relevant equations
    Cos[itex]\gamma[/itex] = F[itex]\underline{}z[/itex]/F


    3. The attempt at a solution
    The first part was to find magnitude of F[itex]\underline{}b[/itex]. The next part was finding the other coordinate directions. I found all but gamma. I don't know how gamma is wrong but the rest are correct. My correct numbers (rounded from the website):

    F[itex]\underline{}B[/itex] = 78.9 lb
    [itex]\alpha[/itex] = 35.4°
    [itex]\beta[/itex] = 68.8°

    Why won't the same method work for finding [itex]\gamma[/itex] (the method being the equation up there)? I keep getting 63.1°.

    Components of F[itex]\underline{}B[/itex]: <69.30 , 28.55 , 35.75>
     
  2. jcsd
  3. Aug 26, 2014 #2

    Zondrina

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    Your ##\alpha## and ##\beta## angles make sense. That is not the correct magnitude of the force at ##B##.

    I get ##F_B = 110 lb##. Could you show some of your working so we can see where you went wrong?
     
  4. Aug 26, 2014 #3
    That's weird because, except for rounding, it said my answers were correct (btw, gamma was 117° and I don't know how . . . ).

    uF[itex]_{}c[/itex] = (6/14)i + (12/14)j + (4/14)k
    F[itex]_{}c[/itex] * 50 (the force) = 21.45i + 42.85j + 14.30k

    F[itex]_{}a[/itex] I did the same, but got this overall vector:
    42.85i - 14.30j + 21.45k It had a negative j because it's left of the point "B" in the picture, on the y axis.

    Then I added those to get the resultant F[itex]_{}B[/itex] vector, 64.30i + 28.55j +35.75k. Finding the magnitude (root of the squares) gets me 78.92, which is correct on the website . . .
     
  5. Aug 26, 2014 #4

    Zondrina

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    Hmm i seem to be getting :

    ##F_B = \{90 \hat i + 40 \hat j - 50 \hat k \} N##

    How are you getting ##\vec F_A## and ##\vec F_C##? You should be finding position vectors ##\vec r_A## and ##\vec r_C##, which you then normalize to obtain unit vectors ##\vec u_A## and ##\vec u_C##.

    You can then express the forces ##\vec F_A## and ##\vec F_C## in terms of the unit vectors.

    Also, you should list the question as stated in the book with all the relevant variables. I'm sure magnitudes for ##F_A## and ##F_C## have already been provided.
     
  6. Aug 26, 2014 #5
    Yes they are 50 lb. I thought I was posting a continuation of a problem. Forgot it was different!

    I'm using the coordinates in the picture. You can see in my Fc equation what my position and magnitude were based solely off the picture.
     
  7. Aug 26, 2014 #6

    Zondrina

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    Well If you have them as ##50 lb## each:

    ##\vec F_A = F_A \vec u_A = \{ \frac{300}{7} \hat i + \frac{100}{7} \hat j - \frac{150}{7} \hat k \} N##

    ##\vec F_C = F_C \vec u_C = \{ \frac{150}{7} \hat i + \frac{300}{7} \hat j - \frac{100}{7} \hat k \} N##

    I get the same vectors as you. So I'm assuming your ##F_B## is correct now that we have the magnitudes of ##F_A## and ##F_C##.

    All you would need to do is use the coordinate angle relationship now to find the three angles.
     
    Last edited: Aug 26, 2014
  8. Aug 26, 2014 #7
    Right. And I did to get alpha and beta. Gamma gave me 63 I believe. Gamma was 117 and I don't know how they calculated that since I used the same numbers and same procedure to get alpha and beta (which were correct).

    Is there a special way to know when the angle you get needs to be subtracted from 180 (I got 63, the website got 117, 180 - 63 = 117)?
     
  9. Aug 26, 2014 #8

    Zondrina

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    Ask yourself which way ##\vec F_{B_z}## points. You should find this changes your answer to ##117 °##.
     
  10. Aug 26, 2014 #9
    I could ask myself that and still not know . . . I can't imagine gamma being obtuse until maybe I see an example probably . . .
     
  11. Aug 26, 2014 #10

    Zondrina

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    What do you mean? Just observing the drawing indicates a net downward force in the ##z## direction at ##B##.
     
  12. Aug 26, 2014 #11
    So if it's going down the Z axis, it's going to be an obtuse angle for gamma?
     
  13. Aug 26, 2014 #12

    Zondrina

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    I wind up computing ##\cos^{-1}(\frac{- 35.71}{78.9})##. The ##z## component of ##B## points downwards.
     
  14. Aug 26, 2014 #13
    Alright, I think I see where I went wrong. This has been way more helpful than class.
     
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