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Resolution of Components Problem

  1. Sep 7, 2005 #1
    Okay, here's a problem from my Statics class. Just dealing with 2 vectors, their direction and magnitude given--need to find the same for the resultant vector. If I break it down into X and Y components, I can do it quite easily....but that's not the way he wants it done.

    It has to be found using ONLY Sine and Cosine Laws, which I get, but am somehow getting stuck on the fact that the given angles are based upon a vertical component that is different from the resultant R after completing the parallelogram....SIMPLE, but throwing me for some reason....

    Here's the problem:
    http://img394.imageshack.us/img394/5334/hw2paper1jt.jpg

    How do I work the angles so I can apply them properly in the Sine and Cosine laws and get R figured out?
     
    Last edited: Sep 8, 2005
  2. jcsd
  3. Sep 7, 2005 #2

    Doc Al

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    Staff: Mentor

    Realize that when you slide one vector to construct the parallelogram, the angle it makes with the vertical (which is given) does not change. That, and the fact that the angles of a triangle add to 180 degrees, should allow you to find all the angles needed to apply the sine and cosine laws.

    For example: To add the two vectors B-P and B-Q, slide a copy of B-Q so that its tail is at P. The original vector B-P and the copy of B-Q form two sides of the triangle you will analyze.
     
  4. Sep 7, 2005 #3
    Okay, that makes sense, but I'm still a little confused for some reason. If you look at the diagram I made in the first one, I DID already complete the parallelogram, but I'm having trouble understanding, would I still use the angles of 40 and 60? I get sliding a copy of B-Q over to touch the tail of B-P, but then what are the angles I plug in? Would I still use the 40deg. on that side as one?

    Forgive my stupidness....something in the image for this problem just isn't working for me.... :frown:
     
  5. Sep 7, 2005 #4

    Doc Al

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    What you need to determine is the angle B-A-D in your diagram. Figure it out like this: The side B-A makes what angle with the vertical? The side A-D makes what angle with the vertical? (Those angles are given, so this should be easy!) Now imagine a vertical line that hits point A. Use those angles to figure out angle B-A-D. (Realize that angles that add to a straight line must add to 180 degrees.)
     
  6. Sep 7, 2005 #5
    Okay, well that would make sense, except for one part--the vertical line shown in the first given diagram is NOT the same line as R which needs to be found. From what I'm getting out of what you said, it would make sense, but only if I were treating the vertical as R. Isn't the R I want to find angled just to the right of the vertical line?

    Am I missing something or just REALLY thick?
     
    Last edited: Sep 7, 2005
  7. Sep 7, 2005 #6

    Doc Al

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    The only purpose of the imaginary vertical line (intersecting point A) is to help you figure out the angle B-A-D. You are correct: It is not to be confused with side R.

    Do this: Actually draw that vertical line. Then mark the three angles:
    -vertical with line B-A (this is given)
    -angle B-A-D (this is the one you want)
    -vertical with line A-D (this is given)

    Realize that these three angles must add to 180 degrees.
     
  8. Sep 7, 2005 #7
    Okay. And then the other side, with angle B-C-D is the same? I guess there, the first at 60deg. is give (correct?), then the other's get based on the left triangle? I would think so, since the vertical doesn't intersect that side anywhere?

    I think I got it now....thanks!
     
    Last edited: Sep 7, 2005
  9. Sep 8, 2005 #8
    Yet again, I'm confused. As far as determining the magnitude of R, I think I have that down. But now I'm stuck on how to find the angle of R.

    Where does it start from/based? The vertical? How? I have to use the Sine Law to find it, which makes sense, except for which components I use where...plus, though my Magnitude answer seems right, something with how I showed the angles still looks wrong...

    http://img389.imageshack.us/img389/9085/hw2paper23oh.jpg
     
    Last edited: Sep 8, 2005
  10. Sep 8, 2005 #9
    Basically, can anyone help me along in getting the angle of R using only the Sine law? I kinda get it, but I've never really used it before, and with this problem it's confusing me...
     
  11. Sep 8, 2005 #10

    Doc Al

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    You have all the sides and one angle (of triangle B-A-D), so you can use the Sine law to find any one of the other angles. Use it to find angle A-B-D of your triangle.

    That's not your final answer, of course. But you can then easily figure out what angle R makes with the vertical.
     
  12. Sep 8, 2005 #11
    Okay, thanks again, I'll see how it goes....
     
  13. Sep 8, 2005 #12
    Done with those...now just confused with the 2nd half of my 3rd problem. Taking a look at this sheet, what does (b) mean to you?

    http://img382.imageshack.us/img382/209/hw2paper19wb.jpg

    The first part is basically just a carbon copy of what I just got done doing, but what does "(b) Determine the components along the u and v axes of F1 and F2" mean?
     
  14. Sep 8, 2005 #13
    Pretend that u and v are your choice of coordinate axes and find the projections of F1 and F2 onto each. If you're not familiar with arbitrary projections, figure out what it is you're doing when you find x and y axis components of a force vector, and apply it to this scenario.
     
  15. Sep 8, 2005 #14
    So that means...um...basically like the U and V components for each, F1 and F2?

    I guess that would make sense, except for the fact that V is angled at 45deg. How do I go about finding the components using only Sine/Cosine laws?
     
  16. Sep 8, 2005 #15
    The x-component of a vector is the length of the leg of the right triangle that the vector forms with a line parallel to the x-axis and a line perpendicular to the x-axis. The leg in question is taken as the one parallel to the x-axis. Draw this out to see the trigonometry required. Use a similar construction to get a vector resolved on an arbitrary axis.
     
  17. Sep 8, 2005 #16
    Thanks...got it!
     
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