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Projected component of F along a line in cartesian question.

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data
    2d98bk2.png


    Determine the magnitude of the projected component of F along AC. Express this component as a Cartesian vector.


    2. Relevant equations
    1scls5.png

    Information I'm working with.


    3. The attempt at a solution

    Most of my attempts have proven fruitless.
    I understand that in order to find a component of F along AC we'd need to have the Uac but I'm just having difficulty finding the cartesian coordiantes for segment AC.

    Cosine law comes to mind but how would you apply cosine law in cartesian for a 3D image?

    Also, do the signs for the cartesian coordinates of AC become times by a (-1) as the force is protruding outwards from point C?

    Any help is much appreciated. I've been stuck on this question for a good 5-6 hours now.

    Thanks!
     
  2. jcsd
  3. Jan 27, 2014 #2

    Simon Bridge

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    If A is on the origin, they are just given to you.
    i.e. AB=(3,4,0)ft so |AB|=5ft. (spot the 3-4-5 triangle?)

    You need to break it down into 2D traingles.
    But I don't think you need this so long as you know how to do a dot product using components of vectors.

    No. A vector can only point outwards from a point because a point has no inside.
    You are given the vector components - use them.
     
  4. Jan 27, 2014 #3
    Appreciate the quick reply Simon, I see AB's triangle but I'm having trouble finding a relation between AB and AC.

    The answer at the back of the book yields Fac = -25.87 lb ( -18.0i -15.4j + 10.3k)

    I don't really understand how they can come up with an answer that has i and j being negative values.
    * = dot product
    Fca = F * Uca
    Where Uca = Cartesian of CA / Sqrt( each cartesian coordinates of ac^2 of)

    Now my question narrows down to: How do I incorporate the cartesians given to yield the cartian coordinates of CA?

    Initially I had thought that

    Fac = (F* Ucb) * Uab

    But that wasn't the case. :(
     
  5. Jan 27, 2014 #4
    Can you see from the figure what the x, y, and z components of the position vector from point A to point C are? In other words, what are the coordinates of point C? For example, from the figure, I can see that the z coordinate of point C is -4 ft.

    Chet
     
  6. Jan 27, 2014 #5

    Simon Bridge

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    You are over-thinking things.
    You don't need the relationship between AB and AC.
    Stop thinking about the equations for a bit.

    AC is the vector pointing from point A to point C.
    Thus AC = C - A. This is true no matter what the coordinate system is.

    You are all-but given the cartesian coordinates of A and C.
    Look carefully at all those measurements in feet that the diagram is peppered with.
    [edit] chet has given you a hint there...
     
  7. Jan 27, 2014 #6

    haruspex

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    Just above C there's an intersection of three black construction lines. What are its coordinates in the XY plane? what, then, are the coordinates of C?
    Edit: A triple response in the same 6 minute window, and in violent agreement.
     
  8. Jan 27, 2014 #7
    Hey Chet, thanks for the reply.

    Yea I do. The coordinates for point C are { 7i + 6j -4k}

    I've tried applying the dot product of point C's coordinates to the force but didn't come to the correct answers.
    My work:

    Fca = F {30i -45k +50k} * Uca ({7i +6j -4k}/ Sqrt(7^2 + 6^2 +4^2))

    Doesnt give me the negative i and j that I'm supposed to be getting.

    Edit: Hahah, I really do appreciate all this help you guys. I've been head-desking over this question for the longest time.
     
    Last edited: Jan 27, 2014
  9. Jan 27, 2014 #8
    It looks like you have the right idea. This is the way I would have done it. Is it just the overall sign that's wrong, or are all the individual components wrong. Maybe they wanted you to dot it with the unit vector in the direction from C to A, rather than from A to C. That would flip the sign.

    Chet
     
  10. Jan 27, 2014 #9
    Not only are the values incorrect but the signs are as well Chet.

    Following through with the calculation using C-A (-7,-6, +4) I end up with : {-20.895i + 26.866j + 19.9007k}
    which is a ways off from the ( -18.0i -15.4j + 10.3k) answer.

    A-C Yields (+, -, -) values
     
  11. Jan 27, 2014 #10

    Simon Bridge

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    C=(7,6,-4), A=(0,0,0) then AC=C-A=(7,6,-4) is the vector from A to C.
    The vector from C to A is CA=-AC=(-7,-6,4)

    The direction of AC is the unit vector AC/|AC|

    The component of F in the direction of AC is the dot product of F with the unit vector.
     
  12. Jan 28, 2014 #11
    OK. But consider this: The unit vector in the direction of the "correct answer" is the same as the unit vector in the C-A direction, while the unit vector in your answer is not. That is:
    18.01/7 = 15.4/6 = 10.3/4 = 2.57. My conclusion is that you must have made a mistake in arithmetic. What did you get for the dot product of F with the unit vector in the direction of CA? It should have been 25.8 N.

    Chet
     
  13. Jan 28, 2014 #12
    I dot producted the force and the unit vector ac (7/10 i , 4/5 j, -2/5k) and got 46.22N
     
  14. Jan 28, 2014 #13
    The sum of the squares of the components of AC is ~101, so the length of AC is ~10.
    So the unit vector in the direction of CA is -0.7 i -0.6 j + 0.4k. So now, please try again with that dot product.

    Chet
     
  15. Jan 28, 2014 #14
    The sum of the squares of the components of AC is ~101, so the length of AC is ~10.
    So the unit vector in the direction of CA is -0.7 i -0.6 j + 0.4k. So now, please try again with that dot product.

    Chet
     
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