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Difference between Force components using triangle rule and dotproduct

  1. Mar 1, 2014 #1
    Hello everyone,
    I have here one important abstract question which makes up some perplexity to my understanding. Attached to this post is one pic of F and two new introduced axes ( u and v ) . Let us for instance not consider them as perpendicular to each other. Now, if I am asked to resolve these forces along u and v USING geometry, I will use the parallelogram rule which will outcome results rather than F cos theta as for the u axis for example. Now, if we want to compute the scalar projection of F along u, we simply can say it is F. unit vector f u which will be F cos theta. However, written in some books, this magnitude is NOT the same as the component of F along u. Why?? To sum up my question, what is the difference between the component of F along any axis using the parallelogram rule and the dot product definition. Sorry for elongating and Thanks to whoever gives me a kind hand.
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2014 #2

    tiny-tim

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    hello ehabmozart! :smile:

    for the usual coordinates i and j,

    F = Fxi + Fyj

    where
    Fx = F·i

    Fy = F·j

    as you've noticed, that doesn't translate into the case of non-perpendicular axes

    so instead of "dotting" with the same coordinate, let's "cross" with the other coordinate …
    Fxixj = Fxj

    Fyjxi = Fxi

    (these are the same as saying that F - Fxi is parallel to j, and F - Fyj is parallel to i, which is what we want)

    if you use a general u and v instead of i and j, the formulas now work :wink:
     
  4. Mar 1, 2014 #3
    [QUOTE
    so instead of "dotting" with the same coordinate, let's "cross" with the other coordinate …
    Fxixj = Fxj

    Fyjxi = Fxi

    (these are the same as saying that F - Fxi is parallel to j, and F - Fyj is parallel to i, which is what we want)

    if you use a general u and v instead of i and j, the formulas now work :wink:[/QUOTE]

    I am sorry my man, but I honestly understood nothing from these statements. What do you mean for instance by Fyjxi = Fxi[/INDENT] ... Thanks for your time and effort, after all :)
     
  5. Mar 1, 2014 #4

    tiny-tim

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    i and j are the usual symbols for the unit vectors in the x and y directions

    so Fx and Fy are Fcosθ and Fsinθ

    and Fxi will be the vector with magnitude Fsinθ
     
  6. Mar 1, 2014 #5
    Let's suppose that u and v are not perpendicular, and [itex]\vec{i}_u[/itex] and [itex]\vec{i}_v[/itex] are the unit vectors along the u and v axes respectively. Then the force [itex]\vec{F}[/itex] can be represented by:
    [tex]\vec{F}=F_u\vec{i}_u+F_v\vec{i}_v[/tex]
    If we take the dot product of [itex]\vec{F}[/itex] with respect to [itex]\vec{i}_u[/itex] and [itex]\vec{i}_v[/itex]
    we get
    [tex]\vec{F}\centerdot \vec{i}_u =F_u+F_v(\vec{i}_v\centerdot \vec{i}_u)[/tex]
    [tex]\vec{F}\centerdot \vec{i}_v =F_u(\vec{i}_v\centerdot \vec{i}_u)+F_v[/tex]

    If you solve for the components of F in the u-v system, you get:
    [tex]F_u=\frac{(\vec{F}\centerdot \vec{i}_u)-(\vec{F}\centerdot \vec{i}_v)(\vec{i}_v\centerdot \vec{i}_u)}{1-(\vec{i}_v\centerdot \vec{i}_u)^2}[/tex]
    [tex]F_v=\frac{(\vec{F}\centerdot \vec{i}_v)-(\vec{F}\centerdot \vec{i}_u)(\vec{i}_v\centerdot \vec{i}_u)}{1-(\vec{i}_v\centerdot \vec{i}_u)^2}[/tex]

    This is the same result you would get using the parallelogram method.

    Chet
     
    Last edited: Mar 1, 2014
  7. Mar 2, 2014 #6
    This is a continuation of my previous post. Let α be the angle between the force and the u axis, and let β be the angle between the force and the v axis. Prove that the results I presented are consistent with the law of sines for the components of the force in the triangle formed by Fu, Fv, and the force (in the parallelogram rule setup).
     
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