Difference between Force components using triangle rule and dotproduct

[ehabmozart]

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.

 

[tiny-tim]

hello ehabmozart!

for the usual coordinates i and j,

F = F_xi + F_yj

where

F_x = F·i

F_y = 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 …

F_xixj = Fxj

F_yjxi = Fxi​

(these are the same as saying that F - F_xi is parallel to j, and F - F_yj 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

 

[ehabmozart]

[QUOTE
so instead of "dotting" with the same coordinate, let's "cross" with the other coordinate …

F_xixj = Fxj

F_yjxi = Fxi​

(these are the same as saying that F - F_xi is parallel to j, and F - F_yj 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 [/QUOTE]

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

 

[tiny-tim]

What do you mean for instance by F_yjxi = Fxi[/INDENT] ... Thanks for your time and effort, after all :)

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

so F_x and F_y are Fcosθ and Fsinθ

and Fxi will be the vector with magnitude Fsinθ

 

[Chestermiller]

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

 

[Chestermiller]

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 F_u, F_v, and the force (in the parallelogram rule setup).