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Determine the angle between the force and the line

by frozenguy
Tags: angle, determine, force, line
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frozenguy
#1
Feb10-10, 01:10 AM
P: 190
1. The problem statement, all variables and given/known data
Just have to determine the angle between the force and the line "OC".



2. Relevant equations
[tex]\vec{P}[/tex] [tex]\bullet[/tex] [tex]\vec{Q}=PQcos(\alpha)[/tex] ??


3. The attempt at a solution
I tried using different triangles to find the angle but couldn't come up with one.

I "moved" the 12X24 box down the x-axis so OC starts from the tail of the vector.

Am I supposed to use Fxy some way?

Thanks for your help, I've been working on this for a couple hours now..
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tiny-tim
#2
Feb10-10, 07:47 AM
Sci Advisor
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tiny-tim's Avatar
P: 26,157
Hi frozenguy!

Forget triangles, use coordinates
what are the coordinates of the vectors AB and OC ?
frozenguy
#3
Feb10-10, 11:22 AM
P: 190
Quote Quote by tiny-tim View Post
Hi frozenguy!

Forget triangles, use coordinates
what are the coordinates of the vectors AB and OC ?
Hi tiny-tim!
Thanks for taking a look!

Ok, so for AB, the coordinates are (12,0,0), (0,24,8)
OC=> (0,0,0), (12,24,0)

Then I can say that the components of the lines are AB: <-12,24,8> OC:<12,24,0>

Doting all of those gets me 432, and the product [tex]\left|AB\right|[/tex][tex]\left|OC\right|=751[/tex]

Therefore, [tex]\theta=cos^{-1}(\frac{432}{751})[/tex]
or, [tex]\theta=54.96[/tex] which is what the answer says!

Is this what you were referring to when you mentioned coordinates?

Also, this probably is really elementary, but what is the relation between coordinates and forces? Because sometimes that 200lb would be considered the length of the hypotenuse, but it's really only 28 in this case.

Thanks again for your help,

Frozenguy


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