# complicated geometry in mechanics

• Jan 16th 2009, 10:42 AM
Sebastian de Vries
complicated geometry in mechanics
https://mirw.kuleuven.be/vragentromm...080025_1_2.JPG
I have to determine the distance along the Z-axis from O to where the upper line crosses the Z-axis and makes angle $\displaystyle \beta$ between the Z-axis and the line.
The lower line which length equals L starts at the origin and makes an angle $\displaystyle \alpha$ between the Z-axis and the line itself (so the angle between the line and the X-axis is $\displaystyle \frac{1}{2}\pi-\alpha$)

This distance along the Z-axis equals (Lcos$\displaystyle \alpha$ + Lsin$\displaystyle \alpha$cotan$\displaystyle \beta$)ez
ez is the unit vector and there should be an arrow above the letter e.
I don't understand why you can express this distance like this.

Because I'm not certain if the rest of you can see this picture, since it's posted on a secured website (login), I have posted the same picture as an attachment.
• Jan 16th 2009, 11:53 AM
running-gag
Hi

See drawing attached

$\displaystyle z = z_0 + z_1$

$\displaystyle z_0 = L\: cos \alpha$

$\displaystyle tan \beta = \frac{L \:sin \alpha}{z_1}$

Therefore $\displaystyle z = L \:cos \alpha + \frac{L \:sin \alpha}{tan \beta}$

ImageShack - Image Hosting :: drawingue6.jpg
• Jan 16th 2009, 12:30 PM
Sebastian de Vries
Great, I understand it. :)

Quote:

Originally Posted by running-gag
Hi

See drawing attached

$\displaystyle z = z_0 + z_1$

$\displaystyle z_0 = L\: cos \alpha$

This part I understood.

Quote:

$\displaystyle tan \beta = \frac{L \:sin \alpha}{z_1}$

Therefore $\displaystyle z = L \:cos \alpha + \frac{L \:sin \alpha}{tan \beta}$
This was my problem because I was confused about that cotan, I tried to solve it by using the geometric circle.
• Jan 16th 2009, 12:38 PM
running-gag
OK (Wink)

Cotangent is only the inverse of tangent

$\displaystyle cotan \theta = \frac{cos \theta}{sin \theta} = \frac {adjacent \:side}{opposite \:side}$

List of trigonometric identities - Wikipedia, the free encyclopedia
• Jan 16th 2009, 04:17 PM
Sebastian de Vries
That wasn't the problem. ;)
I'll attach another file so you can see which figure I used, then you might understand my confusion.
It's a words-file but it just contains 1 picture (copied from a PDF-file).
I assume you can open this file, otherwise just let me know and I'll try to use Imageschack.
As you can see an arbitrary line which makes an arbitrary angle crosses Y=1 (cartesian graph with X-axis and Y-axis, unit circle with a diameter of 1) at X=cotan angle.