3D Force System

• August 4th 2008, 06:41 AM
monkeybars
3D Force System
Hello all. Im so thankfull to have found this site and hope some one can help me. I have exams coming up and am basically completely stuck on these 3D force systems for some reason I cannot get them through my head.

Would anyone be able to walk me through these problems?

http://i35.tinypic.com/15d5fte.jpg
• August 5th 2008, 06:27 AM
earboth
Quote:

Originally Posted by monkeybars
Hello all. Im so thankfull to have found this site and hope some one can help me. I have exams coming up and am basically completely stuck on these 3D force systems for some reason I cannot get them through my head.

Would anyone be able to walk me through these problems?

The direction of the force is

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (-4, -4, 2)$

The length of this vector is: $|(-4, -4, 2)| = \sqrt{16+16+4} = 6$

The length correspond with magnitude $F = 30 kN = k \cdot |(-4, -4, 2)|~\implies~ k = 5$ and therefore $\vec F = (-20, -20, 10)$

• August 6th 2008, 04:00 AM
monkeybars
Thanks a lot!! Any ideas with the other 3?
• August 6th 2008, 04:36 AM
earboth
Quote:

Originally Posted by monkeybars
Thanks a lot!! Any ideas with the other 3?

to Q13:

Keep in mind that the x-axis has the direction vector (1,0,0). Then use the formula

$\cos(\alpha) = \frac{(-4,-4,2) \cdot (1,0,0)}{6 \cdot 1}$

to calculate the angle between $\vec F$ and the x-axis.

The other two angles are calculated similarly. (For your confirmation only: If I didn't make a mistake it is answer a))
• August 6th 2008, 05:44 AM
monkeybars
Thank u earboth!!