3D Force System

**monkeybars** · August 4th 2008, 06:41 AM

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?

Thanks in advance

**earboth** · August 5th 2008, 06:27 AM

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?

Thanks in advance

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 $\vec F = (-20, -20, 10)$

So it is answer b)

**monkeybars** · August 6th 2008, 04:00 AM

Thanks a lot!! Any ideas with the other 3?

**earboth** · August 6th 2008, 04:36 AM

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))

**monkeybars** · August 6th 2008, 05:44 AM

Thank u earboth!!

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