Determine the moment produced by force F_C about point O. Express the result as a Cartesian vector.

#### Solution:

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**This question is the same as the previous question. Steps will be skipped as they are the same. Please refer to the previous question if a section is confusing.**

Let us express force F_C in Cartesian vector form. To do so, we need to write a position vector from A to C.

r_{AC}=\left\{(2-0)i+(-3-0)j+(0-6)k\right\}

r_{AC}=\left\{2i-3j-6j\right\}

r_{AC}=\left\{2i-3j-6j\right\}

The magnitude of this position vector is:

magnitude of r_{AB}\,=\,\sqrt{(2)^2+(-3)^2+(-6)^2}=7

The unit vector is:

u_{AB}\,=\,\left(\dfrac{2}{7}i-\dfrac{3}{7}j-\dfrac{6}{7}k\right)

Force F_C can now be expressed in Cartesian vector form.

F_C=420\left(\dfrac{2}{7}i-\dfrac{3}{7}j-\dfrac{6}{7}k\right)

F_C=\left\{120i-180j-360k\right\} N

F_C=\left\{120i-180j-360k\right\} N

We now need to express a position vector from O to A.

r_{OA}=\left\{0i+0j+6k\right\}

To find the moment at O created by force F_C, we need to take the cross product ofÂ the position vector r_{OA} and force F_C.

M_O=r_{OA}\times F_C

M_O=\begin{bmatrix}\bold i&\bold j&\bold k\\0&0&6\\120&-180&-360\end{bmatrix}

M_O=\left\{1080i+720j\right\}N\cdot m

M_O=\begin{bmatrix}\bold i&\bold j&\bold k\\0&0&6\\120&-180&-360\end{bmatrix}

M_O=\left\{1080i+720j\right\}N\cdot m

#### Final Answer:

M_O=\left\{1080i+720j\right\}N\cdot m