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 of 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