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 If  \(\mathbf{F}_{1}=\{100 \mathbf{i}-120 \mathbf{j}+75 \mathbf{k}\} \mathrm{N}\)  and  \(\mathbf{F}_{2} = \{-200 \mathbf{i} +250 \mathbf{j}+100 \mathbf{k} \} \mathrm{N},\) determine the resultant moment  produced by these forces about point  \(O\) . Express the result  as a Cartesian vector.

\( \overline M_{O} =\overrightarrow r_{OA} \times \overline F_{1}+ +\vec{r}_{O A} \times \vec{F}_{2} \)

\( =\overrightarrow {r}_{O A} \times\left(F_{1}+F_{2}\right) \)

\( F_{1}=\left(100 i-120 j+75 k\right) \)

\( +F_{2}=(-200 i+250 j+100 k] \)

\( \rightarrow F_{1}+F_{2}=(-100 i+130  j+175 k) \)

\( =\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0.8} & {1} & {0.6} \\ {-100} & {130} & {175}\end{array}\right| \)

\( =(175-78) i+(-60-140) j+(104-(-100) k \)

\( =((97) i-200 j+204 k) N \cdot m \)

Determine the resultant moment produced by  forces \(\mathbf{F}_{B}\) and  \(\mathbf{F}_{C}\)  about point  \(O\) . Express the result as a  Cartesian vector. 

\( \vec{r}_{A B}=\vec{r}_{B}-\vec{r}_{A} \)

\( =(0 i+2.5 j+0 k)-(o i+b j+6 k) \)

\( =0 i+2.5 j-6 k \rightarrow\left|r_{A 8}\right|=6.5 m \)

\( \overrightarrow F_{\mathrm{AB}}=F \cdot \overrightarrow U_{\mathrm{AB}} =780 . \frac{0 i+2.5 j-6 k}{6.5} =(0 i+300 j-720 k) N \)

\( \vec{r}_{A C}=\vec{r}_{C}-\vec r_{A} =(2 i-3 j+0 k)-(0 i+0 j+6 k)=(2 i-3 j-6 k) m \)

\( \longrightarrow |r_{A C}|=7 m \)

\( \overrightarrow F_{A C}=F \cdot \vec{U}_{A C}=420. \left(\frac{2 i-3 j-6 k}{7}\right)=(120 i-180 j-360 k) N \)

\( \vec{M}_{O}=\vec r_{O A} \times \overrightarrow F_{A B}+ \vec{r}_{O A}× \vec{F}_{A C} \)

\( =\left|\begin{array}{lll}{i} & {j} & {k} \\ {0} & {0} & {6} \\ {0} & {300} & {-720}\end{array}\right|+ \left|\begin{array}{ccc}{i} & {j} & {k} \\ {0} & {0} & {6} \\ {120} & {-180-360}\end{array}\right| \)

\( =(-1800+1080) i-(-720) j+0 k=(-720 i+720 j+0 k) N \cdot m \)

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