# A slab is subjected to two forces $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ of same magnitude Fas shown in the figure . Force $\overrightarrow{F}_{2}$ in XY - plane while force $\overrightarrow{F}_{1}$ acts along z- axis at the point $\left [ 2\overrightarrow{i}+3\overrightarrow{j} \right ]$. The moment these forces about point O will be:Option 1) $\left ( 3\widehat{i}-2\widehat{j}+3\widehat{k} \right )F$Option 2) $\left ( 3\widehat{i}+2\widehat{j}+3\widehat{k} \right )F$Option 3) $\left ( 3\widehat{i}-2\widehat{j}-3\widehat{k} \right )F$Option 4) $\left ( 3\widehat{i}+2\widehat{j}-3\widehat{k} \right )F$

Torque -

$\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}$

- wherein

This can be calculated by using either  $\tau=r_{1}F\; or\; \tau=r\cdot F_{1}$

$r_{1}$ = perpendicular distance from origin to the line of force.

$F_{1}$ = component of force perpendicular to line joining force.

$\vec{F_{1}}=\frac{-F}{2}\widehat{i}-\frac{\sqrt{3}}{2}F\widehat{j}$

$\vec{r_{1}}=6\widehat{j}$

$\vec{T_{F1}}=\vec{r_{1}}\times \vec{F_{1}}=3F\widehat{k}$

$\vec{F_{2}}=F\widehat{k} & \vec{r_{2}}=2\widehat{i}+3\widehat{j}$

$\vec{T_{F2}}=\vec{r_{2}}\times \vec{F_{2}}=3F\widehat{i}-2F\widehat{j}$

$\widehat{T}.....=T_{r}+T = 3F\widehat{i}-2F\widehat{j}+3F\widehat{K}$

Option 1)

$\left ( 3\widehat{i}-2\widehat{j}+3\widehat{k} \right )F$

Option 2)

$\left ( 3\widehat{i}+2\widehat{j}+3\widehat{k} \right )F$

Option 3)

$\left ( 3\widehat{i}-2\widehat{j}-3\widehat{k} \right )F$

Option 4)

$\left ( 3\widehat{i}+2\widehat{j}-3\widehat{k} \right )F$

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