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

Answers (1)
A admin

 

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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