A slab is subjected to two forces $\overrightarrow{\mathrm{F}_{1}}$ and $\overrightarrow{\mathrm{F}_{2}}$ of…
A slab is subjected to two forces $\overrightarrow{\mathrm{F}_{1}}$ and $\overrightarrow{\mathrm{F}_{2}}$ of same
magnitude $F$ as shown in the figure. Force $\overrightarrow{\mathrm{F}_{2}}$ is in XY-
plane while force $\mathrm{F}_{1}$ acts along $z$ -axis at the point
$(2 \vec{i}+3 \vec{j})$. The moment of these forces about point $\mathrm{O}$ will be:
$(3 \hat{i}-2 \hat{j}+3 \hat{k})$ F
$(3 \hat{i}-2 \hat{j}-3 \hat{k}) \mathrm{F}$
$(3 \hat{i}+2 \hat{j}-3 \hat{k}) \mathrm{F}$
$(3 \hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{F}$
Solution
Given, $\overrightarrow{\mathrm{F}}_{1}=\frac{\mathrm{F}}{2}(-\hat{\mathrm{i}})+\frac{\mathrm{F} \sqrt{3}}{2}(-\hat{\mathrm{j}})$
$\overrightarrow{\mathrm{r}}_{1}=0 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}$
Torque due to $\mathrm{F}_{1}$ force
$
\vec{\tau}_{\mathrm{F}_{1}}=\overrightarrow{\mathrm{r}}_{1} \times \overrightarrow{\mathrm{F}}_{1}=6 \hat{\mathrm{j}} \times\left(\frac{\mathrm{F}}{2}(-\hat{\mathrm{i}})+\frac{\mathrm{F} \sqrt{3}}{2}(-\hat{\mathrm{j}})\right)=3 \mathrm{~F}(\hat{\mathrm{k}})
$
Torque due to $\mathrm{F}_{2}$ force
$
\begin{aligned}
\vec{\tau}_{\mathrm{F}_{2}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times & \mathrm{F} \hat{\mathrm{k}}=3 \mathrm{~F} \hat{\mathrm{i}}+2 \mathrm{~F}(-\hat{\mathrm{j}}) \\
\vec{\tau}_{\mathrm{net}}=\vec{\tau}_{\mathrm{F}_{1}}+\vec{\tau}_{\mathrm{F}_{2}} &=3 \mathrm{Fi}+2 \mathrm{~F}(-\hat{\mathrm{j}})+3 \mathrm{~F}(\hat{\mathrm{k}}) \\
&=(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \mathrm{F}
\end{aligned}
$
