A slab is subjected to two forces ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ of same magnitude F as shown in the figure. Force ${\vec{F}}_{2}$ is in XY-plane while force $F_{1}$ acts along z-axis at the point $(2 \vec{i} + 3 \vec{j}) .$ The moment of these forces about point O will be

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$(3 \hat{i} - 2 \hat{j} - 3 \hat{k}) F$

$(3 \hat{i} + 2 \hat{j} - 3 \hat{k}) F$

$(3 \hat{i} - 2 \hat{j} + 3 \hat{k}) F$

$(3 \hat{i} + 2 \hat{j} + 3 \hat{k}) F$

answer is D.

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Detailed Solution

From the question,

$\vec{F_{1}} = \frac{F}{2} (- \hat{i}) + \frac{F \sqrt{3}}{2} (- \hat{j})$

$\vec{r_{1}} = 0 \hat{i} + 6 \hat{j}$

$\vec{τ} = \vec{r_{1}} \times \vec{F_{1}} = 3 F \hat{k}$

Here, torques for $F_{2}$ force

$\vec{F_{2}} = F \hat{k}$

$\vec{r_{2}} = 2 \hat{i} + 3 \hat{j}$

$\vec{τ} = \vec{r_{2}} \times \vec{F_{2}} = 3 F \hat{i} + 2 f (- \hat{j})$

$τ_{net} = τ_{F_{1}} + τ_{F_{2}} = (3 \hat{i} - 2 \hat{j} + 3 \hat{k}) F$

Hence the correct answer is $(3 \hat{i} - 2 \hat{j} + 3 \hat{k}) F .$

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