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 $\vec{F_1}$ acts along $z$-axis at the point $(2\hat{i} + 3\hat{j})$. The moment of these forces about point O will be:
Answer: (A) $(3\hat{i} - 2\hat{j} + 3\hat{k})F$
Torque of $\vec{F_1} = F\hat{k}$ acting at $(2\hat{i} + 3\hat{j})$:
$$\vec{\tau}_1 = (2\hat{i} + 3\hat{j})\times F\hat{k} = F(-2\hat{j} + 3\hat{i})$$
$\vec{F_2}$ acts at the corner $6\hat{j}$ and makes $30^\circ$ with the $y$-axis, pointing towards the corner as drawn:
$$\vec{F_2} = -F\sin30^\circ\,\hat{i} - F\cos30^\circ\,\hat{j}$$
$$\vec{\tau}_2 = 6\hat{j}\times\left(-\frac F2\hat{i}\right) = 3F\hat{k}$$
(the $\hat{j}$ component of $\vec{F_2}$ is parallel to the position vector and gives no torque).
$$\vec{\tau} = (3\hat{i} - 2\hat{j} + 3\hat{k})F$$
Solution by Sreeraj P, M.Sc Physics