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:

$$\text{For force } \vec{F}_1:$$ $$\text{Position vector: } \vec{r}_1 = 2\hat{i} + 3\hat{j}$$

$$\text{Force vector (acting along z-axis): } \vec{F}_1 = F\hat{k}$$

$$\vec{\tau}_1 = \vec{r}_1 \times \vec{F}_1 = (2\hat{i} + 3\hat{j}) \times (F\hat{k}) = 2F(-\hat{j}) + 3F(\hat{i}) = (3\hat{i} - 2\hat{j})F$$

$$\text{For force } \vec{F}_2:$$ $$\text{From the geometry of the slab, the point of application is on the y-axis at } y = 6\text{ m}:$$ $$\vec{r}_2 = 6\hat{j}$$

$$\vec{F}_2 = -F\sin(30^\circ)\hat{i} - F\cos(30^\circ)\hat{j} = -\frac{F}{2}\hat{i} - \frac{F\sqrt{3}}{2}\hat{j}$$

$$\vec{\tau}_2 = \vec{r}_2 \times \vec{F}_2 = (6\hat{j}) \times \left(-\frac{F}{2}\hat{i} - \frac{F\sqrt{3}}{2}\hat{j}\right) = -3F(\hat{j} \times \hat{i}) - 3\sqrt{3}F(\hat{j} \times \hat{j})$$

$$\text{Using the unit vector properties } \hat{j} \times \hat{i} = -\hat{k} \text{ and } \hat{j} \times \hat{j} = 0:$$ $$\vec{\tau}_2 = 3F\hat{k}$$

$$\text{Total moment about point O:}$$

$$\vec{\tau}_{\text{net}} = \vec{\tau}_1 + \vec{\tau}_2 = (3\hat{i} - 2\hat{j})F + 3F\hat{k} = (3\hat{i} - 2\hat{j} + 3\hat{k})F$$