1. **State the problem:** We have three forces $\mathbf{F}_1, \mathbf{F}_2, \mathbf{F}_3$ acting at points $(1,2), (2,1), (3,2)$ respectively. The forces are given by: \[\mathbf{F}_1 = \hat{i} \cos \theta + \hat{j} \sin \theta, \quad \mathbf{F}_2 = \hat{i} \sin \theta - 2 \hat{j} \cos \theta, \quad \mathbf{F}_3 = \hat{i} \cos \theta\] We want to find the resultant force $\mathbf{F}$ at the origin and the couple moment $G$. 2. **Formula and rules:** - The resultant force $\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3$. - The moment $G$ about the origin due to a force $\mathbf{F}$ at position $\mathbf{r} = x \hat{i} + y \hat{j}$ is given by the scalar (in 2D): $$G = x F_y - y F_x$$ where $F_x, F_y$ are components of the force. 3. **Calculate resultant force $\mathbf{F}$:** \[ \mathbf{F} = (\cos \theta + \sin \theta + \cos \theta) \hat{i} + (\sin \theta - 2 \cos \theta + 0) \hat{j} = (2 \cos \theta + \sin \theta) \hat{i} + (\sin \theta - 2 \cos \theta) \hat{j} \] 4. **Calculate moments $G$ from each force:** - For $\mathbf{F}_1$ at $(1,2)$: $$G_1 = 1 \times \sin \theta - 2 \times \cos \theta = \sin \theta - 2 \cos \theta$$ - For $\mathbf{F}_2$ at $(2,1)$: $$G_2 = 2 \times (-2 \cos \theta) - 1 \times \sin \theta = -4 \cos \theta - \sin \theta$$ - For $\mathbf{F}_3$ at $(3,2)$: $$G_3 = 3 \times 0 - 2 \times \cos \theta = -2 \cos \theta$$ 5. **Sum moments to get total couple moment $G$:** \[ G = G_1 + G_2 + G_3 = (\sin \theta - 2 \cos \theta) + (-4 \cos \theta - \sin \theta) + (-2 \cos \theta) = -8 \cos \theta \] 6. **Final answers:** \[ \mathbf{F} = (2 \cos \theta + \sin \theta) \hat{i} + (\sin \theta - 2 \cos \theta) \hat{j}, \quad G = -8 \cos \theta \] This means the system reduces to a single force $\mathbf{F}$ at the origin and a couple moment $G$ as above.