1. **Problem Statement:** We have a horizontal rod with points A, B, and C. A force $F$ is applied perpendicular to the rod at point B, and the same force $F$ is applied at point C but inclined at a 30-degree angle to the rod. We want to compare the algebraic moments $M_1$ and $M_2$ of these forces about point A. 2. **Formula for Moment:** The moment $M$ of a force about a point is given by: $$M = F \times d \times \sin(\theta)$$ where: - $F$ is the magnitude of the force, - $d$ is the perpendicular distance from the point to the line of action of the force, - $\theta$ is the angle between the force and the lever arm. 3. **Applying to $M_1$:** - Force $F$ at B is perpendicular to the rod, so $\theta = 90^\circ$. - Distance from A to B is $d_{AB}$. - Therefore, $$M_1 = F \times d_{AB} \times \sin(90^\circ) = F \times d_{AB} \times 1 = F d_{AB}$$ 4. **Applying to $M_2$:** - Force $F$ at C is inclined at $30^\circ$ to the rod. - Distance from A to C is $d_{AC}$. - The angle between force and lever arm is $30^\circ$, so $$M_2 = F \times d_{AC} \times \sin(30^\circ) = F \times d_{AC} \times \frac{1}{2} = \frac{F d_{AC}}{2}$$ 5. **Comparing $M_1$ and $M_2$:** Since $d_{AC} > d_{AB}$ (C is farther from A than B), but $M_2$ is multiplied by $\frac{1}{2}$ due to the angle, we compare: $$M_1 = F d_{AB}$$ $$M_2 = \frac{F d_{AC}}{2}$$ If $d_{AC} < 2 d_{AB}$, then $M_1 > M_2$. If $d_{AC} = 2 d_{AB}$, then $M_1 = M_2$. If $d_{AC} > 2 d_{AB}$, then $M_1 < M_2$. 6. **Conclusion:** Without exact distances, the general relation depends on the lengths. However, typically, since $d_{AC}$ is about twice $d_{AB}$ or less, and the force at C is at $30^\circ$, the moment $M_1$ is usually greater than $M_2$. **Final answer:** $M_1 > M_2$ (option b).