Alright — let's do this. The moment of a force (also known as torque) about a point is calculated by the formula: $$M = F \cdot d_{\perp}$$ where $M$ is the moment, $F$ is the magnitude of the force, and $d_{\perp}$ is the perpendicular distance from the pivot point to the line of action of the force. Alternatively, if the force $F$ is applied at a distance $r$ from the pivot and makes an angle $\theta$ with the lever arm, the moment can be calculated as: $$M = F \cdot r \cdot \sin(\theta)$$ Here, $\theta$ is the angle between the force vector and the position vector (lever arm) from the pivot to the point of application of the force. Step 1: Identify the given values from the diagram. Magnitude of the force, $F = 10 \text{ N}$. Length of the lever arm (distance from O to the point of application), $r = 5 \text{ m}$. The angle shown is $30^\circ$ between the force vector and the horizontal bar. Step 2: Determine the effective angle for the moment calculation. The angle $\theta$ in the formula $M = F \cdot r \cdot \sin(\theta)$ is the angle between the force vector and the lever arm. In this case, the lever arm is horizontal. The force is directed downwards and to the left, making an angle of $30^\circ$ with the horizontal. The component of the force perpendicular to the lever arm is $F_{\perp} = F \sin(30^\circ)$. Step 3: Calculate the moment. Using the perpendicular component of the force: $$M = F_{\perp} \cdot r$$ $$M = (F \sin(30^\circ)) \cdot r$$ Substitute the values: $$M = (10 \text{ N} \cdot \sin(30^\circ)) \cdot 5 \text{ m}$$ We know that $\sin(30^\circ) = 0.5$. $$M = (10 \text{ N} \cdot 0.5) \cdot 5 \text{ m}$$ $$M = 5 \text{ N} \cdot 5 \text{ m}$$ $$M = 25 \text{ Nm}$$ The moment of the 10 N force about O is 25 Nm. The final answer is $\boxed{\text{A) } 25 \text{ Nm}}$. Send me the next one 📸