1. **Problem Statement:** Determine the moment of the force about point O for the first problem where a 100 N force is applied at the top end of a vertical section 5 m above point O, with the force vector components forming a right triangle with sides 3 (horizontal) and 4 (vertical). 2. **Formula for Moment:** The moment $M$ about a point is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$: $$M = rF\sin(\theta)$$ where $r$ is the perpendicular distance from point O to the line of action of the force, $F$ is the magnitude of the force, and $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. 3. **Determine the angle of the force:** The force vector forms a right triangle with sides 3 (horizontal) and 4 (vertical), so the angle $\phi$ of the force with respect to the horizontal is: $$\phi = \tan^{-1}\left(\frac{4}{3}\right)$$ 4. **Calculate the moment arm:** The force is applied 5 m above point O vertically. The moment arm is the horizontal distance from O to the line of action of the force. Since the force vector points downward-left, the horizontal component of the force is 3 units left, vertical 4 units down. 5. **Calculate the perpendicular distance $r$:** The force acts at the top of the vertical section 5 m above O. The horizontal offset of the force line of action from O is: $$r = 5 \times \frac{3}{5} = 3 \text{ m}$$ because the force vector components 3 and 4 form a 5-unit hypotenuse. 6. **Calculate the moment:** The magnitude of the force is 100 N, and the perpendicular distance is 3 m, so: $$M = rF = 3 \times 100 = 300 \text{ N}\cdot\text{m}$$ 7. **Direction of the moment:** The force tends to rotate the beam clockwise about point O, so the moment is negative if counterclockwise is positive. **Final answer:** $$M = -300 \text{ N}\cdot\text{m}$$