1. **Problem statement:** Determine the magnitude of the force $\vec{F}$ (in kN) such that the resultant couple moment about point A is 10 kN.m clockwise, given $\theta = 60^\circ$. 2. **Given data and setup:** - Angle $\theta = 60^\circ$ - Resultant couple moment $M = 10$ kN.m clockwise - Vertical member length segments: 2 m above force application point, 4 m below - Two external horizontal forces of 6 kN each acting left at mid and bottom of vertical member 3. **Understanding moments:** The moment $M$ caused by a force $F$ about a point is given by: $$M = F \times d \times \sin(\alpha)$$ where $d$ is the perpendicular distance from the point to the line of action of the force, and $\alpha$ is the angle between the force and the lever arm. 4. **Calculate lever arm distances and force components:** - The force $F$ is applied at the midpoint of the vertical member, which is 4 m below point A. - The force $F$ acts at $60^\circ$ from the horizontal, so its vertical and horizontal components are: $$F_x = F \cos 60^\circ = \frac{F}{2}$$ $$F_y = F \sin 60^\circ = \frac{\sqrt{3}}{2}F$$ 5. **Calculate moments caused by $F$ about point A:** - Horizontal component $F_x$ creates a moment about A due to vertical distance 4 m: $$M_x = F_x \times 4 = \frac{F}{2} \times 4 = 2F$$ - Vertical component $F_y$ creates a moment about A due to horizontal distance 0 (since force acts along vertical member), so no moment from $F_y$. 6. **Moments from other forces:** - Two 6 kN forces act horizontally left at 2 m and 6 m below A. - Moments from these forces about A: $$M_{6kN1} = 6 \times 2 = 12 \text{ kN.m (clockwise)}$$ $$M_{6kN2} = 6 \times 6 = 36 \text{ kN.m (clockwise)}$$ - Total moment from these forces: $$M_{others} = 12 + 36 = 48 \text{ kN.m (clockwise)}$$ 7. **Resultant moment condition:** The total moment about A must be 10 kN.m clockwise, so: $$M_{total} = M_{others} + M_x = 10$$ Substitute values: $$48 + 2F = 10$$ 8. **Solve for $F$:** $$2F = 10 - 48 = -38$$ $$F = \frac{-38}{2} = -19$$ 9. **Interpretation:** Negative sign indicates force direction opposite assumed. Magnitude is: $$|F| = 19 \text{ kN}$$ **Final answer:** $$\boxed{19 \text{ kN}}$$