This problem involves calculating the resultant couple (moment) from the forces acting on a rectangle. Two couples act on the rectangle, each with forces $F$ and $G$, given the rectangle's dimensions of 3 meters (base) and 9 meters (height).

Step-by-Step Solution:

1. Identify the geometry and force components:

• The forces $F = 80 \, \text{N}$ and $G = 70 \, \text{N}$ are applied at an angle of $80^\circ$ to the sides of the rectangle.
• The base of the rectangle is $3 \, \text{m}$ and the height is $9 \, \text{m}$.

2. Break forces into components:

The force $F$ can be broken into horizontal and vertical components as follows: $F_x = F \cos(80^\circ), \quad F_y = F \sin(80^\circ)$ Similarly, the force $G$ can be broken into: $G_x = G \cos(80^\circ), \quad G_y = G \sin(80^\circ)$

Plugging in values for $F = 80 \, \text{N}$ and $G = 70 \, \text{N}$: $F_x = 80 \cos(80^\circ), \quad F_y = 80 \sin(80^\circ)$ $G_x = 70 \cos(80^\circ), \quad G_y = 70 \sin(80^\circ)$ Using a calculator: $F_x \approx 80 \times 0.1736 = 13.888 \, \text{N}, \quad F_y \approx 80 \times 0.9848 = 78.784 \, \text{N}$ $G_x \approx 70 \times 0.1736 = 12.152 \, \text{N}, \quad G_y \approx 70 \times 0.9848 = 68.936 \, \text{N}$

3. Calculate moments:

The moment caused by each force about the center of the rectangle depends on the perpendicular distance from the line of action of the force to the center. For this case:

• The moment due to $F$ (vertical distance is 9 m): $M_F = F_x \times 9$ $M_F \approx 13.888 \times 9 = 125.0 \, \text{Nm}$

• The moment due to $G$ (horizontal distance is 3 m): $M_G = G_y \times 3$ $M_G \approx 68.936 \times 3 = 206.8 \, \text{Nm}$

4. Resultant moment:

The resultant moment $M_R$ is the vector sum of the moments due to $F$ and $G$. Since these moments are perpendicular to each other, we use the Pythagorean theorem: $M_R = \sqrt{M_F^2 + M_G^2}$ $M_R = \sqrt{(125.0)^2 + (206.8)^2} \approx \sqrt{15625 + 42767.04} \approx \sqrt{58392.04} \approx 241.6 \, \text{Nm}$

5. Direction of the resultant moment:

The direction $\theta$ of the resultant moment can be found using: $\theta = \tan^{-1} \left( \frac{M_G}{M_F} \right)$ $\theta = \tan^{-1} \left( \frac{206.8}{125.0} \right) \approx \tan^{-1}(1.6544) \approx 58.1^\circ$

Final Answer:

The magnitude of the resultant moment is approximately 241.6 Nm, and its direction is 58.1° from the horizontal.

Would you like further clarification or details on any part of this solution? Here are some additional questions you might consider:

1. How does the orientation of the rectangle influence the couple moments?
2. What is the physical significance of a couple in mechanics?
3. How would the result change if the angles of the forces were different?
4. Can the resultant moment ever be zero? Under what conditions?
5. How would the calculation differ if the dimensions of the rectangle were not given?

Tip: When calculating moments, always ensure that the distances used are perpendicular to the force's line of action for accuracy.