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To calculate the moment about point **A** due to the two forces \( F_1 \) and \( F_2 \), we need to consider both the magnitude of the forces and the perpendicular distance from point **A** to the line of action of each force.

### Step-by-Step Solution:

1. **Force \( F_1 = 500 \, \text{N} \):**
   - \( F_1 \) is vertical, and the distance from point A to the point of application of \( F_1 \) is **4 m**.
   - The moment generated by \( F_1 \) around point A is:
     \[
     M_{F_1} = F_1 \times d_1 = 500 \, \text{N} \times 4 \, \text{m} = 2000 \, \text{Nm}
     \]
     The direction of the moment (counterclockwise or clockwise) is based on the right-hand rule. Since the force is pushing upwards, the moment will be **counterclockwise**.

2. **Force \( F_2 = 600 \, \text{N} \) (at a 45° angle):**
   - The distance from point **A** to the point where \( F_2 \) is applied is a combination of the horizontal and vertical components.
   - The vertical component of the distance is given as **3 m**, and the horizontal distance is **1 m**.
   - To find the perpendicular distance from A to the line of action of \( F_2 \), we project \( F_2 \) into its horizontal and vertical components:
     \[
     F_{2_x} = F_2 \cos(45^\circ) = 600 \, \text{N} \times \cos(45^\circ) = 600 \times 0.7071 \approx 424.3 \, \text{N}
     \]
     \[
     F_{2_y} = F_2 \sin(45^\circ) = 600 \, \text{N} \times \sin(45^\circ) = 600 \times 0.7071 \approx 424.3 \, \text{N}
     \]
   - Now calculate the moments caused by each component:
     - The horizontal component \( F_{2_x} \) causes a moment based on the vertical distance **3 m**:
       \[
       M_{F_{2_x}} = F_{2_x} \times 3 \, \text{m} = 424.3 \, \text{N} \times 3 \, \text{m} = 1272.9 \, \text{Nm}
       \]
       This moment is **clockwise**.
     - The vertical component \( F_{2_y} \) causes a moment based on the horizontal distance **1 m**:
       \[
       M_{F_{2_y}} = F_{2_y} \times 1 \, \text{m} = 424.3 \, \text{N} \times 1 \, \text{m} = 424.3 \, \text{Nm}
       \]
       This moment is **counterclockwise**.

### Total Moment About Point A:
- Add up the moments considering their directions:
  \[
  M_{\text{total}} = M_{F_1} - M_{F_{2_x}} + M_{F_{2_y}} = 2000 \, \text{Nm} - 1272.9 \, \text{Nm} + 424.3 \, \text{Nm}
  \]
  \[
  M_{\text{total}} = 1151.4 \, \text{Nm}
  \]
  The total moment is **1151.4 Nm counterclockwise**.

Would you like further details on any part of this solution? 

Here are five related questions for further exploration:
1. What is the effect of changing the angle of \( F_2 \) on the total moment?
2. How would the moment change if \( F_1 \) were applied at a different point?
3. What would be the moment about point A if the distance of \( F_1 \) were increased to 5 meters?
4. How does the distribution of forces affect the overall equilibrium of the structure?
5. What is the significance of clockwise vs. counterclockwise moments in engineering problems?

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

Determine the moment with respect to point A for the forces F1 = 500 N and F2 = 600 N at a 45° angle.

This problem demonstrates how to calculate the moment with respect to point A for forces F1 = 500 N and F2 = 600 N applied at a 45° angle. Using trigonometry to break F2 into components and applying the principle of moments, we calculate a total moment of 1151.4 Nm counterclockwise.

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