Thank you for the correction. Let's recalculate the moment with the correct force value \( F = 1600 \, \text{N} \).

### Step-by-Step Solution:

1. **Identify the Force and its Components:**
   - The force \( F \) is now \( 1600 \, \text{N} \).
   - The force \( F \) is acting at a \( 45^\circ \) angle to the horizontal.

2. **Break Down the Force into Components:**
   - The force \( F \) has two components:
     - \( F_x = F \cos(45^\circ) = 1600 \times \cos(45^\circ) = 1600 \times \frac{\sqrt{2}}{2} = 1131.37 \, \text{N} \)
     - \( F_y = F \sin(45^\circ) = 1600 \times \sin(45^\circ) = 1600 \times \frac{\sqrt{2}}{2} = 1131.37 \, \text{N} \)

3. **Determine the Distances:**
   - Horizontal distance from point \( O \) to where the force is applied is \( 3 \, \text{m} + 1 \, \text{m} = 4 \, \text{m} \).
   - Vertical distance from point \( O \) to where the force is applied is \( 1 \, \text{m} \).

4. **Calculate the Moment Contributions:**
   - Moment due to \( F_x \) (horizontal component of force):
     \[
     M_{F_x} = F_x \times 1 \, \text{m} = 1131.37 \, \text{N} \times 1 \, \text{m} = 1131.37 \, \text{Nm}
     \]
   - Moment due to \( F_y \) (vertical component of force):
     \[
     M_{F_y} = F_y \times 4 \, \text{m} = 1131.37 \, \text{N} \times 4 \, \text{m} = 4525.49 \, \text{Nm}
     \]

5. **Direction of the Moments:**
   - The moment due to \( F_x \) causes a clockwise rotation about \( O \), and the moment due to \( F_y \) causes a counterclockwise rotation about \( O \).

6. **Calculate the Resultant Moment:**
   - The resultant moment \( M_O \) about point \( O \) is the algebraic sum of the moments:
     \[
     M_O = M_{F_y} - M_{F_x} = 4525.49 \, \text{Nm} - 1131.37 \, \text{Nm} = 3394.12 \, \text{Nm}
     \]

### Final Answer:
The magnitude of the resultant moment about point \( O \) is \( 3394.12 \, \text{Nm} \).

Would you like more details on any of these steps or have any additional questions?

### Related Questions:
1. What impact does increasing the force have on the magnitude of the moment?
2. How does the angle of the force influence the moment about a point?
3. What is the effect of changing the length of the moment arm on the resulting moment?
4. How would the problem change if the force were acting vertically downward instead of at an angle?
5. Can you explain how the sign of the moment relates to the direction of rotation?
6. How would the moment change if the force acted closer to point \( O \)?
7. What is the relationship between the force components and the distances when calculating moments?
8. How do you determine which moment direction (clockwise or counterclockwise) to use in calculations?

**Tip:** Always double-check the force value and ensure it is correctly used in both components to avoid calculation errors.

Learn how to calculate the moment of a force of 1600 N applied at a 45-degree angle correctly. Step-by-step solution breaking down the force into horizontal and vertical components and determining the resultant moment. Suitable for advanced high school students studying mechanics and vectors.

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