its a 2150 N force at the bottom end, 0.3m from point O
At point O 100 N force at 45 degrees 
the moment is not given


Thanks for the clarification. We have:
1. A 2150 N force at the bottom end, 0.3 m from point \( O \).
2. A 100 N force at point \( O \) acting at 45 degrees.
3. The moment due to a 250 N force acting at 0.6 m from point \( O \).

### Calculating Moments

1. **Moment due to the 2150 N Force:**

The 2150 N force acts horizontally at a distance of 0.3 m from point \( O \):
\[ M_1 = 2150 \text{ N} \times 0.3 \text{ m} = 645 \text{ Nm} \quad \text{(clockwise)} \]

2. **Moment due to the 100 N Force:**

The 100 N force acts at 45 degrees. To find the moment, we need to resolve the force into its horizontal and vertical components:
\[ F_{x} = 100 \cos(45^\circ) \text{ N} = 100 \times \frac{\sqrt{2}}{2} \text{ N} = 70.71 \text{ N} \]
\[ F_{y} = 100 \sin(45^\circ) \text{ N} = 100 \times \frac{\sqrt{2}}{2} \text{ N} = 70.71 \text{ N} \]

Since this force acts at point \( O \), its moment arm is zero, and thus it does not contribute to the moment about \( O \):
\[ M_2 = 0 \text{ Nm} \]

3. **Moment due to the 250 N Force:**

The 250 N force acts vertically at a distance of 0.6 m from point \( O \):
\[ M_3 = 250 \text{ N} \times 0.6 \text{ m} = 150 \text{ Nm} \quad \text{(clockwise)} \]

### Summing the Moments

Sum the moments considering their directions (clockwise moments are positive, and counterclockwise moments are negative):
\[ M_{\text{total}} = 645 \text{ Nm} + 0 \text{ Nm} + 150 \text{ Nm} = 795 \text{ Nm} \]

Thus, the magnitude of the resultant moment about point \( O \) is:
\[ \boxed{795 \text{ Nm}} \]

Do you want details or have any questions? Here are 8 related questions for further exploration:

1. What would be the resultant moment if the 2150 N force was moved closer to point \( O \)?
2. How would the calculation change if the 100 N force was applied 0.5 m away from \( O \)?
3. What is the significance of the direction of forces in calculating moments?
4. How would you calculate the moment if the forces were acting at different angles other than perpendicular?
5. What happens to the moment if the 250 N force was removed?
6. How does the principle of moments apply in real-world engineering structures?
7. Can you derive the perpendicular distances for forces acting at different angles?
8. What tools or methods can be used to visualize and calculate moments in complex structures?

**Tip:** Always ensure to resolve forces into their perpendicular components when calculating moments, especially if they are not acting along standard axes.

Learn how to calculate moments and forces involving a 2150 N force and a 100 N force at point O. Understand the concept of moments, including their direction and calculation methods in statics and engineering.

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