No you got the distance wrong the distance for both 10KN are 5m each

Thank you for the clarification! It seems you're referring to a problem involving forces, likely using the moment of forces (torque) or equilibrium analysis.

To proceed:

- Two forces of **10 kN** each are applied, and the distance of each force from the point of interest (pivot) is **5 meters**.

If you're working on a specific problem (e.g., torque, balancing forces, or something else), can you share more details or confirm the scenario? Here's how I interpret the situation:

### Example Scenario:
If we are calculating torque (moment of force), we use:

\[
\text{Torque} = \text{Force} \times \text{Distance}
\]

For each force of 10 kN applied 5 meters from the pivot:
\[
\text{Torque} = 10 \, \text{kN} \times 5 \, \text{m} = 50 \, \text{kN}\cdot\text{m}
\]

So each force would generate a torque of 50 kN·m about the pivot point.

If both forces are applied in the same direction (clockwise or counterclockwise), you sum the torques. If opposite, they might cancel out or reduce the total torque.

Feel free to upload an image or provide more specifics if you'd like me to walk through a more detailed solution. 

### Would you like more details or have any questions?
#### Here are 5 related questions for further understanding:
1. Are the forces acting in the same or opposite directions?
2. Is the problem about rotational equilibrium?
3. Are there any other forces or moments acting on the system?
4. What is the pivot point in the system (if any)?
5. Does the problem involve static or dynamic analysis?

**Tip**: When calculating torque, always ensure you're measuring the perpendicular distance from the line of action of the force to the pivot point.

Explore how to calculate torque generated by two 10 kN forces applied at 5 meters each from a pivot point. This problem involves understanding key concepts in physics such as torque, moment of forces, and equilibrium. Suitable for high school students in Grades 10-12.

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