The equation you provided looks like a moment equilibrium equation, commonly used in statics to sum the moments about a point to zero. The equation:

\[
R \cdot 3m - 40kN \cdot 27m - 69.3kN \cdot 15m = 0
\]

Let's break it down:

- \( R \cdot 3m \): A reaction force \( R \) acting at a distance of 3 meters from the point where moments are summed.
- \( -40kN \cdot 27m \): A force of 40 kN acting at a distance of 27 meters from the point where moments are summed, creating a negative moment (clockwise direction).
- \( -69.3kN \cdot 15m \): A force of 69.3 kN acting at a distance of 15 meters from the point where moments are summed, also creating a negative moment.

The goal is to solve for the reaction force \( R \), so we rearrange the equation:

\[
R \cdot 3m = 40kN \cdot 27m + 69.3kN \cdot 15m
\]

Now let's calculate the right-hand side:

\[
40kN \cdot 27m = 1080 kN \cdot m
\]
\[
69.3kN \cdot 15m = 1039.5 kN \cdot m
\]

Summing them together:

\[
R \cdot 3m = 1080 + 1039.5 = 2119.5 \, kN \cdot m
\]

Now, divide both sides by 3 to solve for \( R \):

\[
R = \frac{2119.5 \, kN \cdot m}{3m} = 706.5 \, kN
\]

So, the reaction force \( R \) is:

\[
R = 706.5 \, kN
\]

Would you like further details on this solution?

Here are five related questions:
1. What is the principle behind moment equilibrium in statics?
2. How do you determine the direction of moments (clockwise vs. counterclockwise)?
3. How would the reaction force change if the distances were different?
4. Can this method be applied to non-uniformly distributed loads?
5. How can we consider moments in three-dimensional problems?

**Tip:** Always double-check the units when working with forces and distances to ensure consistency in moment calculations.

Explore how to solve the moment equilibrium equation R·3m - 40kN·27m - 69.3kN·15m = 0. This problem involves calculating the reaction force R using statics principles. Suitable for Grades 10-12, this solution includes detailed steps on balancing moments in a system.

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