The problem shown in the image asks to find the reactions at supports $a$ and $b$. It shows a vertical force $F = 36 \text{ kN}$ acting downward, with given distances from the supports.

Let's break down the problem step by step:

Step 1: Identify the known values

• Force $F = 36 \text{ kN}$ acting downward.
• Distance from force to point $a$ is $3 \text{ m}$.
• Distance between supports $a$ and $b$ is $2 \text{ m}$.

Step 2: Write the equations of equilibrium

Since the beam is in static equilibrium, the sum of forces and the sum of moments about any point must be zero.

1. Sum of vertical forces: $R_a + R_b - F = 0$ where $R_a$ and $R_b$ are the reactions at supports $a$ and $b$, respectively.

2. Sum of moments about point $a$: Taking moments about point $a$: $R_b \times 2 \text{ m} - F \times 3 \text{ m} = 0$ Substituting $F = 36 \text{ kN}$: $2R_b = 108 \text{ kN} \cdot \text{m}$ $R_b = 54 \text{ kN}$

Step 3: Calculate the reaction at $a$

Substitute $R_b = 54 \text{ kN}$ back into the sum of vertical forces equation: $R_a + 54 \text{ kN} - 36 \text{ kN} = 0$ $R_a = 36 \text{ kN} - 54 \text{ kN}$ $R_a = -18 \text{ kN}$

Conclusion

• $R_a = -18 \text{ kN}$ (negative indicates the direction is opposite to assumed)
• $R_b = 54 \text{ kN}$

These are the reactions at the supports $a$ and $b$.

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

Here are five related questions to consider:

1. How would the solution change if the distances were different?
2. What if the force $F$ acted at an angle instead of vertically?
3. How do we handle multiple forces acting on the beam?
4. What happens if the beam is not simply supported but fixed at one end?
5. How does the length of the beam between supports affect the reactions?

Tip: Always check the sign of your reaction forces, as a negative sign can indicate the direction of the force is opposite to what was initially assumed.