To calculate the reaction forces in the beam, we need to apply static equilibrium principles. From the diagram and given information, the beam is supported at two points: Point A (a pinned support) and Point B (a roller). There are two external forces acting on the beam: 20N and 30N, spaced 1 meter and 2 meters apart from each other.

For a beam to be in equilibrium, the following conditions must hold:

1. The sum of all vertical forces must be zero: $\sum F_y = 0$.
2. The sum of all moments about any point must be zero: $\sum M = 0$.

Let’s define the forces at the supports as:

• $R_A$: Reaction at point A (vertical force).
• $R_B$: Reaction at point B (vertical force).

Step 1: Moment about point A

Taking moments about point A (to eliminate $R_A$): $\sum M_A = 0 \implies R_B \times 3\,\text{m} - 30\,\text{N} \times 2\,\text{m} - 20\,\text{N} \times 1\,\text{m} = 0$ $3R_B = 60 + 20 = 80$ $R_B = \frac{80}{3} \approx 26.67\,\text{N}$

Step 2: Vertical equilibrium

Now, using the condition that the sum of vertical forces must be zero: $R_A + R_B = 20\,\text{N} + 30\,\text{N} = 50\,\text{N}$ $R_A = 50 - 26.67 = 23.33\,\text{N}$

Final Answer:

• Reaction at Point A, $R_A = 23.33\,\text{N}$
• Reaction at Point B, $R_B = 26.67\,\text{N}$

Would you like more details or have any questions on these calculations?

Here are 5 related questions to expand the topic:

1. How would the reaction forces change if the load positions were altered?
2. What would be the effect if point A were a roller and point B were pinned?
3. How does the presence of a moment affect the reaction forces in a beam?
4. What if an additional horizontal force was applied to the beam at point A—how would you calculate the reactions?
5. How do we apply equilibrium equations to a beam with more than two supports?

Tip: When solving for reactions in beams, always start with the moment equation to eliminate one unknown, making the process simpler.