To determine the vertical reaction at support A, we need to analyze the distributed loads acting on the beam and solve for equilibrium. Let me guide you step by step.

Step 1: Represent the distributed loads as equivalent point loads

1. 20 lb/ft load over 4 ft:

   • Total load: $20 \, \text{lb/ft} \times 4 \, \text{ft} = 80 \, \text{lb}$.
   • Acts at the center of the 4-ft span, i.e., 2 ft from A.

2. 35 lb/ft load over 8 ft:

   • Total load: $35 \, \text{lb/ft} \times 8 \, \text{ft} = 280 \, \text{lb}$.
   • Acts at the center of the 8-ft span, i.e., 4 ft from the end of the distributed load (12 ft from A).

Step 2: Apply equilibrium equations

• Sum of Vertical Forces ($\Sigma F_y = 0$): R_A + R_B - 80 - 280 = 0 \tag{1}

• Sum of Moments about A ($\Sigma M_A = 0$): Taking clockwise moments as positive: -80 \times 2 - 280 \times 10 + R_B \times 12 = 0 \tag{2}

Step 3: Solve for $R_B$ and $R_A$

1. Solve Equation (2) for $R_B$: $-160 - 2800 + 12R_B = 0$ $12R_B = 2960$ $R_B = \frac{2960}{12} = 246.67 \, \text{lb}$

2. Substitute $R_B$ into Equation (1) to find $R_A$: $R_A + 246.67 - 80 - 280 = 0$ $R_A = 113.33 \, \text{lb}$

Final Answer:

The vertical reaction at support A is: $\boxed{R_A = 113.33 \, \text{lb}}$

Let me know if you'd like a diagram of the calculations or further explanation!

Follow-up Questions:

1. How is the centroid of a distributed load determined?
2. What assumptions are made for reactions at supports?
3. Can this method be used for beams with point loads and moments?
4. What happens to reactions if one of the supports is removed?
5. How does changing the distributed load affect $R_A$ and $R_B$?

Tip: Always double-check your moment calculations to avoid small placement errors!