## Step-by-Step Solution Let's analyze the shaft subjected to the given forces. We'll determine the **reactions at supports A and B**. ### 1. **Draw the Free Body Diagram** Let's denote: - \( A_y \): Vertical reaction at A (upward) - \( B_y \): Vertical reaction at B (upward) - Two downward forces: 500 lb and 100 lb #### Position of Forces: - 500 lb force: 3 in from A - 100 lb force: 3 in from B (or 13 in from A, since total length = 3 + 4 + 4 + 2 = 13 in) ### 2. **Summing Vertical Forces** \[ \sum F_y = \] \[ A_y + B_y - 500 - 100 = \] \[ A_y + B_y = 150 \quad \text{(Eq 1)} \] ### 3. **Summing Moments About A** Take moments about point A (counterclockwise positive): \[ \sum M_A = \] \[ -500 \times 3 - 100 \times 11 + B_y \times 13 = \] - 500 lb force acts 3 in from A - 100 lb force acts at (3 + 4 + 4 = 11 in) from A - B_y acts at the far right, which is 13 in from A \[ -150 - 11000 + 13B_y = \] \[ 13B_y = 12500 \] \[ B_y = \frac{12500}{13} \approx 961.54 \text{ lb} \] ### 4. **Find \(A_y\)** From Eq 1: \[ A_y = 150 - B_y = 150 - 961.54 = 538.46 \text{ lb} \] --- ## **Final Answers** ### **a) Reaction at A** \[ A_y = 538.46 \text{ lb (upward)} \] ### **b) Reaction at B** \[ B_y = 961.54 \text{ lb (upward)} \] --- ## **Summary Table** | Reaction | Value (lb) | Direction | |----------|------------|-----------| | \(A_y\) | 538.46 | Upward | | \(B_y\) | 961.54 | Upward | --- **Let me know if you need the bending moment diagram or other calculations!**