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This problem requires finding the reaction forces at points \(A\) and \(B\) using the principles of static equilibrium. Let's solve it step by step.

### Given:
- Force \( X = 42 \, \text{lb} \) (acting horizontally to the left)
- Force \( Y = 50 \, \text{lb} \) (acting vertically downward)
- Distance from force application to point \( A \) (horizontal) = 4 inches = 0.33 feet
- Total length between points \( A \) and \( B \) = 20 feet

### Assumptions:
- The beam is in static equilibrium, so the sum of forces in both the horizontal and vertical directions, and the sum of moments, must all be zero.

### Step 1: Sum of Moments about Point \( A \)

Take the sum of moments about point \( A \), which eliminates the reaction at \( A \). Moments are calculated using the perpendicular distance from the point to the line of action of the force.

- The horizontal force \( X \) creates a counterclockwise moment:
  \[
  M_X = X \times 10 \, \text{ft} = 42 \, \text{lb} \times 10 \, \text{ft} = 420 \, \text{lb-ft}
  \]

- The vertical force \( Y \) creates a clockwise moment:
  \[
  M_Y = Y \times 4/12 \, \text{ft} = 50 \, \text{lb} \times 0.33 \, \text{ft} = 16.67 \, \text{lb-ft}
  \]

- The reaction force at \( B \), \( B_y \), also creates a clockwise moment about point \( A \):
  \[
  M_B = B_y \times 20 \, \text{ft}
  \]

For equilibrium, the sum of moments about point \( A \) is zero:
\[
M_X - M_Y - M_B = 0
\]
Substitute the values:
\[
420 \, \text{lb-ft} - 16.67 \, \text{lb-ft} - B_y \times 20 \, \text{ft} = 0
\]
Solving for \( B_y \):
\[
B_y = \frac{420 - 16.67}{20} = \frac{403.33}{20} = 20.17 \, \text{lb}
\]

### Step 2: Sum of Vertical Forces

For vertical forces, the sum of the forces in the vertical direction must be zero:
\[
A_y + B_y - Y = 0
\]
Substitute the known value of \( B_y \):
\[
A_y + 20.17 \, \text{lb} - 50 \, \text{lb} = 0
\]
Solving for \( A_y \):
\[
A_y = 50 - 20.17 = 29.83 \, \text{lb}
\]

### Step 3: Sum of Horizontal Forces

The sum of horizontal forces must also be zero:
\[
A_x - X = 0
\]
Thus, \( A_x = X = 42 \, \text{lb} \).

### Final Reactions:

- Reaction at \( B \) (vertical component) is \( B_y = 20.17 \, \text{lb} \).
- Reaction at \( A \):
  - Vertical component: \( A_y = 29.83 \, \text{lb} \)
  - Horizontal component: \( A_x = 42 \, \text{lb} \)

Would you like more detailed explanations on any part, or do you have other questions? Here are some related questions:

1. What are the conditions for static equilibrium?
2. How do you calculate the moment of a force?
3. What is the importance of reaction forces in structural analysis?
4. How would the reaction forces change if the force \( Y \) were applied closer to point \( A \)?
5. How can this analysis be extended to a three-dimensional structure?

**Tip**: Always check your unit consistency, especially when dealing with inches and feet, as unit errors can lead to incorrect solutions.

Determine the reactions at A and B. Given: X = 42 lb and Y = 50 lb.

Learn how to determine the reactions at points A and B for a beam subjected to horizontal and vertical forces (X = 42 lb, Y = 50 lb). The solution includes calculating moments, vertical and horizontal reactions, and the application of static equilibrium conditions. Suitable for engineering students.

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