### Calculating Support Reactions for a Loaded BeamIn this example, we will calculate the support reactions at points \(A\) and \(B\) for the loaded beam.#### Given Data:- Force \(P = 143 \text{ lb}\)- Distributed Load \(w = 92 \text{ lb/ft}\)- Distance \(a = 3.2 \text{ ft}\)- Distance \(b = 7.0 \text{ ft}\)#### Diagram Explanation:The diagram provided illustrates a beam supported at two points, \(A\) (fixed support) and \(B\) (roller support). The following elements are depicted:- **Support at A:** This support can provide reactions in both the horizontal (\(A_x\)) and vertical (\(A_y\)) directions.- **Support at B:** This is a roller support that provides a vertical reaction (\(B_y\)).- **Point Load \(P\)** applied vertically downward at a distance \(a\) from support \(A\).- **Distributed Load \(w\):** The load varies linearly across the span from point \(A\) to point \(B\).- **Coordinates:** \(x\) represents the horizontal axis, while \(y\) represents the vertical axis.#### Calculation Objective:We need to compute the reactions at the supports \(A_x\), \(A_y\), and \(B_y\).#### Calculation Steps:1. **Sum of Horizontal Forces:** \[ \Sigma F_x = 0 \] \[ A_x = 0 \] 2. **Sum of Vertical Forces:** \[ \Sigma F_y = 0 \] \[ A_y + B_y - P - w \times (a + b) = 0 \] 3. **Moment Around Point A:** \[ \Sigma M_A = 0 \] \[ -P \times a - \left( \frac{w \times (a + b) \times (a + b/2)}{2} \right) + B_y \times (a + b) = 0 \]By solving these equations, the values for \(A_y\) and \(B_y\) can be determined. Insert these computed values:#### Answers:- \( A_x = \)- \( A_y = \)-

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Answered: Calculate the support reactions at A…

Calculate the support reactions at A and B for the loaded beam. Assume P = 143 lb, w = 92 lb/ft, a = 3.2 ft, and b = 7.0 ft. L A _a. P b W B

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Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freedom under the influence of that force. Here, three degrees of freedom are from the translational motion while the other three are from the rotation motion. However, sometimes the bodies are not allowed to undergo any motion and are fixed, under such conditions the body is said to be constrained.

Question

### Calculating Support Reactions for a Loaded Beam

In this example, we will calculate the support reactions at points \(A\) and \(B\) for the loaded beam.

#### Given Data:
- Force \(P = 143 \text{ lb}\)
- Distributed Load \(w = 92 \text{ lb/ft}\)
- Distance \(a = 3.2 \text{ ft}\)
- Distance \(b = 7.0 \text{ ft}\)

#### Diagram Explanation:
The diagram provided illustrates a beam supported at two points, \(A\) (fixed support) and \(B\) (roller support). The following elements are depicted:

- **Support at A:** This support can provide reactions in both the horizontal (\(A_x\)) and vertical (\(A_y\)) directions.
- **Support at B:** This is a roller support that provides a vertical reaction (\(B_y\)).
- **Point Load \(P\)** applied vertically downward at a distance \(a\) from support \(A\).
- **Distributed Load \(w\):** The load varies linearly across the span from point \(A\) to point \(B\).
- **Coordinates:** \(x\) represents the horizontal axis, while \(y\) represents the vertical axis.

#### Calculation Objective:
We need to compute the reactions at the supports \(A_x\), \(A_y\), and \(B_y\).

#### Calculation Steps:
1. **Sum of Horizontal Forces:**
\[ \Sigma F_x = 0 \]
\[ A_x = 0 \]

2. **Sum of Vertical Forces:**
\[ \Sigma F_y = 0 \]
\[ A_y + B_y - P - w \times (a + b) = 0 \]

3. **Moment Around Point A:**
\[ \Sigma M_A = 0 \]
\[ -P \times a - \left( \frac{w \times (a + b) \times (a + b/2)}{2} \right) + B_y \times (a + b) = 0 \]

By solving these equations, the values for \(A_y\) and \(B_y\) can be determined. Insert these computed values:

#### Answers:
- \( A_x = \)
- \( A_y = \)
-

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