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FROM THE FIGURE SHOWN, SOLVE FOR THE REACTIONS. THEN CHECK THE EQUILIBRIUM

To solve for the reactions in the given beam, we need to use the principles of static equilibrium. The equilibrium conditions are:

1. The sum of vertical forces must be zero: \(\sum F_y = 0\)
2. The sum of horizontal forces must be zero: \(\sum F_x = 0\)
3. The sum of moments about any point must be zero: \(\sum M = 0\)

### Step 1: Identify Loads and Reactions

- **Distributed Load (3 KN/m)** over 3 meters (to the left of point A): 
  - Total load = \(3 \, \text{KN/m} \times 3 \, \text{m} = 9 \, \text{KN}\), acting at its centroid, 1.5 m from point A.

- **Point Load (6 KN)** acting 2 m from point A.

- **Point Load (10 KN)** acting 4 m from point A.

- **Point Load (10 KN)** acting 6 m from point A.

- **Triangular Load (5 KN/m)** over 3 m (to the right of point B):
  - Total load of the triangle = \(\frac{1}{2} \times 5 \, \text{KN/m} \times 3 \, \text{m} = 7.5 \, \text{KN}\), acting at \(1/3\) the base of the triangle from point B (1 m from point B).

### Step 2: Assume Reactions at Supports

- Let \( R_A \) and \( R_B \) be the vertical reactions at supports A and B, respectively. Assume there is no horizontal force, so \( R_{Ax} = 0 \).

### Step 3: Apply Equilibrium Equations

#### 1. Sum of vertical forces \(\sum F_y = 0\):

\[
R_A + R_B - 9 - 6 - 10 - 10 - 7.5 = 0
\]

\[
R_A + R_B = 42.5 \, \text{KN}
\]

#### 2. Sum of moments about point A \(\sum M_A = 0\):

Taking moments about point A to eliminate \( R_A \):

\[
(9 \, \text{KN}) \times (1.5 \, \text{m}) + (6 \, \text{KN}) \times (2 \, \text{m}) + (10 \, \text{KN}) \times (4 \, \text{m}) + (10 \, \text{KN}) \times (6 \, \text{m}) + (7.5 \, \text{KN}) \times (8 + 1) \, \text{m} = R_B \times 8 \, \text{m}
\]

Simplifying:

\[
(9 \times 1.5) + (6 \times 2) + (10 \times 4) + (10 \times 6) + (7.5 \times 9) = R_B \times 8
\]

\[
13.5 + 12 + 40 + 60 + 67.5 = 8 R_B
\]

\[
193 = 8 R_B
\]

\[
R_B = 24.125 \, \text{KN}
\]

#### 3. Calculate \(R_A\):

\[
R_A + 24.125 = 42.5
\]

\[
R_A = 42.5 - 24.125 = 18.375 \, \text{KN}
\]

### Step 4: Check Equilibrium

To verify, check the sum of moments about point B:

\[
R_A \times 8 = (9 \times 6.5) + (6 \times 6) + (10 \times 4) + (10 \times 2) + (7.5 \times 1)
\]

\[
18.375 \times 8 = (58.5) + (36) + (40) + (20) + (7.5)
\]

\[
147 = 147
\]

The moments are balanced, so the system is in equilibrium.

### Final Answer:
- \(R_A = 18.375 \, \text{KN}\)
- \(R_B = 24.125 \, \text{KN}\)

Would you like further details or have any questions?

Here are 5 related questions:
1. How would the reactions change if the point loads increased?
2. What if the triangular load's height increased, but the base remained the same?
3. How do we find the centroid of distributed loads in beams?
4. How would adding a hinge at point B affect the reactions?
5. Can you explain how triangular and trapezoidal loads differ in terms of resultant force?

**Tip**: Always start solving equilibrium problems by summing moments about a point where you can eliminate one reaction. This simplifies your calculations.

Solve for the reactions in a beam under distributed and point loads using static equilibrium principles. This detailed solution includes moment calculations and force summations to verify equilibrium. Suitable for engineering students.

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