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Using the method of sections, find the forces in members CD, FD, and EF. Let P1 = 5 kN, P2 = 14 kN and P3 = 12 kN.

To find the forces in members \( CD \), \( FD \), and \( EF \) of the truss using the method of sections, we will go through the following steps:

### Step 1: Support Reactions
First, we need to determine the support reactions at points \( A \) and \( E \).

1. **Sum of Moments about Point A:**
   Since \( E \) is a roller support, it only has a vertical reaction \( E_y \). Point \( A \) is a pin support, so it has both horizontal and vertical reactions, but only the vertical component \( A_y \) is needed here as no horizontal forces are applied.
   
   \[
   \sum M_A = 0: \quad -P_1 \times 1.5 \text{ m} - P_2 \times 3 \text{ m} - P_3 \times 4.5 \text{ m} + E_y \times 6 \text{ m} = 0
   \]
   Substituting \( P_1 = 5 \text{ kN} \), \( P_2 = 14 \text{ kN} \), and \( P_3 = 12 \text{ kN} \):
   
   \[
   -(5)(1.5) - (14)(3) - (12)(4.5) + E_y \times 6 = 0
   \]
   \[
   -7.5 - 42 - 54 + E_y \times 6 = 0
   \]
   \[
   E_y = \frac{103.5}{6} = 17.25 \text{ kN}
   \]

2. **Sum of Vertical Forces:**
   Now, we can find \( A_y \) by summing the vertical forces.
   
   \[
   \sum F_y = 0: \quad A_y + E_y - P_1 - P_2 - P_3 = 0
   \]
   \[
   A_y + 17.25 - 5 - 14 - 12 = 0
   \]
   \[
   A_y = 13.75 \text{ kN}
   \]

### Step 2: Method of Sections
Now, let's cut through the truss to isolate members \( CD \), \( FD \), and \( EF \). The section will pass through points \( C \), \( D \), and \( F \). We can analyze the right side of the truss after making the cut.

#### Sum of Moments about Point F (for Member \( CD \)):
This helps to eliminate the forces in members \( FD \) and \( EF \), leaving us with only the force in \( CD \).

1. Let \( F_{CD} \) be the force in member \( CD \).
2. Assume the direction of \( F_{CD} \) as tensile (pulling away from the joint).
   
\[
\sum M_F = 0: \quad -P_3 \times 1.5 + F_{CD} \times 2 = 0
\]
\[
-12 \times 1.5 + F_{CD} \times 2 = 0
\]
\[
F_{CD} = \frac{18}{2} = 9 \text{ kN (tensile)}
\]

#### Sum of Moments about Point C (for Member \( EF \)):
This helps to eliminate the forces in members \( CD \) and \( FD \), leaving us with only the force in \( EF \).

1. Let \( F_{EF} \) be the force in member \( EF \).
   
\[
\sum M_C = 0: \quad -P_2 \times 1.5 + F_{EF} \times 2 = 0
\]
\[
-14 \times 1.5 + F_{EF} \times 2 = 0
\]
\[
F_{EF} = \frac{21}{2} = 10.5 \text{ kN (tensile)}
\]

#### Sum of Vertical Forces (for Member \( FD \)):
We can find \( F_{FD} \) by summing the vertical forces on the right section.
   
\[
\sum F_y = 0: \quad F_{FD} - E_y + P_3 = 0
\]
\[
F_{FD} = E_y - P_3
\]
\[
F_{FD} = 17.25 - 12 = 5.25 \text{ kN (tensile)}
\]

### Final Answers
- \( F_{CD} = 9 \text{ kN (tensile)} \)
- \( F_{FD} = 5.25 \text{ kN (tensile)} \)
- \( F_{EF} = 10.5 \text{ kN (tensile)} \)

Would you like a detailed step-by-step explanation of the process, or is there any specific part you'd like to discuss further?

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**Relative Questions:**
1. How would the forces change if \( P_1 \), \( P_2 \), and \( P_3 \) were different values?
2. What would happen to the truss if member \( CD \) were removed?
3. How can you verify these force calculations using a different method?
4. What impact does the height of the truss (2 m) have on the forces in members?
5. How do the reactions at the supports contribute to the overall equilibrium of the truss?

**Tip:** Always double-check the directions (tensile or compressive) assumed for member forces; correcting assumptions can be crucial for the final answer.

Using the method of sections, find the forces in members CD, FD, and EF. Let P1 = 5 kN, P2 = 14 kN and P3 = 12 kN.

Learn how to calculate the forces in truss members CD, FD, and EF using the method of sections. This structural analysis problem involves forces P1 = 5 kN, P2 = 14 kN, and P3 = 12 kN, along with equilibrium conditions. Suitable for students in Grades 10-12.

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