Follow up question use reduced row ehelon form to obtain the equivalent system that solve the problem

To solve this problem using the Reduced Row Echelon Form (RREF) method, we first need to set up the system of linear equations representing the equilibrium of each joint in the truss. We'll use the support reactions and loads calculated in the previous part (c) where $P1 = 9 \text{ kN}$ and $P2 = 15 \text{ kN}$.

Support Reactions:
$A_x = 9 \text{ kN}$ (right)
$A_y = 1.5 \text{ kN}$ (up)
$C_y = 13.5 \text{ kN}$ (up)

Member Angles:
For members FB and CE, which are diagonal:
$\cos \theta = \frac{3}{5} = 0.6$
$\sin \theta = \frac{4}{5} = 0.8$

We will assume all member forces are in tension (pulling away from the joint). A negative result will indicate compression. Let the unknown forces be $F_{AB}, F_{AF}, F_{BC}, F_{CD}, F_{DE}, F_{EF}, F_{FB}, F_{BE}, F_{CE}$.

System of Equations (Equilibrium at each joint):

Joint A (0,0):

1. $\sum F_x = 0: A_x + F_{AB} = 0 \implies 9 + F_{AB} = 0$
2. $\sum F_y = 0: A_y + F_{AF} = 0 \implies 1.5 + F_{AF} = 0$

Joint B (3,0): 3. $\sum F_x = 0: -F_{AB} + F_{BC} + F_{FB} \cos \theta = 0 \implies -F_{AB} + F_{BC} + 0.6 F_{FB} = 0$ 4. $\sum F_y = 0: -P2 + F_{BE} + F_{FB} \sin \theta = 0 \implies -15 + F_{BE} + 0.8 F_{FB} = 0$

Joint C (6,0): 5. $\sum F_x = 0: -F_{BC} + F_{CE} \cos \theta = 0 \implies -F_{BC} + 0.6 F_{CE} = 0$ 6. $\sum F_y = 0: C_y + F_{CD} + F_{CE} \sin \theta = 0 \implies 13.5 + F_{CD} + 0.8 F_{CE} = 0$

Joint D (6,4): 7. $\sum F_x = 0: -F_{DE} = 0$ (Member CD is vertical, so no x-component) 8. $\sum F_y = 0: -F_{CD} = 0$ (Member DE is horizontal, so no y-component)

Joint E (3,4): 9. $\sum F_x = 0: -F_{EF} + F_{DE} - F_{CE} \cos \theta = 0 \implies -F_{EF} + F_{DE} - 0.6 F_{CE} = 0$ 10. $\sum F_y = 0: -F_{BE} - F_{CE} \sin \theta = 0 \implies -F_{BE} - 0.8 F_{CE} = 0$

Joint F (0,4): 11. $\sum F_x = 0: -P1 + F_{EF} - F_{FB} \cos \theta = 0 \implies -9 + F_{EF} - 0.6 F_{FB} = 0$ 12. $\sum F_y = 0: -F_{AF} - F_{FB} \sin \theta = 0 \implies -F_{AF} - 0.8 F_{FB} = 0$

Let's write these equations in matrix form $[A | b]$ where the variables are ordered as $F_{AB}, F_{AF}, F_{BC}, F_{CD}, F_{DE}, F_{EF}, F_{FB}, F_{BE}, F_{CE}$.

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & -9 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & -1.5 \\ -1 & 0 & 1 & 0 & 0 & 0 & 0.6 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0.8 & 1 & 0 & | & 15 \\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0.6 & | & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0.8 & | & -13.5 \\ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & -0.6 & | & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -0.8 & | & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & -0.6 & 0 & 0 & | & 9 \\ 0 & -1 & 0 & 0 & 0 & 0 & -0.8 & 0 & 0 & | & 0 \end{bmatrix}$$

Step 1: Perform row operations to get the matrix into Reduced Row Echelon Form.
The first two rows already give $F_{AB}$ and $F_{AF}$.
$R_1 \to F_{AB} = -9$
$R_2 \to F_{AF} = -1.5$

From $R_7$: $F_{DE} = 0$
From $R_8$: $F_{CD} = 0$

Substitute $F_{CD}=0$ into $R_6$:
$F_{CD} + 0.8 F_{CE} = -13.5 \implies 0 + 0.8 F_{CE} = -13.5 \implies F_{CE} = \frac{-13.5}{0.8} = -16.875$

Substitute $F_{CE}=-16.875$ into $R_5$:
$-F_{BC} + 0.6 F_{CE} = 0 \implies -F_{BC} + 0.6(-16.875) = 0 \implies -F_{BC} - 10.125 = 0 \implies F_{BC} = -10.125$

Substitute $F_{AF}=-1.5$ into $R_{12}$:
$-F_{AF} - 0.8 F_{FB} = 0 \implies -(-1.5) - 0.8 F_{FB} = 0 \implies 1.5 - 0.8 F_{FB} = 0 \implies F_{FB} = \frac{1.5}{0.8} = 1.875$

Substitute $F_{FB}=1.875$ into $R_4$:
$0.8 F_{FB} + F_{BE} = 15 \implies 0.8(1.875) + F_{BE} = 15 \implies 1.5 + F_{BE} = 15 \implies F_{BE} = 13.5$

Substitute $F_{CE}=-16.875$ into $R_{10}$:
$-F_{BE} - 0.8 F_{CE} = 0 \implies -F_{BE} - 0.8(-16.875) = 0 \implies -F_{BE} + 13.5 = 0 \implies F_{BE} = 13.5$ (Consistent)

Substitute $F_{DE}=0$ and $F_{CE}=-16.875$ into $R_9$:
$-F_{EF} + F_{DE} - 0.6 F_{CE} = 0 \implies -F_{EF} + 0 - 0.6(-16.875) = 0 \implies -F_{EF} + 10.125 = 0 \implies F_{EF} = 10.125$

Substitute $F_{FB}=1.875$ and $F_{EF}=10.125$ into $R_{11}$:
$-9 + F_{EF} - 0.6 F_{FB} = 0 \implies -9 + 10.125 - 0.6(1.875) = 0 \implies 1.125 - 1.125 = 0$ (Consistent)

Substitute $F_{AB}=-9$, $F_{BC}=-10.125$, $F_{FB}=1.875$ into $R_3$:
$-F_{AB} + F_{BC} + 0.6 F_{FB} = 0 \implies -(-9) + (-10.125) + 0.6(1.875) = 0 \implies 9 - 10.125 + 1.125 = 0$ (Consistent)

The RREF of the augmented matrix would directly yield these values. Since many variables are directly solved or easily substituted, the RREF process would essentially perform these substitutions.

The equivalent system in RREF form (or the solution obtained by RREF) is:

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & -9 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & -1.5 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & | & -10.125 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & | & 10.125 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & | & 1.875 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & | & 13.5 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & | & -16.875 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}$$

The solution for each member force is:

• $F_{AB} = \boxed{-9 \text{ kN (Compression)}}$
• $F_{AF} = \boxed{-1.5 \text{ kN (Compression)}}$
• $F_{BC} = \boxed{-10.13 \text{ kN (Compression)}}$
• $F_{CD} = \boxed{0 \text{ kN}}$
• $F_{DE} = \boxed{0 \text{ kN}}$
• $F_{EF} = \boxed{10.13 \text{ kN (Tension)}}$
• $F_{FB} = \boxed{1.88 \text{ kN (Tension)}}$
• $F_{BE} = \boxed{13.5 \text{ kN (Tension)}}$
• $F_{CE} = \boxed{-16.88 \text{ kN (Compression)}}$