Question: 6. For the truss shown in the figure, determine the force (in kN) acting in the member BC, and indicate whether Tension or Compression. Answer: 400 kN 3 m 3 @ 3 m = 9 m A, B, C, D are the bottom joints from left to right, with E on the upper-left sloping member, F at the top center directly under the 400 kN downward load, and G on the upper-right sloping member; position_hint: center

1. **Problem Statement:** Determine the force in member $BC$ of the truss under a $400$ kN downward load at joint $F$. 2. **Given Data:** - Load at $F$: $400$ kN downward - Lengths: $AB = BC = CD = 3$ m, total bottom length $= 9$ m - Joints: $A, B, C, D$ (bottom), $E$ (upper-left), $F$ (top center), $G$ (upper-right) 3. **Approach:** Use the method of joints or sections to find the force in member $BC$. 4. **Step 1: Calculate support reactions** Assuming supports at $A$ and $D$, sum of vertical forces and moments: $$\sum M_A = 0 = 400 \times 6 - R_D \times 9 \implies R_D = \frac{400 \times 6}{9} = 266.67\text{ kN}$$ $$\sum F_y = 0 = R_A + R_D - 400 \implies R_A = 400 - 266.67 = 133.33\text{ kN}$$ 5. **Step 2: Analyze joint $B$** At joint $B$, members $AB$, $BC$, and $BE$ meet. Assume tension positive. 6. **Step 3: Geometry of member $BE$** Member $BE$ is diagonal; horizontal length $= 3$ m, vertical height $= 3$ m. Slope length $= \sqrt{3^2 + 3^2} = \sqrt{18} = 4.24$ m. 7. **Step 4: Equilibrium at joint $B$** Sum forces in $x$ and $y$ directions: $$\sum F_x = 0: F_{BC} - F_{BE} \times \frac{3}{4.24} = 0$$ $$\sum F_y = 0: R_A - F_{BE} \times \frac{3}{4.24} = 0$$ 8. **Step 5: Solve for $F_{BE}$** From vertical equilibrium: $$F_{BE} = R_A \times \frac{4.24}{3} = 133.33 \times 1.414 = 188.57\text{ kN}$$ 9. **Step 6: Solve for $F_{BC}$** From horizontal equilibrium: $$F_{BC} = F_{BE} \times \frac{3}{4.24} = 188.57 \times 0.707 = 133.33\text{ kN}$$ 10. **Step 7: Check force direction** Since $F_{BC}$ is positive in assumed tension direction, member $BC$ is in tension. **Final answer:** $$\boxed{F_{BC} = 133.33\text{ kN (Tension)}}$$ Note: The given answer $400$ kN likely refers to the load, not the member force. The calculated force in $BC$ is $133.33$ kN tension based on static equilibrium.