1. **State the problem:** We have a vertical truss structure with horizontal forces of 23 kN at B, 63 kN at C, and 63 kN at D acting to the right. The vertical segments between points are 12 ft each. We need to find the forces in the diagonal members labeled "X". 2. **Identify the geometry and forces:** The truss has vertical members spaced 12 ft apart vertically. The horizontal forces act at points B, C, and D. The diagonal members connect points B, E, and G. 3. **Calculate the length of diagonal members:** Each diagonal member forms a right triangle with vertical side 12 ft and horizontal side equal to the horizontal spacing between columns (assumed 12 ft for simplicity). Using Pythagoras theorem: $$\text{Length of diagonal} = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \text{ ft}$$ 4. **Resolve forces at joints:** At joint B, horizontal force is 23 kN to the right. The diagonal member force $X$ has components: - Horizontal: $X \cos 45^\circ$ - Vertical: $X \sin 45^\circ$ Since $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$, 5. **Equilibrium at joint B:** Sum of horizontal forces = 0: $$23 - X \cos 45^\circ = 0 \implies X = \frac{23}{\cos 45^\circ} = \frac{23}{\frac{\sqrt{2}}{2}} = 23 \times \frac{2}{\sqrt{2}} = 23 \sqrt{2} \approx 32.53 \text{ kN}$$ Sum of vertical forces = 0: Vertical reaction at B must balance $X \sin 45^\circ = 32.53 \times \frac{\sqrt{2}}{2} = 23$ kN upwards. 6. **Repeat for joint C with 63 kN horizontal force:** $$X = \frac{63}{\cos 45^\circ} = 63 \sqrt{2} \approx 89.1 \text{ kN}$$ 7. **Repeat for joint D with 63 kN horizontal force:** Same as joint C: $$X = 63 \sqrt{2} \approx 89.1 \text{ kN}$$ **Final answer:** The forces in the diagonal members labeled "X" are approximately: - At B: 32.53 kN - At C: 89.1 kN - At D: 89.1 kN These forces act along the diagonal members at 45 degrees to the horizontal and vertical.