1. **Problem Statement:** Determine the tension $T$ in the cable and the reaction at point A for the given structure with a 500 N force applied at point C. 2. **Given Data:** - Horizontal rod length $= 200$ mm - Vertical distance from A to base $= 250$ mm - Force at C $= 500$ N (horizontal to the right) - Angles: $30^\circ$, $60^\circ$, $120^\circ$ 3. **Step 1: Define the forces and components** - Tension $T$ has components: - Horizontal: $T \cos 60^\circ = \frac{T}{2}$ - Vertical: $T \sin 60^\circ = \frac{\sqrt{3}}{2} T$ 4. **Step 2: Apply equilibrium equations** - Sum of horizontal forces $\sum F_x = 0$: $$ R_x + T \cos 60^\circ - 500 = 0 $$ where $R_x$ is the horizontal reaction at A. - Sum of vertical forces $\sum F_y = 0$: $$ R_y - T \sin 60^\circ = 0 $$ where $R_y$ is the vertical reaction at A. 5. **Step 3: Moment equilibrium about point A** - Taking moments about A (counterclockwise positive): $$ 500 \times 0.25 - T \sin 60^\circ \times 0.2 = 0 $$ (Note: distances converted to meters: 250 mm = 0.25 m, 200 mm = 0.2 m) 6. **Step 4: Solve for $T$** $$ 500 \times 0.25 = T \sin 60^\circ \times 0.2 $$ $$ 125 = T \times \frac{\sqrt{3}}{2} \times 0.2 $$ $$ 125 = T \times 0.1732 $$ $$ T = \frac{125}{0.1732} \approx 721.7 \text{ N} $$ 7. **Step 5: Calculate reactions at A** - Horizontal reaction: $$ R_x = 500 - T \cos 60^\circ = 500 - 721.7 \times 0.5 = 500 - 360.85 = 139.15 \text{ N} $$ - Vertical reaction: $$ R_y = T \sin 60^\circ = 721.7 \times 0.866 = 624.9 \text{ N} $$ **Final answers:** - Tension in cable $T \approx 722$ N - Reaction at A: $R_x \approx 139$ N (horizontal), $R_y \approx 625$ N (vertical)