1. **Problem Statement:** Calculate the tension in cables AB and AC supporting a 633 kg luminous sign, with gravity $g=9.8\,m/s^2$. 2. **Known Data:** - Mass $m=633\,kg$ - Gravity $g=9.8\,m/s^2$ - Horizontal distances: $AB=3.5\,m$, $AC=4.5\,m$ - Height of walls: $3.0\,m$ 3. **Step 1: Calculate the weight (force) of the sign:** $$W = m \times g = 633 \times 9.8 = 6203.4\,N$$ 4. **Step 2: Geometry of cables:** - Length vertical component for both cables is $3.0\,m$ - Horizontal components: $AB=3.5\,m$, $AC=4.5\,m$ 5. **Step 3: Calculate the angles of cables with the horizontal:** $$\theta_{AB} = \arctan\left(\frac{3.0}{3.5}\right) = \arctan(0.857) \approx 40.6^\circ$$ $$\theta_{AC} = \arctan\left(\frac{3.0}{4.5}\right) = \arctan(0.667) \approx 33.7^\circ$$ 6. **Step 4: Set equilibrium equations at point A:** - Vertical equilibrium: $$T_{AB} \sin\theta_{AB} + T_{AC} \sin\theta_{AC} = W$$ - Horizontal equilibrium: $$T_{AB} \cos\theta_{AB} = T_{AC} \cos\theta_{AC}$$ 7. **Step 5: Express $T_{AB}$ in terms of $T_{AC}$ from horizontal equilibrium:** $$T_{AB} = T_{AC} \frac{\cos\theta_{AC}}{\cos\theta_{AB}} = T_{AC} \frac{\cos 33.7^\circ}{\cos 40.6^\circ} = T_{AC} \frac{0.832}{0.759} = 1.096 T_{AC}$$ 8. **Step 6: Substitute into vertical equilibrium:** $$1.096 T_{AC} \sin 40.6^\circ + T_{AC} \sin 33.7^\circ = 6203.4$$ $$T_{AC} (1.096 \times 0.650 + 0.555) = 6203.4$$ $$T_{AC} (0.712 + 0.555) = 6203.4$$ $$T_{AC} \times 1.267 = 6203.4$$ $$T_{AC} = \frac{6203.4}{1.267} \approx 4897.3\,N$$ 9. **Step 7: Calculate $T_{AB}$:** $$T_{AB} = 1.096 \times 4897.3 = 5367.3\,N$$ **Final answer:** - Tension in cable AB: $5367.3\,N$ - Tension in cable AC: $4897.3\,N$