1. **State the problem:** We have a 40 kg weight suspended by three ropes. One rope pulls at 30 degrees from the x-axis, another pulls horizontally along the x-axis, and the third pulls vertically downward along the negative y-axis. We want to find the tension forces in each rope. 2. **Identify forces and set up equations:** The weight exerts a downward force due to gravity: $$F_g = mg = 40 \times 9.8 = 392\,\text{N}$$. 3. **Define tensions:** Let the tensions be: - $$T_1$$ in the rope at 30 degrees from the x-axis, - $$T_2$$ in the rope pulling along the positive x-axis, - $$T_3$$ in the rope pulling along the negative y-axis. 4. **Resolve $$T_1$$ into components:** - $$T_{1x} = T_1 \cos 30^\circ = T_1 \times \frac{\sqrt{3}}{2}$$ - $$T_{1y} = T_1 \sin 30^\circ = T_1 \times \frac{1}{2}$$ 5. **Write equilibrium equations:** Since the system is in equilibrium, sum of forces in x and y directions are zero. - In x-direction: $$T_{1x} + T_2 = 0 \implies T_1 \frac{\sqrt{3}}{2} + T_2 = 0$$ - In y-direction: $$T_{1y} - T_3 - F_g = 0 \implies T_1 \frac{1}{2} - T_3 - 392 = 0$$ 6. **Solve for $$T_2$$:** $$T_2 = - T_1 \frac{\sqrt{3}}{2}$$ 7. **Solve for $$T_3$$:** $$T_3 = T_1 \frac{1}{2} - 392$$ 8. **Interpretation:** Without additional constraints or values for one tension, we cannot find unique numeric values for all tensions. The tensions depend on $$T_1$$. 9. **Summary:** The tensions are related by: $$T_2 = - T_1 \frac{\sqrt{3}}{2}$$ $$T_3 = T_1 \frac{1}{2} - 392$$ This expresses the tension forces in terms of $$T_1$$, the tension in the rope at 30 degrees.