1. **State the problem:** We have a 40 kg weight suspended by three ropes. One rope pulls at 30 degrees from the positive x-axis, the second pulls along the negative x-axis, and the third pulls 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: - $$T_1$$ for the rope at 30 degrees from the positive x-axis, - $$T_2$$ for the rope along the negative x-axis, - $$T_3$$ for the rope 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: $$T_{1x} - T_2 = 0 \implies T_1 \frac{\sqrt{3}}{2} = T_2$$ - In y-direction: $$T_{1y} - T_3 - F_g = 0 \implies T_1 \frac{1}{2} - T_3 = 392$$ 6. **Express $$T_2$$ and $$T_3$$ in terms of $$T_1$$:** $$T_2 = T_1 \frac{\sqrt{3}}{2}$$ $$T_3 = T_1 \frac{1}{2} - 392$$ 7. **Additional condition:** The ropes must be taut, so tensions are positive. We can solve for $$T_1$$ by considering the geometry or additional constraints. Since the problem does not provide more info, assume the system is balanced vertically and horizontally. 8. **Check vertical equilibrium:** The vertical components must balance the weight: $$T_1 \frac{1}{2} = T_3 + 392$$ Since $$T_3$$ pulls downward (negative y), it must balance the vertical component of $$T_1$$ minus the weight. 9. **Assuming $$T_3$$ supports the weight alone:** If $$T_1$$ and $$T_2$$ balance horizontally, then $$T_3 = 392$$ N. From horizontal equilibrium: $$T_1 \frac{\sqrt{3}}{2} = T_2$$ Assuming $$T_2$$ balances $$T_1$$ horizontally, tensions can be found by substituting values. 10. **Final tensions:** - $$T_3 = 392$$ N - $$T_2 = T_1 \frac{\sqrt{3}}{2}$$ - $$T_1$$ is unknown without more info, but if $$T_2$$ balances $$T_1$$ horizontally, then $$T_2 = T_1 \frac{\sqrt{3}}{2}$$. **Summary:** $$T_3 = 392$$ N $$T_2 = T_1 \frac{\sqrt{3}}{2}$$ $$T_1$$ unknown without further constraints. If you provide more info, we can solve for $$T_1$$ and $$T_2$$ explicitly.