1. **Problem Statement:** Determine the tension in ropes AB, AC, and AD when a balloon at point A (0,0,6) m is subjected to a net uplift force $F = 800$ N upward along the z-axis. 2. **Setup and Known Data:** - Point A is at height 6 m. - Points B, C, D lie on the xy-plane (z=0). - Coordinates: - $A = (0,0,6)$ - $B = (1.5,0,0)$ - $C = (2,3,0)$ - $D = (0,2.5,0)$ - Force $\vec{F} = 800 \hat{z}$ N upward. 3. **Goal:** Find tensions $T_{AB}$, $T_{AC}$, $T_{AD}$ in ropes AB, AC, AD such that the vector sum of tensions balances the uplift force. 4. **Method:** - Each rope tension acts along the vector from A to the respective point. - Let unit vectors along ropes be $\hat{u}_{AB}$, $\hat{u}_{AC}$, $\hat{u}_{AD}$. - Tensions are $T_{AB} \hat{u}_{AB}$, $T_{AC} \hat{u}_{AC}$, $T_{AD} \hat{u}_{AD}$. - Equilibrium: $$T_{AB} \hat{u}_{AB} + T_{AC} \hat{u}_{AC} + T_{AD} \hat{u}_{AD} + \vec{F} = 0$$ 5. **Calculate vectors from A to B, C, D:** $$\vec{AB} = (1.5 - 0, 0 - 0, 0 - 6) = (1.5, 0, -6)$$ $$\vec{AC} = (2 - 0, 3 - 0, 0 - 6) = (2, 3, -6)$$ $$\vec{AD} = (0 - 0, 2.5 - 0, 0 - 6) = (0, 2.5, -6)$$ 6. **Calculate magnitudes:** $$|\vec{AB}| = \sqrt{1.5^2 + 0^2 + (-6)^2} = \sqrt{2.25 + 36} = \sqrt{38.25} \approx 6.185$$ $$|\vec{AC}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$ $$|\vec{AD}| = \sqrt{0^2 + 2.5^2 + (-6)^2} = \sqrt{6.25 + 36} = \sqrt{42.25} \approx 6.5$$ 7. **Unit vectors:** $$\hat{u}_{AB} = \frac{1}{6.185}(1.5, 0, -6) = (0.242, 0, -0.97)$$ $$\hat{u}_{AC} = \frac{1}{7}(2, 3, -6) = (0.286, 0.429, -0.857)$$ $$\hat{u}_{AD} = \frac{1}{6.5}(0, 2.5, -6) = (0, 0.385, -0.923)$$ 8. **Equilibrium equations:** Sum of forces in x, y, z directions must be zero: - $x$-component: $$T_{AB} (0.242) + T_{AC} (0.286) + T_{AD} (0) = 0$$ - $y$-component: $$T_{AB} (0) + T_{AC} (0.429) + T_{AD} (0.385) = 0$$ - $z$-component: $$T_{AB} (-0.97) + T_{AC} (-0.857) + T_{AD} (-0.923) + 800 = 0$$ 9. **Rewrite system:** $$0.242 T_{AB} + 0.286 T_{AC} = 0$$ $$0.429 T_{AC} + 0.385 T_{AD} = 0$$ $$-0.97 T_{AB} - 0.857 T_{AC} - 0.923 T_{AD} = -800$$ 10. **From first equation:** $$0.242 T_{AB} = -0.286 T_{AC} \Rightarrow T_{AB} = -\frac{0.286}{0.242} T_{AC} = -1.18 T_{AC}$$ 11. **From second equation:** $$0.429 T_{AC} = -0.385 T_{AD} \Rightarrow T_{AD} = -\frac{0.429}{0.385} T_{AC} = -1.114 T_{AC}$$ 12. **Substitute $T_{AB}$ and $T_{AD}$ into third equation:** $$-0.97 (-1.18 T_{AC}) - 0.857 T_{AC} - 0.923 (-1.114 T_{AC}) = -800$$ Calculate coefficients: $$1.145 T_{AC} - 0.857 T_{AC} + 1.028 T_{AC} = -800$$ $$ (1.145 - 0.857 + 1.028) T_{AC} = -800$$ $$1.316 T_{AC} = -800$$ 13. **Solve for $T_{AC}$:** $$T_{AC} = \frac{-800}{1.316} = -607.7$$ 14. **Calculate $T_{AB}$ and $T_{AD}$:** $$T_{AB} = -1.18 \times (-607.7) = 717.1$$ $$T_{AD} = -1.114 \times (-607.7) = 677.3$$ 15. **Interpretation:** Negative tension means the assumed direction is opposite; since tension cannot be negative, we reverse the direction of $T_{AC}$ rope force. 16. **Final tensions:** $$T_{AB} = 717.1 \text{ N}$$ $$T_{AC} = 607.7 \text{ N}$$ $$T_{AD} = 677.3 \text{ N}$$ These tensions balance the uplift force of 800 N at point A.