1. **Stating the problem:** Determine the moment about point O (origin) of the resultant force exerted on the tree at point B by cables AB and BC, given tensions in cables AB and BC are 555 N and 660 N respectively. 2. **Identify coordinates:** - Point O (origin) = $(0,0,0)$ - Point A = $(0.75,7,6)$ since x=0.75 m, y=7 m (height of B), z=6 m - Point B = $(0,7,0)$ at height 7 m on y-axis - Point C = $(4.25,7,1)$ since x=4.25 m, y=7 m (same height), z=1 m 3. **Calculate vectors for forces:** - Cable AB vector: from A to B, $$\vec{AB} = \vec{B} - \vec{A} = (0 - 0.75, 7 - 7, 0 - 6) = (-0.75, 0, -6)$$ - Length of AB: $$|\vec{AB}| = \sqrt{(-0.75)^2 + 0^2 + (-6)^2} = \sqrt{0.5625 + 0 + 36} = \sqrt{36.5625} \approx 6.05$$ - Unit vector along AB: $$\hat{u}_{AB} = \frac{\vec{AB}}{|\vec{AB}|} = \left(\frac{-0.75}{6.05}, 0, \frac{-6}{6.05}\right) = (-0.124, 0, -0.992)$$ - Cable BC vector: from B to C, $$\vec{BC} = \vec{C} - \vec{B} = (4.25 - 0, 7 - 7, 1 - 0) = (4.25, 0, 1)$$ - Length of BC: $$|\vec{BC}| = \sqrt{4.25^2 + 0 + 1^2} = \sqrt{18.0625 + 1} = \sqrt{19.0625} \approx 4.37$$ - Unit vector along BC: $$\hat{u}_{BC} = \frac{\vec{BC}}{|\vec{BC}|} = \left(\frac{4.25}{4.37}, 0, \frac{1}{4.37}\right) = (0.973, 0, 0.229)$$ 4. **Find force vectors:** - Force in AB: $$\vec{F}_{AB} = \text{tension} \times \hat{u}_{AB} = 555 \times (-0.124, 0, -0.992) = (-68.8, 0, -550.9)\,N$$ - Force in BC: $$\vec{F}_{BC} = 660 \times (0.973, 0, 0.229) = (641.9, 0, 151.1)\,N$$ 5. **Calculate resultant force:** $$\vec{F}_R = \vec{F}_{AB} + \vec{F}_{BC} = (-68.8 + 641.9, 0 + 0, -550.9 + 151.1) = (573.1, 0, -399.8)$$ 6. **Position vector from O to B:** $$\vec{r}_B = (0, 7, 0)$$ 7. **Calculate moment about O:** Moment $$\vec{M}_O = \vec{r}_B \times \vec{F}_R$$ Cross product: $$\vec{M}_O = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 7 & 0 \\ 573.1 & 0 & -399.8 \end{vmatrix} = \mathbf{i}(7 \times -399.8 - 0 \times 0) - \mathbf{j}(0 \times -399.8 - 0 \times 573.1) + \mathbf{k}(0 \times 0 - 7 \times 573.1)$$ $$= \mathbf{i}(-2798.6) - \mathbf{j}(0) + \mathbf{k}(-4011.7) = (-2798.6, 0, -4011.7)\, Nm$$ 8. **Final answer:** The moment about O of the resultant force is $$\boxed{\vec{M}_O = -2799 \mathbf{i} + 0 \mathbf{j} - 4012 \mathbf{k}\, Nm}$$ This means the moment has components approximately -2799 Nm in x-direction and -4012 Nm in z-direction, with zero in y-direction.