1. **Problem statement:** Determine the vector expression for the tension $\mathbf{T}$ in cable AB acting on member AD, given the tension magnitude is 2.2 kN. Also find the magnitude of the projection of $\mathbf{T}$ along line AC. 2. **Given data:** - Tension magnitude $|\mathbf{T}| = 2.2$ kN - Coordinates (from description): - $A = (0,1.6,5.9)$ m - $B = (0,0,0)$ m (assuming origin for B for simplicity) - $C = (3.8,0,0)$ m - $D$ is connected to $A$ horizontally (coordinates not needed for tension vector in AB) 3. **Find vector $\mathbf{AB}$:** $$\mathbf{AB} = \mathbf{B} - \mathbf{A} = (0-0, 0-1.6, 0-5.9) = (0, -1.6, -5.9)$$ 4. **Calculate unit vector along AB:** $$|\mathbf{AB}| = \sqrt{0^2 + (-1.6)^2 + (-5.9)^2} = \sqrt{2.56 + 34.81} = \sqrt{37.37} \approx 6.11$$ $$\hat{u}_{AB} = \frac{1}{6.11}(0, -1.6, -5.9) = (0, -0.262, -0.966)$$ 5. **Vector expression for tension $\mathbf{T}$:** $$\mathbf{T} = |\mathbf{T}| \hat{u}_{AB} = 2.2 \times (0, -0.262, -0.966) = (0, -0.576, -2.125) \text{ kN}$$ 6. **Find vector $\mathbf{AC}$:** $$\mathbf{AC} = \mathbf{C} - \mathbf{A} = (3.8-0, 0-1.6, 0-5.9) = (3.8, -1.6, -5.9)$$ 7. **Calculate unit vector along AC:** $$|\mathbf{AC}| = \sqrt{3.8^2 + (-1.6)^2 + (-5.9)^2} = \sqrt{14.44 + 2.56 + 34.81} = \sqrt{51.81} \approx 7.20$$ $$\hat{u}_{AC} = \frac{1}{7.20}(3.8, -1.6, -5.9) = (0.528, -0.222, -0.819)$$ 8. **Projection of $\mathbf{T}$ along $\mathbf{AC}$:** $$\text{proj}_{AC} \mathbf{T} = \mathbf{T} \cdot \hat{u}_{AC} = (0)(0.528) + (-0.576)(-0.222) + (-2.125)(-0.819)$$ $$= 0 + 0.128 + 1.740 = 1.868 \text{ kN}$$ **Final answers:** - Vector tension on member AD: $$\mathbf{T} = (0, -0.576, -2.125) \text{ kN}$$ - Magnitude of projection of $\mathbf{T}$ along AC: $$1.868 \text{ kN}$$