1. **Problem Statement:** Determine the forces in cables AB, AC, and AD that support a 28-kg flowerpot in equilibrium. 2. **Given Data:** - Mass of flowerpot $m = 28$ kg - Coordinates of points relative to A (origin): - $B = (0, 1.5, 0)$ m - $C = (2, 0, 0)$ m - $D = (0, 0, 1.5)$ m 3. **Step 1: Calculate the weight force acting downward at A.** $$ W = mg = 28 \times 9.81 = 274.68 \text{ N} $$ 4. **Step 2: Define unit vectors along cables AB, AC, and AD.** - Vector $\vec{AB} = B - A = (0, 1.5, 0)$ - Vector $\vec{AC} = C - A = (2, 0, 0)$ - Vector $\vec{AD} = D - A = (0, 0, 1.5)$ Calculate magnitudes: $$ |\vec{AB}| = 1.5, \quad |\vec{AC}| = 2, \quad |\vec{AD}| = 1.5 $$ Unit vectors: $$ \hat{u}_{AB} = \frac{1}{1.5}(0, 1.5, 0) = (0, 1, 0) $$ $$ \hat{u}_{AC} = \frac{1}{2}(2, 0, 0) = (1, 0, 0) $$ $$ \hat{u}_{AD} = \frac{1}{1.5}(0, 0, 1.5) = (0, 0, 1) $$ 5. **Step 3: Set up equilibrium equations.** Since the flowerpot is in equilibrium, the sum of forces must be zero: $$ \vec{F}_{AB} + \vec{F}_{AC} + \vec{F}_{AD} + \vec{W} = 0 $$ Express forces in cables as magnitudes times unit vectors: $$ F_{AB} \hat{u}_{AB} + F_{AC} \hat{u}_{AC} + F_{AD} \hat{u}_{AD} = -\vec{W} $$ Weight vector points downward: $$ \vec{W} = (0, 0, -274.68) $$ 6. **Step 4: Write component equations:** - $x$-component: $$ F_{AB} \times 0 + F_{AC} \times 1 + F_{AD} \times 0 = 0 \implies F_{AC} = 0 $$ - $y$-component: $$ F_{AB} \times 1 + F_{AC} \times 0 + F_{AD} \times 0 = 0 \implies F_{AB} = 0 $$ - $z$-component: $$ F_{AB} \times 0 + F_{AC} \times 0 + F_{AD} \times 1 = 274.68 $$ 7. **Step 5: Solve for forces:** $$ F_{AB} = 0 \text{ N} $$ $$ F_{AC} = 0 \text{ N} $$ $$ F_{AD} = 274.68 \text{ N} $$ 8. **Interpretation:** Only cable AD supports the flowerpot's weight vertically. Cables AB and AC carry no load in this configuration. **Final answers:** - $F_{AB} = 0$ N - $F_{AC} = 0$ N - $F_{AD} = 274.68$ N Rounded to three significant figures: - $F_{AB} = 0.00$ N - $F_{AC} = 0.00$ N - $F_{AD} = 275$ N