Let's solve the problem step by step: --- ### **Step 1: Understand the Problem** A plant of mass \( m = 10\, \text{kg} \) is hung by three ropes: \( AB \), \( AC \), and \( AD \). - \( Y = 1.2\, \text{m} \) - \( X = 3.2\, \text{m} \) Points: - \( O \): Attachment of plant. - \( A \): At \((0, 0, 0)\) - \( B \): At \((0, 0, 4)\) - \( C \): At \((3, 0, 0)\) - \( D \): At \((0, 3.2, 1.2)\) --- ### **Step 2: Assign Coordinates** - \( O = (0, 0, 0) \) - \( B = (0, 0, 4) \) - \( C = (3, 0, 0) \) - \( D = (0, 3.2, 1.2) \) --- ### **Step 3: Write Rope Vectors** Write vectors from \( O \) to the points: - \( \vec{OB} = (0, 0, 4) \) - \( \vec{OC} = (3, 0, 0) \) - \( \vec{OD} = (0, 3.2, 1.2) \) --- ### **Step 4: Find Unit Vectors** #### \( \vec{e}_{AB} \) (from O to B): \[ \vec{e}_{AB} = \frac{\vec{OB}}{|\vec{OB}|} = \frac{(0, 0, 4)}{4} = (0, 0, 1) \] #### \( \vec{e}_{AC} \): \[ |\vec{OC}| = \sqrt{3^2 + 0^2 + 0^2} = 3 \] \[ \vec{e}_{AC} = \frac{(3, 0, 0)}{3} = (1, 0, 0) \] #### \( \vec{e}_{AD} \): \[ |\vec{OD}| = \sqrt{0^2 + 3.2^2 + 1.2^2} = \sqrt{0 + 10.24 + 1.44} = \sqrt{11.68} \approx 3.418 \] \[ \vec{e}_{AD} = \frac{(0, 3.2, 1.2)}{3.418} \approx (0, 0.936, 0.351) \] --- ### **Step 5: Forces in the Ropes** Let the tensions be: - \( T_{AB} \) in \( AB \) - \( T_{AC} \) in \( AC \) - \( T_{AD} \) in \( AD \) Total force at \( O \) (sum of all tensions and the weight) must be zero: \[ T_{AB} \vec{e}_{AB} + T_{AC} \vec{e}_{AC} + T_{AD} \vec{e}_{AD} + \vec{W} = 0 \] The weight: \[ \vec{W} = (0, 0, -mg) = (0, 0, -98) \text{ N} \] --- ### **Step 6: Write Equilibrium Equations** \[ T_{AB} (0,0,1) + T_{AC} (1,0,0) + T_{AD} (0,0.936,0.351) + (0,0,-98) = (0,0,0) \] So: #### **x-components:** \[ T_{AC} = 0 \] But this can't be right. Let's check the geometry: If C is at (3,0,0), then the rope is horizontal. But the plant can't be in equilibrium if the AC rope is only horizontal. Let's proceed with these equations. #### **y-components:** \[ 0.936 T_{AD} = 0 \implies T_{AD} = 0 \] Again, this seems off. Let's check the equations: Let's let the plant hang at (0, 0, 0), the ropes pull to B (0, 0, 4), C (3, 0, 0), D (0, 3.2, 1.2). So, the tension in each rope is along the direction from O to the respective point. Let the equilibrium equation be: \[ T_{AB} \frac{\vec{OB}}{|\vec{OB}|} + T_{AC} \frac{\vec{OC}}{|\vec{OC}|} + T_{AD} \frac{\vec{OD}}{|\vec{OD}|} = -mg \hat{k} \] --- ### **Step 7: Write Component Equations** \[ T_{AB}(0,0,1) + T_{AC}(1,0,0) + T_{AD}(0,0.936,0.351) = (0,0,98) \] So, #### x: \( T_{AC} = 0 \) #### y: \( 0.936T_{AD} = 0 \implies T_{AD} = 0 \) #### z: \( T_{AB} + 0.351T_{AD} = 98 \) But this would mean only AB is holding the weight, which can't be. #### **Correction:** The plant is not at (0,0,0) if it's held in equilibrium by three ropes. The actual position O needs to be found, given the geometry. Let \( O \) be at \( (x, y, z) \), and the ropes connect from O to B, C, D. But let's assume the plant hangs directly below the point where three ropes meet (let’s call that O), and use vector equilibrium. --- ### **Step 8: Let’s Set Up the Vector Equation** Let’s denote the position of the plant as point \( O \). For the plant to be in equilibrium, the sum of the force vectors from all three ropes must equal the weight vector, i.e., \[ \vec{F}_{AB} + \vec{F}_{AC} + \vec{F}_{AD} + (0, 0, -mg) = 0 \] Let’s calculate the direction cosines for each rope, based on the coordinates of their endpoints. Let’s assume the plant's position is at the origin, and the ropes are attached at: - \( B(0,0,4) \) - \( C(3,0,0) \) - \( D(0,3.2,1.2) \) So, the direction vector of each rope is: - \( \vec{r}_{OB} = (0,0,4) \) - \( \vec{r}_{OC} = (3,0,0) \) - \( \vec{r}_{OD} = (0,3.2,1.2) \) The unit vectors: - \( \vec{e}_{OB} = (0,0,1) \) - \( \vec{e}_{OC} = (1,0,0) \) - \( \vec{e}_{OD} = (0,0.936,0.351) \) Let the tension in each rope be \( T_{AB}, T_{AC}, T_{AD} \) respectively. So, \[ T_{AB}(0,0,1) + T_{AC}(1,0,0) + T_{AD}(0,0.936,0.351) = (0,0,98) \] From which: - x: \( T_{AC} = 0 \) - y: \( 0.936T_{AD} = 0 \implies T_{AD} = 0 \) - z: \( T_{AB} + 0.351T_{AD} = 98 \) But again, this suggests only AB supports the plant, which is not correct. --- ## **Final Step: Realize the Plant is Suspended by Three Non-coplanar Cables** This is a classic statics problem: the plant is at a point in space where three ropes, attached to B, C, and D, support it. Let’s call the plant's position \( O(x,y,z) \). Let’s find \( x, y, z \) such that the three ropes from B(0,0,4), C(3,0,0), D(0,3.2,1.2) meet at O. But the problem does not provide the exact position of the plant, only the anchor points. Since the plant is hanging, its position is at the intersection of the three ropes. --- ### **Summary of Solution Steps** 1. **Write the force equilibrium equation in vector form.** 2. **Express tension in each rope as a force vector along the direction of the rope.** 3. **Set up a system of equations using the components.** 4. **Solve for the unknown tensions.** --- ### **Let’s Use the Coordinates as Given:** Given the data, the plant is at the origin, and the ropes go to the points given. - AB: (0,0,4) - AC: (3,0,0) - AD: (0,3.2,1.2) Force in each rope: - \( \vec{F}_{AB} = T_{AB} (0,0,1) \) - \( \vec{F}_{AC} = T_{AC} (1,0,0) \) - \( \vec{F}_{AD} = T_{AD} (0,0.936,0.351) \) Equilibrium: - \( x: T_{AC} = 0 \) - \( y: 0.936 T_{AD} = 0 \implies T_{AD} = 0 \) - \( z: T_{AB} + 0.351 T_{AD} = 98 \) --- ### **Final Answer** Given these calculations, it seems the plant's weight is supported only by rope AB (vertical). But in a real-world scenario, the plant should be at a point where all three ropes are under tension. **Based on the numbers given and the equilibrium equations:** \[ T_{AC} = 0 \\ T_{AD} = 0 \\ T_{AB} = 98 \, \text{N} \] --- **But if you want to find the force in AC (which is 0 N), enter 0.0 N.** --- #### **If you have more information or the plant's actual position, please provide it for a more accurate answer.** --- **Summary Table:** | Rope | Force (N) | |-------|-----------| | AB | 98 | | AC | 0 | | AD | 0 | --- **Force in AC = 0.0 N** (to the nearest 0.1 N). --- Let me know if you have more details or if you want to discuss a different scenario!

Let's analyze the problem step-by-step to find the tension in the ropes, focusing on the forces and equilibrium conditions. --- ### **Given Data:** - Mass of the plant, \( m = 10\, \text{kg} \) - Weight, \( W = mg = 10 \times 9.8 = 98\, \text{N} \) - Coordinates of attachment points: - \( B = (0, 0, 4) \) - \( C = (3, 0, 0) \) - \( D = (0, 3.2, 1.2) \) --- ### **Assumptions:** - The plant hangs at the point \( O \), where the three ropes meet. - The tension in each rope acts along the line connecting \( O \) to each attachment point. - The system is in static equilibrium: the sum of all forces is zero. --- ### **Step 1: Find the position of the plant \( O \).** Because the problem involves three ropes supporting the plant, the point \( O \) is the intersection point of the three lines from the attachment points. **In many such problems, the plant's position is given or can be inferred.** **Alternatively,** if the plant is hanging in equilibrium, the tension forces in the ropes balance the weight. --- ### **Step 2: Express the tension vectors** Let: - \( T_{AB} \): tension in the rope from \( O \) to \( B \) - \( T_{AC} \): tension in the rope from \( O \) to \( C \) - \( T_{AD} \): tension in the rope from \( O \) to \( D \) The force balance at \( O \): \[ \boxed{ T_{AB} \hat{r}_{B} + T_{AC} \hat{r}_{C} + T_{AD} \hat{r}_{D} + \vec{W} = 0 } \] Where \( \hat{r}_X \) is the unit vector from \( O \) to attachment point \( X \), and \( \vec{W} = (0,0,-98\, \text{N}) \). --- ### **Step 3: Determine the unit vectors** Assuming \( O \) is at the origin \( (0,0,0) \): - \( \hat{r}_B = \frac{\vec{r}_B}{|\vec{r}_B|} = \frac{(0,0,4)}{4} = (0,0,1) \) - \( \hat{r}_C = \frac{(3,0,0)}{3} = (1,0,0) \) - \( \hat{r}_D = \frac{(0,3.2,1.2)}{\sqrt{0^2 + 3.2^2 + 1.2^2}} \) Calculate \( |\vec{r}_D| \): \[ |\vec{r}_D| = \sqrt{0^2 + 3.2^2 + 1.2^2} = \sqrt{0 + 10.24 + 1.44} = \sqrt{11.68} \approx 3.418 \] So, \[ \hat{r}_D = \left( 0, \frac{3.2}{3.418}, \frac{1.2}{3.418} \right) \approx (0, 0.936, 0.351) \] --- ### **Step 4: Write equilibrium equations** The force balance in component form: \[ \begin{cases} T_{AB} \times 0 + T_{AC} \times 1 + T_{AD} \times 0 = 0 \quad \Rightarrow \quad T_{AC} = 0 \\ T_{AB} \times 0 + T_{AC} \times 0 + T_{AD} \times 0.936 = 0 \quad \Rightarrow \quad T_{AD} \times 0.936 = 0 \Rightarrow T_{AD} = 0 \\ T_{AB} \times 1 + T_{AC} \times 0 + T_{AD} \times 0.351 = 98 \end{cases} \] From the first two equations: \[ T_{AC} = 0,\quad T_{AD} = 0 \] From the third: \[ T_{AB} = 98\, \text{N} \] --- ### **Interpretation:** - The tensions in ropes \( AC \) and \( AD \) are zero, meaning they are slack or not supporting the load. - The entire weight is supported by the rope \( AB \) with tension: \[ \boxed{ \textbf{Tension in } AB = 98\, \text{N} } \] --- ### **Final Answer:** | Rope | Tension (N) | |---------|--------------| | \( AB \) | **98** | | \( AC \) | **0** | | \( AD \) | **0** | --- ### **Summary:** - The plant is supported solely by the rope \( AB \) in this configuration. - Tensions in the other two ropes are zero, indicating they are slack. --- **Note:** If the problem intends the plant to be suspended by all three ropes actively, then the position of the plant would be somewhere between the attachment points, and a more detailed 3D equilibrium calculation would be necessary. But based on the given data and typical static equilibrium assumptions, the above is the straightforward solution.