1. **Problem Statement:** Determine the position of the centre of gravity (centroid) of a composite body made of a titanium plate with a circular lead-filled hole. 2. **Given Data:** - Titanium plate thickness = 30 mm - Circular hole diameter = 56 mm - Density of titanium $\rho_{Ti} = 4500$ kg/m³ - Density of lead $\rho_{Pb} = 11300$ kg/m³ - Rectangle width = 160 mm, height = 140 mm - Slanting right side drops 60 mm from top-right corner - Hole center is 40 mm from top edge and 40 mm from right edge - Origin $O$ at bottom-left corner 3. **Approach:** Use the table method to find the centroid of composite shapes: - Calculate area and centroid coordinates of each part - Calculate mass of each part using density and thickness - Use the formula for composite centroid: $$x_c = \frac{\sum m_i x_i}{\sum m_i}, \quad y_c = \frac{\sum m_i y_i}{\sum m_i}$$ 4. **Step 1: Define parts** - Part 1: Titanium plate with polygon shape (rectangle + slant) - Part 2: Lead circular hole (negative area in titanium, positive mass in lead) 5. **Step 2: Calculate areas and centroids** - Titanium polygon area: - Rectangle base = 160 mm, height = 140 mm - Slant right side reduces height by 60 mm over 40 mm horizontal (from 160 to 120 mm base) - Polygon vertices (in mm): (0,0), (120,0), (160,80), (160,140), (0,140) - Calculate polygon area using shoelace formula: $$A_{Ti} = \frac{1}{2} |x_1y_2 + x_2y_3 + ... - (y_1x_2 + y_2x_3 + ...)|$$ Substitute: $$= \frac{1}{2} |0\times0 + 120\times80 + 160\times140 + 160\times140 + 0\times0 - (0\times120 + 0\times160 + 80\times160 + 140\times0 + 140\times0)|$$ Calculate carefully: $$= \frac{1}{2} |0 + 9600 + 22400 + 22400 + 0 - (0 + 0 + 12800 + 0 + 0)| = \frac{1}{2} |54400 - 12800| = \frac{1}{2} \times 41600 = 20800 \text{ mm}^2$$ - Titanium polygon centroid $(x_{Ti}, y_{Ti})$: Use polygon centroid formula: $$C_x = \frac{1}{6A} \sum (x_i + x_{i+1})(x_i y_{i+1} - x_{i+1} y_i)$$ $$C_y = \frac{1}{6A} \sum (y_i + y_{i+1})(x_i y_{i+1} - x_{i+1} y_i)$$ Calculate terms: - Edges: 1: (0,0)-(120,0): $x_i y_{i+1} - x_{i+1} y_i = 0$ 2: (120,0)-(160,80): $120\times80 - 160\times0 = 9600$ 3: (160,80)-(160,140): $160\times140 - 160\times80 = 9600$ 4: (160,140)-(0,140): $160\times140 - 0\times140 = 22400$ 5: (0,140)-(0,0): $0\times0 - 0\times140 = 0$ Sum of cross products = 0 + 9600 + 9600 + 22400 + 0 = 41600 Calculate $C_x$: $$= \frac{1}{6 \times 20800} [(0+120)\times0 + (120+160)\times9600 + (160+160)\times9600 + (160+0)\times22400 + (0+0)\times0]$$ $$= \frac{1}{124800} [0 + 280\times9600 + 320\times9600 + 160\times22400 + 0]$$ $$= \frac{1}{124800} [0 + 2688000 + 3072000 + 3584000 + 0] = \frac{9344000}{124800} \approx 74.87 \text{ mm}$$ Calculate $C_y$: $$= \frac{1}{6 \times 20800} [(0+0)\times0 + (0+80)\times9600 + (80+140)\times9600 + (140+140)\times22400 + (140+0)\times0]$$ $$= \frac{1}{124800} [0 + 80\times9600 + 220\times9600 + 280\times22400 + 0]$$ $$= \frac{1}{124800} [0 + 768000 + 2112000 + 6272000 + 0] = \frac{9152000}{124800} \approx 73.34 \text{ mm}$$ - Titanium polygon thickness = 30 mm - Titanium polygon volume $V_{Ti} = A_{Ti} \times thickness = 20800 \times 30 = 624000 \text{ mm}^3 = 6.24 \times 10^{-4} \text{ m}^3$ - Titanium polygon mass: $$m_{Ti} = \rho_{Ti} \times V_{Ti} = 4500 \times 6.24 \times 10^{-4} = 2.808 \text{ kg}$$ 6. **Step 3: Circular hole (lead filled)** - Diameter $d = 56$ mm, radius $r = 28$ mm - Area $A_{hole} = \pi r^2 = \pi \times 28^2 = 2463.01 \text{ mm}^2$ - Hole centroid coordinates: - From top edge: 40 mm - From right edge: 40 mm - Rectangle width = 160 mm, height = 140 mm - So hole center $x_{hole} = 160 - 40 = 120$ mm - Hole center $y_{hole} = 140 - 40 = 100$ mm - Hole thickness = 30 mm - Hole volume $V_{hole} = A_{hole} \times thickness = 2463.01 \times 30 = 73890.3 \text{ mm}^3 = 7.389 \times 10^{-5} \text{ m}^3$ - Lead mass: $$m_{Pb} = \rho_{Pb} \times V_{hole} = 11300 \times 7.389 \times 10^{-5} = 0.835 \text{ kg}$$ - Titanium mass removed by hole: $$m_{hole,Ti} = \rho_{Ti} \times V_{hole} = 4500 \times 7.389 \times 10^{-5} = 0.3325 \text{ kg}$$ 7. **Step 4: Composite mass and centroid** - Total mass: $$m_{total} = m_{Ti} - m_{hole,Ti} + m_{Pb} = 2.808 - 0.3325 + 0.835 = 3.3105 \text{ kg}$$ - Composite centroid coordinates: $$x_c = \frac{m_{Ti} x_{Ti} - m_{hole,Ti} x_{hole} + m_{Pb} x_{hole}}{m_{total}} = \frac{2.808 \times 74.87 - 0.3325 \times 120 + 0.835 \times 120}{3.3105}$$ Calculate numerator: $$= 210.1 - 39.9 + 100.2 = 270.4$$ $$x_c = \frac{270.4}{3.3105} \approx 81.66 \text{ mm}$$ $$y_c = \frac{m_{Ti} y_{Ti} - m_{hole,Ti} y_{hole} + m_{Pb} y_{hole}}{m_{total}} = \frac{2.808 \times 73.34 - 0.3325 \times 100 + 0.835 \times 100}{3.3105}$$ Calculate numerator: $$= 205.9 - 33.25 + 83.5 = 256.15$$ $$y_c = \frac{256.15}{3.3105} \approx 77.39 \text{ mm}$$ 8. **Final answer:** The centre of gravity of the composite body with respect to origin $O$ is approximately: $$\boxed{(x_c, y_c) = (81.7 \text{ mm}, 77.4 \text{ mm})}$$