Let's solve the problem step by step: --- ## Given Data - **Density, ρ = 2.70 Mg/m³ = 2,700 kg/m³** - All dimensions are in mm (convert to meters for calculations). --- ## (a) **Determine the Mass of the Bracket** ### **Step 1: Break the Bracket into Simpler Parts** Let’s divide the bracket into 3 main parts: 1. **Base plate** 2. **Vertical plate (with hole)** 3. **Triangular support rib** #### **Subtract the cylindrical hole from the vertical plate.** --- ### **Step 2: Calculate Volumes** #### **1. Base Plate** - Dimensions: \( 100 \times 62 \times 12 \) mm \[ V_{\text{base}} = 100 \times 62 \times 12 = 74,400 \text{ mm}^3 \] Convert to m³: \[ V_{\text{base}} = 74,400 \times 10^{-9} = 7.44 \times 10^{-5} \text{ m}^3 \] --- #### **2. Vertical Plate** - Dimensions: \( 100 \times 40 \times 12 \) mm \[ V_{\text{vertical}} = 100 \times 40 \times 12 = 48,000 \text{ mm}^3 \] Convert to m³: \[ V_{\text{vertical}} = 48,000 \times 10^{-9} = 4.8 \times 10^{-5} \text{ m}^3 \] --- #### **3. Triangular Rib** - Right triangle base: \(55 \times 40\) mm - Thickness: \(10\) mm Volume of triangular prism: \[ \text{Area of triangle} = \frac{1}{2} \times 55 \times 40 = 1,100 \text{ mm}^2 \] \[ V_{\text{rib}} = 1,100 \times 10 = 11,000 \text{ mm}^3 \] Convert to m³: \[ V_{\text{rib}} = 11,000 \times 10^{-9} = 1.1 \times 10^{-5} \text{ m}^3 \] --- #### **4. Subtract the Volume of the Hole** - Diameter: 30 mm, thickness: 12 mm \[ \text{Radius } r = 15 \text{ mm} \] \[ V_{\text{hole}} = \pi r^2 h = \pi \times (15^2) \times 12 = \pi \times 225 \times 12 = 8,482 \text{ mm}^3 \] Convert to m³: \[ V_{\text{hole}} = 8,482 \times 10^{-9} = 8.482 \times 10^{-6} \text{ m}^3 \] --- ### **Step 3: Total Volume** \[ V_{\text{total}} = V_{\text{base}} + V_{\text{vertical}} + V_{\text{rib}} - V_{\text{hole}} \] \[ V_{\text{total}} = (7.44 \times 10^{-5}) + (4.8 \times 10^{-5}) + (1.1 \times 10^{-5}) - (8.482 \times 10^{-6}) \] \[ V_{\text{total}} = (7.44 + 4.8 + 1.1 - .8482) \times 10^{-5} \] \[ V_{\text{total}} = (13.34 - .8482) \times 10^{-5} \] \[ V_{\text{total}} = 12.4918 \times 10^{-5} \] \[ V_{\text{total}} = 1.249 \times 10^{-4} \text{ m}^3 \] --- ### **Step 4: Mass Calculation** \[ m = \rho \cdot V_{\text{total}} \] \[ m = 2,700 \times 1.249 \times 10^{-4} \] \[ m \approx .337 \text{ kg} \] --- ### **Final Mass** \[ \boxed{.34 \text{ kg}} \] (Rounded to two decimal places) --- ## (b) **Locate the Center of Mass of the Bracket** This part involves finding the centroid of the composite body. ### **Step 1: Find Centroid of Each Part** Let’s denote the centroid of each part (in mm): - **Base plate:** Center at \((50, 31, 6)\) - **Vertical plate:** Center at \((50, 40, 68)\) - **Rib:** Use centroid for a right triangular prism: - \(x_{\text{rib}} = \frac{1}{3} \times 55 = 18.33\) mm from left - \(y_{\text{rib}} = 40/2 = 20\) mm from bottom - \(z_{\text{rib}} = 10/2 = 5\) mm from base plate (since 10 mm thick, centered) - Plus base offsets for correct global position. - **Hole:** Its centroid is the same as the centroid of the removed cylinder. ### **Step 2: Weighted Average of Centroids** Use the formula: \[ \bar{x} = \frac{\sum x_i V_i}{V_{\text{total}}} \] (Do similarly for \(y\) and \(z\).) #### **Let’s Approximate (Values in mm):** | Part | Volume (mm³) | \(x\) (mm) | \(y\) (mm) | \(z\) (mm) | |--------------|--------------|------------|------------|------------| | Base | 74,400 | 50 | 31 | 6 | | Vertical | 48,000 | 50 | 40 | 68 | | Rib | 11,000 | 18.3 | 20 | 5 | | Hole (-) | 8,482 | 85 | 20 | 68 | Now, compute the weighted average: \[ \bar{x} = \frac{(74,400 \times 50) + (48,000 \times 50) + (11,000 \times 18.3) - (8,482 \times 85)}{V_{\text{total, mm}^3}} \] \[ V_{\text{total, mm}^3} = 74,400 + 48,000 + 11,000 - 8,482 = 124,918 \] Calculate: - \(74,400 \times 50 = 3,720,000\) - \(48,000 \times 50 = 2,400,000\) - \(11,000 \times 18.3 = 201,300\) - \(8,482 \times 85 = 720,970\) Sum: \[ N_x = 3,720,000 + 2,400,000 + 201,300 - 720,970 = 5,600,330 \] \[ \bar{x} = \frac{5,600,330}{124,918} \approx 44.83 \text{ mm} \] Repeat for \(y\): - \(74,400 \times 31 = 2,306,400\) - \(48,000 \times 40 = 1,920,000\) - \(11,000 \times 20 = 220,000\) - \(8,482 \times 20 = 169,640\) \[ N_y = 2,306,400 + 1,920,000 + 220,000 - 169,640 = 4,276,760 \] \[ \bar{y} = \frac{4,276,760}{124,918} \approx 34.24 \text{ mm} \] For \(z\): - \(74,400 \times 6 = 446,400\) - \(48,000 \times 68 = 3,264,000\) - \(11,000 \times 5 = 55,000\) - \(8,482 \times 68 = 576,776\) \[ N_z = 446,400 + 3,264,000 + 55,000 - 576,776 = 3,188,624 \] \[ \bar{z} = \frac{3,188,624}{124,918} \approx 25.54 \text{ mm} \] --- ### **Final Answer** #### **(a) Mass:** \[ \boxed{.34 \text{ kg}} \] #### **(b) Center of Mass (from the origin, in mm):** \[ \boxed{ \begin{align*} \bar{x} &\approx 44.8\ \text{mm} \\ \bar{y} &\approx 34.2\ \text{mm} \\ \bar{z} &\approx 25.5\ \text{mm} \\ \end{align*} } \] --- **Summary Table:** | Part | Volume (mm³) | \(x\) (mm) | \(y\) (mm) | \(z\) (mm) | |----------------|--------------|------------|------------|------------| | Base | 74,400 | 50 | 31 | 6 | | Vertical Plate | 48,000 | 50 | 40 | 68 | | Rib | 11,000 | 18.3 | 20 | 5 | | Hole (remove) | -8,482 | 85 | 20 | 68 | --- ### **Boxed Final Answers** \[ \boxed{ \begin{aligned} \text{(a) Mass of the bracket:} & \quad .34\ \text{kg} \\ \text{(b) Center of mass:} & \quad (44.8,\ 34.2,\ 25.5)\ \text{mm} \end{aligned} } \]