Let's solve the step-by-step--- ## ** Data:** **Box dimensions 5 ft 3 ft 2 ft - **Weight of the box:** 124 lb - **Material when filled:** Homogeneous, weight 240 lb (excluding the box) - **Goal:** Find the increase in moment of inertia about the x-axis when filled with the material. --- ## **Part a: Find the moment of inertia about the x-axis** ### Step 1: Break the box into parts - The box is divided into **front and back panels**, **top and bottom panels**, and **side panels**. - **Ratios of their areas** will help us find the **mass of each part**. ### Step 2: Calculate the **total surface area** (for ratio analysis) - **Total surface area of the box:** \[ A_{total} = 2(lw + lh + wh) \] where: - \( l = 5\, \text{ft} \) - \( w = 3\, \text{ft} \) - \( h = 2\, \text{ft} \) \[ A_{total} = 2(5 \times 3 + 5 \times 2 + 3 \times 2) = 2(15 + 10 + 6) = 2(31) = 62\, \text{sq ft} \] ### Step 3: Determine areas of each panel - **Front and back panels:** \( 5 \times 3 = 15\, \text{sq ft} \) each - **Top and bottom panels:** \( 5 \times 2 = 10\, \text{sq ft} \) each - **Side panels:** \( 3 \times 2 = 6\, \text{sq ft} \) each ### Step 4: Find the **mass of each panel** based on area ratios - **Total mass of the panels (assuming negligible thickness):** Since the box is hollow and weight is 124 lb, the mass ratio is proportional to their areas: \[ \text{Mass of each panel} \propto \text{Area of the panel} \] - Total **panel area sum**: \[ 2 \times 15 + 2 \times 10 + 2 \times 6 = 30 + 20 + 12 = 62\, \text{sq ft} \] which matches the total surface area, so the **mass of each panel**: \[ m_{panel} = \frac{\text{area of panel}}{\text{total area}} \times 124\, \text{lb} \] - **Front/back panels (each):** \[ m_{FB} = \frac{15}{62} \times 124 \approx 30\, \text{lb} \] - **Top/bottom panels (each):** \[ m_{TB} = \frac{10}{62} \times 124 \approx 20\, \text{lb} \] - **Side panels (each):** \[ m_{SP} = \frac{6}{62} \times 124 \approx 12\, \text{lb} \] ### Step 5: Moment of inertia about the x-axis - The **x-axis** runs along the length \( l = 5\, \text{ft} \). - The **moment of inertia** for each panel about the x-axis depends on its position and shape. - Since the panels are thin plates, the **moment of inertia** of each panel is: \[ I_{x} = \sum m_i r_i^2 \] where \( r_i \) is the distance from the x-axis to the mass element. - **Approximate approach:** For each panel, treat it as a thin rectangle, and calculate the moment of inertia about the x-axis considering its distance from the axis. --- ## **Part a: Final calculation of \( I_x \)** ### Step 6: Use parallel axis theorem and approximate distances - **Front and back panels:** located at \( x = \) and \( x = 5\, \text{ft} \). - **Top and bottom panels:** at \( z = \) and \( z = 2\, \text{ft} \). - **Side panels:** at \( y = \) and \( y = 3\, \text{ft} \). - For simplicity, approximate their centers and calculate their moment of inertia contributions. ### **Approximate contributions:** - **Front panel (at \( x= \)):** \[ I_{x,front} \approx m_{front} \times \left(\frac{w}{2}\right)^2 = 30 \times (1.5)^2 = 30 \times 2.25 = 67.5\, \text{lb·ft}^2 \] - **Back panel (at \( x=5\, \text{ft} \)):** \[ I_{x,back} \approx 30 \times (1.5)^2 = 67.5\, \text{lb·ft}^2 \] - **Top panel:** \[ I_{x,top} \approx 20 \times \left( \frac{l}{2} \right)^2 = 20 \times (2.5)^2 = 20 \times 6.25 = 125\, \text{lb·ft}^2 \] - **Bottom panel:** \[ I_{x,bottom} \approx 125\, \text{lb·ft}^2 \] - **Side panels:** \[ I_{x,side} \approx 12 \times \left( \frac{h}{2} \right)^2 = 12 \times (1)^2 = 12\, \text{lb·ft}^2 \] (For both sides, so multiply by 2): \[ 2 \times 12 = 24\, \text{lb·ft}^2 \] ### **Total moment of inertia about x-axis:** \[ I_{x, total} = 67.5 + 67.5 + 125 + 125 + 24 = 409\, \text{lb·ft}^2 \] --- ## **Part b: Increase in moment of inertia when filled** ### Step 7: Find the **mass of the filled material** - **Material weight:** 240 lb (excluding the box). - The **mass of the material**: \[ m_{material} = \frac{240}{32.2} \approx 7.46\, \text{slugs} \] (Using \( g \approx 32.2\, \text{ft/sec}^2 \), or keep in lb for simplicity). - Alternatively, for moment of inertia calculations, since we're working in lb·ft², consider the mass as 240 lb. ### Step 8: Distribution of the material - The material fills the box uniformly. - **Mass density:** \[ \rho = \frac{\text{mass}}{\text{volume}} = \frac{240\, lb}{5 \times 3 \times 2} = \frac{240}{30} = 8\, lb/ft^3 \] - **Mass of the material:** \[ m_{material} = 240\, lb \] - **Moment of inertia of the filled material:** \[ I_{x, material} = m_{material} \times \left(\text{average squared distance from x-axis}\right) \] - **Average squared distance:** Since the material fills the entire box uniformly: \[ \text{Center of mass of material} \approx \left(\frac{l}{2}, \frac{w}{2}, \frac{h}{2}\right) = (2.5, 1.5, 1) \] - The **distance from the x-axis** (along length): \[ r = \text{distance from x-axis} = \text{average of } y^2 + z^2 \] - **Average of \( y^2 \):** \[ \left(\frac{w}{2}\right)^2 = 1.5^2 = 2.25 \] - **Average of \( z^2 \):** \[ \left(\frac{h}{2}\right)^2 = 1^2 = 1 \] - Total: \[ r_{avg}^2 = y^2 + z^2 = 2.25 + 1 = 3.25 \] - **Moment of inertia of the filled material:** \[ I_{x, fill} = m_{material} \times r_{avg}^2 = 240 \times 3.25 = 780\, \text{lb·ft}^2 \] --- ## **Final answer:** ### **Increase in moment of inertia:** \[ \boxed{ \Delta I_x = I_{x, fill} \approx 780\, \text{lb·ft}^2 } \] **The moment of inertia about the x-axis increases by approximately 780 lb·ft² when the box is filled with the homogeneous material.**