Let's break down the problem step by step: --- ## Problem Restatement A car door (1.1 m × .9 m) is submerged vertically in water, with the top edge 10 m below the surface. **Find:** - The net force on the door (normal to its surface) - The location of the center of pressure For two cases: (a) The car is well-sealed, contains air at atmospheric pressure (b) The car is filled with water --- ## **Step 1: Data and Assumptions** - Height of door, \( h = 1.1\,\text{m} \) - Width of door, \( w = .9\,\text{m} \) - Top edge depth, \( d_{\text{top}} = 10\,\text{m} \) - Area, \( A = h \times w = 1.1 \times .9 = .99\,\text{m}^2 \) - Density of water, \( \rho = 100\,\text{kg/m}^3 \) - Atmospheric pressure, \( P_{\text{atm}} = 101,325\,\text{Pa} \) - Gravitational acceleration, \( g = 9.81\,\text{m/s}^2 \) --- ## **Step 2: Hydrostatic Pressure** ### Pressure at depth \( d \) below surface: \[ P = P_{\text{atm}} + \rho g d \] ### For the door: - Top edge: \( d_{\text{top}} = 10\,\text{m} \) - Bottom edge: \( d_{\text{bot}} = d_{\text{top}} + h = 10 + 1.1 = 11.1\,\text{m} \) --- ## **Step 3: Net Force on the Door** ### **Case (a): Car is well-sealed (air inside at atmospheric pressure)** - Outside: Water pressure - Inside: Atmospheric pressure - **Net pressure:** \( (P_{\text{atm}} + \rho g d) - P_{\text{atm}} = \rho g d \) #### **Force on door (integrate pressure over area):** Let \( y \) be measured downward from the water surface (\( y= \) at surface). - Top of door: \( y_1 = 10\,\text{m} \) - Bottom of door: \( y_2 = 11.1\,\text{m} \) At depth \( y \): \( P(y) = \rho g y \) Total force: \[ F = \int_{y_1}^{y_2} \rho g y \cdot w\,dy \] \[ F = \rho g w \int_{y_1}^{y_2} y\,dy = \rho g w \left[ \frac{y^2}{2} \right]_{y_1}^{y_2} \] \[ F = \rho g w \left( \frac{y_2^2 - y_1^2}{2} \right) \] Plug in values: - \( w = .9 \) - \( y_1 = 10 \) - \( y_2 = 11.1 \) \[ y_2^2 = (11.1)^2 = 123.21 \] \[ y_1^2 = (10)^2 = 100 \] \[ y_2^2 - y_1^2 = 123.21 - 100 = 23.21 \] \[ F = (100)(9.81)(.9) \left( \frac{23.21}{2} \right) \] \[ F = 100 \times 9.81 \times .9 \times 11.605 \] \[ F = 100 \times 9.81 \times 10.4445 \] \[ F = 100 \times 102.499845 \] \[ F \approx 102,500\,\text{N} \] --- ### **Center of Pressure** \[ y_{cp} = \frac{\int y P(y) dA}{\int P(y) dA} \] For a vertical surface: \[ y_{cp} = \frac{1}{A \bar{P}} \int_{A} y P(y)\, dA \] But for a vertical rectangle submerged from \( y_1 \) to \( y_2 \): \[ y_{cp} = \frac{\int_{y_1}^{y_2} y (\rho g y) w\, dy}{F} = \frac{\rho g w \int_{y_1}^{y_2} y^2 dy}{F} = \frac{\rho g w \left[ \frac{y^3}{3} \right]_{y_1}^{y_2}}{F} \] \[ \left[ \frac{y^3}{3} \right]_{y_1}^{y_2} = \frac{y_2^3 - y_1^3}{3} \] \[ y_2^3 = (11.1)^3 = 1369.221 \] \[ y_1^3 = (10)^3 = 100 \] \[ y_2^3 - y_1^3 = 369.221 \] \[ \frac{y_2^3 - y_1^3}{3} = 123.07 \] So: \[ y_{cp} = \frac{\rho g w (123.07)}{F} \] Plug in the values: \[ \rho g w \times 123.07 = 100 \times 9.81 \times .9 \times 123.07 = 100 \times 9.81 \times 110.763 \] \[ = 100 \times 1086.55 = 1,086,550 \] Recall \( F \approx 102,500 \): \[ y_{cp} = \frac{1,086,550}{102,500} \approx 10.6\,\text{m} \] --- ### **Case (b): Car is filled with water** Inside and outside pressure are both hydrostatic and equal. **Net force = ** (since pressure cancels at every point). --- ## **Final Answers** ### **(a) Car is well-sealed (air inside at atmospheric pressure):** - **Net force on door:** \(\boxed{102,500\,\text{N}}\) (toward inside) - **Center of pressure:** \(\boxed{10.6\,\text{m}}\) below the water surface ### **(b) Car is filled with water:** - **Net force:** \(\boxed{\,\text{N}}\) - **(Center of pressure is irrelevant, as net force is zero.)** --- ### **Summary Table** | Case | Net Force (N) | Center of Pressure (m below surface) | |------|---------------|--------------------------------------| | (a) | 102,500 | 10.6 | | (b) | | -- |