# Problem Restatement A **quarter-cylinder** (radius \( a \), depth into page \( d \)) is anchored at one tip. - Water is on the **left side** (exerts hydrostatic pressure). - **Right side** is open to air (zero gauge pressure). - The cylinder's height is \( a \) and the submerged height is \( h_1 + h_2 = a \). - **Find:** The **torque** required at the anchor to keep the structure upright. --- # Step 1: Hydrostatic Force on Curved Surface ## Hydrostatic Pressure The pressure at depth \( h \) in water is: \[ P = \rho g h \] where - \( \rho \) = density of water, - \( g \) = gravitational acceleration, - \( h \) = depth below water surface. ## Surface Area The **quarter-cylinder** has: - **Curved surface area:** \( \frac{1}{4} \) of a cylinder's side area - For full cylinder: \( A_\text{cyl} = 2\pi a \cdot d \) - For **quarter**: \( A = \frac{1}{4} (2\pi a d) = \frac{\pi a d}{2} \) --- # Step 2: Resultant Hydrostatic Force The **total force** on a curved surface is equal to the force on its vertical projection (perpendicular to direction of pressure). ## Vertical Projection The vertical projection is a **square** of side \( a \), so area \( = a \cdot d \). ## Total Force The **average pressure** on the vertical projection (from to \( a \)): \[ \bar{P} = \frac{\rho g}{a} \int_^a h\,dh = \frac{\rho g}{a} \cdot \frac{a^2}{2} = \frac{1}{2} \rho g a \] Total force: \[ F = \text{average pressure} \times \text{area} = \left( \frac{1}{2} \rho g a \right) (a d) = \frac{1}{2} \rho g a^2 d \] --- # Step 3: Center of Pressure The **center of pressure** for a vertical surface from to \( a \) is at \[ h_{cp} = \frac{\int_^a h \cdot h\, dh}{\int_^a h\,dh} = \frac{\int_^a h^2\,dh}{\int_^a h\,dh} = \frac{\frac{a^3}{3}}{\frac{a^2}{2}} = \frac{2}{3} a \] below the water surface. --- # Step 4: Calculation of Torque The **torque** about the anchor is: \[ \tau = F \times r \] where - \( F = \frac{1}{2} \rho g a^2 d \) - \( r = \frac{2}{3} a \) (distance from anchor to center of pressure) \[ \tau = \left(\frac{1}{2} \rho g a^2 d\right) \left(\frac{2}{3} a\right) = \frac{1}{3} \rho g a^3 d \] --- # Step 5: Final Answer \[ \boxed{ \tau = \frac{1}{3} \rho g a^3 d } \] where - \( \rho \): density of water, - \( g \): gravitational acceleration, - \( a \): radius of cylinder, - \( d \): depth into the page. --- **Summary:** The torque required by the anchor to keep the quarter-cylinder upright is \[ \boxed{\tau = \frac{1}{3} \rho g a^3 d} \]