Let's break down and solve the problem step by step: --- ## **Given Data** - **Steel sphere diameter (\(d_{sphere}\))**: .3 m - **Density of steel (\(\rho_{steel}\))**: 785 kg/m³ - **Density of plastic shell (\(\rho_{plastic}\))**: 940 kg/m³ - **Density of polystyrene foam (\(\rho_{foam}\))**: 80 kg/m³ - **Density of seawater (\(\rho_{sea}\))**: 1025 kg/m³ - **Metacentric height (\(GM\))**: 1.4 m --- ### **Step 1: Find the Weight of the Steel Sphere** #### **Volume of Sphere** \[ V_{sphere} = \frac{4}{3} \pi r^3 \] \[ r = \frac{.3}{2} = .15 \text{ m} \] \[ V_{sphere} = \frac{4}{3} \pi (.15)^3 = \frac{4}{3} \pi (.003375) \approx .01414\ \text{m}^3 \] #### **Mass of Sphere** \[ m_{sphere} = \rho_{steel} \cdot V_{sphere} = 785 \times .01414 \approx 111.1\ \text{kg} \] #### **Weight of Sphere** \[ W_{sphere} = m_{sphere} \cdot g = 111.1 \times 9.81 \approx 109\ \text{N} \] --- ### **Step 2: Draught of the Buoyant Cylinder** Let's denote: - \(d\) = draught (depth of immersion of buoy) - \(D\) = diameter of buoy (unknown, as not provided; we’ll keep in terms of \(D\)) - \(H\) = height of buoy (also not given; if needed, keep in terms of \(H\)) #### **Buoyant Force = Total Weight (Buoy + Sphere)** Assume the weight of rod is negligible and the buoy's mass is much less than the sphere’s. \[ \text{Buoyant force} = W_{sphere} \] \[ \text{Buoyant force} = \text{Volume submerged} \times \rho_{sea} \times g \] \[ \text{Volume submerged} = \pi \left(\frac{D}{2}\right)^2 d \] \[ \text{Buoyant force} = \pi \left(\frac{D}{2}\right)^2 d \times \rho_{sea} \times g \] Set equal to sphere weight: \[ W_{sphere} = \pi \left(\frac{D}{2}\right)^2 d \times \rho_{sea} \times g \] \[ 109 = \pi \left(\frac{D}{2}\right)^2 d \times 1025 \times 9.81 \] \[ 109 = \pi \left(\frac{D}{2}\right)^2 d \times 10055.25 \] \[ \boxed{ d = \frac{109}{\pi \left(\frac{D}{2}\right)^2 10055.25} } \] **You need the diameter of the buoy to find a numerical answer for the draught. If you have the value of \(D\), substitute to get \(d\).** --- ### **Step 3: Metacentric Height Condition** The metacentric height \(GM\) is given by: \[ GM = BM - BG \] where: - \(BM = \frac{I}{V}\), \(I\) = second moment of area of waterplane, \(V\) = submerged volume - \(BG\) = distance from center of buoyancy to center of gravity #### **For a Cylinder:** - Waterplane area: \(A = \pi \left(\frac{D}{2}\right)^2\) - \(I = \frac{\pi}{64} D^4\) - Submerged volume: \(V = \pi \left(\frac{D}{2}\right)^2 d\) \[ BM = \frac{I}{V} = \frac{\frac{\pi}{64} D^4}{\pi \left(\frac{D}{2}\right)^2 d} = \frac{D^2}{16d} \] So, \[ GM = \frac{D^2}{16d} - BG \] Given \(GM = 1.4\,m\): \[ 1.4 = \frac{D^2}{16d} - BG \] --- ### **Step 4: Find BG (Buoyancy to Center of Gravity)** Assume the center of gravity is below waterline by a distance \(h\). For a floating system with a heavy weight below, the center of gravity is lowered, increasing stability. Let’s denote: - \(h\) = depth below waterline to center of gravity (to be found in part ii) \[ BG = d - h \] So, \[ 1.4 = \frac{D^2}{16d} - (d - h) \] \[ 1.4 = \frac{D^2}{16d} - d + h \] \[ h = 1.4 + d - \frac{D^2}{16d} \] --- ### **Step 5: Center of Gravity Calculation** The center of gravity is found using the moments: \[ h = \frac{m_{buoy} \cdot h_{buoy} + m_{sphere} \cdot h_{sphere}}{m_{buoy} + m_{sphere}} \] Assume the buoy's mass is negligible: \[ h \approx h_{sphere} \] If the sphere is suspended at depth \(L\) below the waterline: \[ \text{Depth below mean sea level} = L + r \] where \(r = .15\,m\) (center of sphere is one radius below its top). --- ## **Summary of Steps and Final Answers** ### **(i) Draught of Buoyant Cylinder** \[ d = \frac{109}{\pi \left(\frac{D}{2}\right)^2 10055.25} \] *You need the actual buoy diameter \(D\) to get a numerical value.* ### **(ii) Depth of Center of Gravity Below Mean Sea Level** \[ h = 1.4 + d - \frac{D^2}{16d} \] *You need \(D\) and \(d\) to calculate this value. If the weight of the buoy is negligible, \(h \approx\) the vertical distance from the waterline to the sphere’s center.* --- ### **If Additional Data Provided** If the diameter of the buoy (\(D\)) is provided from part (a) or elsewhere, plug in that value to get explicit numbers for draught and center of gravity. --- ### **Key Formulas Used** - **Sphere volume:** \(V_{sphere} = \frac{4}{3}\pi r^3\) - **Buoyant force:** \(F_B = \rho_{sea} g V_{sub}\) - **Moment of inertia (cylinder):** \(I = \frac{\pi}{64} D^4\) - **Metacentric height:** \(GM = \frac{D^2}{16d} - BG\) --- **Let me know the diameter of the buoy if you want explicit numerical answers!**