To determine the maximum height of the center of gravity of the load for the cylindrical buoy to remain in equilibrium, we can follow these steps:

Given Data:

• Diameter of the buoy (d) = 1.8 m → Radius (r) = 0.9 m
• Height of the buoy (h) = 1.2 m
• Weight of the buoy (W_buoy) = 10 kN
• Weight of the load (W_load) = 3 kN
• Density of salt water (ρ) = 1025 kg/m³
• Gravitational acceleration (g) = 9.81 m/s²

Step 1: Calculate the Volume of the Buoy

The volume $V$ of the cylinder is given by the formula: $V = \pi r^2 h$ Substituting the values: $V = \pi (0.9^2)(1.2) \approx 3.194 \, \text{m}^3$

Step 2: Calculate the Weight of the Displaced Water

The weight of the displaced water $W_{water}$ can be calculated using the formula: $W_{water} = V \cdot \rho \cdot g$ $W_{water} = 3.194 \, \text{m}^3 \cdot 1025 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 \approx 32.0 \, \text{kN}$

Step 3: Determine the Height of the Center of Buoyancy

The center of buoyancy (B) is located at the centroid of the submerged volume, which is at half the height of the buoy: $h_B = \frac{1.2}{2} = 0.6 \, \text{m} \text{ from the bottom}$

Step 4: Apply the Conditions for Equilibrium

For the buoy to be in equilibrium, the moments about the point of contact with the water surface must be balanced.

The total weight acting downwards: $W_{total} = W_buoy + W_load = 10 \, \text{kN} + 3 \, \text{kN} = 13 \, \text{kN}$

Step 5: Calculate the New Height of the Center of Gravity

Let $h_G$ be the height of the center of gravity of the total system (buoy + load) from the bottom. It can be expressed as: $h_G = \frac{W_buoy \cdot h_{CG_{buoy}} + W_load \cdot h_{CG_{load}}}{W_{total}}$ Where:

• $h_{CG_{buoy}} = 0.5 \, \text{m}$ (height of the center of gravity of the buoy from the bottom)
• $h_{CG_{load}}$ is the height of the center of gravity of the load from the bottom, which we want to determine.

Setting up the equation: $h_G = \frac{10 \cdot 0.5 + 3 \cdot h_{CG_{load}}}{13}$ The center of gravity $h_G$ must also equal the center of buoyancy $h_B$ to maintain equilibrium: $\frac{10 \cdot 0.5 + 3 \cdot h_{CG_{load}}}{13} = 0.6$

Step 6: Solve for $h_{CG_{load}}$

Multiply through by 13: $10 \cdot 0.5 + 3 \cdot h_{CG_{load}} = 7.8$ $5 + 3 \cdot h_{CG_{load}} = 7.8$ $3 \cdot h_{CG_{load}} = 2.8$ $h_{CG_{load}} = \frac{2.8}{3} \approx 0.933 \, \text{m}$

Step 7: Calculate the Maximum Height of the Load’s Center of Gravity

The maximum height of the center of gravity of the load from the bottom of the buoy is: $\text{Maximum height from the bottom} = 0.933 + 1.2 = 2.133 \, \text{m}$

Conclusion

The maximum height of the center of gravity of the load from the bottom of the buoy for it to remain in equilibrium is approximately 2.133 m.

Further Questions:

1. What would happen if the buoy were placed in fresh water instead?
2. How would increasing the height of the buoy affect stability?
3. What other factors could impact the equilibrium of the buoy?
4. How does the shape of the buoy affect its stability in water?
5. What methods can be used to calculate the stability of floating objects?

Tip:

Understanding buoyancy and stability involves balancing forces and torques, so always visualize the forces acting on the object!