To solve this problem, we need to apply principles from hydrostatics and buoyancy. Specifically, we'll use Archimedes' principle, which states that the buoyant force is equal to the weight of the displaced liquid.

Step-by-Step Solution

1. Given Information:

• Diameter of the spherical buoy, $D = 1.8 \, \text{m}$.
• Specific gravity of the liquid, $SG = 1.5$.
• Density of lead, $\rho_{\text{lead}} = 7220 \, \text{kg/m}^3$.

We are tasked with finding the weight of the lead anchor needed to fully submerge the buoy.

2. Volume of the Buoy:

The volume $V_{\text{buoy}}$ of the sphere is calculated using the formula for the volume of a sphere: $V_{\text{buoy}} = \frac{4}{3} \pi \left(\frac{D}{2}\right)^3$ Substituting $D = 1.8 \, \text{m}$: $V_{\text{buoy}} = \frac{4}{3} \pi \left(\frac{1.8}{2}\right)^3 = \frac{4}{3} \pi (0.9)^3$ $V_{\text{buoy}} \approx 3.05 \, \text{m}^3$

3. Buoyant Force:

The buoyant force $F_b$ is equal to the weight of the liquid displaced. The weight of the displaced liquid depends on the volume of the buoy and the density of the liquid.

The density of the liquid $\rho_{\text{liquid}}$ can be calculated using its specific gravity: $\rho_{\text{liquid}} = SG \times \rho_{\text{water}} = 1.5 \times 1000 \, \text{kg/m}^3 = 1500 \, \text{kg/m}^3$

Now, the buoyant force $F_b$ is: $F_b = \rho_{\text{liquid}} \times V_{\text{buoy}} \times g$ where $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity).

Substitute the values: $F_b = 1500 \times 3.05 \times 9.81 \approx 44,900 \, \text{N}$

4. Weight of the Buoy:

The buoy is half-submerged, so the current buoyant force is balancing half its weight. To submerge it completely, we need to add weight.

The total weight of the buoy, $W_{\text{buoy}}$, is: $W_{\text{buoy}} = \frac{F_b}{2} = \frac{44,900}{2} \approx 22,450 \, \text{N}$

The mass of the buoy $m_{\text{buoy}}$ is: $m_{\text{buoy}} = \frac{W_{\text{buoy}}}{g} = \frac{22,450}{9.81} \approx 2,288 \, \text{kg}$

5. Required Weight of Lead Anchor:

To completely submerge the buoy, the total downward force (buoy's weight plus the lead anchor's weight) must be equal to the buoyant force.

Let the weight of the lead anchor be $W_{\text{anchor}}$. We need: $W_{\text{buoy}} + W_{\text{anchor}} = F_b$ $W_{\text{anchor}} = F_b - W_{\text{buoy}} = 44,900 - 22,450 = 22,450 \, \text{N}$

The mass of the lead anchor $m_{\text{anchor}}$ is: $m_{\text{anchor}} = \frac{W_{\text{anchor}}}{g} = \frac{22,450}{9.81} \approx 2,288 \, \text{kg}$

6. Volume of Lead Anchor:

Finally, the volume of the lead anchor $V_{\text{anchor}}$ can be calculated using the formula: $V_{\text{anchor}} = \frac{m_{\text{anchor}}}{\rho_{\text{lead}}} = \frac{2,288}{7220} \approx 0.317 \, \text{m}^3$

Answer:

The minimum weight of the lead anchor required to completely submerge the buoy is approximately $2,288 \, \text{kg}$.

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Do you want further details or have any questions?
Here are some related questions:

1. What is Archimedes' principle and how does it apply to buoyancy?
2. How would the result change if the liquid had a lower specific gravity?
3. What happens if the buoy were less than half-submerged initially?
4. How would we calculate the forces if the shape of the buoy were different (e.g., a cylinder)?
5. What is the effect of gravity variations on the buoyant force?

Tip: When solving hydrostatic problems, ensure that units are consistent throughout the calculations.