To solve this problem, we will break it down into several steps, involving both buoyancy and weight calculations.

Step 1: Volume of the Spherical Buoy

The diameter of the spherical buoy is given as 1.8 meters. The volume $V$ of a sphere is calculated using the formula:

$V = \frac{4}{3} \pi r^3$

where $r$ is the radius of the sphere, which is half the diameter.

$r = \frac{1.8}{2} = 0.9 \text{ meters}$

Thus, the volume of the buoy is:

$V = \frac{4}{3} \pi (0.9)^3 = \frac{4}{3} \pi \times 0.729 = 3.054 \text{ m}^3$

Step 2: Buoyant Force

The buoyant force $F_b$ is equal to the weight of the liquid displaced by the fully submerged buoy. The buoy is floating in a liquid with a specific gravity of 1.5. Since the specific gravity is 1.5, the density of the liquid $\rho_{\text{liquid}}$ is:

$\rho_{\text{liquid}} = 1.5 \times 1000 = 1500 \text{ kg/m}^3$

The buoyant force is given by:

$F_b = \rho_{\text{liquid}} \times V_{\text{buoy}} \times g$

where:

• $V_{\text{buoy}}$ is the volume of the buoy,
• $\rho_{\text{liquid}}$ is the density of the liquid,
• $g$ is the acceleration due to gravity (approximated as $9.81 \, \text{m/s}^2$).

Substituting the values:

$F_b = 1500 \times 3.054 \times 9.81 = 44,969.61 \, \text{N}$

This is the buoyant force acting on the buoy when it is fully submerged.

Step 3: Weight of the Buoy

Since the buoy floats half-submerged, its weight is half the buoyant force when it is fully submerged. Therefore, the weight $W_{\text{buoy}}$ of the buoy is:

$W_{\text{buoy}} = \frac{1}{2} \times F_b = \frac{1}{2} \times 44,969.61 = 22,484.81 \, \text{N}$

Step 4: Additional Force to Submerge the Buoy Completely

To submerge the buoy completely, the additional downward force needed is the difference between the buoyant force and the weight of the buoy:

$F_{\text{extra}} = F_b - W_{\text{buoy}} = 44,969.61 - 22,484.81 = 22,484.81 \, \text{N}$

Step 5: Weight of the Lead Anchor

The weight of the lead anchor required to provide this additional force must equal the extra force needed to submerge the buoy completely. The density of lead is given as $7220 \, \text{kg/m}^3$.

The weight of an object is given by:

$W = \rho \times V \times g$

Rearranging this to solve for the volume $V$ of lead required:

$V_{\text{lead}} = \frac{F_{\text{extra}}}{\rho_{\text{lead}} \times g} = \frac{22,484.81}{7220 \times 9.81} = 0.317 \, \text{m}^3$

Thus, the volume of lead required is $0.317 \, \text{m}^3$.

Finally, the weight of the lead anchor is:

$W_{\text{lead}} = \rho_{\text{lead}} \times V_{\text{lead}} = 7220 \times 0.317 = 2287.74 \, \text{kg}$

Conclusion:

The minimum weight of the lead anchor required to completely submerge the buoy is approximately 2287.74 kg.

Do you want more details or have any questions?

Here are some related questions:

1. How would the result change if the specific gravity of the liquid were different?
2. What is the significance of specific gravity in determining buoyancy?
3. How would the required anchor weight change if the buoy had a different shape (e.g., a cube)?
4. How does the concept of Archimedes' principle apply to this problem?
5. Can the anchor's material be replaced with something lighter, and how would that affect the calculations?

Tip: Always double-check the units in buoyancy and force calculations to ensure consistency across all terms.