Answer :
To solve this problem, we'll apply principles of buoyancy, Archimedes' principle, and some basic calculations in physics.
a) The volume of liquid displaced:
When a body floats in a fluid, according to Archimedes' principle, the weight of the liquid displaced is equal to the weight of the body. The specific gravity of the wood block is 0.64, meaning its density is 0.64 times the density of water (which is approximately $1000 \text{ kg/m}^3$).
First, find the density of the wooden block:
[tex]\text{Density of wood} = 0.64 \times 1000 \text{ kg/m}^3 = 640 \text{ kg/m}^3[/tex]
Next, calculate the volume of the block:
[tex]\text{Volume of the block} = 4 \times 1.25 \times 2 = 10 \text{ m}^3[/tex]
Now, find the weight of the block:
[tex]\text{Weight of the block} = \text{Volume} \times \text{Density of wood} = 10 \times 640 = 6400 \text{ kg}[/tex]
Since the block floats, it displaces a volume of water equal to its own weight. Therefore, the volume of displaced water, which equals the weight of the block divided by the density of water, is:
[tex]\text{Volume of water displaced} = \frac{6400}{1000} = 6.4 \text{ m}^3[/tex]
b) The position of center of buoyancy:
The center of buoyancy is the centroid of the displaced volume of fluid. Since the block is floating horizontally and symmetrically, the center of buoyancy will be at the geometric center of the displaced volume underwater. The block's submerged height can be determined by:
[tex]\frac{\text{Submerged volume}}{\text{Block area}} = \frac{6.4 \text{ m}^3}{4 \times 1.25 \text{ m}^2} = 1.28 \text{ m}[/tex]
Therefore, the center of buoyancy is 1.28 m below the top surface, or equivalently, 2 - 1.28 = 0.72 m above the bottom of the block.
c) The metacentric height:
The metacentric height (GM) can be calculated using the formula:
[tex]GM = BM - BG[/tex]
where BM, the distance between the center of buoyancy and the metacenter, is given by:
[tex]BM = \frac{I}{V}[/tex][tex]I[/tex] is the second moment of area about the axis through the center of buoyancy, and [tex]V[/tex] is the submerged volume.
For a rectangular block:
[tex]I = \frac{1}{12} \times \text{length} \times \text{width}^3 = \frac{1}{12} \times 4 \times (1.25)^3 \approx 0.6510 \text{ m}^4[/tex]
Therefore,
[tex]BM = \frac{0.6510}{6.4} \approx 0.1017 \text{ m}[/tex]
BG, the distance from the center of gravity (1.5 m) to the center of buoyancy (0.72 m):
[tex]BG = 1.5 - 0.72 = 0.78 \text{ m}[/tex]
Finally, calculate GM:
[tex]GM = 0.1017 - 0.78 = -0.6783 \text{ m}[/tex]
This negative GM value indicates the block is unstable in this floating position. It means if the block is slightly tilted, it will not return to its original position without external intervention.
