15.8
Design Example: Application of Archimedes' Principle
Consider a wooden block with dimensions 1.25 meters wide, 2 meters deep, and 4 meters long floating in water held in a rectangular container.
The specific gravity of wood is 0.64, and the weight of water is 1025 kg(f)/m3.
The objective is to determine the volume of liquid displaced by the block and locate the block's center of buoyancy.
The weight of the liquid displaced by the block is equal to the weight of the block, as per Archimedes' principle.
The volume of water displaced can be calculated by dividing the block's weight by the unit weight of water.
Once the submerged block's depth is known, the position of the center of buoyancy can be calculated.
This depth can be obtained by equating the volume of the block's immersed portion to the volume of the water displaced by the block.
The center of buoyancy is located at the center of gravity of the submerged part of the block, which is at half the depth of the submerged portion.
Archimedes' principle is fundamental in analyzing the buoyant force and stability of floating bodies. In this example, a wooden block with a rectangular section floats in seawater. Based on the block's dimensions, its specific gravity and the specific weight of seawater are used to find the volume of water displaced and the center of buoyancy.
The volume of seawater displaced by the block is determined by first calculating the block's weight. This is done by multiplying the block's volume by its specific gravity. According to Archimedes' principle, the weight of displaced water is equal to the weight of the block. The volume of displaced water is then found by dividing the block's weight by the unit weight of seawater.
The center of buoyancy, which is the centroid of the displaced water volume, indicates where the buoyant force acts. To locate this point, the immersed depth is calculated by equating the submerged portion's volume to the volume of displaced water. The center of buoyancy is positioned at half the immersed depth from the bottom of the block, representing the equilibrium point where the buoyant force balances the weight of the block.
Consider a wooden block with dimensions 1.25 meters wide, 2 meters deep, and 4 meters long floating in water held in a rectangular container.
The specific gravity of wood is 0.64, and the weight of water is 1025 kg(f)/m3.
The objective is to determine the volume of liquid displaced by the block and locate the block's center of buoyancy.
The weight of the liquid displaced by the block is equal to the weight of the block, as per Archimedes' principle.
The volume of water displaced can be calculated by dividing the block's weight by the unit weight of water.
Once the submerged block's depth is known, the position of the center of buoyancy can be calculated.
This depth can be obtained by equating the volume of the block's immersed portion to the volume of the water displaced by the block.
The center of buoyancy is located at the center of gravity of the submerged part of the block, which is at half the depth of the submerged portion.
From Chapter 15:
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15.8 : Design Example: Application of Archimedes' Principle
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