15.8
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 rectangul…
Copyright © 2026 MyJoVE Corporation. All rights reserved.