15.8
Design Example: Application of Archimedes' Principle
Archimedes' principle is fundamental in analyzing the buoyant force and stability of floating bodies. In this example, a wooden block with a rectangul…
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.
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Q1: How do you calculate the volume of water displaced by a floating wooden block?
The volume of water displaced equals the block's weight divided by the unit weight of water. First, calculate the block's weight by multiplying its total volume by its specific gravity. According to Archimedes' principle, the weight of displaced water equals the block's weight. Dividing this weight by seawater's unit weight (1025 kg(f)/m³) gives the displaced volume.
Q2: What is the relationship between a floating block's weight and the water it displaces?
Archimedes' principle states that the weight of liquid displaced by a floating block equals the block's weight. This equilibrium condition ensures the block floats stably. For a wooden block with specific gravity 0.64 floating in seawater, the buoyant force balances the block's weight, determining how deeply it submerges and establishing buoyancy and stability for submerged and floating bodies.
Q3: Where is the center of buoyancy located on a floating block?
The center of buoyancy is located at the centroid of the submerged portion of the block. It is positioned at half the immersed depth from the bottom of the block. This point represents where the buoyant force acts and indicates the equilibrium location where the upward buoyant force balances the downward weight of the block.
Q4: How do you determine the immersed depth of a floating wooden block?
The immersed depth is found by equating the volume of the submerged portion to the volume of water displaced. Once you know the displaced volume from dividing the block's weight by water's unit weight, divide this by the block's cross-sectional area. For a 1.25 m wide by 2 m deep block, this calculation yields the depth to which the block sinks into the water.
Q5: Why is specific gravity important when analyzing floating bodies?
Specific gravity determines how much of a block's volume must be submerged to achieve equilibrium. A wooden block with specific gravity 0.64 displaces a volume of water equal to 64% of its total volume. This ratio directly controls the immersed depth and the position of the center of buoyancy, making it essential for predicting floating body behavior.
Q6: What role does the unit weight of seawater play in buoyancy calculations?
The unit weight of seawater (1025 kg(f)/m³) is the denominator used to convert the block's weight into the volume of water displaced. A higher unit weight means less volume is needed to displace the same weight, causing the block to float higher. This parameter is critical for determining both the displaced volume and the immersed depth in floating body analysis.
Q7: How do block dimensions affect the calculation of buoyant force?
Block dimensions determine its total volume and cross-sectional area, which are used to calculate weight and immersed depth. A 1.25 m × 2 m × 4 m block's volume multiplied by its specific gravity yields its weight. The cross-sectional area (1.25 m × 2 m) then divides the displaced volume to find immersed depth, showing how geometry directly influences buoyant force distribution.