Understanding the Calculation of Thrust Using Archimedes’ Principle
Thrust calculation via Archimedes’ principle quantifies buoyant force on submerged objects. This article explores detailed methodologies and applications.
Discover comprehensive formulas, variable explanations, and real-world examples to master thrust calculation in fluid mechanics.
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- Calculate the thrust on a steel cube submerged in freshwater at 20°C.
- Determine the buoyant force on a helium balloon in air at sea level.
- Compute the thrust acting on a submarine hull at 100 meters depth in seawater.
- Evaluate the buoyant force on a wooden log floating partially submerged in river water.
Comprehensive Tables of Common Values for Thrust Calculation
Accurate thrust calculation requires knowledge of fluid densities, gravitational acceleration, and typical object volumes. The following tables provide essential reference values for these variables under standard conditions.
| Fluid | Density (kg/m³) | Temperature (°C) | Gravitational Acceleration (m/s²) | Notes |
|---|---|---|---|---|
| Freshwater | 998.2 | 20 | 9.80665 | Standard freshwater density at room temperature |
| Seawater | 1025 | 15 | 9.80665 | Average ocean water density |
| Air (dry, sea level) | 1.225 | 15 | 9.80665 | Standard atmospheric conditions |
| Mercury | 13546 | 20 | 9.80665 | High-density fluid used in barometers |
| Oil (crude) | 850 | 25 | 9.80665 | Typical crude oil density |
| Liquid Nitrogen | 807 | -196 | 9.80665 | Cryogenic fluid |
| Object Material | Typical Density (kg/m³) | Common Volume Range (m³) | Application Examples |
|---|---|---|---|
| Steel | 7850 | 0.001 – 10 | Ship hulls, structural components |
| Wood (Oak) | 700 | 0.01 – 5 | Boat construction, floating logs |
| Aluminum | 2700 | 0.001 – 2 | Aerospace, lightweight structures |
| Plastic (Polyethylene) | 950 | 0.001 – 1 | Buoy markers, containers |
| Helium Gas | 0.1786 (kg/m³ at STP) | Variable | Balloons, airships |
Fundamental Formulas for Thrust Calculation Based on Archimedes’ Principle
Archimedes’ principle states that the buoyant force (thrust) on a submerged object equals the weight of the fluid displaced by the object. This principle is mathematically expressed as follows:
Thrust (Buoyant Force), Fb = ρfluid × Vdisplaced × g
- Fb: Buoyant force or thrust (Newtons, N)
- ρfluid: Density of the fluid (kg/m³)
- Vdisplaced: Volume of fluid displaced by the object (m³)
- g: Acceleration due to gravity (m/s²), typically 9.80665 m/s²
Each variable plays a critical role in determining the magnitude of the thrust:
- Density (ρfluid): Varies with fluid type, temperature, and salinity. For example, seawater is denser than freshwater, increasing buoyant force.
- Displaced Volume (Vdisplaced): Equal to the volume of the submerged portion of the object. For fully submerged objects, this equals the object’s volume.
- Gravity (g): Standard gravitational acceleration on Earth’s surface; slight variations occur with altitude and latitude.
For partially submerged objects, the volume displaced is less than the total volume, requiring precise measurement or calculation of the submerged fraction.
Additional Relevant Formulas
To fully analyze thrust and related forces, the following formulas are also essential:
Weight of Object, W = ρobject × Vobject × g
Net Force, Fnet = Fb – W
- ρobject: Density of the object (kg/m³)
- Vobject: Total volume of the object (m³)
- Fnet: Net upward or downward force (N)
The net force determines whether the object sinks, floats, or remains neutrally buoyant:
- If Fnet > 0, the object experiences upward thrust and tends to float.
- If Fnet = 0, the object is neutrally buoyant.
- If Fnet < 0, the object sinks.
Detailed Explanation of Variables and Their Typical Values
Density of Fluid (ρfluid): This is a critical parameter that depends on the fluid’s composition and temperature. For example, freshwater at 20°C has a density of approximately 998.2 kg/m³, while seawater averages around 1025 kg/m³ due to dissolved salts. Air density at sea level is much lower, about 1.225 kg/m³, which significantly affects buoyancy calculations for gases and balloons.
Volume of Displaced Fluid (Vdisplaced): This volume corresponds to the portion of the object submerged in the fluid. For fully submerged objects, it equals the total volume. For floating objects, it is the submerged fraction, which can be calculated by balancing forces or measured experimentally.
Acceleration Due to Gravity (g): Standard gravity is 9.80665 m/s², but it can vary slightly depending on geographic location. For most engineering calculations, this standard value is used unless high precision is required.
Density of Object (ρobject): The object’s density determines its weight and interaction with the fluid. Materials like steel (7850 kg/m³) are much denser than water, causing them to sink unless shaped to displace sufficient fluid volume. Conversely, wood (around 700 kg/m³) is less dense and typically floats.
Real-World Applications and Case Studies
Case Study 1: Buoyant Force on a Submerged Steel Cube in Freshwater
Consider a steel cube with a side length of 0.1 m fully submerged in freshwater at 20°C. Calculate the buoyant force acting on the cube.
- Density of freshwater, ρfluid = 998.2 kg/m³
- Density of steel, ρobject = 7850 kg/m³
- Volume of cube, V = (0.1 m)³ = 0.001 m³
- Gravity, g = 9.80665 m/s²
Using the formula for buoyant force:
Fb = ρfluid × V × g = 998.2 × 0.001 × 9.80665 = 9.79 N
Weight of the steel cube:
W = ρobject × V × g = 7850 × 0.001 × 9.80665 = 76.96 N
Net force:
Fnet = Fb – W = 9.79 – 76.96 = -67.17 N
The negative net force indicates the cube sinks, as expected for steel in water.
Case Study 2: Buoyant Force on a Helium Balloon in Air
Calculate the thrust acting on a helium balloon with a volume of 0.5 m³ at sea level (15°C, 1 atm).
- Density of air, ρfluid = 1.225 kg/m³
- Density of helium, ρobject = 0.1786 kg/m³
- Volume of balloon, V = 0.5 m³
- Gravity, g = 9.80665 m/s²
Buoyant force:
Fb = ρfluid × V × g = 1.225 × 0.5 × 9.80665 = 6.00 N
Weight of helium inside balloon:
W = ρobject × V × g = 0.1786 × 0.5 × 9.80665 = 0.875 N
Net upward force (thrust):
Fnet = Fb – W = 6.00 – 0.875 = 5.125 N
This net force is the effective thrust lifting the balloon upward.
Additional Considerations for Accurate Thrust Calculations
- Temperature Effects: Fluid density varies with temperature, affecting buoyant force. For precision, use temperature-corrected density values.
- Salinity and Pressure: In seawater, salinity and depth pressure alter density. Deep-sea applications require pressure-corrected densities.
- Partial Submersion: For floating bodies, the submerged volume must be calculated using equilibrium conditions or measured directly.
- Dynamic Fluids: In moving fluids, additional forces such as drag and lift may influence net thrust.
Recommended External Resources for Further Study
- Engineering Toolbox: Archimedes’ Principle – Comprehensive resource on buoyancy and fluid properties.
- NASA Glenn Research Center: Buoyancy and Archimedes’ Principle – Educational material with practical examples.
- Encyclopedia Britannica: Archimedes’ Principle – Historical and scientific background.
- ScienceDirect Topics: Archimedes’ Principle – Technical articles and research papers.