Buoyant Force Equation: Formula, 7 Forms, and Worked Examples

The buoyant force equation is F_b = ρ_fluid × g × V_displaced, which gives the upward force of a fluid on an immersed object. The force equals the weight of the displaced fluid and is expressed in newtons (N). The 7 forms cover displaced volume, displaced mass, apparent weight, floating objects, net force, specific gravity, and gases.

For a reliable reference, see OpenStax explains Archimedes principle and the weight of displaced fluid.

Both problems use force balance; compare the setup with Inclined Plane Physics: Forces, Formulas, Friction, and Worked Examples.

What Is the Buoyant Force Equation and What Are Examples of It?

The buoyant force equation is the formula that calculates the upward force of a fluid on an object from the fluid density, the displaced volume, and gravitational acceleration. The principle behind it is called Archimedes’ principle. According to OpenStax University Physics Volume 1, the buoyant force on an object equals the weight of the fluid the object displaces. Archimedes of Syracuse described the principle in On Floating Bodies in the 3rd century BC.

Buoyant force is a vector quantity. It acts upward, opposite to gravity, through the center of the displaced fluid volume. Other names include upthrust and flotation force.

Fluids include liquids and gases. Examples of buoyancy include ships floating on seawater, icebergs, submarines at depth, swimmers, and helium balloons in air.

The table below defines each symbol in the equation F_b = ρ_fluid × g × V_displaced. The last column gives a typical value on Earth.

What Are the 7 Forms of the Buoyant Force Equation?

The 7 forms use displaced volume, displaced mass, apparent weight, object weight, density difference, specific gravity, or gas densities. The table below lists the equation, the known quantities, and the condition for each form.

Symbols in the table: m_f is the mass of displaced fluid (kg), W is the weight of the object in air (N), W_app is the apparent weight in the fluid (N), ρ_o is the object density (kg/m³), SG_f is the specific gravity of the fluid (no unit), and ρ_water is 1,000 kg/m³.

How Do You Calculate Buoyant Force?

Buoyant force is calculated by multiplying the fluid density, the gravitational acceleration, and the displaced volume. The 6 steps are:

1. Identify the fluid and find its density in kg/m³.
2. Identify the displaced volume, which equals the submerged volume of the object.
3. Convert the volume to m³.
4. Use g = 9.8 m/s².
5. Multiply ρ_fluid × g × V_displaced.
6. Express the result in newtons, with the direction upward.

Example: A block with a volume of 0.002 m³ sits fully submerged in freshwater. The buoyant force is 1,000 × 9.8 × 0.002 = 19.6 N.

The same block in seawater experiences 1,025 × 9.8 × 0.002 = 20.1 N, which is 2.5% larger.

How Is the Buoyant Force Equation Derived?

The equation is derived from the difference between the hydrostatic pressure on the bottom face and the top face of a submerged object. Fluid pressure at depth h is P = P₀ + ρgh, where P₀ is the pressure at the surface.

The 5 steps for a cylinder with a cross-sectional area A, a top face at depth h₁, and a bottom face at depth h₂ are:

1. The fluid pushes down on the top face with a force of (P₀ + ρgh₁)A.
2. The fluid pushes up on the bottom face with a force of (P₀ + ρgh₂)A.
3. The net upward force is ρg(h₂ − h₁)A, because P₀ cancels.
4. The volume of the cylinder is V = (h₂ − h₁)A.
5. The buoyant force is F_b = ρgV.

Forces on the side faces cancel in pairs. Example: A cylinder with A = 0.01 m² and a height of 0.10 m has a volume of 0.001 m³. The pressure difference is 1,000 × 9.8 × 0.10 = 980 Pa, and the force is 980 × 0.01 = 9.8 N.

What Does the Buoyant Force Depend On?

Buoyant force depends on 3 factors: the fluid density, the displaced volume, and the gravitational acceleration. The mass of the object, the material of the object, and the depth of a fully submerged object do not change the force. The table below lists 6 changes and the effect of each on the buoyant force.

Fluid density changes with temperature. Water has a density of about 1,000 kg/m³ at 4 °C, 998.2 kg/m³ at 20 °C, and 958.4 kg/m³ at 100 °C. Water at 100 °C supports 4.0% less force per unit volume than water at 20 °C.

Compressible objects such as air-filled lungs and foam change volume with depth, so their buoyant force decreases as the pressure increases.

What Are the Densities of Common Fluids and the Buoyant Force on 1 m³?

Fluid density sets the buoyant force per unit volume. The table below lists 7 fluids with the approximate density at about 20 °C and 1 atm, and the buoyant force on 1 m³ of displaced fluid, calculated with g = 9.8 m/s². Freshwater is rounded to 1,000 kg/m³.

Seawater has an average salinity of about 35 g/kg, which raises its density above that of freshwater.

How Do You Find Apparent Weight With the Buoyant Force Equation?

Apparent weight is found by subtracting the buoyant force from the weight of the object in air (W_app = W − F_b). Apparent weight is the reading of a spring scale that holds a submerged object. The equivalent form is W_app = (ρ_o − ρ_fluid) g V.

Example: An aluminum block has a volume of 0.001 m³ and a density of 2,700 kg/m³.

1. The mass is 2,700 × 0.001 = 2.7 kg.
2. The weight in air is 2.7 × 9.8 = 26.46 N.
3. The buoyant force in freshwater is 1,000 × 9.8 × 0.001 = 9.8 N.
4. The apparent weight is 26.46 − 9.8 = 16.66 N.

The table below lists 8 solids in freshwater with a density of 1,000 kg/m³. The columns give the density, the result, the percent of volume that is submerged, and the apparent weight as a percent of the weight in air, calculated as 1 − 1,000 / ρ_o for sinking objects.

How Do You Know if an Object Sinks, Floats, or Stays Suspended?

An object sinks if its density exceeds the fluid density, floats if its density is lower, and stays suspended if the 2 densities are equal. The table below compares the object density with the fluid density, the buoyant force with the weight for a fully submerged object, and the result.

Average density decides the result for hollow and composite objects. Example: A hollow steel box has a mass of 1,000 kg and an enclosed volume of 2.0 m³. The average density is 500 kg/m³, so the box floats with 50% of its volume submerged.

How Do You Find the Submerged Fraction of a Floating Object?

The submerged fraction of a floating object equals the object density divided by the fluid density (V_sub / V_total = ρ_o / ρ_fluid). The relationship follows from F_b = W, because ρ_fluid g V_sub = ρ_o g V_total.

Example for a block: A block 0.20 m tall has a density of 600 kg/m³ and a base area of 0.05 m². It floats in freshwater.

1. The submerged fraction is 600 / 1,000 = 0.60.
2. The submerged depth is 0.20 × 0.60 = 0.12 m.
3. The submerged volume is 0.05 × 0.12 = 0.006 m³.
4. The buoyant force is 1,000 × 9.8 × 0.006 = 58.8 N.
5. The weight is 600 × 0.01 × 9.8 = 58.8 N, which matches.

Example for ice: Ice has a density of 917 kg/m³. In freshwater, 91.7% of the volume is submerged. In seawater, 917 / 1,025 gives 89.5% submerged and 10.5% above the surface.

How Do You Find Density or Volume With the Buoyant Force Equation?

Density is found with ρ_o = ρ_fluid × W / (W − W_app), which uses the weight in air and the apparent weight in a fluid. The formula follows from F_b = W − W_app = ρ_fluid g V and W = ρ_o g V. The volume equals V = F_b / (ρ_fluid g).

Vitruvius recorded the story of Archimedes and the golden crown in the 1st century BC. The story describes a method that finds the density of an irregular object.

Example: A crown weighs 9.8 N in air and 9.0 N in freshwater.

1. The buoyant force is 9.8 − 9.0 = 0.8 N.
2. The volume is 0.8 / (1,000 × 9.8) = 8.16 × 10⁻⁵ m³.
3. The density is 1,000 × 9.8 / 0.8 = 12,250 kg/m³.

Pure gold has a density of 19,300 kg/m³ and gives an apparent weight of 9.29 N for a 9.8 N sample. The measured density of 12,250 kg/m³ indicates a metal mixture with a lower average density.

How Do You Calculate Buoyant Force in Gases and Balloons?

Net lift of a balloon is found with F_lift = (ρ_air − ρ_gas) × g × V. Air acts as the fluid, and the balloon displaces its own volume of air. The lift is available to support the balloon structure and the payload.

Example for a helium balloon: Air has a density of 1.204 kg/m³ and helium has a density of 0.166 kg/m³ at 20 °C. A balloon with a volume of 1 m³ has a net lift of (1.204 − 0.166) × 9.8 × 1 = 10.2 N. The lift supports about 1.04 kg.

Example for a hot air balloon: Air inside the envelope at 100 °C has a density of 0.946 kg/m³. The envelope volume is 2,800 m³. The net lift is (1.204 − 0.946) × 9.8 × 2,800 = 7,080 N. The lift supports about 722 kg of envelope, basket, fuel, and passengers.

How Does the Buoyant Force Equation Apply to Ships and Submarines?

A ship floats when the buoyant force equals the weight of the ship, so the ship displaces a mass of water equal to its own mass. The shape of the hull spreads the mass over a large displaced volume.

Example for a ship: A ship has a mass of 50,000,000 kg.

1. The buoyant force is 50,000,000 × 9.8 = 4.9 × 10⁸ N.
2. The displaced volume in seawater is 50,000,000 / 1,025 = 48,780 m³.
3. The displaced volume in freshwater is 50,000,000 / 1,000 = 50,000 m³.

The ship sits 2.5% deeper in freshwater, in terms of displaced volume. The Merchant Shipping Act of 1876 in the United Kingdom required the Plimsoll mark, which shows the maximum safe loading depth.

A submarine controls buoyancy with ballast tanks. Water in the tanks increases the mass, and compressed air that expels the water decreases the mass.

Example for a submarine: A submarine has an outer volume of 3,000 m³. The buoyant force in seawater is 1,025 × 9.8 × 3,000 = 3.01 × 10⁷ N. Neutral buoyancy occurs at a total mass of 1,025 × 3,000 = 3,075,000 kg.

How Does Gravity Change the Buoyant Force?

Buoyant force is proportional to the gravitational acceleration g. Weight is also proportional to g, so the outcomes of floating and sinking stay the same on every body. The table below lists the surface gravity of 3 bodies and the buoyant force on 1 m³ of displaced water.

What Determines the Stability of a Floating Object?

A floating object is stable when the metacenter lies above the center of gravity. 3 points define the stability of a floating body:

• Center of gravity (G): the point where the weight acts.
• Center of buoyancy (B): the centroid of the displaced fluid volume, where the buoyant force acts.
• Metacenter (M): the point where the line of action of the buoyant force crosses the vertical centerline after a small tilt.

The distance GM is the metacentric height. A positive GM produces a restoring torque that returns the object to its upright position. A negative GM causes the object to capsize. Heavy cargo placed low lowers G and increases GM.

What Is the Difference Between Buoyant Force, Weight, Density, and Pressure?

Buoyant force is the upward force of the fluid, weight is the downward force of gravity, density is mass per unit volume, and pressure is force per unit area. The table below compares 5 related quantities by definition, equation, unit, and role in buoyancy.

What Are the Units of the Buoyant Force Equation and How Are They Converted?

The SI unit of buoyant force is the newton (N), which equals 1 kg·m/s². NIST defines standard gravity as exactly 9.80665 m/s². The table below lists 5 quantities with the symbol, the SI unit, and the conversions.

The specific weight form is F_b = γ V. Example: A displaced volume of 3 ft³ of freshwater gives a buoyant force of 62.4 × 3 = 187.2 lbf.

Example conversion: A 500 cm³ object equals 5.0 × 10⁻⁴ m³. The buoyant force in water is 1,000 × 9.8 × 5.0 × 10⁻⁴ = 4.9 N.

How Do You Measure Buoyant Force Experimentally?

Buoyant force is measured by comparing the weight of an object in air with its apparent weight in a fluid, using a spring scale. An overflow can collects the displaced fluid for a check of Archimedes’ principle. The 6 steps are:

1. Weigh the object in air and record W.
2. Fill an overflow can with water up to the spout.
3. Lower the object into the water while it hangs from the scale.
4. Collect the overflow water in a cup.
5. Record the apparent weight W_app and calculate F_b = W − W_app.
6. Weigh the collected water and compare m_f g with F_b.

Example: An object weighs 4.0 N in air and 3.0 N in water. The buoyant force is 1.0 N. The collected water has a mass of 1.0 / 9.8 = 0.102 kg, which equals 102 g. The density of the object is 1,000 × 4.0 / 1.0 = 4,000 kg/m³.

Repeating the measurement in 3 trials reduces random error.

What Are the Limits of the Buoyant Force Equation?

The buoyant force equation applies to an object fully or partly immersed in a fluid at rest, with fluid contacting the submerged surfaces. 6 conditions limit the equation:

• Moving fluids add drag and dynamic forces to the buoyant force.
• An object sealed against the container bottom, with no fluid underneath, experiences no buoyant force.
• Surface tension supports small objects such as water striders and needles in addition to buoyancy.
• Density changes with depth, temperature, and salinity, so the equation uses an average density or an integral.
• Compressible objects change volume with depth.
• Accelerating containers replace g with the effective gravitational acceleration.

What Are the Common Mistakes When Using the Buoyant Force Equation?

7 common mistakes cause incorrect buoyant force results. Loverude, Kautz, and Heron published a study in the American Journal of Physics in 2003 and found that many university students have difficulty applying Archimedes’ principle and relating buoyancy to fluid pressure.

The mistakes include:

• Using the total volume of a floating object instead of the submerged volume.
• Using the object density instead of the fluid density.
• Mixing units, such as cm³ for volume with kg/m³ for density.
• Treating mass in kilograms as force in newtons.
• Assuming the buoyant force increases with depth for a fully submerged object.
• Assuming heavy objects always sink.
• Applying the freshwater density to seawater.

Example of the first mistake: A floating block has a total volume of 0.01 m³ and a submerged volume of 0.006 m³. Using 0.01 m³ gives 98 N, while the correct value is 58.8 N.

What Are Real-Life Applications of the Buoyant Force Equation?

The buoyant force equation applies to 7 technologies and natural systems. Examples include:

• Ships and boats, which float by displacing water equal to their mass.
• Submarines, which adjust ballast water to reach neutral buoyancy.
• Hot air balloons and airships, which displace air with a lower density gas.
• Life jackets, which add low-density volume to a swimmer.
• Hydrometers, which measure liquid density from the depth of flotation.
• Hydrostatic weighing, which measures body density from weight in air and weight underwater.
• Fish, which adjust the gas volume of the swim bladder to control buoyancy.

What Do People Also Ask About the Buoyant Force Equation?

What is the buoyant force equation?

The buoyant force equation is F_b = ρ_fluid × g × V_displaced.

What is Archimedes’ principle?

Archimedes’ principle states that the buoyant force equals the weight of the fluid displaced by the object.

What is the unit of buoyant force?

The SI unit of buoyant force is the newton (N).

How do you calculate buoyant force without volume?

Use F_b = W − W_app, which needs the weight in air and the apparent weight in the fluid.

How do you calculate buoyant force from mass?

Divide the mass by the object density to find the volume, then multiply by ρ_fluid and g, if the object is fully submerged.

What is the buoyant force on a floating object?

The buoyant force on a floating object equals the weight of the object (F_b = m g).

Does buoyant force depend on depth?

Buoyant force does not depend on depth for a fully submerged object in an incompressible fluid.

Does buoyant force depend on the mass of the object?

Buoyant force depends on the displaced volume, not on the mass of the object.

Is buoyant force greater in salt water?

Seawater (1,025 kg/m³) produces a 2.5% larger buoyant force than freshwater (1,000 kg/m³) for the same displaced volume.

Why does ice float on water?

Ice has a density of 917 kg/m³, which is lower than the density of water, so 91.7% of its volume is submerged in freshwater.

Can the buoyant force be greater than the weight?

Buoyant force exceeds weight for a fully submerged object with a lower density than the fluid, and the object rises until the 2 forces are equal.

How do you find buoyant force from apparent weight?

Subtract the apparent weight in the fluid from the weight in air.

Is buoyant force a vector or a scalar?

Buoyant force is a vector, and it points upward.

Is upthrust the same as buoyant force?

Upthrust is another name for buoyant force.

What Is the Buoyant Force Equation and How Is It Applied?

The buoyant force equation is F_b = ρ_fluid × g × V_displaced, which gives the upward force of a fluid in newtons. The 7 forms use displaced volume, displaced mass, apparent weight, object weight, density difference, specific gravity, or gas densities. An object sinks when ρ_object exceeds ρ_fluid, floats when it is lower, and stays suspended when the 2 values are equal. A floating object displaces a volume fraction equal to ρ_object / ρ_fluid. The force depends on fluid density, displaced volume, and gravitational acceleration, and does not depend on depth for a fully submerged object. Applications include ships, submarines, balloons, life jackets, hydrometers, and hydrostatic weighing.

Title: Statistics Formulas: Definitions, 9 Categories, and Worked Examples

Maurice Cotterell

MEd., BSc. QTS/OCT – Physics Teacher at Davenant Foundation School
The University of the West Indies

My passion as a teacher and leader in education is to contribute to students’ enhanced learning by creation of a positive teaching/learning environment, where they are motivated to develop strong critical thinking, analytical and problem-solving skills.
As acting Vice-Principal and head of Physics, I have amassed five years of experience in leading meetings, implementing strategies to maintain discipline and school ethos, liaise with stakeholders in planning for future development of the institution; successfully negotiated the implementation of various projects for student development as well as supervising effective execution of such. For the past 19 years, I have incorporated discovery learning within the classroom through practical activities, problem solving and cooperative learning and have realized that students retention in Physics and Mathematics have increased significantly.

I am always cognizant that students have different learning styles, and as such teaching strategies during planning and delivery must take this crucial factor into consideration and be tailored to meet such differences.