# Buoyancy

Buoyancy is a vertical force that acts on any object immersed in a fluid. This force is known as Archimedes’ Principle.

**buoyancy will appear on the object, which is exerted by the fluid and has a vertical direction and an upward direction.**

The image above shows a spherical object partially immersed in a fluid of density ρ . Two forces act on the object: weight and buoyancy.

**How does the thrust arise?**

Imagine an object completely immersed in water. The pressure exerted by the liquid acts on all points of the object. As the body sinks, the pressure in its lower points becomes greater than the pressure in its upper parts. The pressure difference generates the force called buoyancy.

**How was thrust discovered?**

The discovery of this force is attributed to the Greek Archimedes , who defined greatness as follows:

*“Any body immersed in a liquid experiences a force called buoyancy that corresponds to the weight of the volume of liquid displaced”.*

The buoyancy is treated in several literatures as Archimedes’ Principle.

**How is thrust calculated?**

The buoyancy corresponds to the weight of the volume of liquid displaced by the body immersed in a fluid. Knowing that the weight is the product of mass and gravity and calling the mass of displaced liquid m _{DES , we have:}

**E = m _{DES} . g**

The volumetric density is defined as the ratio between the mass and the volume of the substance, therefore, for the volume of liquid displaced (V _{DES} ), we have:

**ρ = m _{DES} ÷ V _{DES}**

**mDES = ρ _{. }V _{DES}**

**E = ρ . V _{DES} _ g**

NOTE: The volume of liquid displaced corresponds to the immersed volume of the body immersed in the fluid.

**flotation of ships**

Have you ever wondered how huge cargo ships or ocean liners float in the sea being so heavy? The answer is in the thrust!

When a non-massive object is deposited on a fluid, its weight acts vertically downwards. As the object descends, the amount of fluid displaced increases and the buoyancy also increases. The moment the buoyancy becomes equal to the weight force, the object will remain in a state of static equilibrium and will float on the surface of the liquid.

**exercise example**

An object of density 300 kg/m ^{3} and mass 15,000 kg floats in the calm waters of a lake. If the density of the fluid is 1000 kg/m ^{3} , determine the surface volume of the body.

**Resolution:**

As the object floats in the water, we can say that the buoyancy is equal to the weight:

E = p

ρ . V _{DES} _ g = m. g

ρ . V _{DES} = m

1000 V _{DES} = 15,000

V _{DES} = 15 m ^{3}

From the definition of density, we can find the total volume of the body:

ρ = m ÷ V

V = m ÷ ρ

V = 15,000 ÷ 300

V = 50 m ^{3}

As the total volume of the body is 50 m ^{3} and the volume immersed in water is 15 m ^{3} , we can conclude that the volume emerged is 35 m ^{3} .