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Buoyancy
  • Last Updated : 12 May, 2021

Buoyancy is a phenomenon due to buoyant force that causes an object to float. When you put an object in a liquid, an upward force is exerted on the object by the liquid this force is equal to the weight of the liquid that has been displaced. The amount of liquid that has been displaced depends upon the density and the volume of the object immersed in the liquid.

Have you ever wondered why does an iron nail sinks in water but a ship made up of iron floats? And why does an iron ball sink but a plastic ball of the same size floats in water? These wonders happen due to the phenomenon known as Buoyancy.

Buoyancy can be defined as an upward force exerted on an object that is completely or partially submerged in liquid. The unit of buoyant force is Newton.

Buoyancy force depends upon two factors:

  1. Amount (Volume) of liquid displaced by object
  2. The density of the object.

In the first example, iron nail is having less volume displaces very less amount of water so less buoyant force (upward force) therefore sinks. Whereas ship have more volume displaces more water and therefore greater buoyant force (upward force) by water and it floats.



In the second example, the iron ball has more density and therefore greater gravitational force (downward force) than the buoyant force applied by liquid, hence it sinks. Whereas plastic ball is hollow and very less dense, so smaller gravitational force (downward force) than buoyant force (upward force) applied by liquid and it floats.

Boat, submarines, lifeboat, life jacket, and swimming works on the principle of buoyancy.

Archimedes’ Principle

Physical law of buoyancy, was given by Greek mathematician and inventor Archimedes. Archimedes’ principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces.

In the image given above the thrust produced by fluid, opposes gravity and actual gravitational forces decreases and therefore given by:

Apparent weight = Weight of object (in the air) – Thrust force (buoyancy)

Formula for Archimedes’ Principle

Archimedes’ law mathematically written as:

Fb = ρ x g x V



Where,

Fb is the buoyant force.

ρ is the density the fluid.

V is the submerged volume.

g is the acceleration due to gravity.

Derivation of the Formula

After rearranging the formula ρV is the density of the displaced fluid multiplied by the volume of the displaced fluid, and we know the mass, density and volume relation.

m = ρV

That means the term ρV corresponds to the mass of the displaced fluid.

Fb = ρ x g x V    

Fb = mg i.e. The mass of the displaced fluid times the magnitude of the acceleration due to gravity is just the weight of the displaced fluid.



Fb = W            

The above equation when stated in words called Archimedes’ principle.          

Assuming Archimedes’ principle to be reformulated

Apparent immersed weight = weight – Weight of displaced fluid

Then inserted into the quotient of weights, which has been explained by the mutual volume.

\frac{Density}{Density \ of \ fuild} = \frac{Weight}{Weight \ of \ displaced \ fluid}

The density of the immersed object relative to the density of the fluid can be calculated:

 \frac{Density \ of \ object}{Density \ of \ fuild} = \frac{Weight}{Weight - Apparant \ immersed \ weight}

Fb = PA = g ρ V = ρ g h A       —-   (i)

P = pressure

Fb = force of buoyancy in Newton,

A = Area in meter square,

g = acceleration due to gravity,

h = Height at which force acts taken from the surface

p = density of the fluid,

V = volume of the object inserted into the fluid.

Fb = Wa – Wf             —–(ii)

Fb = force of buoyancy

Wa = Weight of the object in the air

Wf  = Weight of the object when it is immersed in the fluid

Using (l) and (ll),

g ρ V = Wa – Wf   ———-(iii)

If the object is not sinking then Fg – Fb

Mg = g ρ V

Sample Problems on Archimedes’ Principle

Question 1. Find volume of immersed object if M is 10000 kg.

Solution:

ρ=997 kg/m3

10000 kg x g = g x 997 kg/m3 V

1000 kg = 997 kg/ m3 V

V = \frac{10000 kg }{997 kg/m3}=10.03 m3

Question 2. A block of wood with length = 3.2 m, width = 0.8 m and height = 0.6 m. The density of water is 1000 kg/m3. If the block is placed in the water, what is the buoyant force? Acceleration due to gravity is 10 N/kg.

Solution:

Volume of the block (V) = length x width x height = 3.2 x 0.8 x 0.6 = 1.53 m3

Density = 1000 kg/m3

Gravity = 10 N/Kg

F = density x gravity x volume

F = 1000 x 10 x 1.53 = 15300N

Ans : F = 15300N

Question 3. Weight of an object in air is 108 N. The object is placed in a liquid. Increase in volume of liquid is 1.8 m3. If specific weight of the liquid is 10 N/m3, what is the weight of the object in liquid?

Solution:

Object’s weight in liquid = object’s weight in air – buoyant force

Object’s weight in liquid = 108 N – buoyant force

F = ρ g V

The density of liquid is 1 kg/m3

F = ρ g V = (1 kg/m^{3})(10 m/s^{2})(1.8 m^{3}) = 18kg m/s^{2} = 18 N

Object’s weight in fluid = 100 N – 18 N = 82N

Ans: 82 N

Question 4. An object weighs 12N in air, when immersed fully in water, it weighs only 9N. What would be the weight of the liquid displaced by the object?

Solution:

According to Archimedes’s law:

Apparent weight = Weight of object (in the air) – Thrust force (buoyant force)

Apparent weight = 9N

Weight of object (in the air) = 12N

Thrust force (buoyant force) = ?

Apparent weight = 12N – Fb

Fb = 12N – 9N = 3N

Ans: 3N

Question 5. Volume of an object is 50 cm3  and the mass in 20 g. Density of water is 1 gcm-3.Will the object float on water or sink in water?

Solution:

Density = \frac{Mass}{Volume}=\frac{20}{50}= 0.4 g/cm^{3}

Ans: The density of object is less than the water therefore object will float.

A floating object is stable if it tends to restore itself to an equilibrium position after a small displacement. For example, floating objects will generally have vertical stability, as if the object is pushed down slightly, this will create a greater buoyancy force, which, unbalanced by the weight force, will push the object back up.

Given a small angular displacement, the vessel may return to its original position (stable), move away from its original position (unstable) Or remain where it is (neutral). Rotational stability depends on the relative lines of action of forces on an object.

The upward buoyancy force on an object acts through the centre of buoyancy (CB), being the centroid of the displaced volume of fluid. The weight force on the object acts through its centre of gravity (CG).

Buoyant object will be stable if the centre of gravity is beneath the centre of buoyancy because any angular displacement will then produce a ‘righting moment’. The stability of a buoyant object at the surface is more complex, and it may remain stable even if the centre of gravity is above the centre of buoyancy, provided that when disturbed from the equilibrium position, the centre of buoyancy moves further to the same side that the centre of gravity moves, thus providing a positive righting moment.

What is Upthrust?

Any object placed in a fluid receives an upward force called Upthrust. The object is pushed in an upward direction, therefore, named Upthrust. An object that is partly, or completely, submerged experiences a higher pressure on bottom surface than on its top surface. This force is called upthrust. When we put something on the surface of the water body it displaces some of the fluid. The upthrust force is equal to the weight of the fluid displaced by the object. The unit of upthrust is Newton.

As upthrust is the force which is applied by the water and one other force is also acting on the body due to mass of body which is know as gravitational force. Gravitational force is acting to the opposite direction of upthrust force, Upthrust force acting in vertically upward direction and Gravitational force acting on vertically downward direction.

Types of Buoyancy

There are three types of buoyancy:

1. Positive buoyancy: When the weight of an object is lighter than the fluid it displaces is called positive buoyancy.

For example, a boat that weighs (3000 kg) but displaces (4500 kg) of water will easily float

2. Negative buoyancy: When the weight of an object is greater than the fluid it displaces is called positive buoyancy.

For example, an iron nail may weigh 27 grams, but if it only displaces 17 grams of water, it will sink.

3. Natural buoyancy: When the weight of an object is equal to the fluid it displaces.

For example, a submarine can adjust its weight by adding or expelling water in special tanks called ballast tanks is an example of natural buoyancy.

Application of Buoyancy

1. The boat is designed in a way so that the shape of the boat is hollow. Due to the hollow shape, the density becomes less than the density of the sea. The volume of water displaced by the boat is equal to the weight of the boat and this helps the ship to float.

2. Fishes stay buoyant underwater. Fishes have a special organ called a swim bladder that is usually filled with gases thus making the body lighter. This enables the fishes to go up.

3. Floatation of submarines in the water.

4. To find the volume of the body.

5. To keeps swimmers on top of the water body.

Sample Problems

Question 1. A block of wood with length = 3.2 m, width = 0.8 m and height = 0.6 m. The density of water is 1000 kg/m3. If the block is placed in the water, what is the buoyant force? Acceleration due to gravity is 10 N/kg.

Solution:

F = ρ g V

F = 1000 x 10 x 1.58 = 15840N

Ans : F = 15840N

Question 2. Weight of an object in air is 108 N. The object is placed in a liquid. Increase in volume of liquid is 1.8 m3. If specific weight of the liquid is 10 N/m3, what is the weight of the object in liquid?

Solution:

Object’s weight in liquid = object’s weight in air – buoyant force

Object’s weight in liquid = 108 N – buoyant force

FA = ρ g V

The density of liquid is 1 kg/m3

FA = ρgV = (1kg/m3)(10m/s2)(1.8m3) = 18kgm/s3 = 18N

Object’s weight in fluid = 100 N – 18 N = 82N

Ans: 82 N

Question 3. A piece of marble tile weighs 285 g in air. If its density is 3.5 g/cc, what will be its weight in water?

Solution:

Weight in air = 285 g

Volume = 285g /(3.5 g/cc) = 81.4 cc

Weight in water = 285 g – 81.4g = 203.6 g

Ans: 203.6 g

Question 4. Volume of an object is 50 cm3 and the mass is 30 g. Density of water is 1 gcm-3. Will the object float on water or sink in water?

Answer: 

Object will float because the density of object is less than that of water.

Question 5. Volume of an object is 20 cm3  and the mass in 10 g. Density of water is 1 gcm-3.Will the object float on water or sink in water?

Solution:

Density = \frac{Mass}{Volume}=\frac{10}{20}= 0.5 g/cm^{3}

Ans: The density of object is less than the water therefore object will float.

Question 6. An object weighs 15N in air, when immersed fully in water, it weighs only 10N. What would be the weight of the liquid displaced by the object?

Solution:

According to Archimedes’s law:

Apparent weight = Weight of object (in the air) – Thrust force (buoyant force)

Apparent weight = 10N

Weight of object (in the air) = 15N

Thrust force (buoyant force) = ?

Apparent weight = 15N – Fb

Fb = 15N – 10N = 5N

Ans: 5N

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