Archimedes Principle

Archimedes Principle is the foundation of fluid mechanics and explains why objects float or sink in a fluid (liquid or gas). It serves as the basis for many phenomena related to flotation and buoyancy designs.

1.0Archimedes Principle Statement

Archimedes' Principle states: "When a body is partially or completely immersed in a fluid (liquid or gas), it experiences an upward buoyant force equal to the weight of the fluid displaced by the body."

Here the weight of the submerged object in fluid will be less than its actual weight in the air. The decreased weight is also known as the apparent weight of the object. 

The principle is named after the Greek scientist Archimedes, who discovered it about 250 BCE. According to Archimedes Principle, objects experience an upward push when immersed in a fluid.

2.0Archimedes Principle Explained 

This principle can be described using the concept of pressure. The fluid will exert pressure on all sides of an immersed object. It is evident that pressure increases with depth. The bottom side of an object would have greater pressure than its top side. This is due to a pressure difference that causes a buoyant force to move upward.

Archimedes Principle and Buoyancy 

Buoyancy is the vertical upward force exerted by a fluid upon a submerged object. The concept of buoyancy is fundamentally related to Archimedes Principle, which explains why an object feels lighter when it is plunged into a fluid.

Archimedes Principle and Buoyant Force

Buoyancy is a fluid's upward vertical force (that depends upon the density of the fluid) acting on an immersed body. Archimedes Principle and Buoyant Force are inherently connected, describing why a body experiences the sensation of weight reduction upon being submerged in a fluid. If the buoyant force is more, then the object will float or sink if the buoyant force is less. 

Mathematically, Archimedes principle can be written by: 

$F_B=p.V.g$

Here, 

• $F_B$ the Buoyant force
• p represents the density of the fluid. 
• V represents the volume of the displaced fluid. 
• g represents the acceleration due to gravity. 

Archimedes Principle and Floatation 

The Principle of Floatation is a special application of Archimedes Principle. According to this principle, if the weight of the fluid displaced by a body is equal to or greater than the weight of the body itself, it will float on the fluid. To float, an object must experience an upward buoyant force that is equal to the object’s weight.

Experiment to Verify Archimedes Principle

For verification of Archimedes principle in physics, we can do the following Archimedes Principle Experiment:

• Take an object (say, a metal sphere) and measure its weight in air with a spring balance.
• Fill a graduated cylinder with water and record the initial water level.
• Immerse the object completely in the water and observe the displacement of the water level.
• In this case, the volume of the water displaced equals the volume of the object.
• Calculate the buoyant force by finding the weight of the water displaced. This should be equal to the buoyant force acting on the object.

3.0Archimedes Principle Examples 

Example 1: Buoyant Force on a Submerged Object 

Problem: A metal block with a volume of 0.5 m³ is completely submerged in water. Take the value of water's density as 1000 kg/m³. Calculate the buoyant force acting on the metal block.

Solution: Volume of the metal block, say V = 0.5 m³

Density of water, = 1000kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Applying Archimedes principle formula for buoyant force, 

$F_B=\rho .V.g$

$F_B=1000\times 0.5\times 9.8=4900N$

Hence, the force that is acting on the metal block is 4900N. 

Example 2: Apparent Loss of Weight

Problem: A rock weighs 100 N in air. When it is fully submerged in water, its weight reduces to 70 N. Calculate the buoyant force and the volume of the rock. The density of water is 1000 kg/m³.

Solution: Weight of the rock in air, say W_air = 100N 

Weight of the rock in water, W_water = 70 N 

We know that the buoyant force F_B is equal to the apparent weight loss when the given object is submerged in any fluid. This can be mathematically represented as: 

$F_B=W_{air}-W_{water}=100-70=30N$

Now, with the help of buoyant force formula: 

$F_B=\rho .V.g$

$30N=1000\times V\times 9.8$

$V=\frac{30}{1000\times 9.8}=0.00306m^3$

Problem: A solid spherical ball of metal with a radius of 10 cm is fully submerged in water. The ball has a density of 8.5 g/cm³. Find the buoyant force acting on the ball when it is submerged in water, and determine whether the ball will float or sink.

Given: 

• The radius of the ball, r = 10cm 
• The density of the ball, ${\rho }_{ball}$=$8.5g/c{m}^3$
• Density of water, ${\rho }_{water}$=1g/cm3
• g = 9.8m/s²

Solution: The Volume of the spherical ball: 

$V=\frac{4}{3}\pi r^3$

$V=\frac{4}{3}\times 3.1416\times 10^3$

$V\approx \frac{4000}{3}\times 3.1416=4188.79c{m}^3$

The Buoyant force

$F_B={\rho }_{water}\times V_{displaced}\times g$

Here ${\rho }_{water}=1g/c{m}^3$, and $V_{displaced}=V.$

$1c{m}^3=1\times 10^{-6}{m}^3$

$V_{displaced}=4188.79c{m}^3=4188.79\times 10^{-6}{m}^3=0.00418879m^3$

$F_B=1000\times 0.00418879\times 9.8=41.1N$

Weight of the ball: 

$W={\rho }_{ball}\times V_{ball}\times g$

$W=8500\times 0.00418879\times 9.8$

$W=348.8N$

For the ball to float or sink: 

If the buoyant force (F_B) is greater than the weight of the ball (W), it will float. Otherwise, it will sink in the case of F_B < W.

Here, 

F_B<W, the ball will sink.