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# Archimedes Principle and Buoyant Force, Law of Floatation, Apparent Weight

A ship is able to float on water because it displaces some water equal to its own weight. The weight of the body acting downwards is balanced by the buoyant force exerted by the liquid. One can imagine buoyant force as the upward force exerted by the liquid on the body; it is also known as upthrust. Archimedes came up with this principle when he got into his bathtub. He observed that the water level rose when he got into it; he climbed out shouting, “Eureka!Eureka!”

## Archimedes' principle

It states that an object partially or fully submerged in a liquid experiences the force in upward direction called buoyant force (FB) equal to the weight of the displaced fluid.

This principle is true for the body of any shape and for partially or fully submerged bodies.

Let’s understand Archimedes’ principle using the following diagram. A container filled with water having a hole just above the water level. Now when you place a block on the water surface it experiences a buoyant force (FB). As an object gets submerged in the water, water level will increase and extra water will come out of hole as shown. This water is collected in a beaker and measured. So according to Archimedes' principle buoyant force is equal to the weight of displaced water which is collected in the beaker. So FB=Wf

## Buoyant force

When an object is submerged in water or any other fluid, we observe that the object is subjected to a downward force opposing gravitational attraction, which causes the object's weight to drop. The weight of an object immersed in a fluid is counteracted by the upward force exerted by the fluid. The pressure in a fluid column increases with depth, as we all know. As a result, the pressure at the bottom of a submerged object is larger than at the top. The difference in pressure causes the object to experience a net upward force, which we call buoyancy. There will be no force of buoyancy if gravity does not exist.

Let us consider a body of mass m and density floating in a liquid of density . Let V be the volume of the body that remains immersed in the liquid.

Fig shows a body of weight W floating in a liquid of density

Then weight of the body W acting downwards is,

W=mg=Vg.

If Fb denotes the upthrust exerted by the liquid; then

Fb=Vg=Weight of displaced liquid

If A denotes the cross sectional area of the body and x is the depth to with it is immersed in the liquid,

Volume of immersed object is, V=Ax

So Fb=Vg=Axg

The point where the buoyant force is applied or the point on the object where the force acts is termed the Center of Buoyancy, which is at the center of gravity of the displaced fluid.

## Laws of Floatation

If s indicates the density of the solid body and L indicates the density of the liquid in which it is floating, then weight of the solid Ws=Vsg.

Case1: Floating

As shown in the diagram if buoyant force is equal to the weight of the object then it will float on the liquid surface. Consider the area of the cube is A and the height as H of which x is submerged in the liquid.

So FB=Ws

If we balance the forces on the object in vertical direction,

FB=Weight of displaced liquid

FB=(Ax)Lg

Ws=AHsg

As FB=Ws

(Ax)Lg=AHsg

x=HsL

Immersed Volume of CubeTotal Volume=AxAH=xH=sL

Case2: Sinking

If buoyant force is less than the weight of the object it will sink in the liquid. So FB<WS. That means s>L. In this case buoyant force will not be able to balance the weight of the object. So it will start sinking.

## Apparent Weight

The weight of a body, when immersed in a liquid, weighs lesser than the actual weight of the body due to the buoyant force acting on the body. The Apparent Force (Wapp) is defined as the net force on the body inside water.

So

Wapp=Wactual-FB=sVg-LVg

Wapp=(s-L)Vg

Note: If the body is floating then weight is balanced by buoyant force. So in this case apparent weight is zero.

## Practice Problems

Q1. Ice is floating in water of density 1000 kgm-3. Find the fraction of volume of ice outside the water. (Given,density of ice is 900 kgm-3)?
Answer. Given,ice is floating on water

Weight=Upthrust

V1 is the volume of ice inside water

V is the volume of ice

Vsg=V1lg ;s=900 kgm-3

l=1000 kgm-3

V1V=sl=900100=0.9Fraction outside water =1-0.9=0.1

That is 10% of volume is outside water.

Q2. A body floats with 25th of its volume outside water. The same body floats with 35th of its volume outside oil. The relative density of oil is?
Answer. Let V be the volume of the body; then weight of the body= Upthrust

Since 25th of its volume floats outside water; 35th floats inside water.

Vsg=35wg ; w=1000

It floats with 25th of its volume inside oil;

Vsg=25oilg

Dividing both equations;

oilw=5235=1.5

Q3. An ornament weighing 36 g in air, weighs 32 g in water. Assuming that copper is mixed with gold to make the ornament, find the amount of copper in it? Given, density of gold =19.3 gcm-3 and density of copper=8.9 gcm-3
Answer. Density of gold Au=19.3 gcm-3

Density of copper Cu=8.6 gcm-3.

Weight of the ornament in the air is 36 g and in the water it is 34 g. So 36-32=4 g is the buoyant force.

FB=LVg

4g=1Vg

So V=4 cm3

Let mass of the copper mCu=x g

Then mass of the gold mAu=(36 -x )g

Volume of copper Vcu=mCucu=x8.9 cm3

Volume of the gold VAu=mAuAu=36-x19.3 cm3

Let V=Vcu+VAu

4=x8.9+36-x19.3

Solving, we get x=2.225 g

Q4. A block of volume 2m3 is immersed in a fluid of density 800 kgm-3. Calculate the upthrust (in N) acting on the body?

## FAQs

Question 1. What is the use of Archimedes' principle?
Answer. Archimedes’s principle is used to calculate the volume of objects having irregular shapes.

Question 2. What is the condition for a body to float in a liquid?
Answer. If the average density of the body is lesser than the density of the liquid,.i.e s<L Then the body floats in the liquid.

Question 3. What is the condition for a body to remain immersed in a liquid?
Answer. If the density of the object is greater than the density of the liquid then the object sinks in the liquid.

Question 4. What will be the buoyant force in gravity free space?
Answer. Buoyancy is caused due to the gravitational force. So in gravity free space buoyant force will not exist.

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