# Physics Tutorial: Buoyancy. Archimedes' Principle

In this Physics tutorial, you will learn:

• The meaning of buoyancy
• What caused the buoyancy phenomenon?
• How to calculate the buoyant force?
• What does Archimedes' Principle say on buoyancy?
• How to calculate the maximum load a floating object can hold without sinking?
• How does air buoyancy act on flying objects?
• What is the difference between buoyancy in liquids and gases?
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9.5Buoyancy. Archimedes' Principle

## Introduction

Do you think it is easier to learn swimming in seawater or in lake water (in salty or sweet water)? Why?

What happens if you are in water and hold your breath? What happens if you then release your breath?

Why is it difficult to keep a ball completely immersed water?

Why balloons fly so high? Why a plastic balloon filled with helium raises in the sky while another balloon filled with air falls on the ground?

Why you should not fill a vessel completely with water if you want to put other things in it?

All these questions will be discussed and explained in the current tutorial, which deals with one of the most important topics in Hydrostatics i.e. Buoyancy.

## What is Buoyancy?

Buoyancy is the lifting ability of fluids on objects immersed in them. In this sense, we can define buoyancy as an opposing force to gravity caused by the resistance of fluid against any molecular shift caused by the downward tendency of objects' motion. This resistance comes due to the medium's inertia (here, medium is represented by the fluid). Look at the figure. In the figure, you can see that the falling object compresses the molecules below it and leaves for the moment empty spaces above it, until these empty spaces refill by other molecules coming from other places. This difference in the density of fluid (i.e. of medium's molecules) produces a resistive effect towards motion through the fluid, which we call buoyancy. Even though the object may continue fall down, we say buoyancy causes a lifting effect because it prevents a quick fall of the object due to gravity.

## Buoyant Force

The resistive effect caused by the fluid we mentioned above, causes a resistive force, which we call "buoyant force". In other words, buoyant force is the upward force exerted by a fluid on an object that is completely or partially immersed in it.

Buoyant force is symbolically expressed as Fb and like all the other forces, it is measured in Newton.

Since the direction of buoyant force is vertically upwards, the equation for the resultant vertical force acting on an object immersed in a fluid is

FR = Fg - Fb

where Fg is the gravitational force acting on the object. Look at the figure. There are three possible cases of an object immersed in a fluid:

1- The object is not at rest because Fg > Fb and therefore, FR = Fg - Fb > 0. As a result, the object is falling down (sinking). 2- When buoyant force is numerically equal to the gravitational force, we have

FR = Fg - Fb = 0
As a result, the object stays unmoveable, completely immersed inside the fluid, as shown below 3- When buoyant force is numerically greater than gravitational force (when we force an object to stay immersed in fluid), initially we have

FR = Fg - Fb(initial) < 0

As a result, when released the object raises up, until a part of it is pushed out of fluid. This is why air bubbles rise in water until they go to the water surface and "explode", i.e. they move to the air. In floating objects, the equilibrium of forces is re-established when they go to the liquid surface and then objects float at rest on liquids as shown in the figure below. When the equilibrium is established, we have

FR = Fg - Fb (final) = 0

Or

Fg = Fb (final)

Let's use the above explanation to obtain a formula for the buoyant force Fb. From experience, we know that heavy objects such as metals, stones, etc., sink when immersed in water (case 1). This is because they have a greater density than water (i.e. the medium). On the other hand, light objects such as wood, polystyrene, etc., float on water (case 3). This is because their density is smaller than the density of water.

When objects have a density close to the density of water, they usually stay completely immersed in water, neither sinking nor floating (case 2). As an example, we can mention the human body. It is known that about 70% of the human body is made by water. The rest are mostly ashes, flesh and air, whose average density is close to the density of water (1000 kg/m3) as well. Therefore, when we hold our breath, we float on water as the extra air we have on our lungs reduces the density of our body, making it slightly smaller than the density of water. On the other hand, when we release the breath from our lungs, the density of body increases becoming temporarily slightly greater than the density of water. As a result, we sink. In normal condition, we stay completely immersed in water (divers for example). This is because the density of our body is very close to that of water.

Said this, we can use the help of the second case (object completely immersed in liquid) to derive a formula for the buoyant force. As stated above, in this case, we have

FR = Fg-Fb = 0

or (numerically):

Fg = Fb

Since

Fg = mobj × g
= ρobj × Vobj × g

we obtain after substitutions,

ρobj × Vobj × g = Fb

Since in the second case ρobject = ρliquid, we obtain for the buoyant force Fb:

Fb = ρliq × Vobj × g

The same formula is used for the other two cases as well. The only thing to keep in mind is that in the third case (in floating objects), we must consider only the part of object's volume immersed in liquid because (as discussed earlier) the new buoyant force is smaller than the original one, when the object was forced to stay completely immersed in liquid. Thus, the new (reduced) buoyant force is auto-regulated to equal the force of gravity by sending out of liquid the unnecessary part of the volume. As in the other two cases the total volume of the object complies with that of the immersed part, we can generalize as

Fb = ρliq × Vimmersed part of object × g

The above formula is the standard formula of buoyant force.

Now, it is understandable why we can learn more easily swimming in seawater than in lake water. As the density of seawater is greater than that of lake water, the buoyant force in seawater (salty water) is greater than the buoyant force in lake (sweet or pure) water. This helps keep us floating on water.

### Example 1

A

n object, which normally weighs 45 N (figure a), is hanged on a spring balance and then immersed in water. Now, the spring balance shown the value 30 N (figure b). What is the density of the object? Take g ≈ 10 m/s2 and ρwater = 1000 kg/m3. ### Solution 1

From the first figure, we can find the mass of object. We have:

W = Fg = mobj × g
⇒ mobj = W/g
= 45 N/10 m/s2
= 4.5 kg

The buoyant force exerted by the water is obtained by calculating the difference between the two readings of spring balance. Thus,

Fb = Fg - Win water
= 45 N - 30 N
= 15 N

This result helps in calculating the volume of object through the other formula of buoyant force, i.e. from the formula

Fb = ρliq × Vimmersed part of object × g

we can individualize the volume of the immersed part of object which corresponds to its total volume. Thus,

Vimmersed part of object = Vobj
= Fb/ρliq × g
= 30 N/1000 kg/m3 × 10 m/s2
= 0.003 m3

Therefore, the object's density is

ρobj = mobj/Vobj
= 4.5 kg/0.003 m3
= 1500 kg/m3

## Archimedes' Principle

Buoyancy as a phenomenon, was first discussed and explained by the famous ancient Greek scientist Archimedes, who was able to calculate the density of an irregular object (the famous crown donated to the King by a jeweller). Thus, by measuring the volume of the displaced water, he calculated indirectly the volume of the immersed object (the crown). Then, he used the equation of density (density = mass / volume) to find whether the donated crown was made of pure gold or not (the density of gold was already known at that time). In the first figure, water is poured in the container up to the hose level. After putting the object (the crown) inside the container, Archimedes used a bucket to collect the spilled water due to the rise in water level. (It is known that two things cannot occupy the same space at the same time. As a result, the water level rises up when the crown is placed inside the vessel because its place now is occupied by the crown). Then he used the procedure mentioned above to calculate the crown density.

However, the calculation of crown density, i.e. the calculation of irregular objects density was not Archimedes' major achievement in this experiment. He accidentally discovered that:

"The buoyant force acting on an object immersed in a liquid is numerically equal to the weight of the displaced liquid."

The above statement is known as Archimedes Principle" and it is one of the founding principles of Hydrostatics.

Mathematically, we can write the Archimedes principle as

Fb = Wdisplaced liquid

Given that

Vobject = Vdisplaced liquid

Archimedes managed to obtain the formula of buoyant force we discussed earlier, because

Fb = Wdisplaced liquid
Fb = mdisplaced liquid × g
= ρliquid × Vdisplaced liquid × g

which is the same formula we found earlier for the buoyant force because

Vdisplaced liquid = Vimmersed object

### Example 2

How many m3 of water spill out of a bathtub when a 80 kg man enters in it, if the bathtub initially was filled to the brim with water? Take the density of human body equal to that of water, i.e. 1000 kg/m3. Also, take the gravity g ≈ 10 m/s2.

### Solution 2

From Archimedes' Principle we have

Fb = Wdisplaced liquid

Here, the buoyant force Fb is equal to the man's weight as both the man and water have the same density. Thus,

Fb = Wman
= mman × g
= 80 kg × 10 m/s2
= 800 N

From the new equation of buoyant force (based on Archimedes' Principle), we have

Fb = ρwater × Vdisplaced water × g

Therefore,

Vdisplaced water = Fb/(ρwater × g)
= 800 N/1000 kg/m3 × 10 m/s2
= 0.08 m3
= 80 dm3
= 80 L

Archimedes' Principle is very important in daily life as its applications include a wide range of situation, where the most notorious is the ships construction. Thus, we can use Archimedes' Principle to calculate how much weight a boat or ship can hold without sinking. Let's see an example in this regard.

### Example 3

A rectangular raft of dimensions 5 m × 4 m × 20 cm is made of wood (density of wood is 600 kg/m3).

1. What is the maximum load (in kilograms) the raft can hold if the load is very precious and it must be prevented from getting wet?
2. What part of the raft is initially immersed in water?
Take g = 10 m/s2. ### Solution 3

a In the first figure (before putting the extra object on it), the raft is floating while in the second figure (after putting the extra object on the raft), it is at limits of floating and completely immersed. The equation for the first figure is

Fb1 (water) = Fg (raft)

and for the second figure, it is

Fb2 (water) = Fg (raft) + W(extra object)

From the first equation, we obtain for the initial buoyant force cause by water

Fb1 (water) = mraft × g
= ρwood × Vraft × g
= 600 × (5 × 4 × 0.2) × 10
= 24 000 N

When the raft is completely immersed due to the downward push of the extra object (Vimmersed part = Vtotal), we have for the new buoyant force caused by the water:

Fb2 (water) = ρwater × Vimmersed part × g
= ρwater × Vtotal × g
= 1000 × (5 × 4 × 0.2) × 10
= 40 000 N

The weight of the extra object can be obtained by subtracting the two buoyant forces found above, as they represent the total weight of the system without and with load respectively. Thus,

Wextra object = Fb2 - Fb1
= 40 000 N - 24 000 N
= 16 000 N

Hence, the maximum mass of the extra load the raft can hold is

mextra object = Wextra object/g
= 16 000 N/10 m/s2
= 1600 kg

b The part of volume initially immersed in water is obtained by dividing the volume that was initially immersed by the total volume of the raft. Thus,

% immersed = V1 (immersed)/Vtotal × 100%

The initially immersed volume can be obtained by the formula of buoyant force, i.e.

Fb1 (water) = ρwater × V1 (immersed) × g

Thus,

V1 (immersed) = Fb1 (water)/ρwater × g
= 24 000/1000 × 10
= 2.4 m3

The raft's total volume is calculated by multiplying its dimensions provided in the clues. Thus,

Vtotal = 5m × 4m × 0.2 m
= 4 m3

Hence, the part of the raft that was initially immersed in water expressed in percentage is

% immersed = 2.4 m3/4 m3 × 100%
= 60%

Remark! Besides the ratio of volumes found above, the immersed part of the object can be found through two other ratios as well.

1 The ratio of buoyant forces, i.e.

% immersed = Fb1/Fb2 × 100%
= 24 000 N/40 000 N × 100%
= 60%

and

2 The ratio of densities (of object and of liquid respectively), i.e.

% immersed = ρwood/ρwater × 100%
= 600 kg/m3/1000 kg/m3 × 100%
= 60%

## Buoyancy in Gases

We can say the same words for buoyancy in gases as well. The only difference is that in gases there is no floating; we consider the object completely immersed in gas (especially in air). The third case mentioned above when buoyancy in liquids was explain (i.e. when density of object is smaller than the density of gas), is discussed only to see whether the object is moving up or not. Thus, the three cases explained in the buoyancy of liquids adapted for buoyancy in gases are:

1. When Fb < Wobject the object falls down on the ground. In this case, ρgas < ρobject
2. When Fb = Wobject the object stays unmoveable, suspended in gas. In this case, ρgas = ρobject
3. When Fb > Wobject the object rises up. In this case, ρgas > ρobject

### Example 4

A balloon weighs 12 000 N when empty (all people and equipment are included in this value). When the balloon is flying, it has a volume of 1500 m3. What is happening with the balloon (is it rising up, saying unmoveable in air or falling down) if the balloon is filled with hot air (ρhot air = 0.8 kg/m3) and f the surrounding (cold) air has a density of 1.3 kg/m3? Take g = 10 m/s2. ### Solution 4

The solution is simple. We must calculate the buoyant force caused by the cold air and compare it with the weight of empty balloon plus the weight of hot air inside the balloon. We have:

Fb (cold air) = ρcold air × Vballoon × g
= 1.3 × 1500 × 10
= 19 500 N

Weight of the hot air inside the balloon is

Whot air = mhot air × g
= ρhot air × Vballoon × g
= 0.8 × 1500 × 10
= 12 000 N

Therefore, the total downward force is

Fdown = Wempty balloon + Whot air
= 12 000 N + 12 000 N
= 24 000 N

Since this force is greater than the buoyant force (lifting force) of the surrounding (cold) air, (24 000 N > 19 500 N), the balloon is falling down at 24 000 N - 19 500 N = 4500 N of resultant force.

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