SINKING AND FLOATING

Archimedes’ Principle

• Consider a piece of wood that is held below the surface of a

liquid and then released .The wood comes to the surface

immediately.

• When a piece of wood is immersed in a fluid, then it floats due to

the buoyant force or upthrust.

• Upthrust is the upward force that enables the object to float or at

least seem lighter

• The upthrust is greater than the weight of the wood, that is why

the wood is pushed to the surface

Archimedes Principle (the law of buoyancy)

It states that:

“When a body is partially or totally immersed in a fluid it

experiences an upthrust which is equal to the weight of the fluid

displaced”

Relationship between Real Weight and Apparent Weight

• Consider the diagram of the mass (weight) of the object below

• Real Weight

• Apparent weight

is the weight of an object in air

is the weight of an object in fluid

• ∴ Apparent loss in weight = Real weight – apparent weight of a

body in liquid

∴ 푼 = 푹 − 푨

Example

1. Given that the weight of a body in air is 10.10N while the weight

of the body when immersed in water is 9.2N. Find the up thrust.

Solution:

Weight in air (R) = 10.10 N

Weight in water (A) = 9.2 N

∴ 푼 = 푹 − 푨 = ퟏퟎ.ퟏퟎ − ퟗ. ퟐ = ퟎ. ퟗ 푵

Relative Density by using Archimedes Principle

 The relative density of a substance can be expressed as:

Example

1. A piece of glass weighs 5 N in air and 3 N when completely

immersed in water calculate its.

(b) Density of glass Solution:

Given: R = 5 N

(a) Relative density

A = 3 N

From:

kgm^-3

∴ The density of the piece of glass is 2400 kgm^-3

Relative Density of other liquid from water by solid

substance in Archimedes Principle

 When a solid immersed in a liquid and water the relative density

is given by liquid displaced over water displaced

Mathematically

R.D =

R.D

Example

In an experiment to determine the relative density of a liquid, a solid X

weighs as follows:

Weight of X in air, W_A(R) = 8.6N

Weight of X in water, W_A= 6.0N

Weight of X in liquid, W_L= 5.4N

Solution:

From: R.D

R.D

Sinking

 Sinking is the tendency of an object to fall or drop to lower levels

in a fluid

Conditions for Sinking

• The upthrust exerted by the fluid must be less than the weight of

an object

• The density of the object should be greater than that of the fluid

Floating

• Floating is the tendency of an object to remain on the surface of

a fluid due to the upthrust.

• The ability of an object to float in a fluid is called Buoyancy

Conditions for Floating

• The upthrust due to the liquid must be equal to the total weight

of the object

• The density of the body must be less than that of fluid.

• The Volume of submerged object must be large enough to displace

a lot of fluid.

Difference between floating and sinking

Floating

Sinking

The body stays at the surface of

the liquid

The body drops to the bottom of

the liquid

Takes place when the upward Takes place when the upward

force is greater than the weight force is less than the weight of

of the body

the object

Takes place when the density Takes place when the density of

of the body is less than that of the body is greater than that of

the liquid

The following conditions can be used to determine the

position of the object in the fluid:-

• If W > U, there is downward movement of the body (known as

sinking)

• If W < U, there is upward movement of the body

• If W = U, the body is in equilibrium under the action of two equal

and opposite force. (Thus the body floats)

The Law of Flotation

• It states that:

“A floating body displaces its own weight of the fluid in

which it floats”

• For a substance to float

(a)

(b)

Upthrust (U) = weight of displaced fluid (W)

Real weight(R) = weight of displaced fluid (W)

• But: Apparent weight = Real weight – Upthrust (weight of

displaced fluid)

• For a body to float Apparent weight = 0 N

Therefore: R = U = W

 To find percentage of submerged substance consider

the equation

Real weight of a substance (R) = Weight of displaced fluid

(W)

But now: R = mass of substance (m_s) x g

R = density (흆_s) × volume of substance (v_s) × 품

Also:

U = mass of fluid (m_f) x g = density (흆_f) × volume of

fluid (v_f) × 품

•

But: Volume of fluid displaced = % of object submerged (S) x

Volume of object (Vs)

V_f= % Sub x V_s---------------------- (i)

∴ Percentage submerged (%S)^풗=^{풐풍풖풎풆 풔풖풃풎풆풓품풆풅}

_{풗풐풍풖풎풆 풐풇 풔풖풃풔풕풂풏풄풆}x 100%

•

Since: R = U (when substance floats)

Then: ρs x vs x g = ρf x vf x g ---------------- (ii)

Substitute equation (i) into equation (ii)

ρs x Vs x g = ρ_fx %Sub x V_Sx g

흆

풔

∴

=

× ퟏퟎퟎ%

% Sub

흆

풇

Application of Flotation

The Law of flotation is applicable in various

objects like:-  Applied when Filling Balloons

(Hot air balloons)

• Submarines

• Ships

• Hydrometer

Balloons (Hot air balloons)

• A balloon is a light bag filled with hydrogen or helium gas.

• These gases are less denser than air. An air ship is a large

balloon with a motor to make it and fins to steer it.

• The downward force in a balloon is equal to the weight of the

bag plus the weight of a gas in it.

• The balloon rises if the upthrust is greater than the down ward

force. That is

NB:

• As a balloon rises, the atmospheric pressure on it becomes less.

The gas in the balloon tends to expands

. Therefore, the

gas – bag must not be filled completely when the balloon is

on the ground.

• Consider the diagram below

• If the balloon is filled with some gas of known density. Then the

volume of gas required just to lift the balloon into the air is given

by

Example

1. A balloon and the gas in it has a mass of 450 g and its volume is

500 cm³. What is the maximum load it can lift in air of density 1.3 g/cm³

Solution:

From: The Principle of floatation,

Mass of balloon + Load = Mass of air displaced

= Volume of gas x density of air

= 500 x 1.3 = 650 g

∴ 푀푎푥푖푚푢푚 푙표푎푑 = (풎풂풔풔 풐풇 풃풂풍풍풐풐풏 + 풍풐풂풅) − 푴풂풔풔 풐풇 풃풂풍풍풐풐풏

= ퟔퟓퟎ − ퟒퟓퟎ = ퟐퟎퟎ 품

Sub Marines

• A Submarine is a watercraft capable of independent operation

underwater

• A submarine has ballast tanks which can be filled with water or

air

• When full of water ,the average density of the submarine is

slightly greater than the density of the sea – water and it sinks

• When air is pumped into the tanks the average density of the

submarine falls until it is the same or slightly less than that of the

water around it

• The submarine therefore stays at one depth or rises to the

surface

Ships

• A ship is a large watercraft that travels the world’s oceans and

other sufficiently deep waterways

• It is used to carry passengers or goods or in supporting

specialized missions such as defense, research and fishing

• A Ship is made of steel and is expected to sink due to its weight.

it contains hollow which increases the volume of ship which

helps on making less denser than water

• So for safety loading of the ship under different sea conditions

plimsol lines (plimsol marks) are provided

• Plimsoll lines: Are lines which show maximum height of the

ship that should

be under water

• Plimsoll lines are also referred as plimsol marks. See the figure

below

Whereby:

F =

W =

Fresh water

Winter Sea

S =

TF =

Summer sea

Tropical

T =

fresh water WNA = Winter North Atlantic

Tropical seas

Hydrometer

• Is a floating instrument used for measuring the densities of

liquids. OR

• Is an instrument used for determining the relative density of

liquids.

Parts (Structure) of Hydrometer

• Heavy sinker (bulb): containing mercury or lead shots that keeps

the hydrometer upright when it floats

• Air bulb: it increases volume of displaced liquid and overcomes

the weight of the sinker

• Stem: stem is thin so that small changes in density (height) gives

large difference in reading

• Scale: Inside stem graduated in densities

• It is made up of glass to prevent soaking of the liquid

Mode of action of hydrometer

• A hydrometer is made to float in the liquid whose relative density

is to be measured

• It sinks to different levels depending on the relative density of the

liquid in which it floats

• The liquid whose relative density is to be determined is poured

into a tall jar and the hydrometer is gently lowered into the liquid

until it floats freely.

• The point where the surface of the liquid touches the stem of the

hydrometer indicates the relative density of the liquid

• Example, the hydrometer sinks more in methanol than in water.

This indicates that water is denser than methanol

NB:

• The greater the density of the liquid the shorter the stem of

hydrometer immersed

• Hydrometer works on the principle of Archimedes

Relative Density of Liquid by Hydrometer

• When hydrometer floats over water the weight of hydrometer

(w_h) must be equal to the weight of water displaced (w_w)

That

is w_h= w_w

• When hydrometer floats over liquid the weight of hydrometer (w_h)

must be equal to the weight of liquid displaced (w_L)

That is w_h

= w_L

• Since the relative density of liquid is given by ratio of density of

liquid (ρ_L) to the

But

Thus:

, Since m_L= m_w

= m_h

• Since cross-section area of the hydrometer is uniform, the

volume of water and that of liquid displaced are proportional to

the lengths immersed in them

∴ R.D =

Consider the diagram below

Whereby:

Stem volume, V₁= Ah

Bulb volume, V₂= V

Total volume, V_T= V₁+ V₂= Ah + V

But: R = U (Weight of hydrometer = Up thrust of liquid)

Whereby:

ρ_mn= minimum

density ρ_mx=

maximum density

U = V_tx ρ_mnx g

R = V x ρ_mxx g

Then:

V_tx ρ_mnx g = V x ρ_mxx g

(Ah + V) x ρ_mnx g = V x ρ_mxx g

(Ah + V) x ρ_mn= V x ρ_mx

Ah x ρ_mn+ V x ρ_mn= V x ρ_mx

subject

--------------- make V₂the

V x ρ_mx– V x ρ_mn= Ah x ρ_mn

V x (ρ_mx– ρmn) = Ah x ρ_mn

푨풓풆풂 풐풇 풔풕풆풎 ×풉풆풊품풉풕 풐풇 풔풕풆풎 ×풎풂풏풊풎풖풎 풅풆풏풔풊풕풚

푨풉 × 흆풎풏

∴

=

V_Bulb=

풎풂풙풊풎풖풎 풅풆풏풔풊풕풚 − 풎풊풏풊풎풖풎 풅풆풏풔풊풕풚

흆풎풙 – 흆풎풏

Examples

2. Consider the diagram below used to measure density of liquid

between 1g/cm³to 0.81g/cm³(The area of cross section area

of stem is 0.5 cm²). Find the volume of hydrometer below 1.0

g/cm³graduated

Data given

Cross section area of stem, A = 0.5cm²

Height of steam, h = 16 cm

The volume of steam, V₁= Ah = 8 cm³

Total volume, V_T= (8 + V₂) cm³

Minimum density, ρmn = 0.8 g/cm³

Maximum density, ρmx = 1.0 g/cm³

The volume of bulb, V₂=?

Solution

From: V_Bulb

V_Bulb

3. The diagram below shows on form of man hydrometer used to

measure the densities of liquid over the range of 0.8 to 1.00

g/cm³. If the area of cross section of the stem is 0.5 cm²and

the distance between the 0.80 and 100 division is 18cm

determine

(a)The volume of hydrometer below 1.00 graduated

(b)The position of the 0.90 graduation

Solution:

Given

Cross section area of stem, A = 0.5cm²

Height of stem, h = 18 cm

The volume of stem, V₁= Ah = 9 cm³

Total volume, Vt = (9 + V₂) cm³

Minimum density, ρmn = 0.8 g/cm³

Maximum density, ρmx = 1.0g/cm³

The volume of bulb, V₂=?

(a)The volume of bulb, V₂= ?

푨풉

From:

VBulb

(b)What height, h₂of hydrometer when shifted to measure 0.9 g/cm³

푨풉

From:

V_Bulb

Daily Application of the Law of Floatation

(a)

It is used in transportation by water ways: (By ships,

submarines and ferry boats)

(b)

It is used in transportation by air ways:(By hot air

balloons and air ships)

(c)

It is used in decoration: (Balloons of different colors and

shapes are filled with lighter gas so that will float in air)

(d)

It is used in determination of specific gravity of liquids:

(By hydrometer)