ARCHEMEDIS PRINCIPLES

Archimedes’s Principle and Law of Flotation

Archimedes’s Principle

Consider the diagram below

When piece of wood immersed in fluid then floats due to the buoyant force or upthrust. Where by upthrust force is the upward force enable the object to float or at least seem higher

Archimedes principle also called the law of buoyancy which state 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”

U = W

Relationship between Real Weight and Apparent Weight

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

Real Weight is the weight of object in air and apparent weight is the weight of object in fluid

Mathematically

U = R - A

Where:

U = Upthrust or Apparent weight loss

A = Apparent weight

R = Real weight

Nb:

i. fluid normally exerts an Upthrust

ii. Upthrust tends to reduce weight of body

iii. 1g of water = 0.01N of water

iv. 1g of water = 1cm3 = 1ml of water

Example,

Given Weight of body in air is 10.10N weight of body when immersed in water is 9.2N. Find the upthrust.

Data given

Real weight, R =10.10N

Apparent weight, A = 9.2N

Upthrust, U = ?

Solution

From: U = R - A

W = 10.10 – 9.2

W =0.9N

Example,

The weight of a body when totally immersed in a liquid is 4.2N if the weight of the liquid displaced is 2.5N. Find the weight of the body in air.

Data given

Apparent weight, A = 4.2N

Upthrust, U = 2.5N

Real weight, R =?

Solution

From: U = R - A - make W1 subject

R = U + A

R = 2.5 + 4.2

R =6.7N

Example,

When an object is totally immersed in water, its weight is recorded as 3.1N if its weight in air is 4.9N. Find upthrust.

Data given

Real weight, R =4.9N

Apparent weight, A = 3.1N

Upthrust, U = ?

Solution

U = 1.8N

Example,

A body immersed in water displaced 1.1N of the liquid if its weight white in the water is 3.3N. Find they weight in air.

Data given

Apparent weight, A = 1.1N

Upthrust, U = 3.3N

Real weight, R =?

Solution

From: U = R - A - make W1 subject

R = U + A

R = 3.3 + 1.1

R = 4.4N

Relative Density by Using Archimedes Principle

Consider the formula below

R.D = msmw (Vw=Vs)

R.D = Ms x gmw x g

Where:

i. mw x g = upthrust = U

ii. Ms x g = real weight = R

iii. R.D = relative density

iv. ms = mass of substance

v. mw (vw = vw) = Mass of an equal volume of water or mass of water displaced

R.D = 𝑅U

But: U = R – A

R.D = 𝑹R−A

Example,

A piece of glass weight in air 1.2N and 0.7N when completely immersed in water calculate it’s.

(a) Relative density

(b) Density of glass

Data given

Weight of body in air, R = 1.2N

Weight of body in water, A = 0.7N

Density of water, ρw = 10000kg/m3

Relative density of glass, R.D = ?

Density of glass, ρg = ?

Solution

(a) Relative density of glass, R.D = ?

From: R.D = R/(R - A) = 1.2/ (1.2 - 0.7) = 2.4

R.D = 2.4

(b) Density of glass, d = ?

From: R.D = ρg/ρw = make ρg subject

ρg = R.D x dw = 2.4 x 1000

ρg = 2400

ρg = 2400 kg/m3

Relative Density of other liquid from water by solid substance in Archimedes Principle

When solid immersed in liquid and water the relative density is given by liquid displaced over water displaced

Mathematically

R.D = weight of liquid displacedWeight of water displaced

R.D = weight of object in air − weight of object in liquidweight of object in air− weight of object in water

R.D = upthrust on liquidupthrust on water = 𝑈𝑙𝑈𝑤

R.D = 𝑅− 𝐴𝑙𝑅− 𝐴𝑤

Where:

𝑈𝑙= upthrust on liquid

𝑈𝑤 =upthrust on water

𝐴𝑙= Apparent weight on liquid

𝐴𝑤 = Apparent weight on water

Example,

In an experiment to determine the relative density of a liquid, a solid Q weighted as follows:

Weight Q in air, R = 8.6N

Weight Q in water, AW = 6.0N

Weight Q in liquid, AL = 5.4N

Data given

Weight of body Q in air, R = 8.6 N

Weight of body Q in water, AW = 6.0N

Weight of body Q in liquid, AL = 5.4N

Solution

From:

R.D = 𝑅− 𝐴𝑙𝑅− 𝐴𝑤

R.D = 8.6 − 5.4 8.6 – 6.0 = 3.2/2.6 = 1.2

R.D = 1.2

Example,

Using the data shown above and determine the relative density of the liquid

Data given

Weight of body in air, R = 15 N

Weight of body in water, Aw = 11N

Weight of body in liquid, Al = 9N

Relative density, R.D = ?

Solution

From:

R.D = 𝑅− 𝐴𝑙𝑅− 𝐴𝑤

R.D = 15 − 9 18 – 11 = 6/7 = 1.5

R.D = 1.5

Example,

A body weights 0.52N in air. Total immersed in water it weighs only 0.32N while its weight when immersed in another liquid is 0.36N. The density of water is 1000 kg/m3. What is the density of the other liquid?

Data given

Weight of body in air, R = 0.52 N

Weight of body in water, Aw = 0.32N

Weight of body in liquid, Al = 0.36N

Density of water ρw = 1000 kg/m3

Relative density, R.D = ?

Density of liquid, ρl = ?

Solution

From:

R.D = 𝑅− 𝐴𝑙𝑅− 𝐴𝑤

R.D = 0.52−0.360.52−0.32 = 0.16/0.2 = 0.8

R.D = 0.8

But: R.D = ρl/ ρw – make ρl subject

ρl = R.D x ρw = 0.8 x 1000 = 800

ρl = 800 kg/m3

Sinking

Defn: Sinking is the tendency of an object to fall or drop to lower levels in a fluid

Condition for Sinking

i. Upthrust exerted by fluid is less than weight of object

ii. Object denser than fluid means object has great density than fluid

Floating

Defn: Floating is the tendency of an object to be suspended in (remain) on the surface of a fluid due to the upthrust

Nb:

The ability of an object to float is called Buoyancy

Condition for Floating

i. Upthrust exerted by fluid must be equal or greater to the real weight of the object

Nb: apparent weight approximately equal to zero

A ≈ 0 N

ii. The density of body must lees than that of fluid

iii. Volume of submerged object must be large enough to displace a lot of fluid

Law of Flotation

The law states that

“A floating body displaces its own weight of the fluid in which if floats”

Nb:

The real weight of the body can be expressed in terms of its density

For substance to float, R = U

But: R =mS x g

Since: mS = ρS x vS

Therefore: R = ρS x vS x g

Also: U =mf x g

Since: mf = ρf x vf

Therefore: U = ρf x vf x g

But: vf = %S x vs

Then: %S = (vt/vs) x 100

But: R = U

Then: ρs x vs x g = ρf x vf x g

Ρs x vs x g = ρf x %S x vf x g

Ρs = ρf x %S

Where

R = real weight of substance

U = W = weight of fluid displaced

Ms = mass of substance

Ρs = density of substance

Vs = volume of substance

Mf = mass of fluid displaced

Ρf = density of fluid displaced

Vf = volume of fluid displaced

%S= percentage of substance submerged

Vt = volume submerged

Application of Flotation

Law of flotation is applicant in various substance include

i. Filling Balloons

ii. Filling Hot air balloon

iii. Submarines

iv. Ships

v. Hydrometer

Balloons

Consider the diagram below

Since: mf = ρf x vf

Therefore: U = ρf x vf x g

But: vf = %S x vs

Then: %S = (vt/vs) x 100

But: R = U

Then: ρs x vs x g = ρf x vf x g

Ρs x vs x g = ρf x %S x vf x g

Ρs = ρf x %S

Where

R = real weight of substance

U = W = weight of fluid displaced

Ms = mass of substance

Ρs = density of substance

Vs = volume of substance

Mf = mass of fluid displaced

Ρf = density of fluid displaced

Vf = volume of fluid displaced

%S= percentage of substance submerged

Vt = volume submerged

Application of Flotation

Law of flotation is applicant in various substance include

i. Filling Balloons

ii. Filling Hot air balloon

iii. Submarines

iv. Ships

v. Hydrometer

Balloons

Consider the diagram below

Where:

F = force drift up below

U = upthrust

R = weight of balloon

Mechanism

Balloons is filled with a light gas e.g. Helium which displace volume of air equal to its volume. Since filled gas has the light or has low density that than displaced air so the balloon drifted up by a force

Hot - Air Balloon

Mechanism

When air heated in the envelope is increased and gas inside expand so its volume increase and mass of gas inside envelope remains Constant so its density lowered as if you compare to the external gas (cold air). This different in density drift the balloon and it’s passengers into the air

Consider the diagram below

For substance to float, R = U

But: U = Mt x g

U = (MB + Mc + mh) x g

But: mh = vh x ρh

Then: U = (Mt + vh x ρh) x g - - - - - 1

Also: R = mc x g

But: mc = vc x ρc x g

Then: R = vc x ρc x g - - - - - 2

But: vc = vh = v

Then: equation 1 = equation 2 (U = R)

(Mt + vh x ρh) x g = vc x ρc x g

(Mt + vh x ρh) = vc x ρc

Mt + v x ρh = v x ρc ce: mf = ρf x vf

Mt + v x ρh = v x ρc – make v subject

v = Mt/(ρc - ρh)

v = (ml + mb)/Δρ

Where

Total mass = mt = (MB + Mc + mh)

Volume of hot air = vh

Volume cold air = vc

Mass of hot air = mh

Mass cold air = mc

Mass of balloon = mb

Density hot air = ρh

Density cold air = ρc

Example,

A hot air balloon including the envelope, gondola, burner and fuel and one passenger has a total mass of 450kg. Air outside balloon is at 20℃ and has a density of 1.29kg/m3 the air inside at temperature 120℃ has density of 0.90kg/m3. To what volume must the envelope expand to just lift the balloon into the air?

Data given

Total mass, mt = 450kg

Density at 120℃ , ρ2 = 0.90 kg/m3

Density at 20℃ , ρ1 = 1.29 kg/m3

Volume of air displaced, v1 = ?

Solution

From: v = (ml + mb)/Δρ

v = 450kg /(1.29-0.9)

v = 450kg /(1.29-0.9)

v = 450/0.39 = 1.15

v = 1.15m3

Example,

A balloon has a capacity of 20m3 and it is filled with hydrogen. The balloon fabric and the container have a mass of 2.5kg. What mass of instruments can be lifted by the balloon? (Density of hydrogen = 0.089kg/m3 and density of air is 1.29kg/m3)

Data given

Total mass = mt = ml + mb

Volume capacity, v = 20m3

Volume cold air, v1 =20m3

Mass of balloon, mb = 2.5kg

Density at v2, ρ2 = 0.089kg/m3

Density at v1, ρ1 = 1.29kg/m3

Mass of instrument), ml = ?

Solution

From: v = (ml + mb)/Δρ – make ml subject

Ml = (v x Δρ) – mb)

Ml = (20 x (1.29 – 0.089)) – 2.5

Ml = (20 x 1.201) – 2.5

Ml = 24.02 – 2.5

Ml = 21.52kg

Sub Marine

Consider the diagram above

Mechanism

Submarine made with empty space filled with air called ballast in order to increase its volume in order to devise density of submarine and vice versa

When water filled in the ballast the submarine submerged and when balloon admitted to special tank its due and when ballast is filled with air the sub marine floats like other ship

Nb:

When water quantity increased/ filled in the blast is tend to reduce volume hence increase the density of submarine.

Ship

Ship is made of steel and is expected to sink due to its weight. it contains hollow which increase the volume of ship which help on making less dense that the waters

But when load put on ship it tends to increase the density and mass of ship when overloaded the ship sink completely. To check on over wading ships are marked with line or mark called Plimsoll marks.

Plimsoll line

Plimsoll line on a ship used to show minimum heights (maximum density) above different of water types in different sea as shown in the diagram below

 

Where:

F = for fresh water

S = for sea in summer time

W = for sea in water time

TF = tropical fresh water

WNA = winter in Atlantic

T = Tropical

Hydrometer

Defn: hydrometer is an instrument used for measuring the densities of liquids or hydrometer is an instrument used for determine the relative density of liquids.

Structure of Hydrometer

i. Heavy sinker (bulb): containing mercury or lead shots that keep the hydrometer upright when it floats

ii. Air bulb: it increases volume of displaced liquid and overcomes the weight of the sinker

iii. Stem: stem is thin so that small changes in density (height) give large differences in reading

iv. Scale: inside stem graduated in densities

v. Made up of glass to prevent soaking of the liquid

Nb:

The greater the density of the liquid the shorter the stem of hydrometer immersed

Relative Density of Liquid by Hydrometer

i. When hydrometer floats over water the weight of hydrometer (wg) must equal to the weight of water displaced (ww)

wg = ww

ii. When hydrometer floats over liquid the weight of hydrometer (wg) must equal to the weight of liquid displaced (wl)

wg = wl

iii. since relative density of liquid is given by ratio of density of liquid (ρl) to the density of water (ρw)

R.D = ρl/ρw = wl/vl ÷ ww/vw

Where:

vl = volume of liquid displaced

vw = volume of water displaced

wl = ww

Then: R.D = vw/vl

iv. since cross-section area of the hydrometer is uniform, the volume of water and of liquid displaced are proportional to the lengths immersed in them

R.D = lw/ll

Where:

Lw = length of hydrometer immersed in water

Ll = length of hydrometer immersed in liquid

Consider the Diagram Below

 Where:

Steam volume, v1 = Ah

Bulb volume, v2 = v

Total volume, vt = v1 + v = Ah + v2

But: R = U

Where:

U = upthrust in liquid

ρmn = minimum density

ρmx = maximum density

R = weight of hydrometer

U = vt x ρmn x g

R = v x ρmx x g

But: R = U

Then: vt x ρmn x g = v x ρmx x g

(Ah + v) x ρmn x g = v x ρmx x g

(Ah + v) x ρmn = v x ρmx

Ah x ρmn + v x ρmn = v x ρmx – make v2 subject

v x ρmx - v x ρmn = Ah x ρmn

v x (ρmx - ρmn) = Ah x ρmn

v = (Ah x ρmn)/(ρmx – ρmn

v = (Ah x ρmn)/(ρmx - ρmn)

Example,

Consider the diagram below used to measure density of liquid between 1g/cm3 to 0.81g/cm3 (The area of cross section area of stem is 0.5cm2). Find the volume of hydrometer below 1.0 g/cm3 graduated

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, vt = (8 + v2) cm3

Minimum density, ρmn = 0.8 g/cm3

Maximum density, ρmx = 1.0 g/cm3

The volume of bulb, v2 = ?

Solution

The volume of bulb, v2 = ?

From: v2 = (Ah x ρmn)/(ρmx - ρmn)

v2 = (8 x 0.8)/(1 – 0.8)

v2 = 6.4/0.2

v2 = 32 cm3

Example, : NECTA form IV 2012 QN: 4

(a) What does a solid body weight more in air than when immersed in a liquid?

(b) An ordinary hydrometer of mass 27g floats with 4cm of its stem out of water. If cross section area of stem is 0.75cm2 calculate

i. The total volume of stem just under the surface of the liquid

ii. The relative density of the liquid

Data given (b):

Mass of hydrometer, mh = 27g

Mass of water displaced, mw = 27g

Area of stem, A = 0.75cm2

Height of stem, h = 4cm

Density of water, ρw = 1g/cm3

Density of liquid, ρl = ?

Solution

i. Total volume, vt = v1 + v

Volume of stem, v1 = Ah = 3 cm3

Volume of bulb, v = mw x ρw = 27 cm3

vt = v1 + v

vt = Ah + mw x ρw

vt = (3 + 27) cm3

vt = 30 cm3

ii. Relative density, R.D = ?

From: R.D = ρl/ρw

But: ρl = ml/vt = 27/30 = 0.9 g/cm3

Then: R.D = ρl/ρw 0.9/1 = 0.9

R.D = 0.9

Example,

A balloon of volume 2000m3 is filled with hydrogen of density 0.09 kg/m3. If the mass of fabric is 100kg and that of the pilot is 75kg,

i. What will be the greatest mass of equipment that can be carried when operation in air is 1.25kg/m3?

ii. How would this figure change if helium, which has twice the density of hydrogen under the same condition, were to be used?

Solution

i. For hydrogen

Data given

Mt total mass = mt = (ml + mb)

Volume of hydrogen, v2 = 2000m3

Volume surrounding, v1 =20000m3

Mass of balloon, mb = (75+100) = 175kg

Density at v2, ρ2 = 0.09kg/m3

Density at v1, ρ1 = 1.25kg/m3

Density change, Δρ = (ρ2-ρ1) = (1.25 - 0.09) = 1.16kg/m3

Solution

Mass of instrument), ml = ?

From: v = (ml + mb)/Δρ – make ml subject

Ml = (v x Δρ) - mb

Ml = (2000 x 1.16) – 175

Ml = 2320 – 175

Ml = 2145kg

ii. For helium

Data given

Mt total mass = mt = (ml + mb)

Volume of helium, v2 = 2000m3

Volume surrounding, v1 =20000m3

Mass of balloon, mb = (75+100) = 175kg

Density at v2, ρ2 = 0.18kg/m3

Density at v1, ρ1 = 1.25kg/m3

Density change, Δρ = (ρ2-ρ1) = (1.25 - 0.18) = 1.07kg/m3

Solution

Mass of instrument), ml = ?

From:

v = (ml + mb)/Δρ – make ml subject

Ml = (v x Δρ) - mb

Ml = (2000 x 1.07) – 175

Ml = 2140 – 175

Ml = 1965kg

Example,

The mass of a piece of cork (0.25 g/cm3) is 20g. What fraction of the cork is immersed when it floats in water?

Data given

Mass of cork, mc = 20g

Density of cork, ρc = 0.25 g/cm3

Density of water, ρw = 1 g/cm3

Cork Fraction immersed, x = ?

Volume of water displaced, vw = ?

Volume of cork immersed, vc = ?

Solution

From: principle of flotation (vw = vc)

Mc = mass 0f water, mw

20 = vw x 1

Vw = vc = 20 cm3

But: x = vc/vt

Where: Volume of cork, vt = ?

From: ρc = mc/vt

Vt = mc/ ρc = 20/0.25 = 80 cm3

Vt = 80 cm3

X = 20/80 = 1/4

X = 1/4

Example,

The mass of a piece of cork (0.25 g/cm3) is 20g. What fraction of the cork is immersed when it floats in alcohol?(density of alcohol is 0.8 g/cm3)

Data given

Mass of cork, mc = 20g

Density of cork, ρc = 0.25 g/cm3

Density of alcohol, ρa = 0.8 g/cm3

Cork Fraction immersed, x = ?

Volume of alcohol displaced, va = ?

Volume of cork immersed, vc = ?

Solution

From: principle of flotation (va = vc)

Mc = mass 0f water, mw

20 = vw x 0.8

Vw = va = 25 cm3

But: x = vc/vt

Where: Volume of cork, vt = ?

From: ρc = mc/vt

Vt = mc/ ρc = 20/0.25 = 80 cm3

Vt = 80 cm3

X = 25/80 = 5/16

X = 5/16

Example,

A uniform pencil floats upright in water with 8cm of its length immersed. What length is immersed when its floats in glycerol (density of glycerol is 1.3 g/cm3)?

Data given

Length of pencil immersed in water,lw = 8cm

Density of glycerol, ρg = 1.3 g/cm3

Density of water, ρw = 1 g/cm3

Length of pencil immersed in glycerol, lg = ?

Solution

But: R.D of glycerol = lw/lg

Then: lg = lw/R.D = 8/1.3 = 6.2cm

lg = 6.2cm

Example,

A balloon and the gas in it have a mass of 450g and its volume is 500 litres. What is the maximum load it can lift in air of density 1.3 g/cm3?

Data given

Total mass, mt = (ml + mb)

Mass of balloon, mb = 450g

Density of gas, ρg = 1.3 g/cm3

Density change, Δρ = 1.3 g/cm3

Volume of balloon, vb = 500 litres

Maximum load, ml = ?

Solution

Maximum load, ml = ?

From: v = (ml + mb)/Δρ – make ml subject

Ml = (v x Δρ) - mb

Ml = (500 x 1.3) – 450

Ml = 650 – 450

Ml = 200g

Example, : NECTA 2012

The diagram below shows on form of man hydrometer used to me 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 cm2 and 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

Data given

Cross section area of stem, A = 0.5cm2

Height of steam, h = 18 cm

The volume of steam, v1 = Ah = 9 cm3

Total volume, vt = (9 + v2) cm3

Minimum density, ρmn = 0.8 g/cm3

Maximum density, ρmx = 1.0 g/cm3

The volume of bulb, v2 = ?

Solution

(a) The volume of bulb, v2 = ?

From: v2 = (Ah x ρmn)/(ρmx - ρmn)

v2 = (9 x 0.8)/(1 – 0.8)

v2 = 7.2/0.2

v2 = 36 cm3

(b) What height, h2 of hydrometer when shifted to measure 0.9 g/cm3

But: weight of hydrometer never change

V3 x 0.9 x 0.01N = 0.36N

V3 = 40 3

V3 = Ah2

h2 = 8cm


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