1. Archimedes and Up thrust
Archimedes’ principle:-
When a body is wholly or partially immersed in a fluid, there is an
upward force or upthrust on the body equal to the weight of the fluid
displaced by the body.
The principle is also quoted as:-
apparent loss in weight equal to the weight of the fluid displaced by the
body.

2. Archimedes and Upthrust

3. Archimedes and Upthrust

4. Archimedes and Upthrust
The picture below shows a modern submarine breaching the surface during an emergency surface routine. To do this the water is forced from the ballast tanks to reduce the weight of the boat.

5. Pressure at Depth This picture shows a modern
submarine on a rapid dive test.
Air is blown from the ballast tanks
and replaced with water, thus
increasing the weight of the boat.

6. Archimedes and Upthrust
Big Boat:-
Down:-
Dn:
Up:-
up:-

7. Archimedes and Upthrust
Typhoon class boat:-
The biggest yet !
Displacement:
Surfaced: 23,200-24,500 tonnes
Submerged: 33,800-48,000 tonnes
Length:175 metres
Beam:23 metres
Draught:12 metres
Propulsion and power:2×OK-650
pressurized-water nuclear reactors
90 megawatt each
2×VV-type steam turbines
37 megawatt each2 shaft,
7 blades shrouded screws
Speed:Surfaced: 12 knots Submerged: 27 knots
Test depth:400 metres
Complement:163

8. Archimedes and Upthrust
The Kursk – disaster !
At 154m long – and four stories high – it was the largest attack submarine ever built.
The outer hull, made of high-nickel, high-chrome content steel 8.5 mm thick, had exceptionally good
resistance to corrosion and a weak magnetic signature which helped prevent detection by Magnetic
Anomaly Detection (MAD) systems. There was a two-metre gap to the 50.8 mm thick steel inner hull.
The Kursk sailed out to sea to perform an
exercise of firing dummy torpedoes.
On August 12, 2000 the torpedoes were fired,
but soon after there was an explosion on the Kursk.
The chemical explosion blasted with the force
of kg of TNT and registered 2.2 on the Richter scale.
The submarine sank to a depth of 108 metres (350 ft),
about 135km (85 miles) from Severomorsk,

9. Archimedes and Upthrust
If an object (eg solid block of steel) is weighed in air and then weighed
again whilst immersed in water it will ‘appear’ to have lost weight.
However as weight is product of mass and gravity, neither of which will
change by immersion in water, then the apparent weight loss must be
due to another force acting.
It can be shown that the other force acting, termed upthrust, is equal to
the weight of the water displaced by the steel block.

10. Archimedes and Upthrust
Floatation
If an object is seen to float on a liquid then its total weight is supported
by the liquid.
Archimedes’ principle would suggest that the weight of the fluid
displaced by the floating object is equal to the weight of the object.
As weight is mass x gravity, and gravity assumes the constant g (9.81)
then it is acceptable to use mass in place of weight.
Thus:- If a body floats in a fluid then the mass of fluid displaced is equal
to the mass of the body.

11. Archimedes and Upthrust
Floatation
If an object is seen to float then the term upthrust is often replaced with
the term buoyancy.
A block of steel which will sink in water as a block can be made to float
if shaped into a ships hull.
Upthrust calculations are based on previous work involving pressure at
depth, and the equation ρ g h = P
as Force = pressure x area (P A) Then Force on a body at depth is
found from ρ g h x Area . The area is the surface area on which the
upthrust can be considered to act.

12. Archimedes and Upthrust
Floatation
As depth h x area A give the volume of the fluid displaced then:-
Upthrust = ρ g x Volume displaced.
For an object to float this must balance the weight of the object.
For a submarine the ballast tanks are flooded with water or filled with
air to alter the total weight of the boat without changing its volume.
eg. A submarine floats in sea water of density 1030 kg m-3 with ¾
of its volume submerged. Find the mass of water the submarine must
take in to its ballast tanks to allow it to ‘float’ completely submerged.
The total volume of the submarine is 1250 m3.

13. Archimedes and Upthrust
eg. A submarine floats in sea water of density 1030 kg m-3 with ¾ of its volume
submerged. Find the mass of water the submarine must take in to its ballast
tanks to allow it to ‘float’ completely submerged. The total volume of the
submarine is 1250 m3.
Solution:
The weight of the submarine with its tanks empty = weight of water displaced
by ¾ of its volume = ρ g Vol. = x 9.81 x (¾ x 1250) = kN
When completely submerged upthrust of water = ρ g Vol. = 1030x9.81x1250
= kN. This figure must also include the weight of water in the ballast
Thus: kN = kN + Weight of water W
Weight of water W = kN
Mass of water = ( x 1000) N ÷ = kg
Mass of water = tonnes

14. Archimedes and Upthrust
The upthrust concept of ρ x g x Volume can be put to other uses:
eg. A steel casting weighs 32 N in air. When suspended
completely immersed in a tank of oil of relative density 0.8 the weight
reduces to 26.5 N. Find the volume of the casting.
Solution:
Apparent weight = weight in air – upthrust
26.5 N = N ρ x g x Volume
26.5 N = N (0.8 x 1000 kg/m3) x 9.81 x Vol.
Volume = m3
Check this solution!

15. Archimedes and Upthrust
Try:- A plastic floatation cylinder has flat ends 160 mm diameter, the
cylinder has an overall length of 90mm and a total mass of 1.7 kg.
Calculate the length of float above the surface if it is placed in:
a. pure water b. sea water of density 1030 kg/m3
Solution: As the object floats then the weight = upthrust (ρ x g x Vol)
Mass x g = ρ x g x Volume g is a constant on both sides
Mass = ρ x Volume and volume = area x depth = π d2/4 x depth
a. For pure water:- ρ = 1000 kg/m3 , diameter = 0.16m
1.7 kg = 1000 kg/m3 x π 0.162/4 x depth
Depth = 84.5 mm and so the float height above water = = 5.5mm
a. For sea water:- ρ = 1030 kg/m3 , height = 8mm
Re-calculate for oil of density 800 kg/m3 and note your findings!

16. Archimedes and Upthrust
Try:
A weather balloon has a total weight of 9000 N and contains hydrogen
gas with a volume of 2240 m3. If the density of the surrounding air is
1.23 kg m-3 calculate the tension in the cable needed to hold the
balloon down.
Tip:- Cable tension will be the difference between the upthrust from the
air pressure and the weight of the balloon.
ρ x g x Volume N
(1.23 x 9.81 x 2240) – 9000N = 18 kN

17. Archimedes and Upthrust
Try:
A metal sphere, 50mm diameter is suspended in oil of relative density
0.7. The apparent weight of the sphere is 4.9N, find the density of the
sphere metal.
Solution:
Sphere volume = 4/3 π r3 = 4/3 x π x = x 10-6 m3
Weight of sphere in air = ρm x g x Vol. (ρm = metal density)
= ρm x 9.81 x 65.5 x 10-6 m3 = ρm x x 10-6
Upthrust due to oil = ρoil x g x Vol. = 700 x 9.81 x (65.5 x 10-6) = 0.449N
Apparent weight = weight in air - upthrust in oil
4.9N = ρm x x N
Metal density ρm = kg m-3