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.

Archimedes and Upthrust

Archimedes and Upthrust

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.

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.

Archimedes and Upthrust Big Boat:- http://www.youtube.com/watch?v=ykYHN3pP2SA Down:- http://www.youtube.com/watch?v=6pKyRogjlfc Dn:http://www.youtube.com/watch?v=8F5bf6XMleQ&feature=related Up:-http://www.youtube.com/watch?v=YNj2xrschr4&feature=related up:- http://www.youtube.com/watch?v=lxyMzR9K1Dk

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

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 100-250 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,

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.

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.

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.

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.

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. = 1030 x 9.81 x (¾ x 1250) = 9472.8 kN When completely submerged upthrust of water = ρ g Vol. = 1030x9.81x1250 = 12630.4 kN. This figure must also include the weight of water in the ballast Thus:- 12630.5 kN = 9472.8 kN + Weight of water W Weight of water W = 3157.6 kN Mass of water = (3157.6 x 1000) N ÷ 9.81 = 321873 kg Mass of water = 322 tonnes

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 = 32 N - ρ x g x Volume 26.5 N = 32 N - (0.8 x 1000 kg/m3) x 9.81 x Vol. Volume = 0.0007 m3 Check this solution!

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 = 90 - 84.5 = 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!

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 - 9000N (1.23 x 9.81 x 2240) – 9000N = 18 kN

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 0.0253 = 65.5 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 642.6 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 642.6 x 10-6 - 0.449N Metal density ρm = 8325 kg m-3