Buoyancy and Archimedes Principle PDF
Preview Buoyancy and Archimedes Principle
Lecture 8 Buoyancy Archimedes Principle Fluid flow Viscosity Buoyancy and Archimedes Principle Archimedes principle: Any object completely or partially immersed in a fluid experiences an upward or buoyant force equal to the weight of the fluid it displaces. Archimedes (287 BC – 212 BC) Greek Physicist, Mathematician Buoyant force easily explained Pressure is greater at greater depth: F 1 therefore the upward force is greater at the bottom of an object than the downward F 2 force at the top of the object F > F 2 1 Net upward force is the buoyant force is equal to the weight of the displaced fluid. Buoyancy and Archimedes Principle Cube of same fluid Buoyant force F b F = P A - P A b 2 2 1 1 h P A Cube remains stationary 1 1 1 h 2 F = weight of fluid b = weight of fluid mg P A displaced 2 2 P = rgh F = P A - P A = m g 1 1 b 2 2 1 1 f where m =mass of fluid P = rgh f 2 2 Cube of different material (only mass of cube changes) buoyant force remains unchanged (F = m g) b f F = m g =r V g b f fluid fluid • If an object floats then the buoyant force must equal its weight. • If an object sinks then its weight must be greater than the buoyant force. Buoyancy and Archimedes Principle Objects that float Object’s weight is equal to the buoyant force F b w = F object b Applying Archimedes principle F is equal to weight of fluid displaced b therefore w = w (displaced) object fluid True only if object floats F = m g = r V g F = m g = r V g b f fluid fluid b f object object r V g = r V g fluid fluid object object r V volume of object submerged object fluid = = volume of fluid displaced r V fluid object “The fraction of the object that is submerged is equal to the ratio of the density of the object to the density of the fluid.” Buoyancy and Archimedes Principle Applications Why do ships float Average density is less than that of water Ship partially submerges until weight of ship equals weight of water displaced. Hot air (or Helium) balloons Hot air less dense that cold air resulting in a net upward force on the balloons Human brain is immersed in cerebrospinal fluid, density ≈ 1007 kgm-3 average density of brain ≈ 1040 kgm-3 Most of brain’s weight is supported by the buoyant force Buoyancy and Archimedes Principle Objects that sink (totally submerged) Applying Archimedes principle F is equal to weight of fluid displaced b F = m g = (V )(r )g b fluid fluid fluid But volume of fluid displaced = volume of object F = (V )(r )g therefore b object fluid Downward gravitational force on object m g = (r ) (V )g object object object Net force upwards on object = F - w b object = (V )(r )g -(r ) (V )g object fluid object object Net force upwards= (r -r )(V )g fluid object object If r > r object will float fluid object If r < r object will sink fluid object Buoyancy and Archimedes Principle 0 0 w - F object b w object Fluid Apparent weight = net downward force = w - F object b F = weight – apparent weight b F = weight of liquid displaced b Buoyancy and Archimedes Principle Example Determine the density of a liquid if an object of known volume (50cm3) and mass (0.150kg) has an apparent mass of 0.105kg in the liquid. Applying Archimedes principle Buoyant force is equal to weight of liquid displaced Weight – apparent weight =weight of liquid displaced Mass – apparent mass =mass of liquid displaced 0.150kg - 0.105kg =0.045kg is displaced by the object Volume of object =50cm3 = 50x10-6m3 = Volume of liquid displaced Therefore density of liquid = (0.045kg) /(50x10-6m3) = 900kgm-3 Buoyancy and Archimedes Principle Calculate the volume and density of an Irregular shaped object. Example A person has a mass of 75kg in air and an apparent mass of 2kg when submerged in water. Calculate the volume and density of the person. Mass – apparent mass = mass of water displaced 75kg – 2kg = 73kg = mass of water displaced Volume of water displaced =(m )/(r ) water water = 73kg/1000kgm-3 = 73x10-3m3 Therefore volume of person = 73x10-3m3 r =(m )/(volume ) person person person r =(75kg)/(73x10-3m3) = 1027kgm-3 person Buoyancy and Archimedes Principle A 70 kg statue lies at the bottom of the sea (Density 1.025x103 kg m-3). Its volume is 3.0x104cm3. How much force do you need to lift it? Weight - Apparent weight = weight of fluid displaced mg –W = weight of fluid displaced app W = mg –m g app fluid W = mg –r Vg app fluid W = g(m –r V) app fluid W =9.8ms-2(70kg –1.025x103 kg m-3 3.0x10-2m3) app W = 9.8ms-2( 70kg –30.75kg) app 384.65N
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