Pressure and buoyancy in fluids FCQ s for lecture and tutorials will be next week. Buoyancy force today Fluid dynamics on Monday (alon with the loudest demonstration of the semester). Review on Wednesday and Friday next week. Final exam at 7:30am on Tuesday, May 5.
Pressure in a liquid A Consider the bottom of a column of water with depth d and cross sectional area A inside a container open to the atmosphere. Atmospheric pressure pushes down with force of A. m pa A d pa The weiht of the column pushes down with force m. For a liquid with density ρ, m=ρv=ρad. Because the liquid is in static equilibrium, the upward force from pressure, pa, must equal the downward forces pa = A + ρad so p = + ρd (hydrostatic pressure) When you descend in a liquid, the weiht of the liquid above you causes the pressure to increase. 2
Clicker question 1 Set frequency to BA Three vessels are full of the same liquid and open to the same atmosphere. The pressure is measured in each at a distance of 3 m below the surface. What can we say about the pressures? A. only two are the same B. all three are different C. all three are the same 3 m The hydrostatic pressure is p = + ρd It only depends on the pressure on top and the amount of water in a column directly overhead so it is the same for all 3. 3
In the problem, the liquids had the same heiht because they were filled that way. Pressure in a liquid If they were all connected (as shown), the liquid levels would have to be the same. A B Why? Well, assume the first container had a hiher level. Then, since p = + ρd, pressure at A would be reater than at B. Assume a radual decrease from A to B. Then at any point between them, pressure from the left is reater than from the riht resultin in a net force to the riht (not equilibrium). Therefore, fluid will flow to the riht until equilibrium is reached. 4
Some rules for pressure Anywhere in a connected, static, uniform density fluid, the pressure at a iven heiht is the same. Pascal s law: A pressure chane at one point in an incompressible fluid appears undiminished at all points in the fluid. A force F is applied to a piston of area A, increasin the pressure by F/A. F p = + F A d d 2 1 p = = + p F 0 A + ρd 1 p = + ρd 2 5
Clicker question 2 Set frequency to BA Uma Thurman (mass of 60 k) is standin on a piston, connected as shown to another piston on which a 6000 k stretch Hummer is restin. How much bier in area is the piston under the Hummer compared to the one under Uma? Try writin down two expressions for the pressure at the dotted line. A. same B. 10 times C. 100 times D. 1000 times E. 10000 times On the left side we have p L = + F L A L F L = m U Riht side: F R = m H p R = + F R A R Both pressures at same heiht so must be the same. Settin equal: + F L = + F R F L A R = F R A L A R = F R so so A L =100 A L A L A R F L
Buoyancy Take a volume of water with density ρ f and area A extendin from a depth of d 2 to d 1. Summin the forces of the free body diaram ives F y m f A + ρ f Ad 1 d 2 A + ρ f Ad 2 = A + ρ f Ad 1 A ρ f Ad 2 m f = ρ f A(d 1 d 2 ) m f = 0 d 1 A pa If we replace the fluid with an object, the only difference is mass. F y = ρ f A(d 1 d 2 ) m o = 0 Note that A(d 1 -d 2 ) is the volume V f of fluid displaced Upward force ρ f A(d 1 d 2 ) = ρ f V f equals the weiht of displaced fluid
Archimedes principle A body partially or fully immersed in a fluid feels an upward force equal to the weiht of the displaced fluid. This force is called the buoyant force: = ρ f V f As shown, it is due to the increase of pressure with depth in a fluid. If the object is fully immersed then the volume of the displaced fluid is equal to the volume of the object: V f = V o Note that volume is related to mass and density: m o = ρ o V o If an object is only partially submered, the volume of the displaced fluid is less than the volume of the object: V f < V o
Buoyancy example A 2 cm by 2 cm by 2 cm cube of iron (ρ=8 /cm 3 ) is weihed with the iron outside, half in and fully in the water, as shown in the diaram. What is the measured weiht in each case? Iron mass: m o = ρ o V o = 8 /cm 3 8 cm 3 = 64 = 0.064 k Out of the water: T T m o = 0 so m o T = m o = 0.064 k 10 m/s 2 = 0.64 N In the water: ½ in the water: T m o +T m o = 0 so T = m o 3 3 2 = ρf Vf = 1/cm 8cm 10m/s = 0.08N so T = m o = 0.64 N 0.08 N = 0.56 N T m o +T m o = 0 so T = m o = ρ f V f =1 /cm 3 4 cm 3 10 m/s 2 = 0.04 N so T = m o = 0.64 N 0.04 N = 0.60 N
Archimedes crown Archimedes lived from 287 BC to 212 BC. Kin Hiero II ave a oldsmith old to make a crown. When the crown was made, it was found to weih the same as the old iven but the kin suspected that silver had been mixed in so he asked Archimedes to find out (without damain the crown). One day while ettin into his tub, Archimedes noticed water was displaced in an amount equal to his volume and fiure out a way to determine the density of the crown. He ran naked throuh the streets of Syracuse shoutin Eureka, Greek for I found it. What had he found? Balance the weiht of the crown with pure old in air. Submere both in water. If the crown is less dense than the old, it must have a larer volume (in order to have the same weiht). In that case, the crown will displace more water than the old, leadin to a larer buoyant force on the crown, and a smaller apparent weiht.
Clicker question 3 Set frequency to BA An ice cube is floatin in a lass of water. As the ice cube melts, the level of the water A. rises. B. falls. The buoyant force is always C. stays the same. equal to the weiht of the liquid displaced by the object. Since the ice cube is in equilibrium, the weiht of water displaced is equal to the weiht of the ice cube. When the ice cube melts it becomes water with the same weiht as the ice cube so real water takes the place of the displaced water.