Lecture 19 Fluids: density, pressure, Pascal s principle and Buoyancy.

Lecture 19 Water tower Fluids: density, pressure, Pascal s principle and Buoyancy. Hydraulic press Pascal s vases Barometer

What is a fluid? Fluids are substances that flow. substances that take the shape of the container Atoms and molecules must be free to move.. No long range correlation between positions (e.g., not a crystal). Gas or liquid or granular materials (like sand)

Density, pressure Density: m V Ex: Pure water: 1000 kg/m 3 In a fluid: Particles are always moving i.e., hitting surfaces i.e., exerting (perpendicular) forces on surfaces Pressure: p F A Units: Surface of area A F Pascal Pa = 1 N/m 2 psi = lb/in 2 atmosphere 1 atm = 1.013 10 5 Pa bar 1 bar = 10 5 Pa

Atmospheric pressure The atmosphere of Earth is a fluid, so every object in air is subject to some pressure. At the surface of the Earth, the pressure is p atm ~ 1.013 x 10 5 Pa = 1 atm Area of a hand ~ 200 cm 2 = 0.02 m 2 F p A ~ 2000 N on your hand due to air! atm DEMO: Piston and weight

Vacuum gun Sealed tube, air pumped out Ping-pong ball What happens if we punch a little hole on one side? W = Δ KE FL ( p atm R 2 )L = 1 2 m v 2 0 v 2p atm R 2 L m 300 km/s Length of tube L ~ 3 m Mass of ball m ~ 3 g Radius of tube R ~ 2 cm DEMO: Vacuum gun

Pressure vs. depth DEMO: Plastic tube with cover Imaginary box of fluid with density ρ with bases of area A and height h For floating object, net force must be zero! F F mg bottom top mg F top h F bottom P bottom/top = F bottom/top A p p gh bottom m Ah top Example: How deep under water is p = 2 atm? h = p bottom p top ρ fluid g = 1.01 10 5 Pa (10 3 kg/m 3 ) (9.81 m/s 2 ) = 10.3 m (i.e. 1 atm is produced by a 10.3 m high column of water) Called guage pressure

DEMO: Pascal s vases Fluid in an open container Pressure is the same at a given depth, independent of the shape of the container. y p(y) Fluid level is the same everywhere in a connected container (assuming no surface forces) A B If liquid height was higher above A than above B p A > p B Net force Net flow This is not equilibrium!

ACT: U tube Two liquids Y and G separated by a thin, light piston (so they cannot mix) are placed in a U-shaped container. What can you say about their densities? A. ρ G < ρ Y B. ρ G = ρ Y A Y G B C. ρ G > ρ Y

ACT: U tube DEMO: U-tube with water and kerosene Two liquids Y and G separated by a thin, light piston (so they cannot mix) are placed in a U-shaped container. What can you say about their densities? A. ρ G < ρ Y h 1 h 2 Y G h 3 B. ρ G = ρ Y A B C. ρ G > ρ Y Pressure at A and B must be the same: gh gh p gh p Y 1 G 2 atm G 3 atm h h h Y 1 G 3 2 Since h 1 < h 3 h 2 ρ y > ρ G

Water towers Water towers are a common sight in the Midwest because it s so flat! h p p hg house atm water

So physics sucks, but how much? Your physics professor sucks on a long tube that rises out of a bucket of water. He can get the liquid to rise 5.5 m (vertically). What is the pressure in His mouth at this moment? A. 1 atm A. 0.67 atm B. 0.57 atm C. 0.46 atm D. 0 atm DEMO: Sucking through a hose p gh p mouth p p gh mouth 5 3 3 2 10 Pa 10 kg/m 9.8 m/s 5.5 m water atm water atm 46100 Pa 0.46 atm x B x A h

Pascal s principle Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. Pascal s Principle is most often applied to incompressible fluids (liquids): Increasing p at any depth (including the surface) gives the same increase in p at any other depth

Hydraulic chamber F 1 A 1 = F 2 A 2 F 2 = A 2 A 2 F 1 F 2 can be very large No energy is lost: =( W = F 1 d F 1 2 A 1 )( A d A 2 ) 2 2 A 1 =F 2 d 2 Incompressible fluid: A 1 d 1 = A 2 d 2

ACT: Hydraulic chambers In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference d i between the liquid levels. If A 2 = 2A 1, then: d A M A. d A < d B A 1 A 10 B. d A = d B C. d A > d B d B M A 2 A 10

Measuring pressure with fluids Barometer Measures absolute pressure Top of tube evacuated (p = 0) Bottom of tube submerged into pool of mercury open to sample (p) Pressure dependence on depth: h = p ρ Hg g Vacuum vacuum p p=0 = 0 Sample at p=p p 0 atmosphere h Manometer Measures gauge pressure: pressure relative to atmospheric pressure. Pressure dependence on depth: Δ h = p p atm ρ Hg g p p atm p atm h h A unit for pressure 760 mm Hg = 760 torr = 1 atm

ACT: Side tube A sort of barometer is set up with a tube that has a side tube with a tight fitting stopper. What happens when the stopper is removed? vacuum A. Water spurts out of the side tube. stopper B. Air flows in through the side tube. C. Nothing, the system was in equilibrium and remains in equilibrium.

Buoyancy and the Archimedes principle A box of base A and height h is submerged in a liquid of density ρ. Net force by liquid: F F bottom F top Ap bottom Ap top y top y bottom F top A h atm bottom atm top A p gy A p gy F bottom A hg Vg direction up Archimedes principle: The liquid exerts a net force upward called buoyant force whose magnitude is equal to the weight of the displaced liquid.

In-class example: Hollow sphere A hollow sphere of iron (ρ Fe = 7800 kg/m 3 ) has a mass of 5 kg. What is the maximum diameter necessary for this sphere to be completely submerged in water? (ρ water = 1000 kg/m 3 ) A. It will always be submerged. F B B. 0.11 m C. 0.21 m D. 0.42 m E. It will always only float. The sphere sinks if F B < mg 4 ρ water 3 π R3 g < mg R < 3 What if the sphere is solid?! mg 3m = 0.106 m 4π ρ water Maximum diameter 2R 0.21 m max

Density rule DEMO: Frozen helium balloon A hollow sphere of iron (ρ Fe = 7800 kg/m 3 ) has a mass of 5 kg. What is the maximum diameter necessary for this sphere to be fully submerged in water? (ρ water = 1000 kg/m 3 ) Answer: R = 0.106 m. And what is the average density of this sphere? sphere m 3 1000 kg/m 4 3 4 3 R 3 3 5 kg 0.106 m water An object of density ρ object placed in a fluid of density ρ fluid sinks if ρ object > ρ fluid is in equilibrium anywhere in the fluid if ρ object = ρ fluid floats if ρ object < ρ fluid (will not be completely submerged) This is why you cannot sink in the Dead Sea Buoyancy (in the Dead Sea) (ρ Dead Sea water = 1240 kg/m 3, ρ human body = 1062 kg/m 3 )!

ACT: Styrofoam and lead A piece of lead is glued to a slab of Styrofoam. When placed in water, they float as shown. Pb styrofoam What happens if you turn the system upside down? styrofoam styrofoam Pb Pb C. It sinks. A B The displaced volume in both cases must be the same (volume of water whose weight is equal to the weight of the lead+styrofoam system)

ACT: Floating wood Two cups have the same level of water. One of the two cups has a wooden block floating in it. Which cup weighs more? A. Cup 1 B. Cup 2 C. They weigh the same. 1 2

ACT: Floating wood Two cups have the same level of water. One of the two cups has a wooden block floating in it. Which cup weighs more? A. Cup 1 B. Cup 2 C. They weigh the same. 1 2 Cup 2 has less water than cup 1. The weight of the wood is equal to the weight of the missing liquid (= displaced liquid ) in 2. DEMO: Bucket of water with wooden block

ACT: Aluminum and lead Two blocks of aluminum and lead with identical sizes are suspended from the ceiling with strings of different lengths and placed inside a bucket of water as shown. In which case is the buoyant force greater? A. Al B. Pb C. It s the same for both ceiling The displaced volume (= volume of the block) is the same in both cases. Depth or object density do not play any role. The different weight is compensated with a different tension in the strings. Al Pb

ACT: Wooden brick When a uniform wooden brick (1 m x 1 m x 2 m) is placed horizontally on water, it is partially submerged and the height of the brick above the water surface is 0.5 m. If the brick was placed vertically, the height of brick above the water would be: A. 0.5 m B. 1.0 m C. 1.5 m. The displaced volume in both cases needs to be the same (Because the weight of the wood did not change) : half of the volume of the brick. Same volume 0.5 m