Report

Matter 1. Density: m – mass V – volume m V Units: 1kg / m 3 water 1g / cm3 1000kg / m3 Example: Volume of 1.00 kg of iron is 128 cm3. Find density of iron. 3 m 1.00 kg 100 cm 1.00 kg 3 V 128 cm 1m 128 cm 3 6 cm 3 3 3 10 7 . 8 10 kg / m 3 m m 1.00 103 g 3 3 3 7 . 8 g / cm 7 . 8 10 kg / m V 128cm3 Specific gravity (SG): SG water at 4.0 C 1 2. Pressure F P Units: F P A In fluid: F|| 0 F F A P 1Pa 1N / m 2 1atm 1.013 105 Pa Example: You are walking out on a frozen lake and you begin to hear the ice cracking beneath you. What is your best strategy for getting off the ice safely? A) Stand absolutely still and don’t move a muscle B) Jump up and down to lessen your contact time with the ice C) Try to leap in one bound to the bank of the lake D) Shuffle your feet (without lifting them) to move towards shore E) Lie down flat on the ice and crawl toward shore Gauge pressure = (absolute pressure - 1atm) Pg P 1atm 2 Fluid Mechanics What is a fluid? •Any substance that can flow •Liquids or gases (there are tree common phases of mater: solid, liquid, gas) •Fluid can not sustain a force that is tangent it its surface •Fluids conform to the boundaries of any container in which we put them •No crystal structure 3 A. Fluid Statics 1. Pascal’s law (principle) F P0 A y F 0 P0 PA P0 A gAh 0 mg gAh P gh P P0 gh h Incompressible fluid: const F PA ( P0 P) A A change in the pressure applied to an enclosed incompressible fluid is transmitted to every portion of the fluid and to the walls of its container 4 Example: Three containers are filled with water to the same height and have the same surface area at the base, but the total weight of water is different for each. Which container has the greatest total force acting on its base? 1) 2) 3) 4) container 1 container 2 container 3 all three are equal The pressure at the bottom of each container depends only on the height of water above it! This is the same for all the containers. The total force is the product of the pressure times the area of the base, but since the base is also the same for all containers, the total force is the same. 5 2. Applications of Pascal’s law 2a. Hydraulic lift (lever) F1 F2 PA1 PA2 F1 PA1 F2 PA2 W F1x1 PA1x1 PV W F2 x2 PA2 x2 PV F1 A1 F2 A2 W PV F1x1 F2 x2 Similar to solid lever Example: F1=10N A1=0.01m2 A2=1m2 F2 - ? A2 F2 F1 A1 1m 2 F2 10N 1000N 2 0.01m 6 2b. The open tube manometer (measuring gauge pressure) h h2 h1 P0 P gh1 P0 gh2 P h h2 Pg P P0 gh h1 2c. The mercury barometer (measuring atmospheric pressure) P=0 P0 h P0 gh 7 3. Buoyancy. Archimedes’ principle Fb m f g f Vg Fb m g Vg Example: Imagine holding two identical bricks in place under water. Brick 1 is just beneath the surface of the water, while brick 2 is held about 2 feet down. The force needed to hold brick 2 in place is: 1) greater 2) the same 3) smaller 1 2 The force needed to hold the brick in place underwater is: W – Fb. According to Archimedes’ Principle, Fb is equal to the weight of the fluid displaced. Since each brick displaces the same amount of fluid, then Fb is the same in both cases. 8 3a. An object floating in equilibrium Fb m f g f V f g Fg mg Vg Fb Fg f V f g Vg f V f V V f Vf Example: An object floats in water with 3/4 of its volume submerged. What is the ratio of the density of the object to that of water? Vf 3 f V 4 If the ratio of the volume of the displaced water to the volume of the object is 3/4, the object has 3/4 the density of water. 9 Example 1: A 15.0 kg solid gold statue is being raised from a sunken ship. What is the tension in the hosting cable when he statue is at rest and a) completely immersed; b) out of the water? T T = mg - Fb Fb = mfg = ρfVg Fb = mgρf /ρgold T = mg(1 - ρf /ρg) a) T = (15.0kg) (9.8 m/s2)(1 - 1.03 /19.3)=139N mg = ρgoldVg gold 19.3g / cm 3 b) air gold 1.2 103 19.3 6 105 1 T = mg= (15.0kg) (9.8 m/s2)=147N Example 2: You place a container of sea water on a scale and note the reading on the scale. You now suspend the statue of example 1 in the water. Haw does the scale reading change? Water exerts an upward buoyant force, Fb on the statue, so the statue must exert an equal downward force on the water, making the scale reading by Fb grater than the weight of water and container. From example 1: Fb = mgρf /ρg= (15.0kg)( 9.8 m/s2)1.03 /19.3=7.8N 10 Example: The “weight” of a chunk of metal measured in air is 100g, and when the chunk is submerged into water the “weight” is 87.2g. What is the density of the metal. T1 = mg = ρVg T1 T2 Fb T2 = mg - Fb T1 T 2 Fb f Vg mg mg T1 Vg T1 T 2 f Vg f T1 f T1 T2 100g 1.0 g / cm 7.8 g / cm3 100g 87.2 g 3 11 B. Fluid Dynamics. Bernoulli’s equation (Ideal fluid - incompressible and has no viscosity) y A2 v2 A1d1 A2 d 2 V d2 W K U A1 W F1d1 F2 d 2 P1 A1d1 P2 A2 d 2 P1 P2 V v1 d1 K 12 mv22 12 mv12 12 V (v22 v12 ) U mg(h2 h1 ) Vg (h2 h1 ) P1 P2 12 (v22 v12 ) g (h2 h1 ) P1 gh1 12 v12 P2 gh2 12 v22 12 Example: An open water tank has a hole a distance h = 4.9 m below the water surface. The water surface area of the tank is much bigger then the hole’s area. What is the speed of the water emerging from the hole? P0 P1 gh1 12 v12 P2 gh2 12 v22 v0=0 h P0 v-? gh 12 v2 v 2gh 13