Physics 101: Lecture 18, Pg 1 Physics 101: Lecture 18 Fluids II Exam III Textbook Sections 9.6 – 9.8

Physics 101: Lecture 18, Pg 2 Review Static Fluids l Pressure is force exerted by molecules “bouncing” off container P = F/A l Gravity/weight effects pressure  P = P 0 +  gd l Buoyant force is “weight” of displaced fluid.  F =  g V Today include moving fluids! A 1 v 1 = A 2 v 2 P 1 +  gy 1 + ½  v 1 2 = P 2 +  gy 2 + ½  v

Physics 101: Lecture 18, Pg 3 Archimedes’ Principle l Buoyant Force (F B ) è weight of fluid displaced è F B =  fluid Vol displaced g è F g = mg =  object Vol object g è object sinks if  object >  fluid è object floats if  object <  fluid l If object floats… è F B = F g è Therefore:  fluid g Vol displ. =  object g Vol object è Therefore: Vol displ. /Vol object =  object /  fluid 10

Physics 101: Lecture 18, Pg 4 Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up, causing the water to spill out of the glass. 2. Go down. 3. Stay the same. CORRECT Preflight 1 12

Physics 101: Lecture 18, Pg 5 Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up, causing the water to spill out of the glass. 2. Go down. 3. Stay the same. CORRECT Preflight 1 B =  W g Vol displaced W =  ice g Vol ice   W g Vol melted_ice Must be same! 12

Physics 101: Lecture 18, Pg 6 Preflight 2 Which weighs more: 1. A large bathtub filled to the brim with water. 2. A large bathtub filled to the brim with water with a battle-ship floating in it. 3. They will weigh the same. Tub of water Tub of water + ship Overflowed water CORRECT Weight of ship = Buoyant force = Weight of displaced water 15

Physics 101: Lecture 18, Pg 7Continuity A four-lane highway merges down to a two-lane highway. The officer in the police car observes 8 cars passing every second, at 30 mph. How many cars to the officer on the motorcycle observe passing every second? A) 4 B) 8 C) 16 How fast must the cars in the two-lane section be going? A) 15 mph B) 30 mph C) 60 mph 18 All cars stay on the road… Must pass motorcycle too Must go faster, else pile up!

Physics 101: Lecture 18, Pg 8 Continuity of Fluid Flow Watch “plug” of fluid moving through the narrow part of the tube (A 1 ) Time for “plug” to pass point  t = x 1 / v 1 Mass of fluid in “plug” m 1 =  Vol 1 =  A 1 x 1 or m 1 =  A 1 v 1  t Watch “plug” of fluid moving through the wide part of the tube (A 2 ) Time for “plug” to pass point  t = x 2 / v 2 Mass of fluid in “plug” m 2 =  Vol 2 =  A 2 x 2 or m 2 =  A 2 v 2  t Continuity Equation says m 1 = m 2 fluid isn’t building up or disappearing A 1 v 1 = A 2 v 2 22

Physics 101: Lecture 18, Pg 9 Faucet Preflight A stream of water gets narrower as it falls from a faucet (try it & see). Explain this phenomenon using the equation of continuity A1A1 A2A2 V1V1 V2V2 23 As the the water falls, its velocity is increasing. Since the continuity equations states that if density doesn't change...Area1*velocity1=Area2*velocity2. From this equation, we can say that as the velocity of the water increases, its area is going to decrease I tried to do this experiment and my roommate yelled at me for raising the water bill. Adhesion and cohesion…

Physics 101: Lecture 18, Pg 10 Fluid Flow Concepts Mass flow rate:  Av (kg/s) Volume flow rate: Av (m 3 /s) Continuity:  A 1 v 1 =  A 2 v 2 i.e., mass flow rate the same everywhere e.g., flow of river A 1 P 1 A 2 P 2 v1v1 v2v2  24

Physics 101: Lecture 18, Pg 11 Pressure, Flow and Work l Continuity Equation says fluid speeds up going to smaller opening, slows down going to larger opening l Acceleration due to change in pressure. P 1 > P 2 è Smaller tube has faster water and LOWER pressure l Change in pressure does work!  W = P 1 A 1  x 1 - P 2 A 2  x 2 = (P 1 – P 2 )V olume 28 Demo Recall: W=F d = PA d = P Vol

Physics 101: Lecture 18, Pg 12 Pressure ACT l What will happen when I “blow” air between the two plates? A) Move Apart B) Come Together C) Nothing 31 There is air pushing on both sides of plates. If we get rid of the air in the middle, then just have air on the outside pushing them together.

Physics 101: Lecture 18, Pg 13 Bernoulli’s Eqs. And Work l Consider tube where both Area, height change.  W =  K +  U (P 1 -P 2 ) V = ½ m (v 2 2 – v 1 2 ) + mg(y 2 -y 1 ) (P 1 -P 2 ) V = ½  V (v 2 2 – v 1 2 ) +  Vg(y 2 -y 1 ) P 1 +  gy 1 + ½  v 1 2 = P 2 +  gy 2 + ½  v Note: W=F d = PA d = P V

Physics 101: Lecture 18, Pg 14 Bernoulli ACT l Through which hole will the water come out fastest? A B C P 1 +  gy 1 + ½  v 1 2 = P 2 +  gy 2 + ½  v 2 2 Note: All three holes have same pressure P=1 Atmosphere  gy 1 + ½  v 1 2 =  gy 2 + ½  v 2 2 gy 1 + ½ v 1 2 = gy 2 + ½v 2 2 Smaller y gives larger v. Hole C is fastest 37

Physics 101: Lecture 18, Pg 15 Act A large bucket full of water has two drains. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bucket near the top, which bent is downward such that the bottom of this pipe even with the other hole, like in the picture below: Though which drain is the water spraying out with the highest speed? 1. The hole 2. The pipe 3. Same CORRECT 40 Note, the correct height, is where the water reaches the atmosphere, so both are exiting at the same height!

Physics 101: Lecture 18, Pg 16 Example (like HW) A garden hose w/ inner diameter 2 cm, carries water at 2.0 m/s. To spray your friend, you place your thumb over the nozzle giving an effective opening diameter of 0.5 cm. What is the speed of the water exiting the hose? What is the pressure difference between inside the hose and outside? Bernoulli Equation P 1 +  gy 1 + ½  v 1 2 = P 2 +  gy 2 + ½  v 2 2 P 1 – P 2 = ½  (v 2 2 – v 1 2 ) = ½ x (1000 kg/m 3 ) (1020 m 2 /s 2 ) = 5.1x10 5 PA Continuity Equation A 1 v 1 = A 2 v 2 v 2 = v 1 ( A 1 /A 2 ) = v 1 ( r 1 2 / r 2 2 ) = 2 m/s x 16 = 32 m/s

Physics 101: Lecture 18, Pg 17 Lift a House Calculate the net lift on a 15 m x 15 m house when a 30 m/s wind (1.29 kg/m 3 ) blows over the top. P 1 +  gy 1 + ½  v 1 2 = P 2 +  gy 2 + ½  v 2 2 P 1 – P 2 = ½  (v 2 2 – v 1 2 ) = ½  (v 2 2 – v 1 2 ) = ½ (1.29) (30 2 ) N / m 2 = 581 N/ m 2 F = P A = 581 N/ m 2 (15 m)(15 m) = 131,000 N = 29,000 pounds! (note roof weighs 15,000 lbs) 48

Physics 101: Lecture 18, Pg 18 Fluid Flow Summary Mass flow rate:  Av (kg/s) Volume flow rate: Av (m 3 /s) Continuity:  A 1 v 1 =  A 2 v 2 Bernoulli: P / 2  v  gh 1 = P / 2  v  gh 2 A 1 P 1 A 2 P 2 v1v1 v2v2  50

Physics 101: Lecture 18, Pg 19 Practice Problems Chapt. 9, problems 1, 5, 9, 13, 15, 17, 21, 29, 33, 35, 41, 45,47.