Exam III Physics 101 Lecture 18 Fluids II

Exam III Physics 101: Lecture 18 Fluids II Textbook Sections 9. 6 – 9. 8 Physics 101: Lecture 18, Pg 1

Archimedes’ Principle l Determine force of fluid on immersed cube èDraw FBD » FB = F 2 – F 1 » » = P 2 A – P 1 A = (P 2 – P 1)A = gd. A = g. V l Buoyant force is weight of displaced fluid! Physics 101: Lecture 18, Pg 2

Fb Archimedes Example A cube of plastic 4. 0 cm on a side with density = 0. 8 g/cm 3 is floating in the water. When a 9 gram coin is placed on the block, how much does it sink below the water surface? SF=ma Mg mg h Fb – Mg – mg = 0 g Vdisp = (M+m) / h A = (M+m) / M = plastic Vcube = 4 x 4 x 4 x 0. 8 = 51. 2 g h = (M + m)/ ( A) = (51. 2+9)/(1 x 4) = 3. 76 cm Physics 101: Lecture 18, Pg 3

Review Static Fluids l Pressure is force exerted by molecules “bouncing” off container P = F/A l Gravity/weight effects pressure èP = P 0 + gd l Buoyant force is “weight” of displaced fluid. èF = g V Today include moving fluids! A 1 v 1 = A 2 v 2 P 1+ gy 1 + ½ v 12 = P 2+ gy 2 + ½ v 22 Physics 101: Lecture 18, Pg 4

Archimedes’ Principle l Buoyant Force (FB) èweight of fluid displaced èFB = fluid. Voldisplaced g èFg = mg = object Volobject g èobject sinks if object > fluid èobject floats if object < fluid l If object floats… èFB = Fg èTherefore: fluid g Voldispl. = object g Volobject èTherefore: Voldispl. /Volobject = object / fluid Physics 101: Lecture 18, Pg 5

Preflight 1 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. CORRECT 3. Stay the same. B= W g Voldisplaced W = ice g Volice W g Volmelted_ice 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. Tub of water + ship 3. They will weigh the same. CORRECT Tub of water Weight of ship = Buoyant force = Overflowed water Weight of displaced water Physics 101: Lecture 18, Pg 7

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

Faucet Preflight A stream of water gets narrower as it falls from a faucet (try it & see). Explain this phenomenon using the equation of V 1 continuity As the water leaves the faucet, gravity is pulling the water down and causes the velocity to increase. As velocity increases, the area is suppose to decrease, which it does. The stream gets narrower. V 2 A 1 A 2 Post-spring break answers: It's a miracle that has not been explained by modern science. All sinks are fitted with bottom-mounted singularities which draw streams of water directly into a single point of nearly infinite mass from which not even light can escape. i tried it and it didn't do that! I just recently watched A. I. for the first time, please warn the class that it is a terrible movie. Physics 101: Lecture 18, Pg 9

Fluid Flow Concepts r A 1 P 1 v 1 A 2 P 2 v 2 • 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 Physics 101: Lecture 18, Pg 10

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

Pressure ACT l What will happen when I “blow” air between the two plates? A) Move Apart B) Come Together C) Nothing 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 12

Bernoulli’s Eqs. And Work l Consider tube where both Area, height change. èW = DK + DU Note: (P 1 -P 2) V = ½ m (v 22 – v 12) + mg(y 2 -y 1) (P 1 -P 2) V = ½ V (v 22 – v 12) + Vg(y 2 -y 1) P 1+ gy 1 + ½ v 12 = P 2+ gy 2 + ½ v 22 W=F d = PA d =PV Physics 101: Lecture 18, Pg 13

Bernoulli ACT l Through which hole will the water come out fastest? P 1+ gy 1 + ½ v 12 = P 2+ gy 2 + ½ v 22 Note: All three holes have same pressure P=1 Atmosphere A B gy 1 + ½ v 12 = gy 2 + ½ v 22 gy 1 + ½ v 1 = gy 2 + 2 ½v 22 C Smaller y gives larger v. Hole C is fastest Physics 101: Lecture 18, Pg 14

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 Note, the correct height, is where the water reaches the atmosphere, so both are exiting at the same height! Physics 101: Lecture 18, Pg 15

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? Continuity Equation A 1 v 1 = A 2 v 2 = v 1 ( A 1/A 2) = v 1 ( π r 12 / π r 22) = 2 m/s x 16 = 32 m/s Bernoulli Equation P 1+ gy 1 + ½ v 12 = P 2+ gy 2 + ½ v 22 P 1 – P 2 = ½ (v 22 – v 12) = ½ x (1000 kg/m 3) (1020 m 2/s 2) = 5. 1 x 105 PA Physics 101: Lecture 18, Pg 16

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 12 = P 2+ gy 2 + ½ v 22 P 1 – P 2 = ½ (v 22 – v 12) = ½ (1. 29) (302) N / m 2 = 581 N/ m 2 F=PA = 581 N/ m 2 (15 m) = 131, 000 N = 29, 000 pounds! (note roof weighs 15, 000 lbs) Physics 101: Lecture 18, Pg 17 48

Fluid Flow Summary r A 1 P 1 v 1 A 2 P 2 v 2 • Mass flow rate: Av (kg/s) • Volume flow rate: Av (m 3/s) • Continuity: A 1 v 1 = A 2 v 2 • Bernoulli: P 1 + 1/2 v 12 + gh 1 = P 2 + 1/2 v 22 + gh 2 Physics 101: Lecture 18, Pg 18 50

Practice Problems Chapt. 9, blue numbered problems in sections 2, 3, 4, 5, 6, and 7. Physics 101: Lecture 18, Pg 19