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5. (II) A bottle has a mass of 35.00 g when empty and 98.44 g when filled with water. When filled with another fluid, the mass is 88.78 g. What is the specific gravity of this other fluid? 5. Take the ratio of the density of the fluid to that of water, noting that the same volume is used for both liquids (so V cancels). SG 1 ΔP = ρgh = (1.05 x 103 kg/m3)(9.8m/s2)(1.60m) = 1.646 x 104 N/m2 1mmHg × 133N /m 2 = 124 mmHg € 2 9. (I) (a) Calculate the total force of the atmosphere acting on the top of a table that measures (b) What is the total force acting upward on the underside of the table? 9. (a) The total force of the atmosphere on the table will be the air pressure times the area of the table. (b) Since the atmospheric pressure is the same on the underside of the table (the height difference is minimal), the upward force of air pressure is the same as the downward force of air on the top of the table, 3 10.(II) In a movie, Tarzan evades his captors by hiding underwater for many minutes while breathing through a long, thin reed. Assuming the maximum pressure difference his lungs can manage and still breathe is 85 mmHg, calculate the deepest he could have been. 10.The pressure difference on the lungs is the pressure change from the depth of water 4 Atmospheric Pressure • Earth’s atmosphere: A fluid. • ⇒ Change in pressure: ΔP = ρgΔh • At sea level: P0 ≡ 1.013 × 105 N/m2 = 101.3 kPa ≡ 1 atm The cause of (air) pressure at any height in atmosphere is: weight of air above that height! Measurement of Pressure; Gauges and the Barometer There are a number of different types of pressure gauges. This one is an opentube manometer. The pressure in the open end is atmospheric pressure; the pressure being measured will cause the fluid to rise until the pressures on both sides at the same height are equal. 6 Pressure beneath surface of liquid open to the atmosphere Gauge Pressure Initial Pressure (atmospheric pressure) ABSOLUTE PRESSURE Gauge Pressure = The amount by which P exceeds atmospheric pressure Po 7 Gauge Pressure • Pressure gauges (like tire gauges, etc.) measure difference between atmospheric pressure P0 & internal pressure (of tire, for example). • Gauge pressure: PG = P – P0 Measurement of Pressure; Gauges and the Barometer Here are two more devices for measuring pressure: the tire pressure gauge and the aneroid gauge. 9 Measurement of Pressure; Gauges and the Barometer This is a mercury barometer, developed by Torricelli to measure atmospheric pressure. The height of the column of mercury is such that the pressure in the tube at the surface level is 1 atm. Therefore, pressure is often quoted in millimeters (or inches) of mercury. 10 Measurement of Pressure; Gauges and the Barometer Any liquid can serve in a Torricelli-style barometer, but the most dense ones are the most convenient. (Density of Hg is 13.5 g/cm3) This barometer uses water. (Density of H2O is 1 g/cm3) 11 Mercury Barometer • Weather reports: • Barometric pressure (atmospheric pressure): 28-32 inches Hg 76 cm = 760 mm = 29.29 inches When h = 760 mm, P = ρHg gh = 1.013 × 105 N/m2 =1atm • If we use water instead of mercury, P = 1atm = ρH2O gh ⇒ h ≈ 10 m ≈ 30 feet! Example How high would the level be in an alcohol barometer at normal atmospheric pressure? (The density of the alcohol is 0.79 g/cm3) o A closer look at Pressure vs. Depth Depth below surface Initial Pressure – May or MAY NOT be atmospheric pressure ABSOLUTE PRESSURE 14 Example (a) What are the total force and the absolute pressure on the bottom of a swimming pool 22.0 m by 8.5 m whose uniform depth is 2.0 m? The total force is the absolute pressure times the area of the bottom of the pool. (b) What will be the pressure against the side of the pool at the bottom? The pressure against the side of the pool, near the bottom, will be the same as the pressure at the bottom, Pascal’s Law • Experimental fact: An external pressure P applied to a confined fluid increases the pressure throughout a fluid by P ≡ Pascal’s Law • Simple example: Water in a lake (at rest). At depth h below surface, pressure is P = P0 + ρgh (P0 = atmospheric pressure) If P0 changes, P changes evenly at a given height! Pascal’s Law • The volume of liquid pushed down on left must equal volume pushed up on right. So, Pascal’s Law • The volume of liquid pushed down on left must equal volume pushed up on right. So, • Combining this with Pascal’s Law • The volume of liquid pushed down on left must equal volume pushed up on right. So, • Combining this with • And knowing V = AΔx A = V/Δx F1Δx1 F2 Δx 2 = V V Pascal’s Law • The volume of liquid pushed down on left must equal volume pushed up on right. So, • Combining this with • Gives Pascal’s Law • The volume of liquid pushed down on left must equal volume pushed up on right. So, • Combining this with • Gives ⇒ This means that Work1 = Work2 Pascal’s Law • The volume of liquid pushed down on left must equal volume pushed up on right. So, • Combining this with • Gives ⇒ This means that Work1 = Work2 So, Pascal’s Law is consequence of Conservation of Mechanical Energy A closed system If you take a liquid and place it in a system that is CLOSED like plumbing, for example, or a car’s brake line, the PRESSURE is the same everywhere. Since this is true, if you apply a force at one part of the system, the pressure is the same at the other end of the system. The force, on the other hand MAY or MAY NOT equal the initial force applied. It depends on the AREA. You can take advantage of the fact that the pressure is the same in a closed system as it has MANY applications. The idea behind this is called PASCAL’S PRINCIPLE 23 Pascal’s Principle If an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. Pin = Pout Hydraulic press 24 Hydraulic press An applied force F1 can be “amplified”: Examples: hydraulic brakes, forklifts, car lifts, etc. Pascal’s Law, Example #3 • Car lift in a service station. A large output force can be applied by means of a small input force. Volume of liquid pushed down on left must equal volume pushed up on right. • Circular cross section system. On left, r1 = 5 cm = 0.05 m. On right r2 = 15 cm = 0.15 m. Car’s weight mg = 13,300 N. Pascal’s Law, Example • Car lift in a service station. See figure. A large output force can be applied by means of a small input force. Volume of liquid pushed down on left must equal volume pushed up on right. • Circular cross section system. On left r1 = 5 cm = 0.05 m. On right r2 = 15 cm = 0.15 m. Car’s weight mg = 13,300 N. N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N Calculate minimum F1 to lift the car & pressure P in the system. Pascal’s Law, Example • Car lift in a service station. See figure. A large output force can be applied by means of a small input force. Volume of liquid pushed down on left must equal volume pushed up on right. • Circular cross section system. On left r1 = 5.00 cm = 0.0500 m. On right r2 = 15.0 cm = 0.150 m. Car’s weight mg = 13,300 N. N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N Calculate minimum F1 to lift the car & pressure P in the system. Pascal’s Law ⇒ Pascal’s Law, Example • Car lift in a service station. See figure. A large output force can be applied by means of a small input force. Volume of liquid pushed down on left must equal volume pushed up on right. • Circular cross section system. On left r1 = 5.00 cm = 0.0500 m. On right r2 = 15.0 cm = 0.150 m. Car’s weight mg = 13,300 N. N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N Calculate minimum F1 to lift the car & pressure P in the system. Gives Pascal’s Law ⇒ F1 = (A1/A2)F2 Pascal’s Law, Example • Car lift in a service station. See figure. A large output force can be applied by means of a small input force. Volume of liquid pushed down on left must equal volume pushed up on right. • Circular cross section system. On left r1 = 5.00 cm = 0.0500 m. On right r2 = 15.0 cm = 0.150 m. Car’s weight mg = 13,300 N. N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N Calculate minimum F1 to lift the car & pressure P in the system. Gives Pascal’s Law ⇒ F1 = (A1/A2)F2 2 2 A π r r where A=πr2 ... so 1 = 1 2 = 1 2 A2 πr2 r2 Pascal’s Law, Example • Car lift in a service station. See figure. A large output force can be applied by means of a small input force. Volume of liquid pushed down on left must equal volume pushed up on right. • Circular cross section system. On left r1 = 5.00 cm = 0.0500 m. On right r2 = 15.0 cm = 0.150 m. Car’s weight mg = 13,300 N. N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N Calculate minimum F1 to lift the car & pressure P in the system. Gives Pascal’s Law ⇒ F1 = (A1/A2)F2 = 1,476.7 N =1480N Pascal’s Law, Example • Car lift in a service station. See figure. A large output force can be applied by means of a small input force. Volume of liquid pushed down on left must equal volume pushed up on right. • Circular cross section system. On left r1 = 5.00 cm = 0.0500 m. On right r2 = 15.0 cm = 0.150 m. Car’s weight mg = 13,300 N. N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N Calculate minimum F1 to lift the car & pressure P in the system. Gives Pascal’s Law ⇒ F1 = (A1/A2)F2 = 1,476.7 N =1480N P = (F1/A1) = 1.88 × 105 Pa Example: What is the resulting force on the bottom of the jug? F1 F2 = A2 A1 € 10N 2 = (500cm ) 2 5cm 1m 2 2 × = 0.0005m 100 2 cm 2 € (since units on A cancel, no need to convert cm2 to m2) = 1000 N Buoyancy and Archimedes’ Principle This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. The buoyant force is found to be the upward force on the same volume of water: Buoyancy and Archimedes’ Principle This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. The buoyant force is found to be the upward force on the same volume of water: = P2 A - P1 A = ρf gh2 A-ρf gh1 A Buoyancy and Archimedes’ Principle This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. The buoyant force is found to be the upward force on the same volume of water: Buoyancy and Archimedes’ Principle This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. The buoyant force is found to be the upward force on the same volume of water: Buoyancy and Archimedes’ Principle This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. The buoyant force is found to be the upward force on the same volume of water: Archimedes Principle Any object completely or partially submerged in a fluid is buoyed up by a force whose magnitude is equal to the weight of the fluid displaced by the object. 39 Buoyancy and Archimedes’ Principle The new (“apparent”) weight of the object is then the difference between the buoyant force and the gravitational force. Example in extra chapter hand-out, p.284 Buoyancy and Archimedes’ Principle Applied to Floating Objects If the object’s density is less than that of water, there will be an upward buoyant force on it that is greater than the downward force of its weight, and it will rise until it is partially out of the water. 41 Floating objects If the log is fully submerged, it will displace a mass of water mF = ρFV = (1000 kg/m3)(2.0 m3) = 2000kg So the buoyant force on the log will be greater than its weight, and it will float to the surface. It will come to equilibrium when it displaces 1200 kg of water. (At this time, FB has gone down). In general, for a floating object, the buoyant force will equal the weight of the object. FB = w 42 Buoyancy and Archimedes’ Principle This principle also works in the air; this is why hot-air and helium balloons rise. 43 Checking Understanding Two blocks of identical size are submerged in water. One is made of lead (heavy), the other of aluminum (light). Upon which is the buoyant force greater? A. On the lead block. B. On the aluminum block. C. They both experience the same buoyant force. Answer Two blocks of identical size are submerged in water. One is made of lead (heavy), the other of aluminum (light). Upon which is the buoyant force greater? A. On the lead block. B. On the aluminum block. C. They both experience the same buoyant force. Checking Understanding Two blocks are of identical size. One is made of lead, and sits on the bottom of a pond; the other is of wood and floats on top. Upon which is the buoyant force greater? A. On the lead block. B. On the wood block. C. They both experience the same buoyant force. Answer Two blocks are of identical size. One is made of lead, and sits on the bottom of a pond; the other is of wood and floats on top. Upon which is the buoyant force greater? A. On the lead block. B. On the wood block. C. They both experience the same buoyant force. Checking Understanding A barge filled with ore floats in a canal lock. If the ore is tossed overboard into the lock, the water level in the lock will A. rise. B. fall. C. remain the same. Answer A barge filled with ore floats in a canal lock. If the ore is tossed overboard into the lock, the water level in the lock will A. rise. B. fall. C. remain the same. 4913-30 Slide A helicopter lowers a probe into Lake Michigan which is suspended on a cable. The probe has a mass of 500 kg and its average density is 1400 kg/m3. What is the tension in the cable? 1400 N a) mg upward b) ρH20Vg upward c) (ρH20-ρwood) Vg upward a) mg upward b) ρH20Vg upward c) (ρsteel-ρH20) Vg upward 40 cm p. 323/ #54, 57, 69, 72, 74, 75 due Tues Fluids in Motion; Flow Rate and the Equation of Continuity If the flow of a fluid is smooth, it is called streamline or laminar flow (a). Above a certain speed, the flow becomes turbulent (b). Turbulent flow has eddies; the viscosity of the fluid is much greater when eddies are present. 55 Bernoulli’s Principle & Fluid Continuity 56 Principles of Fluid Flow • The speed of a fluid depends on the cross sectional area – This is because the same volume of fluid must pass through the given object per unit of time. • Fluid Continuity A1v1 = A2v2 • When you put your finger over the end of the hose, what happens? – Animation 57 The Equation of Continuity Have you ever used your thumb to control the water flowing from the end of a hose? 58 The Equation of Continuity Have you ever used your thumb to control the water flowing from the end of a hose? When the end of a hose is partially closed off, thus reducing its cross-sectional area, the fluid velocity increases. This kind of fluid behavior is described by the equation of continuity. 59 Equation of Continuity 60 If there are no losses of fluid within a uniform tube, the mass of fluid flowing into the tube in a given time must be equal to the mass flowing out of the tube in the same time. So the mass Δm1 entering the tube during time Δt is: Δm1 = ρ1ΔV1 = ρ1(A1Δx1 ) = ρ1(A1v1Δt) 61 Equation of Continuity 62 Equation of Continuity So the mass entering is Δm1 = ρ1ΔV1 = ρ1(A1Δx1 ) = ρ1(A1v1Δt) And the mass leaving is Δm2 = ρ2ΔV2 = ρ2(A2Δx2 ) = ρ2(A2v2Δt) And since mass is conserved, Δm1 = Δm2 and ρ1A1v1 = ρ2A2v2 And for an incompressible fluid, the density is constant, so A1v1 = A2v2 63 Equation of Continuity 64 Equation of Continuity 65 Fluids in Motion; Flow Rate and the Equation of Continuity The mass flow rate is the mass that passes a given point per unit time. The flow rates at any two points must be equal, as long as no fluid is being added or taken away. This gives us the equation of continuity: 66 Fluids in Motion; Flow Rate and the Equation of Continuity If the density doesn’t change – typical for liquids – this simplifies to . Where the pipe is wider, the flow is slower. 67 Flow Rate Equation • The continuity equation is sometimes called the flow rate equation. Av is called the volume rate of flow and is the volume of fluid that passes by a point in the tube per unit time. • Av has units: m2 m/s = m3/s, volume per time. Flow rate = Q = Av = Vol/time Example: High cholesterol in the blood can cause fatty deposits called plaques to form on the walls of blood vessels. Suppose a plaque reduces the effective radius of an artery by 25%. How does this partial blockage affect the speed of blood through the artery? A1v1 = A2v2 π r12v1 = π r22v2 v2 = (r1 / r2)2v1 Since r2 = .75 r1 , then r1 / r2 = 1 / 0.75 So v2 = (1 / 0.75)2 v1 = 1.8v1 The speed through the clogged artery increases by 80% 69 Example 2: Blood flows at a rate of 5.00 L/min through an aorta with a radius of 1.00 cm. What is the speed of blood flow in the aorta? Flow rate = Av 70 Example 2: Blood flows at a rate of 5.00 L/min through an aorta with a radius of 1.00 cm. What is the speed of blood flow in the aorta? 5L min 1000cm 3 13 m 3 × × × Flow rate = Av = min 60sec L 100 3 cm 3 € 71 Example 2: Blood flows at a rate of 5.00 L/min through an aorta with a radius of 1.00 cm. What is the speed of blood flow in the aorta? 5L min 1000cm 3 13 m 3 × × × Flow rate = Av = min 60sec L 100 3 cm 3 = 8.3 x 10-5 m3/s € 72 Example 2: Blood flows at a rate of 5.00 L/min through an aorta with a radius of 1.00 cm. What is the speed of blood flow in the aorta? 5L min 1000cm 3 13 m 3 × × × Flow rate = Av = min 60sec L 100 3 cm 3 = 8.3 x 10-5 m3/s € v = 8.3 x 10-5 m3/s / A 73 Example 2: Blood flows at a rate of 5.00 L/min through an aorta with a radius of 1.00 cm. What is the speed of blood flow in the aorta? 5L min 1000cm 3 13 m 3 × × × Flow rate = Av = min 60sec L 100 3 cm 3 = 8.3 x 10-5 m3/s € v = 8.3 x 10-5 m3/s / A = 8.3 x 10-5 m3/s / π r 2 74 Example 2: Blood flows at a rate of 5.00 L/min through an aorta with a radius of 1.00 cm. What is the speed of blood flow in the aorta? 5L min 1000cm 3 13 m 3 × × × Flow rate = Av = min 60sec L 100 3 cm 3 = 8.3 x 10-5 m3/s € v = 8.3 x 10-5 m3/s / A = 8.3 x 10-5 m3/s / π r 2 v = 0.265 m/s 75 Bernoulli’s Equation For steady flow, the speed, pressure, and elevation of an incompressible and nonviscous fluid are related by an equation discovered by Daniel Bernoulli (1700–1782). 76 Bernoulli’s Principle • Pressure also changes with velocity. • Bernoulli’s Principle – When the speed of a fluid increases, the pressure within the fluid decreases. • This has to do with conservation of energy. – If the velocity increases, the KE increases. – In order to keep the total energy constant, a quantity must decrease. – Since PE is constant, work must be done. – W = Fd and P = F/A W = PAd = PV • The volume stays the same, therefore the pressure decreases. 77 Bernoulli’s Equation In the steady flow of a nonviscous, incompressible fluid of density ρ, the pressure P, the fluid speed v, and the elevation y at any two points (1 and 2) are related by 78 Bernoulli’s Equation A fluid can also change its height. By looking at the work done as it moves, we find: This is Bernoulli’s equation. One thing it tells us is that as the speed goes up, the pressure 79 goes down. It’s a conservation of energy equation! Applications of Bernoulli • Why during storm might a roof blow off? • There is fast moving air above (bunched up streamlines) the roof. Faster moving air means less pressure above than inside. 81 Applications of Bernoulli • A small jeep has a soft, ragtop roof. When the jeep is at rest the roof is flat. When the jeep is cruising at highway speeds with its windows rolled up, does the roof a. Bow upward b. Remain flat c. Bow downward • A. The roof bows upward – When the jeep is in motion air flows over the top of the roof, while the air inside the jeep is at rest—since the windows are closed. Thus, there is less pressure over the roof than under it. As a result the roof bows upward. 82 Bernoulli’s Equation example Water circulates throughout a house in a hot-water heating system. If the water is pumped at a speed of 0.50 m/s through a 4.0-cm-diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6-cmdiameter pipe on the second floor 5.0 m above? Assume these pipes do not divide into branches. 83 Solution First, calculate the flow speed on the second floor, calling it v2, using the equation of continuity. 84 Solution First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. We can call the basement point 1. 85 Solution First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. We can call the basement point 1. v2A2 = v1A1 86 Solution First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. We can call the basement point 1. v2 = (v1A1) / A2 = (v1π(r1)2) / (π(r2)2) = (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s 87 Bernoulli’s Equation, Is often more conveniently rearranged to solve for P2. And re-labeling the ys as hs, we write: P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2) Solution First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. We can call the basement point 1. v2 = (v1A1) / (A2) = (v1π(r1)2) / (π(r2)2) = (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s To find pressure, we use Bernoulli’s equation: P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2) 89 Solution First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. v2 = (v1A1) / (A2) = (v1π(r1)2) / (π(r2)2) = (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s To find pressure, we use Bernoulli’s equation: P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2) P2 = (3.0 x 105 N/m2) + (1.0 x 103 kg/m3)(9.8 m/s2)(-5.0m) + ½ (1.0 x 103 kg/m3)[ (0.50 m/s)2 – (1.2 m/s)2] 90 Solution First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. v2 = (v1A1) / (A2) = (v1π(r1)2) / (π(r2)2) = (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s To find pressure, we use Bernoulli’s equation: P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2) P2 = (3.0 x 105 N/m2) + (1.0 x 103 kg/m3)(9.8 m/s2)(-5.0m) + ½ (1.0 x 103 kg/m3)[ (0.50 m/s)2 – (1.2 m/s)2] P2 = (3.0 x 105 N/m2) – (4.9 x 104 N/m2) – (6.0 x 102 N/m2) 91 Solution First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. v2 = (v1A1) / (A2) = (v1π(r1)2) / (π(r2)2) = (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s To find pressure, we use Bernoulli’s equation: P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2) P2 = (3.0 x 105 N/m2) + (1.0 x 103 kg/m3)(9.8 m/s2)(-5.0m) + ½ (1.0 x 103 kg/m3)[ (0.50 m/s)2 – (1.2 m/s)2] P2 = (3.0 x 105 N/m2) – (4.9 x 104 N/m2) – (6.0 x 102 N/m2) P2 = 2.5 x 105 N/m2 92 Applications of Bernoulli’s Principle: Using Bernoulli’s principle, we find that the speed of fluid coming from a spigot on an open tank is: This is called Torricelli’s theorem. (Since P1=P2, Bernoulli’s equation becomes ½ ρv12 + ρgy = ρgy2) … solve for v1 93