Fluids Pascal s Principle Measuring Pressure Buoyancy

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1 Fluids Pascal s Principle Measuring Pressure Buoyancy Lana Sheridan De Anza College April 11, 2018

2 Last time shear modulus introduction to static fluids pressure bulk modulus pressure and depth

3 Overview pressure and depth Pascal s principle measurements of pressure buoyancy and Archimedes principle

4 Pressure varies with Depth

5 Pascal s Barrel

6 Density of Water For water: ρ w = kg/m 3 That is ρ w = 1 g/cm 3.

7 Density of Water For water: ρ w = kg/m 3 That is ρ w = 1 g/cm 3. Originally, the gram was defined to be the mass of one cubic centimeter of water at the melting point of water.

8 Questions Calculate the water pressure at the base of the Hoover Dam. The depth of water behind the dam is 220 m. 1 2 Question from Hewitt, Conceptual Physics, 11th ed. 3 See example 14.4, page 422, Serway & Jewett, 9th ed.

9 Questions Calculate the water pressure at the base of the Hoover Dam. The depth of water behind the dam is 220 m. 1 Density of water: ρ w = 1000 kg/m 3 2 Question from Hewitt, Conceptual Physics, 11th ed. 3 See example 14.4, page 422, Serway & Jewett, 9th ed.

10 Questions Calculate the water pressure at the base of the Hoover Dam. The depth of water behind the dam is 220 m. 1 Density of water: ρ w = 1000 kg/m 3 Now consider, if the dam is 380 m long, what is the total force exerted by the water on the dam? 2 2 Question from Hewitt, Conceptual Physics, 11th ed. 3 See example 14.4, page 422, Serway & Jewett, 9th ed.

11 he middle ear. Using this technique equalizes the pressure. Questions Now consider, if the dam is 380 m long, what is the total force exerted by the water on the dam? 14.5). Determine the h y cannot calculate the pressure in the water e dam also increases., we must use integraproblem. H ttom of the dam. We x a distance y above the O on each such strip is Figure (Example 14.4) Water See example 14.4, page 422, Serway & Jewett, 9th ed. w dy y

12 Pressure in a liquid We have this expression for total pressure: P total = P 0 + ρgh What if the pressure at the surface of the liquid, P 0, was increased to P 1. How would we expect this relation to change?

13 Pressure in a liquid We have this expression for total pressure: P total = P 0 + ρgh What if the pressure at the surface of the liquid, P 0, was increased to P 1. How would we expect this relation to change? P total = P 1 + ρgh The differences in pressure between the different layers of liquid remain the same, but the pressure at each depth h increases.

14 Pascal s Law This simple idea is captured by Pascal s Law. Pascal s law applied to confined, incompressible fluids. Pascal s Law A change in pressure applied to one part of an (incompressible) fluid is transmitted undiminished to every point of the fluid.

15 Pascal s Law This simple idea is captured by Pascal s Law. Pascal s law applied to confined, incompressible fluids. Pascal s Law A change in pressure applied to one part of an (incompressible) fluid is transmitted undiminished to every point of the fluid. This does not mean that the pressure is the same at every point in the fluid. It means that if the pressure is increased at one point in the fluid, it increases by the same amount at all other points.

16 Pascal s Principle Since the changes in pressures at the left end and the right end are the same: Since A 2 > A 1, F 2 > F 1. P 1 = P 2 F 1 = F 2 A 1 A 2

17 Hydraulic Lift This has applications: 1 Figure from hyperphysics.phy-astr.gsu.edu.

18 Question If a pair of pistons are connected on either end of a hydraulic tube. The first has area 0.2 m 2 and the second has an area of 4 m 2. A force of 30 N is applied the first piston. What is the force exerted by the second piston on a mass that rests on it?

19 Question If a pair of pistons are connected on either end of a hydraulic tube. The first has area 0.2 m 2 and the second has an area of 4 m 2. A force of 30 N is applied the first piston. What is the force exerted by the second piston on a mass that rests on it? If the first piston is depressed a distance of 1 m by the 30 N force, how far does the second piston rise?

20 Measuring Pressure 418 Chapter 14 Fluid Mechanics Pressure in a fluid could be measured using a device like this: Vacuum A S F Figure 14.2 A simple device for The pressure is measuring proportional the topressure the compression exerted of the spring. by a fluid. 1 Diagram from Serway & Jewett, 9th ed. The pre The device to a spring the piston is balanced measured d force exert the fluid at of the force

21 Barometers Barometers are devices for measuring local atmospheric pressure. Typically, simple barometers are filled with mercury, which is very dense. The weight of the mercury in the tube exerts the same pressure as the surrounding atmosphere. On low pressure days, the level of the mercury drops. On high pressure days it rises.

22 Mercury Barometer The pressure at points A and B is the same. The pressure at B is P 0. h P 0 Above the mercury in the tube is a vacuum, so pressure at A is ρ Hg gh. (ρ Hg = 13.6 kg/m 3 ) A P 0 B Therefore, P 0 = ρ Hg gh. h P 0 Pressure is sometimes quoted in inches of mercury. a 1 Diagrams from Serway & Jewett, 9th ed.

23 mon baromd Manometer at one end. 14.6a). The of the merpoint A, due to the atmold move mererefore, P 0 5 the mercury lumn varies, us determine atm 5 be the pres- The pressure being measured, P, can be compared to atmospheric pressure P 0 by measuring the height of the incompressible fluid in the U-shaped tube. a P A h P 0 B b If h is positive, P > P 0, if negative, P < P m Figure 14.6 Two devices for P P 0 is calledmeasuring the gauge pressure: pressure. (a) a mercury barometer and (b) an open-tube

Barometers Barometers are devices for measuring local atmospheric pressure. Typically, simple barometers are filled with mercury, which is very dense.

24 Barometers Barometers are devices for measuring local atmospheric pressure. Typically, simple barometers are filled with mercury, which is very dense. The weight of the mercury in the tube exerts the same pressure as the surrounding atmosphere. On low pressure days, the level of the mercury drops. On high pressure days it rises.

25 Mercury Barometer The pressure at points A and B is the same. The pressure at B is P 0. h P 0 Above the mercury in the tube is a vacuum, so pressure at A is ρ Hg gh. (ρ Hg = 13.6 kg/m 3 ) A P 0 B Therefore, P 0 = ρ Hg gh. h P 0 Pressure is sometimes quoted in inches of mercury. a 1 Diagrams from Serway & Jewett, 9th ed.

26 mon baromd Manometer at one end. 14.6a). The of the merpoint A, due to the atmold move mererefore, P 0 5 the mercury lumn varies, us determine atm 5 be the pres- The pressure being measured, P, can be compared to atmospheric pressure P 0 by measuring the height of the incompressible fluid in the U-shaped tube. a P A h P 0 B b If h is positive, P > P 0, if negative, P < P m Figure 14.6 Two devices for P P 0 is calledmeasuring the gauge pressure: pressure. (a) a mercury barometer and (b) an open-tube

27 Buoyancy Astronauts training in their spacesuits: The total mass of NASA s EMU (extravehicular mobility unit) is 178 kg. Why does training underwater make maneuvering in the suits easier? 1 Picture from Hubblesite.org.

28 Buoyancy The apparent weight of submerged objects is less than its full weight. For an object that would float, but is held underwater, its apparent weight is negative! There is an upward force on an object in a fluid called the buoyant force.

29 Buoyancy Why does this force exist? Where does it come from?

30 Buoyancy Why does this force exist? Where does it come from? We know pressure depends on depth, so an object that s not completely flat will have different pressure on different parts of its surface. Consider a rectangular object of height h and base area A with its top edge at a depth d.

31 Buoyancy Why does this force exist? Where does it come from? We know pressure depends on depth, so an object that s not completely flat will have different pressure on different parts of its surface. Consider a rectangular object of height h and base area A with its top edge at a depth d. The force on each of the four sides will be equal. The force on the bottom will be (P 0 + ρg(h + d))a. The force on the top will be (P 0 + ρgd)a.

32 Buoyancy Why does this force exist? Where does it come from? We know pressure depends on depth, so an object that s not completely flat will have different pressure on different parts of its surface. Consider a rectangular object of height h and base area A with its top edge at a depth d. The force on each of the four sides will be equal. The force on the bottom will be (P 0 + ρg(h + d))a. The force on the top will be (P 0 + ρgd)a. There will be a net upward force from the pressure difference!

33 Buoyancy How big will the upward force be? F buoy = F up F down

34 Buoyancy How big will the upward force be? F buoy = F up F down = (P 0 A + ρg(h + d)a) (P 0 A + ρgda) = ρgha = ρgv obj because the volume of the submerged block is V obj = ha.

35 Buoyancy How big will the upward force be? F buoy = F up F down = (P 0 A + ρg(h + d)a) (P 0 A + ρgda) = ρgha = ρgv obj because the volume of the submerged block is V obj = ha. Notice that ρv obj = m f, the mass of the displaced fluid. (We are assuming ρ is constant.)

36 Buoyancy and Archimedes Principle Archimedes Principle The buoyant force on an object is equal to the weight of the fluid that the object displaces. Logically, if a brick falls to the bottom of a pool it must push an amount water equal to its volume up and out of the way.

37 Buoyancy and Archimedes Principle For a fully submerged object the buoyant force is: F buoy = ρ f V obj g where ρ f is the mass density of the fluid and V obj is the volume of the object. ρ f V obj is the mass of the water moved aside by the object.

38 Buoyancy and Archimedes Principle An object that floats will displace less fluid than its entire volume. For a floating object: F buoy = ρ f V sub g where V sub is the volume of the part of the object underneath the fluid level only.

39 Sinking and Floating Will a particular object sink or float in a particular fluid? If the object is less dense than the fluid it will float. If the object is more dense than the fluid it will sink. If the object and the fluid have the same density if will neither float or sink, but drift at equilibrium.

40 Sinking and Floating Since the relative density of the object to the fluid determines whether it will sink or float, we sometimes use the notion of specific gravity. The specific gravity of an object relates its density to the density of water (or occasionally other liquids): Specific gravity, SG of a sample is the ratio of its density to that of water. SG = ρ sample ρ water Often referenced in brewing!

41 Sinking and Floating A floating object displaces a mass of fluid equal to its own mass! (Equivalently, a weight of fluid equal to its own weight.) This also means that ρ f V sub = m obj.

42 Summary pressure and depth Pascal s principle measuring pressure buoyancy and Archimedes principle Test Tuesday, April 17, in class. Collected Homework due Monday, April 16. (Uncollected) Homework Serway & Jewett: Ch 14, onward from page 435, OQs: 7; CQs: 5; Probs: 8, 15, 19, 21, (25, 27, 29, 35, 36, 65, 71, 73,) 77 - brackets are buoyancy questions

Fluids Pascal s Principle Measuring Pressure Buoyancy

Fluids Pascal s Principle Measuring Pressure Buoyancy Lana Sheridan De Anza College April 11, 2018 Last time shear modulus s introduction to static fluids pressure bulk modulus pressure and depth re at