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AP Physics B

AP Physics B. Fluid Dynamics. College Board Objectives. . FLUID MECHANICS AND THERMAL PHYSICS Fluid Mechanics Hydrostatic pressure Students should understand the concept of pressure as it applies to fluids, so they can: a) Apply the relationship between pressure, force, and area.

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AP Physics B

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  1. AP Physics B Fluid Dynamics

  2. College Board Objectives . FLUID MECHANICS AND THERMAL PHYSICS • Fluid Mechanics • Hydrostatic pressure Students should understand the concept of pressure as it applies to fluids, so they can: a) Apply the relationship between pressure, force, and area. b) Apply the principle that a fluid exerts pressure in all directions. c) Apply the principle that a fluid at rest exerts pressure perpendicular to any surface that it contacts.

  3. Fluid mechanics, cont. d) Determine locations of equal pressure in a fluid. e) Determine the values of absolute and gauge pressure for a particular situation. f) Apply the relationship between pressure and depth in a liquid, DP = r g Dh

  4. Fluid Mechanics, cont. Buoyancy Students should understand the concept of buoyancy, so they can: a) Determine the forces on an object immersed partly or completely in a liquid. b) Apply Archimedes’ principle to determine buoyant forces and densities of solids and liquids.

  5. Fluid Mechanics, cont. Fluid flow continuity Students should understand the equation of continuity so that they can apply it to fluids in motion. Bernoulli’s equation Students should understand Bernoulli’s equation so that they can apply it to fluids in motion.

  6. Equations : P = F/ A Sp.Gr. = ρ substance / ρwater P = ρhg _____________________________ Ptotal = Patm + P liquid

  7. Homework : Chapter 9 Solids and Fluids Summaries 2 Examples per section 2 End of Chapter 9 problems per section 2 PROBLEMS IN THE COLLEGE BOARD https://apstudent.collegeboard.org/apcourse/ap-physics-b/exam-practice

  8. 10.1&2 Density & Specific Gravity 1. The mass density  of a substance is the mass of the substance divided by the volume it occupies: unit: kg/m3 r for aluminum 2700 kg/m3 or 2.70 g/cm3 mass can be written as m = rV and weight as mg = rVg Specific Gravity: r substance / r water

  9. 2. A fluid - a substance that flows and conforms to the boundaries of its container. 3. A fluid could be a gas or a liquid; however on the AP Physics B exam fluids are typically liquids which are constant in density.

  10. 4. An ideal fluid is assumed • to be incompressible (so that its density does not change), • to flow at a steady rate, • to be nonviscous (no friction between the fluid and the container through which it is flowing), and • flows irrotationally (no swirls or eddies).

  11. 5. Turpentine has a specific gravity of 0.9 . What is its density ?

  12. 5. Turpentine has a specific gravity of 0.9 . What is its density ? Specific Gravity=r substance / r water Ρsubstance = sp.gr. X ρwater = 0.9 X 1000kg/m3 = 900 kg / m3

  13. 6. A cork has volume of 4 cm3 and weighs 10-2 N . What is the specific gravity of cork ?

  14. 6. A cork has volume of 4 cm3 and weighs 10-2 N . What is the specific gravity of cork ? Fg = Weight = mg = 10-2 N = m (10 m/s2) m = 10-2 N / 10m/s2 = 10-3 kg ρcork = m/v= 10-3 kg / 4cm3 ( 100 cm)3 / m3 = 250 kg / m3 sp. Gr. = ρsubstance /ρwater = 250kg /m3 / 1000kg /m3 sp. Gr. = 0.25

  15. 10.3 5. Pressure Any fluid can exert a force perpendicular to its surface on the walls of its container. The force is described in terms of the pressure it exerts, or force per unit area: Units: N/m2 or Pa (1 Pascal*) dynes/cm2 or PSI (lb/in2) 1 atm = 1.013 x 105 Pa or 15 lbs/in2 *One atmosphere is the pressure exerted on us every day by the earth’s atmosphere.

  16. 6. The pressure is the same in every direction in a fluid at a given depth. 7.Pressure varies with depth. P = F/A = mg/A = ρVg/A P = F = rAhg so P = rgh A A

  17. 8. A FLUID AT REST EXERTS PRESSURE PERPENDICULAR TO ANY SURFACE THAT IT CONTACTS. THERE IS NO PARALLEL COMPONENT THAT WOULD CAUSE A FLUID AT REST TO FLOW.

  18. PROBLEM 10-9 • (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? ,

  19. PROBLEM 10-9 • (a) Calculate the total force of the atmosphere acting on the top of a table that measures Patmosphere = 1.013 X 105 N/m2 (b) What is the total force acting upward on the underside of the table? 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,

  20. 10. A vertical column made of cement has a base area of 0.5 m2 If its height is 2m , and the specific gravity of cement is 3 , how much pressure does this column exert on the ground ?

  21. 10. A vertical column made of cement has a base area of 0.5 m2 If its height is 2m , and the specific gravity of cement is 3 , how much pressure does this column exert on the ground ? P = F / A = mg /A = ρVg /A = ρAhg/A = ρhg Sp. Gr. = ρ cement / ρwater ρcement=Sp. Gr. X 1000 kg / m3 = 3 X 1000 kg / m3 = 3000 kg/m3 P = ρh g = 3,000 kg/m3 X 2m X 10m/s2 = 60,000 Pa = 60k Pa

  22. p1 p1 p1 h h h p2 p2 p2 11. Atmospheric Pressure and Gauge Pressure • The pressure p1 on the surface of the water is 1 atm, or 1.013 x 105 Pa. If we go down to a depth h below the surface, the pressure becomes greater by the product of the density of the water , the acceleration due to gravity g, and the depth h. Thus the pressure p2 at this depth is

  23. p1 p1 p1 h h h p2 p2 p2 12. In this case, p2 is called the absolute(total) pressure -- the total static pressure at a certain depth in a fluid, including the pressure at the surface of the fluid 13. The difference in pressure between the surface and the depth h is gauge pressure Note that the pressure at any depth does not depend of the shape of the container, only the pressure at some reference level (like the surface) and the vertical distance below that level.

  24. 14. (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? (b) What will be the pressure against the side of the pool near the bottom? (a)The absolute pressure is given by Eq. 10-3c, and the total force is the absolute pressure times the area of the bottom of the pool.

  25. 14. (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? (b) What will be the pressure against the side of the pool near the bottom? (a)The absolute pressure is given by Eq. 10-3c, and the total force is the absolute pressure times the area of the bottom of the pool.

  26. (b) The pressure against the side of the pool, near the bottom, will be the same as the pressure at the bottom,

  27. CW : Hydrostatic Pressure • 1. What is the hydrostatic gauge pressure at a point 10m below the surface of the ocean ? The specific gravity of seawater is 1.025.

  28. 1. What is the hydrostatic gauge pressure at a point 10m below the surface of the ocean ? The specific gravity of seawater is 1.025. Gauge pressure = ρgh = P2 - P1 Sp. Gr. = ρsubstance / ρwater ρocean = Sp. Gr. X 1000kg/m3 = 1.025 X 1000kg/m3 = = 1025 kg/ m3 Gauge Pressure = ρgh = 1025 kg/m3 ( 10 m/s2) ( 10m) = = 102, 500 Pa

  29. 2. A swimming pool has a depth of 4 m . What is the hydrostatic gauge pressure at a point 1 m below the surface ?

  30. 2. A swimming pool has a depth of 4 m . What is the hydrostatic gauge pressure at a point 1 m below the surface ? Gauge Pressure = ρhg = 1000kg/m3 ( 1m) (10m/s2) = = 10,000 Pa

  31. 15. What happens to the gauge pressure if we double our depth below the surface of the liquid ? What happens to the total pressure ?

  32. 15. What happens to the gauge pressure if we double our depth below the surface of the liquid ? What happens to the total pressure ? Gauge pressure will double. Gauge Pressure = ρgh Total Pressure = Patm + ρgh It will increase based on ρgh and having Patm as constant

  33. 16. A flat piece of wood , of area 0.5m2, is lying at the bottom of a lake . If the depth of the lake is 30 m , what is the force on the wood due to pressure ? ( use Patm= 1 X 10 5 Pa)

  34. 16. A flat piece of wood , of area 0.5m2, is lying at the bottom of a lake . If the depth of the lake is 30 m , what is the force on the wood due to pressure ? ( use Patm= 1 X 10 5 Pa) F= P/A P total = P atm + ρgh = 1 X 105 Pa + (1000 kg /m3) ( 10 m/s2) ( 30m) = 4 X 10 5 Pa Force = P A = 4 X 10 5 Pa X 0.5 m2= 2 X 10 5 N

  35. CW : Hydrostatic Pressure • 3. Consider a closed container , partially filled with a liquid of density ρ = 1200 kg/m3 and a point X that’s 0.5 m below the surface of the liquid . • A. if the space above the surface of the liquid is vacuum , what is the absolute pressure at point X ? • B. if the space above the surface of the liquid is occupied by a gas whose pressure is 2.4 X 104 Pa , What is the absolute pressure at Point X?

  36. CW : Hydrostatic Pressure • Consider a closed container , partially filled with a liquid of density ρ = 1200 kg/m3 and a point X that’s 0.5 m below the surface of the liquid . • A. if the space above the surface of the liquid is vacuum , what is the absolute pressure at point X ? • P absolute = P atm + ρgh = 0 + 1200 kg/m3 (10m/s2)( 0.5m) • = 6 X 10 3 Pa • B. if the space above the surface of the liquid is occupied by a gas whose pressure is 2.4 X 104 Pa , What is the absolute pressure at Point X?

  37. CW : Hydrostatic Pressure • Consider a closed container , partially filled with a liquid of density ρ = 1200 kg/m3 and a point X that’s 0.5 m below the surface of the liquid . • A. if the space above the surface of the liquid is vacuum , what is the absolute pressure at point X ? • P absolute = P atm + ρgh = 0 + 1200 kg/m3 (10m/s2)( 0.5m) • = 6 X 10 3 Pa • B. if the space above the surface of the liquid is occupied by a gas whose pressure is 2.4 X 104 Pa , What is the absolute pressure at Point X? • P absolute = P1 +ρgh = 2.4 X 104 Pa + 6 X 10 3 Pa = 3X 104 Pa

  38. 10.5 Pascal’s Principle • 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. Applications: hydraulic lift and brakes Pout = Pin And since P = F/a Fout = Fin Aout Ain Mechanical Advantage: Fout = Aout Fin Ain

  39. 17. 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. 18. The buoyant force is found to be the upward force on the same volume of water:

  40. 10-7 Buoyancy and Archimedes’ Principle 19. The net force on the object is then the difference between the buoyant force and the gravitational force.

  41. 10-7 Buoyancy and Archimedes’ Principle 20. If the object’s density is less than that of water, there will be an upward net force on it, and it will rise until it is partially out of the water.

  42. 10-7 Buoyancy and Archimedes’ Principle 21. For a floating object, the fraction that is submerged is given by the ratio of the object’s density to that of the fluid.

  43. 10-7 Buoyancy and Archimedes’ Principle This principle also works in the air; this is why hot-air and helium balloons rise.

  44. A geologist finds that a Moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock?

  45. A geologist finds that a Moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock? ρrock = mrock/ Vrock ρwater = mwater / V water Vrock = Vdisplaced water = Mdisplacedwater /ρ water = 9.28kg -6.18 kg / 1000 kg/m3 = 0.0031 m3 ρrock = mrock/ Vrock = 9.28 kg / .0031 m3 ρrock =2.99X 103 kg/m3

  46. 23. A crane lifts the 18,000-kg steel hull of a ship out of the water. Determine (a) the tension in the crane’s cable when the hull is submerged in the water, and (b) the tension when the hull is completely out of the water. Tension (T) Fb = Fwater Buoyant Force mg

  47. 23. A crane lifts the 18,000-kg steel hull of a ship out of the water. Determine (a) the tension in the crane’s cable when the hull is submerged in the water, and (b) the tension when the hull is completely out of the water. When the hull is submerged, both the buoyant force and the tension force act upward on the hull, and so their sum is equal to the weight of the hull. The buoyant force is the weight of the water displaced.

  48. (b)When the hull is completely out of the water, the tension in the crane’s cable must be equal to the weight of the hull.

  49. 24. A 5.25-kg piece of wood floats on water. What minimum mass of lead, hung from the wood by a string, will cause it to sink? For the combination to just barely sink, the total weight of the wood and lead must be equal to the total buoyant force on the wood and the lead.

  50. A 5.25-kg piece of wood floats on water. What minimum mass of lead, hung from the wood by a string, will cause it to sink? 34.For the combination to just barely sink, the total weight of the wood and lead must be equal to the total buoyant force on the wood and the lead.

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