2. Physics 1710—Warm-up Quiz Why does the earth “have more gravity,” i.e. g is larger on earth, than does the moon? A.The earth as an atmosphere. B.The earth has a larger diameter. C.The earth has more mass. D.The earth rotates faster. E.None of the above.

3. Analysis: The earth has an atmosphere because of its larger g, not vis versa. The earth has an atmosphere because of its larger g, not vis versa. The larger diameter means smaller g~1/R 2 The larger diameter means smaller g~1/R 2 The larger mass of the earth m = 4/3πρR 3 makes a larger g=GM/R 2 The larger mass of the earth m = 4/3πρR 3 makes a larger g=GM/R 2 Gravity has nothing to do with the rate of rotation.Gravity has nothing to do with the rate of rotation. Physics 1710—Chapter 14 Fluid Dynamics

4. Summary: Pressure is the force per unit area. P =F/APressure is the force per unit area. P =F/A Unit of pressue [Pacal] = [N]/[m 2 ] Unit of pressue [Pacal] = [N]/[m 2 ] The hydrostatic pressure is P = P o + ρghThe hydrostatic pressure is P = P o + ρgh Archimedes’ Principle: F bouyant = ρ fluid g V Archimedes’ Principle: F bouyant = ρ fluid g V Equation of Continuity: A 1 v 1 = A 2 v 2 Equation of Continuity: A 1 v 1 = A 2 v 2 Bernoulli’s Equation: P + ½ ρv 2 + ρgy = constant. Bernoulli’s Equation: P + ½ ρv 2 + ρgy = constant. Physics 1710—Chapter 14 Fluid Dynamics

5. 80/20 Pressure = Force per unit Area P = F /A Physics 1710—Chapter 14 Fluid Dynamics

6. 80/20 The unit of pressure in SI units is the Pascal [Pa] = 1 Newton/ m 2 1 Newton force 1 square meter area 1 Pascal pressure: Physics 1710—Chapter 14 Fluid Dynamics

7. 80/20 Standard atmosphere = 1.0 atm =101 kPa =101 kPa =14.7 psi Tire pressure A few 100 kPa Physics 1710—Chapter 14 Fluid Dynamics

8. Which applies the greatest pressure on their foot? Physics 1710—Chapter 14 Fluid Dynamics

9. Which applies the greatest pressure on their foot? A.Man. B.Elephant. C.Woman. Physics 1710—Chapter 14 Fluid Dynamics

10. Force = 200 lb = 891 N Force = 4,000 lb = 17,800 N Force = 100 lb = 445 N Foot Area = 36.2 in 2 = 0.023 m 2 Foot Area = 43.5 in 2 = 0.028 m 2 Foot Area = 12.7 in 2 =.0082 m 2 p = F/A =200/(2 x 36.2) = 2.8 psi p = 891/(2 x 0.023) = 19 kPa p = F/A =4000/(4 x 43.5) = 23. psi p = 17800/(4 x 0.028) = 159 kPa p = F/A =100/(2 x 12.7) = 3.9 psi p = 445/(2 x 0.0082) = 27 kPa Physics 1710—Chapter 14 Fluid Dynamics

11. Force = 100 lb = 445 N Heel Area = 0.16 in 2 =.0001 m 2 p = F/A =100/(2 x 0.16) = 313. psi p = 445/(2 x 0.0001) = 2200. kPa p = 19 kPa p = 159 kPa p = 27 kPa Physics 1710—Chapter 14 Fluid Dynamics

12. The Mysterious Swami Seat Physics 1710—Chapter 14 Fluid Dynamics

13. 80/20 Pressure = Force per unit Area p = F /A Same Force Different Area ∴Different Pressure Physics 1710—Chapter 14 Fluid Dynamics

14. Pressure is the hydraulic stress: Hydraulic Stress: Bulk Modulus B Stress = modulus x strain σ = F/A = p = B ε = B ΔV/V Physics 1710—Chapter 14 Fluid Dynamics P V ΔV

15. Pressure is isotropic: Physics 1710—Chapter 14 Fluid Dynamics P V

16. Pressure is isotropic: Physics 1710—Chapter 14 Fluid Dynamics P P PP F 1 = A 1 P; F 2 = A 2 P

17. Hydrostatic Pressure is due to weight of fluid above: Physics 1710—Chapter 14 Fluid Dynamics F = m g = (Ah)ρ fluid g P = F/A = ρ fluid g h h A P F

18. Archimedes Pressure Physics 1710—Chapter 14 Fluid Dynamics F = AΔP - mg= A ρ fluid g Δh – mg F = F bouyant - mg F bouyant = ρ fluid g V P A - mg V P

19. Archimedes Pressure Physics 1710—Chapter 14 Fluid Dynamics F bouyant = ρ fluid g V P - ρ fluid V g V P

20. Continuity Physics 1710—Chapter 14 Fluid Dynamics V constant if fluid is incompressible. A 1 v 1 = A 2 v 2

21. Work done on fluid Physics 1710—Chapter 14 Fluid Dynamics ΔW = ½ m v 2 + mg y dW/dV = ½ ρ v 2 +ρg y = d (Fy)/dV = dF/dA =ΔP ½ ρ v 2 +ρg y =ΔP

22. Summary: Pressure is the force per unit area. P =F/APressure is the force per unit area. P =F/A Unit of pressue [Pacal] = [N]/[m 2 ] Unit of pressue [Pacal] = [N]/[m 2 ] The hydrostatic pressure is P = P o + ghThe hydrostatic pressure is P = P o + gh Archimedes’ Principle: F bouyant = ρ fluid g V Archimedes’ Principle: F bouyant = ρ fluid g V Equation of Continuity: A 1 v 1 = A 2 v 2 Equation of Continuity: A 1 v 1 = A 2 v 2 Bernoulli’s Equation: P + ½ ρv 2 + ρgy = constant. Bernoulli’s Equation: P + ½ ρv 2 + ρgy = constant. Physics 1710—Chapter 14 Fluid Dynamics

23. No Talking! Think! Confer! Peer Instruction Time Why does the platform spin faster when he brings his arms in? Physics 1710—Chapter 11 App: E & E

24. Why does the platform spin faster when he brings his arms in? 10 A.He increases his angular momentum. B.He increase his moment of inertia. C.He decrease his moment of inertia. D.He pushes against the inertia of the weights. E.None of the above Answer Now ! 1234567891011121314151617181920 2122232425262728293031323334353637383940 0% 0 of 1 Physics 1710—Chapter 11 App: E & E

25. Where should the fulcrum be place to balance the teeter-totter?10 A. A. B. B. C. C. Answer Now ! 1234567891011121314151617181920 2122232425262728293031323334353637383940 0% 0 of 1 Physics 1710—Chapter 14 Fluid Dynamics

26. Which way will the torque ladder move?10 A.Clockwise B.Counterclockwise C.Will stay balanced Answer Now ! 1234567891011121314151617181920 2122232425262728293031323334353637383940 0% 0 of 1 Physics 1710—Chapter 14 Fluid Dynamics