2. Oct. 29, 2001 Dr. Larry Dennis, FSU Department of Physics1 Physics 2053C – Fall 2001 Chapter 10 Fluids

3. 2 Pressure and Density Density = Mass/Volume A property of the material. Pressure = Force/Area Depends on the height of the fluid. Same in all directions. Units are: Force/Area = N/m 2. Pascals  1 Pa = 1 N/m 2. Atmosphere  1 atm = 1.013 X 10 5 N/m 2.

4. 3 Buoyant Forces Force exeted by a displaced liquid. F t -F b = B  gAh t -  gAh b = B B = g  A(h t – h b ) = W t - W b = B B =  A(h t – h b ) * g

5. 4 Equation of Continuity Flow 1 = Flow 2 assuming  1 =  2 (same liquid)  1 A 1 v 1 =  2 A 2 v 2 A 1 v 1 = A 2 v 2 so v 2 = x v 1 A1A1 A2A2

6. 5 Bernoulli’s Equation P = Pressure v = velocity  = density of fluid y = height g = acceleration due to gravity P 1 + ½  v 1 2 +  gy 1 = P 2 + ½  v 2 2 +  gy 2 1 2

7. 6 CAPA #1 What is the absolute pressure on the bottom of a swimming pool 20.0 m by 11.60 m whose uniform depth is 1.92 m? P w =  gh = (1.0x10 3 kg/m 3 )(9.8 m/s 2 )(1.92m) = 1.89x10 4 N/m 2 But, we need absolute pressure… P = P w + P atm P = 1.89x10 4 N/m 2 + 1.013x10 5 N/m 2 = 1.20x10 5 N/m 2

8. 7 CAPA #2-3 2. What is the total force on the bottom of that swimming pool? Area = 20.0 m x 11.60 m F = P x A = (1.20x10 5 N/m 2 )(20.0 m)(11.60 m) = 2.79x10 7 N 3. What will be the pressure against the side of the pool near the bottom? The pressure near the bottom is the same as on the bottom P = 1.20x10 5 N/m 2

9. 8 CAPA #4 4. A dentist’s chair of mass 236.0 kg is supported by a hydraulic lift having a large piston of cross-sectional area 1434.0 cm 2. The dentist has a foot pedal attached to a small piston of cross-sectional area 76.0 cm 2. What force must be applied to the small piston to raise the chair? F chair F? F chair = mg = (236.0 kg)(9.80 m/s 2 ) = 2312.8 N

10. 9 CAPA #4 (cont) 4. A dentist’s chair of mass 236.0 kg is supported by a hydraulic lift having a large piston of cross-sectional area 1434.0 cm 2. The dentist has a foot pedal attached to a small piston of cross-sectional area 76.0 cm 2. What force must be applied to the small piston to raise the chair? P 1 = P 2 P = F/A F 1 F 2 A 1 A 2 F 2 = (A 2 /A 1 )F 1 = ((76.0 cm 2 )/(1434.0 cm 2 )) x (2312.8 N) F 2 = 123 N F chair F? ___ = ___

11. 10 Floating Objects Buoyant force must equal the weight if the object is to float. Weight = Mg =  Vg Buoyant Force =  w Vg Apparent Weight = W – B = (  -  w ) Vg

12. 11 Floating Objects How much of an ice cube is above the water line? Mg B F = 0 = B – Mg B = Mg  water gV below = Mg  V below = M/  water =  ice V total /  water In order to float we must have V below < V total Only floats if  ice <  water.

13. 12 Applications of Bernoulli’s Principle Why does an airplane fly? Air moving over the top of the wing is moving faster than the air moving below the wing. The air pressure on the top top of the wing is lower than the pressure on the bottom of the wing. F =  Pressure * Wing Area P top + ½  V 2 top = P bottom + ½  V 2 bottom P bottom – P top = ½  (V 2 top - V 2 bottom )

14. 13 Application to CAPA Problems Airplane wing: F =  P*A P b + ½  v b 2 = P t + ½  v t 2  P = P b – P t = ½  (v t 2 – v b 2 ) Roof in a Hurricane F =  P*A P b + ½  v b 2 = P t + ½  v t 2  P = P b – P t = ½  (v t 2 – v b 2 ) V b = 0 m/s

15. 14 Quiz 7 Density:  = M/V Pressure: P = F/A Buoyant Forces: B =  liquid gV displaced Continuity Eqn: A 1 v 1 = A 2 v 2 Bernoulli’s Principle: P 1 + ½  v 1 2 +  gy 1 = P 2 + ½  v 2 2 +  gy 2

16. 15 Quiz 7 Sample Questions: Chap. 10: 10, 20, 27 Sample Problems: Chap. 10: 23, 29, 43, 71

17. 16 Next Time Quiz 7. Begin Chapter 11. Please see me with any questions or comments. See you Wednesday.