Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 10 Part 1 Pressure, Archimedes Principle, Buoyancy

2. Announcements • Today: chapter 10, sec. 1 – 7 • Next Tuesday: chapter 10, sec. 8 – 10 • Problem set 10A due. • Next Thursday: exam on chapters 7 – 9 (only the sections covered in the homework)

3. Pressure in a Fluid • Pressure : P = F/A. • Imagine an imaginary box of water at depth h. • Static equilibrium: • F2 – F1 = mg = rgV • → P2A – P1A =rgAh. • ThenP2 – P1 = rgh. h F1 A d F2 mg imaginary box

4. Water behind a Dam • Suppose I have two reservoirs, both of the same depth and width, but one holding a lake 2 mileslong, and the other holding a lake 20 mileslong. • Which dam has more force on it?

5. Water Behind a Dam • It doesn’t matter: The pressure on the dam at depth hisrghin either case. • The same would be true if the dam just held back just an inch of water!

6. Different shaped Vessels • Which of these vessels, filled with the same depth of water and with the same base area, has the greater force on the base? A B C D E = all the same

7. Different shaped Vessels • The force on the base is the same in each case, even though the mass of water in each container is different. The pressure only depends on the depth of the water. E = all the same

8. Different shaped Vessels • Which vessel has the greater pressure at the bottom? They are filled to the same depth, and are all the same shape. A B C D E = all the same

9. Atmospheric Pressure • The weight of the air above us produces atmospheric pressure at sea level equal to • 1 atm = 1.013 × 105 N/m2. • Pressure is also measured inPascals: • 1 Pa = 1 N/m2.

10. Gauge Pressure • Pressure gauges are normally set to zero when only atmospheric pressure is present. • Gauge pressure is then the additional pressure beyond that due to the atmosphere. • The total pressure including atmospheric pressure is the absolute pressure.

11. Suction Pump • A negative gauge pressure corresponds to suction. If we produce a negative gauge pressure on a straw, water will be “sucked” up the straw. • Why does the water go up the straw? P < Patm h

12. Suction Pump • What is the highest a suction pump can draw water up a tube? • The absolute pressure can’t be less than zero. • Assume the pressure in the tube is reduced to P = 0. P < Patm h

13. Suction Pump • The water must be supported by atmospheric pressure. • rgh= 1 atm = 1.013 × 105 N/m2 P= 0 h 1.013 × 105 N/m2 (1000 kg/m3)(9.8 m/s2) h = = 10.3 m. P= rgh

14. Hydraulic Lift • A hydraulic lift is a simple machine which uses the fact that any fluid pushed into a cylinder on one end V must push out the same volume into a cylinder at the other end. L2 V L1

15. Hydraulic Lift • Since the volumes are the same, the distances the cylinders move is related to their areas: • A1L1 = V = A2L2. • →L2/L1 = A1/A2. V A2 L2 V A1 L1

16. Hydraulic Lift • The work done by a force on one cylinder will equal the work done by the other cylinder. • F1L1= W = F2L2. F2 V L2 V F1 L1

17. Hydraulic Lift • Mechanical advantage • F2 / F1 = L1 / L2 • = A2/ A1. F2 V A2 L2 V A1 F1 L1

18. Pascal’s Principle • Know: F2/ F1 = A2 / A1 • Rewrite: • F2/A2= F1/A1 • →DP2 = DP1. F2 V A2 V A1 F1

19. Pascal’s Principle • When an force is applied to a closed vessel, the pressure increases by the same amount throughout the vessel. F2 A2 DP A1 F1 DP = F1/A1 = F2/A2

20. Pascal’s Demonstration d • Pascal demonstrated his principle by inserting a thin 12 m long tube of diameter 6 mminto a wine barrel of diameter 40 cm. • Filling the tube with water caused the barrel to burst! • What was the weight of water in the tube? • What was the force on the lid? h D

21. Pascal’s Demonstration d = 0.006 m • What was the weight of the water in the tube? • m = r(pr2) h • = 1000 kg/m3 • × (2.8 × 10-5 m2)× 12 m • = 340 g. h = 12 m r = 0.003 m D • mg = 0.340 kg × 9.8 m/s2 • = 3.33 N (0.75 lb)

22. Pascal’s Demonstration d • What was the force on the lid of the barrel? • The force on the bottom of the tube is • F1 = mg = 0.75 lbon area • A1 = pd2/4(= 2.8 x 10-5 m2) h D

23. Pascal’s Demonstration d • The force on the lid of the barrel is • F2 = mg (A2/A1)with • A2 = pD2/4 • ThenF2 = mg (D/d)2 h D Area is proportional to the length squared.

24. big number! Pascal’s Demonstration d = 0.006 m • The force on the lid of the barrel is • F2 = mg (D/d)2 • = 0.75 lb (0.40 m / 0.006 m)2 • = 0.75 lb × 4400 • = 3300 lb h D = 0.4 m

25. Pressure in a Fluid • Suppose I replace the imaginary box of water by an actual box. • What is the net force due to pressure on the box? • F2– F1= mwg, just as before! • This is the buoyant force. • Archimedes Principle: The buoyant force equals the weight of the water displaced. F1 V F2 submerged box

26. Floating Example • A plastic block with specific gravity 0.3floats in water. What fraction of the block’s volume is under water? • The weight of the block equals the buoyant force.

27. Floating Example V • If the volume of the block is V and the volume under water is Vw, we need to find Vw/V. • Use FB = rw Vwg • = mg = rVg. • Then rw Vw = rV, and • Vw/V = r/rw = SG = 0.3 Vw 30% of the block is under water.

28. Buoyancy Puzzle • Two identical cups have water in them, to the same level. But one also has a plastic block floating in it. • Which is heavier? • Or are they the same weight? A: B:

29. V Buoyancy Puzzle • The total mass in case Ais MA = rV • What is the mass in case B? • Don’t guess – use physics. A: B: v1 v2

30. Buoyancy Puzzle • Remember Archimedes’ Principle: • The weight of the plastic block equals the weight of the water that would have occupied the submerged volume of the block. A: B:

31. V Buoyancy Puzzle • This means the mass of the whole block equals the mass of water that would occupy volume v2. • Therefore the two cups have the same total mass. A: B: v1 v2