Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 10 Part 1 Pressure, Archimedes Principle, Buoyancy

Physics 203 – College Physics I Department of Physics – The Citadel 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)

Physics 203 – College Physics I Department of Physics – The Citadel Pressure in a Fluid Pressure : P = F/A. Imagine an imaginary box of water at depth h. Static equilibrium: F 2 – F 1 = mg =  gV → P 2 A – P 1 A =  gAh. Then P 2 – P 1 =  gh. h F1F1F1F1 F2F2F2F2 d mgmgmgmg A imaginary box

Physics 203 – College Physics I Department of Physics – The Citadel Water behind a Dam Suppose I have two reservoirs, both of the same depth and width, but one holding a lake 2 miles long, and the other holding a lake 20 miles long. Which dam has more force on it?

Physics 203 – College Physics I Department of Physics – The Citadel Water Behind a Dam It doesn’t matter: The pressure on the dam at depth h is  gh in either case. The same would be true if the dam just held back just an inch of water!

Physics 203 – College Physics I Department of Physics – The Citadel 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

Physics 203 – College Physics I Department of Physics – The Citadel 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

Physics 203 – College Physics I Department of Physics – The Citadel 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

Physics 203 – College Physics I Department of Physics – The Citadel Atmospheric Pressure The weight of the air above us produces atmospheric pressure at sea level equal to 1 atm = × 10 5 N/m 2. Pressure is also measured in Pascals: 1 Pa = 1 N/m 2.

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel Suction 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? h P < P atm Pump

Physics 203 – College Physics I Department of Physics – The Citadel Suction 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. h P < P atm Pump

Physics 203 – College Physics I Department of Physics – The Citadel Suction The water must be supported by atmospheric pressure.  gh = 1 atm = × 10 5 N/m × 10 5 N/m 2 (1000 kg/m 3 )(9.8 m/s 2 ) h = = 10.3 m. h P = 0 Pump P =  gh

Physics 203 – College Physics I Department of Physics – The Citadel Hydraulic Lift A hydraulic lift is a simple machine which uses the fact that any fluid pushed into a cylinder on one end V V must push out the same volume into a cylinder at the other end. L1L1 L2L2

Physics 203 – College Physics I Department of Physics – The Citadel Hydraulic Lift Since the volumes are the same, the distances the cylinders move is related to their areas: A 1 L 1 = V = A 2 L 2. → L 2 /L 1 = A 1 /A 2. V V L1L1 L2L2 A1A1 A2A2

Physics 203 – College Physics I Department of Physics – The Citadel Hydraulic Lift The work done by a force on one cylinder will equal the work done by the other cylinder. F 1 L 1 = W = F 2 L 2. V V L1L1 L2L2 F2F2 F1F1

Physics 203 – College Physics I Department of Physics – The Citadel Hydraulic Lift Mechanical advantage F 2 / F 1 = L 1 / L 2 = A 2 / A 1. V V L1L1 L2L2 F2F2 F1F1 A2A2 A1A1

Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Principle Know: F 2 / F 1 = A 2 / A 1 Rewrite: F 2 /A 2 = F 1 /A 1 →  P 2 =  P 1. V V F2F2 F1F1 A2A2 A1A1

Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Principle When an force is applied to a closed vessel, the pressure increases by the same amount throughout the vessel. F2F2 F1F1 PP  P = F 1 /A 1 = F 2 /A 2 A2A2 A1A1

Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration Pascal demonstrated his principle by inserting a thin 12 m long tube of diameter 6 mm into a wine barrel of diameter 40 cm. Filling the tube with water caused the barrel to burst! (a)What was the weight of water in the tube? (b)What was the force on the lid? D h d

Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration What was the weight of the water in the tube? m =  (  r 2 ) h = 1000 kg/m 3 × (2.8 × 10  5 m 2 ) × 12 m = 340 g. D h = 12 m d = m r = m mg = kg × 9.8 m/s 2 = 3.33 N (0.75 lb)

Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration What was the force on the lid of the barrel? The force on the bottom of the tube is F 1 = mg = 0.75 lb on area A 1 =  d 2 /4 (= 2.8 x 10  5 m 2 ) D d h

Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration The force on the lid of the barrel is F 2 = mg (A 2 /A 1 ) with A 2 =  D 2 /4 Then F 2 = mg (D/d) 2 D d h Area is proportional to the length squared.

Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration The force on the lid of the barrel is F 2 = mg (D/d) 2 = 0.75 lb (0.40 m / m) 2 = 0.75 lb × 4400 = 3300 lb D = 0.4 m h d = m big number!

Physics 203 – College Physics I Department of Physics – The Citadel 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? F 2 – F 1 = m w g, just as before! This is the buoyant force. Archimedes Principle: The buoyant force equals the weight of the water displaced. V F1F1F1F1 F1F1F1F1 F2F2F2F2 submerged box

Physics 203 – College Physics I Department of Physics – The Citadel Floating Example A plastic block with specific gravity 0.3 floats in water. What fraction of the block’s volume is under water? The weight of the block equals the buoyant force.

Physics 203 – College Physics I Department of Physics – The Citadel Floating Example If the volume of the block is V and the volume under water is V w, we need to find V w /V. Use F B =  w V w g = mg =  Vg. Then  w V w =  V, and V w /V =  w = SG = 0.3 VwVw V 30% of the block is under water.

Physics 203 – College Physics I Department of Physics – The Citadel 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:

Physics 203 – College Physics I Department of Physics – The Citadel Buoyancy Puzzle The total mass in case A is M A =  V What is the mass in case B? Don’t guess – use physics. V A: B: v2v2 v1v1

Physics 203 – College Physics I Department of Physics – The Citadel 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:

Physics 203 – College Physics I Department of Physics – The Citadel Buoyancy Puzzle This means the mass of the whole block equals the mass of water that would occupy volume v 2. Therefore the two cups have the same total mass. V A: B: v2v2 v1v1