Physics 17 Part H Fluids and Pressure Uni-Directional Pressure The pressure that is applied over an area is the force applied, divided by the area. A 400-N force is applied to a tabletop over a square area with side-length L = 20-cm. What pressure does this portion of the table experience? Area of a square = L 2 A = (0.20 m) 2 = 0.04 m 2 P = 400 N / 0.04 m 2 = 10,000 N/m 2 1.0 pascal (Pa) = 1.0 N/m 2 P = 10,000 Pa Example B: A 100-lb (445 N) person wearing stiletto heels, standing on one leg, rocks back onto the heel. The base of the heel is circular, with a radius of 0.20 cm. What pressure does the heel exert on the floor? 1.0 cm = 1 x 10-2 m (Area of a circle: πr 2 ) A = π (0.20 x 10-2 m) 2 = 1.26 x 10-5 m 2 P = F/A = 445 N /(1.26 x 10-5 m 2 ) = 3.54 x 10 7 N/m 2 = 3.54 x 10 7 Pa Heel pressure is unidirectional : its effect is felt only in one direction. Example C: The chest of an average human can be punctured or crushed by a uni-directional pressure of about six million pascals. What would have to be the radius of the stiletto heal in Example B to kill the person who is subject to the pressure of the single heel? P = F/A = F/P πr 2 = 445/6.0 x 10 6 r = 0.0049 m = -0.49 cm 1
Omni-Directional Pressures Earth s atmosphere is a 40-mile thick blanket of oxygen, nitrogen, and other gases weighing 5.17 x 10 19 N. This blanket of air rests on Earth s surface, exerting a downward force on it. R = radius of Earth = 6.38 x 10 6 m Each square inch on the surface of the empty bottle below experiences a 14.7-pound force directed inward. Why doesn t atmospheric pressure crush it? A = 4πR 2 = 4π (6.38 x 10 6 ) 2 = 5.12 x 10 14 m 2 Po = F/A = 5.17 x 10 19 / 5.12 x 10 14 = 101,000 Pa = 101 kilo-pascal (kpa) Po = 14.7 lb/in 2 (psi) (The subscript zero indicates that this is atmospheric pressure at zero elevation, i.e., sea-level. ) Atmospheric pressure is omni-directional: its effect isn t just downward; it applies a 101,000 Pa pressure in all directions: downward, upward, rightward, leftward, forward, backward. Try this: half-fill the bottle with hot tap-water, put the cap on tightly and place the bottle in a freezer. After 5-10 minutes, the hot air inside will have cooled, causing the air pressure inside to drop below atmospheric pressure which then exerts a greater inwardly-directed force than the outwardly-directed force: The bottle partially collapses. Now heat the bottle under hot faucet water and watch as the bottle re-inflates. 2
What is atmospheric pressure 2000 meters above sea-level? P = 101 (1-2.3 x 10-5 x 2000) 5 = 79.8 kpa Example B: At what altitude is atmospheric pressure 80 kpa? 80 = 101 (1-2.3 x 10-5 h ) 5 h = 1980 m Notice that the pressure at negative elevations (below sealevel) is greater than Po (101,000 Pa). Shower Caddy Example C: Calculate the force in pounds due to atmospheric pressure on one of the shower caddy suction cups, assuming the cups are circular, and have a radius of 0.70 inch. Recall: Po = 14.7 lbs/in 2 (psi) F = PA = 14.7 π(0.70) 2 = 22.6 lbs 3
Absolute Pressure vs Gauge Pressure Gauge Pressure = P - Po The gauge pressure is how much more than 14.7 psi the air pressure is. Example: Air pressure at sea-level is 14.7 lbs/in 2 (psi). The air pressure inside an inflated tire (the absolute pressure ) is 50.0 psi. What is the gauge pressure? Air pressure gauges measure how much greater than atmospheric pressure the pressure inside the tire is. Gauge Pressure = P - Po = 50.0-14.7 = 35.3 psi 4
Density The density of a substance is the mass per volume. ρ = m/v Substance ρ (kg/m 3 ) ρ (g/cm 3 ) Gold 19,300 19.30 Water 1,000 1.00 Ice 920 0.92 Wood 800 0.80 Air 1 ----- The density in units of g/cm 3 equals the density in kg/m 3, divided by 1000. What is the mass of a cylindrical barrel of water whose height is 1.7 meters and whose top and bottom areas are 0.25 m 2? V = 0.25 (1.7) = 0.425 m 3 m = (1000) 0.425 = 425 kg Example B: What must be the radius of a solid gold sphere for it to have a mass of 238 grams? ρv = m 19.3 (4/3) πr 3 = 238 r =1.4 cm Gold sold for $42 per gram on February 28, 2018. How much was this ball of gold worth that day? 5
Omni-Directional Water Pressure The cylinder below is filled with water. The water s weight causes a pressure on the base of the cylinder. The pressure at the base of each of the cylinders (within broken lines) of water below equals the weight of the water inside, divided by the area A of the cylinder s base. F = mg = (ρv) g = ρ(ah)g P = F/A = ρ(ah)g/a = ρgh = 9800h (Assuming the fluid is water.) Water pressure is proportional to depth: P = 9800 h Example B: What is the water pressure 10 meters below water? P = 9800 h = 9800 (10) = 98,000 Pa What is the total fluid pressure at that location? P = Po + 98,000 = 199,000 Pa This pressure is omni-directional : its effect is felt in all directions equally: upward, downward, sideways, forward, backward. The crush pressure for World War II German submarines ( U-boats ) was about 3.0 x 10 6 Pa. How many feet below the surface of the ocean could these submarines dive? (Convert answer in meters to feet, using conversion factor 1.0 meter = 3.28 feet) P = 9800 h 3.0 x 10 6 = 9800 h h = 306.12 m = 306.12 m (3.28 ft/m) = 1004 feet Destroyer depth charges are set to explode at depths less than 1004 feet. Dive! Dive! 6
Buoyancy and Archimedes Principle The underside of the block is farther below the water s surface than is the upper side, so the water pressure below is greater than the water pressure above. Thus, the upward force is greater than the downward force. The difference between these two pressure forces is the buoyant force, B. Archimedes Principle: B = weight of water displaced (no proof provided) 7
A solid spherical ball made of metal whose density is 2800 kg/m 3 and whose radius is 0.30 m is placed in a tank of water, and sinks. Example B: The sinking ball in Example A eventually comes to rest on the bottom of the tank. What is the contact force between the bottom of the tank and the ball? (a) What is the buoyant force acting on the ball? (b) What is the weight of the ball? (c) What is the net force? (d) What is the ball s acceleration? (a) V = (4/3π)(0.30) 3 = 0.1131 m 3 m = ρv (m = mass of water) = 1000 (0.1131) = 113.1 kg B = mg = 113.1 (9.8) = 1108 N Acceleration equals zero, so the net force is likewise zero: B + C - mg = 0 C = mg - B = 3103-1108 = 1995 N (b) m = ρv (m = mass of ball) = 2800 (0.1131) = 316.67 kg w = mg = 316.67 (9.8) = 3103 N (c) F = 1108-3103 = -1995 N (d) a = F/m (m = mass of ball) = -1995/316.67 = -6.30 m/s 2 Acceleration is negative, as expected because the ball is accelerating downward, in the negative direction. 8
Floating Icebergs The fraction of a floating object s volume that is under water equals the ratio of the densities. The block below is floating in water. Only the top 6 cm are above water. What is the block s density? ρo = density of object ρ = density of water Proof: A floating object has zero acceleration, so the net force on it is zero, so the object s weight equals the buoyant force, which is the weight of the water: Weight of Object = Weight of Water V/Vo = 14/20 = 0.70 ρo /ρ = V/Vo = 0.70 ρo = ρ (0.70) = 1000 (0.70 = 700 kg/m 3 or ρovog = ρvg V/Vo = ρo /ρ ρo /ρ = V/Vo Example B: Sea water has a density of 1029 kg/m 3. Ice has a density of 920 kg/m 3. What fraction of an iceberg is above the water line? Fraction Below = 920/1029 = 0.89 Fraction Above = 0.11 Only the tip of the iceberg is seen. 9
Pascal s Principle and the Hydraulic Lift Pressure added to any part of a completely enclosed fluid is transmitted undiminished to all parts of the fluid. --Blaise Pascal In the diagram below, a force F is applied over the area A of the top of a push piston. This adds a pressure to what already existed everywhere else in the liquid. The pressures at three locations are indicated before, and then after the application of a 50-N force on the 0.10 m 2 push piston. Added pressure is 50/0.10 = 500 Pa. 1623-1662 French mathematician, physicist and religious philospher. 10
The Hydraulic Lift What force F must be applied to the top of the push piston to lift a 9800 N automobile? The ratio of the areas is one-hundred, so the force needed is one-hundredth of 9800 N: Ratio of areas is one-hundred. F = 98 N Example B: Solve the problem in Example A more systematically, i.e., the longer way. The added pressure is the same everywhere: whatever it is under the push piston, it s the same under the lift piston: Let P1 and A1 be the added pressure and area for the push piston, and P2 and A2 be the corresponding numbers for the lift piston: F1/A1 = F2 /A2 F1 = (A1/A2) F2 = (0.002/0.200) 9800 = (1/100) 9800 = 98 N The ratio of the areas is one-hundred, so the push-piston force is one-hundredth of 9800 N. 11
Volume Flow Rate A1 = 0.04 m 2 v1 = 2.0 m/s We want the speed at the right end to be 8.0 m/s. What does the area at that end have to be? The figure above shows a portion of a tapered pipe along which water is flowing. The product Av at any point in the pipe is the volume flow rate, R, which has units of cubic meters per second (m 3 /s). R = Av = volume flow rate R is the same everywhere: R2 = R1 Units of R = Av: (m 2 )(m/s) = m 3 /s For example, if the volume flow rate at Point 1 is 30 cubic meters per second, then the flow rate at Point 2 is also 30 cubic meters per second. A2 =? A2v2 = A1v1 A2 = (v1/v2) A1 = (2.0/8.0) 0.04 = 0.01 m 2 Example B: An empty swimming pool 8.0 meters long, 4.0 meters wide, is to be filled to an average depth of 2.0 meters. A hose filling the pool has an inside radius of 4.0 cm. Water exiting the hose has a speed of 2.0 m/s. In hours and minutes, how long will it take to fill the pool? R = Av = [π (0.04 2 ) m 2 ] 2.0 m/s = 0.0101 m 3 /s V = 8.0 (4.0) (2.0) = 64.0 m 3 t = V/R = 64.0 m 3 / 0.0101 m 3 /s = 6337 s = 1.76 hours = one hour and 46 minutes 12