#### Transcript Review of buoyant force

Physics Warm Up: Agenda Copy these assignemts into your binder December 2 WarmUp: Agenda/ Review of buoyant force InClass: Archimedes Worksheet Homework: Read and take notes p358-362 DUE NEXT CLASS December 3-4 WarmUp: Specific Heat of Copper Lab: Specific Heat Homework: Answer q 5p374 DUE NEXT CLASS December 5-6 WarmUp: Heat of Fusion Lab: Heat of Fusion Homework: When the air temperature is 22.2°C many people find it a comfortable temperature, yet the same people often find a swimming in 22.2°C water too cold to be comfortable. Use specific heat to explain the reason for the difference in sensation of temperature of the air and the water. Explain it to somebody and bring in a written copy of your explanation. WHEN YOU FINISH WRITING DOWN THE HOMEWORK, PLEASE GET OUT YOUR BUOYANCY HOMEWORK FROM LAST WEEK, QUESTIONS 1 AND 2 P324 SO THAT WE CAN GO OVER THE QUESTIONS Buoyant force p324 There was some difficulty with problems 1 and 2 of the homework last week. We didn’t go over it in class. Question 1: A piece of metal weighs 50.0N in Air. 36.0N in water, and 41.0N in oil. Find the density of a. The metal Two pages back are the equations used in the sample problem Fnet = (ρfVf – ρoVo) g Fnet = (ρf – ρo) V•g Fg(object) ρo V•g = FB ρf V•g Fg(object) ρf = ρo FB 50.0N • 1.00g/cm3 14.0N where Vnet is the apparent weight, and the subscripts f and o stand for The fluid and the object respectively Because the object is submerged, the volumes are equal and we can simplify Use the ratio (you can find this on p322 if you forgot where it comes from) and solve for the density of the object (ρo). Substitute values from the question = 3.57g/cm3 = 3.57x103kg/m3 Buoyant force p324 There was some difficulty with problems 1 and 2 of the homework last week. We didn’t go over it in class. Question 1: A piece of metal weighs 50.0N in Air. 36.0N in water, and 41.0N in oil. Find the density of Then B asks you to find the density of the oil. Fg(object) FB ρf = = ρo V•g ρf V•g We can use the same simple ratio, but now solve for the density of the fluid, (ρf) ρo FB Fg(object) 3kg/m • 9.0N 3.57x10 3 ρf = 50.0N Substitute values from the question = 640x103kg/m3 Archimedes Principle Question 2 from that homework is easy if you just remember that the buoyant force acting on a submerged object is equal to the weight (force) of the fluid it displaces. The mass of this water Verify it: On the demo table is a can with a spout. Fill the with just enough water to run out of the spout. Weigh the small beaker. When the spout stops dripping, put the beaker under it. Add the 100g weight to the water and collect the displaced water. Weigh the beaker with the water and calculate the weight of the displaced water. Record it on your paper. Weigh the 100g weight in air by hanging it from the hook on the balance. Fill the 250mL beaker and place it on the arm of the balance. Weigh the 100g weight in water and calculate the difference. The difference is the buoyant force. It should be very close to the weight of the water displaced. Should equal the difference in weight Buoyant force p324 2. An empty rubber balloon has a mass of 0.0120kg. The balloon is filled with helium at 0°C, 1atm pressure and a density of 0.181kg/m3. The filled balloon has a radius of 0.500m. a. What is the buoyant force acting on the balloon V= 4/3 πr3 The buoyant force acting on the balloon is equal to V= 4/3 π(0.5m) the weight of the air it displaces (p320). The mass of the displaced air is the volume of the V= .524m3 balloon times the density of the. Multiplying that by Fg(air)= v• ρf • g acceleration due to gravity gives its weight. Fg(air)=0.524m3 • 1.29kg/m3 • 9.8 m/s2 B. The net force will be the difference between the balloon’s weight and the buoyant force. The weight of the balloon is Fg(air)= 6.62kgm/s2 = 6.62N the density of the helium times the volume of the balloon added to the mass of the B. empty balloon. Fg(balloon)=9.8 m/s2 (0.524m3 )(0.181kg/m3)+9.8 m/s2(0.0120kg) Fg(balloon)=1.05N Fnet= Fg(air) - Fg(balloon) Fnet= 6.62N- 1.05N Fnet= 5.57N Pressure & Pascal’s Law Pressure is force per area. p = f/a Liquid Pressure = (depth)(density) Units of pressure: pascal = n/m2 Atmosphere, mm of Hg (in manometer) kg/cm2 (mass units). Common pressures: 1 atm = 100 kPa 760 mm of Hg 10 m of H2O 1 kg/cm2 p = h Blaise Pascal Tap water pressure = 4 atm 400 kPa 4 kg/cm2 Car tire 2 atm Pressure cont’d Total Force = (pressure)(area) TF = pA . Two Sample Problems Solutions are on the next page Find the pressure needed to push water to the top of Tower 2, a height of 8.0 meters. (Use mass units). The density of water is 1.0 g/cm3. Mainmachinery Find the Total Force on the filled school dam whose dimensions are 5.0m by 2.0m. The average depth is 1.0m. Castle attack Solutions Find the pressure needed to push water to the top of Tower 2, a height of 8.0 meters. (Use mass units). The density of water is 1.0 g/cm3. 1 2 3 4 p = h = (8.0m)(100cm/m)(1.0g/cm3 2 ) = 800 g/cm2 Find the Total Force on the filled school dam whose dimensions are 5.0m by 2.0m. The average depth is 1.0m. 1 2 3 4 f = pA = (100g/cm2)(10.0m2)(104cm2/m2) * = 1 X 107 g = 1 X 104 kg or 10 tons! Note: 1kg = 1000g 1 metric ton = 1000kg 1 2 3 4 A = LW = (5.0m)(2.0m) = 10.0 m2 * There are 104 cm2 in a m2. 1 2 3 4 p = h = (1.0m)(100cm/m)(1.0g/cm3 2 ) = 100 g/cm2 Pascal’s Law The pressure on a confined fluid is transmitted in all directions. Pascal’s Vases show pressure depends only on depth & density. Pascal Pressure in Rocket (1) pascalrocket.mov Hiero’s Fountain Demo h2 is higher than h1, so the pressure is greater in the right system which pushes the water up into the fountain. Try it! Add a about 50mL of water to the funnel of the fountain on the demo table! Hydraulic Lift: Demo: Syringes Try this with different sized syringes. The pressure is transmitted undiminished in all directions. Hydraulic Brakes The applied pressure to the master cylinder is transmitted equally to all four brake pistons. Hydraulics lifts a House! (2) Oh, Pascal! Thanks to Mark Shisler hydraulics.mov An Uplifting Experience: Demo A strong rubber balloon inflated beneath a car or truck can lift 15 metric tonnes of load. The pressure is low, but the surface area is large. Total Force = (press)(area). Demo: A garbage bag blown up by a vacuum cleaner can lift a massive person.