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## Ch. 15 Fluids

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**Fluids**• Substances that flow readily, usually from high pressure to low pressure. • They take the shape of the container rather than retain their shape. • We refer both liquids and gases as fluids.**Density**• Is the mass of a substance divided by the volume of that substance.**Density Equation**Mass Density = Volume**m**D V**Density Units**• Metric: kg / m3 or g / cm3 • English: lbm / ft3**Sample Problem: What is the volume of my 2.16 g titanium**wedding band if the density is 4.50g/ cm3?**G: m = 2.16 g,**D = 4.50 g/cm3 U: V = ? E: D = m / V V = m/D S: V = 2.16 / 4.50 g/cm3 S: V = 0.48 cm3**Pressure**• is defined as force per unit of area.**Each book exerts the same force, but which one exerts a**greater pressure?**Equation**Force Pressure = Area**F**A P**Units of Pressure**• English: lb / ft2, lb / in2, or p.s.i. • Metric: N/m2 or Pascal (Pa)**Sample Problem: If an airplane window that is 40 cm by 60 cm**feels a pressure of 40,000 Pa. How much force is acting on the window?**G: L = 40 cm = 0.4 m**w = 60 cm = 0.6 m P = 40,000 Pa U: F = ? E: F = PA**We don’t have area (A),**A = Lw A = (.4)(.6) A = 0.24 m2**F = (40,000)(.24)**F = 9600 N**Atmospheric Pressure**The pressure of the earth’s atmosphere pushing down on you.**Earth's atmosphere is pressing against each square inch of**you with a force of 1 kilogram per square centimeter (14.7 pounds per square inch).**Sample ProblemCalculate the net force on an airplane window**if cabin pressure is 90% of the pressure at sea level, and the external pressure is only 50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.**Since pressure acts equally in all directions, gauge**pressure is used to deal with this. • If you have a flat tire, is there a pressure in the tire or is the pressure zero? • Yes, it is equal to atmospheric pressure.**Gauge Pressure (Pg)**• Is the pressure inside an inflated object, that is above the atmospheric pressure. • Pg = P – Patm • If a tire is inflated to 35 psi, that means it is 35 psi above the atmospheric pressure. So the pressure inside the tire is actually 35 psi + 14.7 psi.**Pressure increases with depth.**The deeper the body of water or you go, the more water there is on top of you per square inch.**Pressure depth relationship**Phyd = rgh r – weight density of liquid**The pressure of a liquid is the same at any given depth**regardless the shape of the container.**Absolute Pressure**• Absolute pressure is obtained by adding the atmospheric pressure to the hydrostatic pressure. • pabs = patm + rgh • p2 = p1 + rgh**Pascal’s Principle**Pressure applied to a fluid in a closed container is transmitted equally**to every point of the fluid and to the walls of the**container.**When the area of the 2 “pistons” are the same the input**force is the same as the output force.**What if you increased the area of the 2nd “piston” by**25? By Pascal’s Principle: P1 = P2**By the definition of Pressure:**F1/A1 = F2/A2 So if the 2nd area is increased by a factor of 25.**In order to maintain equality, the force has to be increased**by the same factor.**Distance each piston moved is different.**F1d1 = F2d2**Sample problem: The small piston on a hydraulic lift has an**area of 0.20 m2. A car weighing 1.2 x 104 N sits on the rack with a large piston. If the area of the large piston is 0.9 m2, how much force must be applied to the small piston to support the car?**G: A1 = 0.2 m2, F2 = 1.2 x 104 N, A2 =0.9 m2**U: F1 = ? E: F1/A1 = F2/A2 F1 = F2A1/A2**S: F1=(1.2 x 104)(0.2)/(0.9)**S: F1 = 2666.7 N**How far will the 2nd piston rise if the 1st piston moves 1.2**m?**The distance the 2nd piston moves is equal to the reciprocal**of the factor of the ↑ or ↓ in area times the distance the 1st piston moves.**Pascal’s Principle applies to all fluids, gases as well**liquids.**Buoyancy**• The apparent loss of weight of an object that is submerged in a fluid.