Exam III Physics 101: Lecture 17 Fluids Physics 101: Lecture 17, Pg 1 States of Matter Solid Hold Volume Hold Shape Liquid Fluids Hold Volume Adapt Shape Gas Adapt Volume Adapt Shape Physics 101: Lecture 17, Pg 2 Qualitative Demonstration of Pressure Force due to molecules of fluid colliding with container. Impulse Fav t = p Average Pressure = F / A y p y average vertical force f y t mv y t Physics 101: Lecture 17, Pg 3 Atmospheric Pressure Basically weight of atmosphere! Air molecules are colliding with you right now! Pressure = 1x105 N/m2 = 14.7 lbs/in2! Example: Sphere with r = 0.1 m Magdeburg Spheres demo A = 4 p r2 = .125 m2 F = 12,000 Newtons (over 2,500 lbs)! Can demo Physics 101: Lecture 17, Pg 4 Pascal’s Principle A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid. Hydraulic Lift P1 = P2 F1/A1 = F2 / A2 F1 = F2 (A1/A2) lift demo Compare the work done by F1 with the work done by F2 A) W1 > W2 B) W1 = W2 C) W1 < W2 Physics 101: Lecture 17, Pg 5 Gravity and Pressure Two identical “light” containers are filled with water. The first is completely full of water, the second container is filled only ½ way. Compare the pressure each container exerts on the table. 1 A) P1 > P2 B) P1 = P2 2 C) P1 < P2 Physics 101: Lecture 17, Pg 6 Pascal’s Principle (Restated) 1. Without gravity: Pressure of a confined fluid is everywhere the same. 2. With gravity: P = Patm + r g h Density r = M/V Pressure of a fluid is everywhere the same at the same depth. [vases demo] In general: in a confined fluid, change in pressure is everywhere the same. Physics 101: Lecture 17, Pg 7 Dam ACT B A A Two dams of equal height prevent water from entering the basin. Compare the net force due to the water on the two dams. A) FA > FB B) FA=FB C) FA< FB Physics 101: Lecture 17, Pg 8 Pressure and Depth Barometer: a way to measure atmospheric pressure For non-moving fluids, pressure depends only on depth. p1=0 p2 = p1 + rgh Patm - 0 = rgh p2=patm h Measure h, determine patm example--Mercury r = 13,600 kg/m3 patm = 1.05 x 105 Pa h = 0.757 m = 757 mm = 29.80” (for 1 atm) Physics 101: Lecture 17, Pg 9 Checkpoint Is it possible to stand on the roof of a five story (50 foot) tall house and drink, using a straw, from a glass on the ground? 1.No 2.Yes Physics 101: Lecture 17, Pg 10 Archimedes’ Principle Determine force of fluid on immersed cube Draw FBD » FB = F2 – F1 » = P 2 A – P1 A » = (P2 – P1)A » =rgdA » =rgV » = (Mfluid/V) g V » = Mfluid g Buoyant force is weight of displaced fluid! Physics 101: Lecture 17, Pg 11 Fb Archimedes Example A cube of plastic 4.0 cm on a side with density = 0.8 g/cm3 is floating in the water. When a 9 gram coin is placed on the block, how much does it sink below the water surface? FNet = m a = 0 Mg mg h Fb – Mg – mg = 0 r g Vdisp = (M+m) g M = rplastic Vcube Vdisp = (M+m) / r = 4x4x4x0.8 h A = (M+m) / r = 51.2 g h = (M + m)/ (r A) = (51.2+9)/(1 x 4 x 4) = 3.76 cm [coke demo] Physics 101: Lecture 17, Pg 12 Summary Pressure is force exerted by molecules “bouncing” off container P = F/A Gravity/weight affects pressure P = P0 + rgd Buoyant force is “weight” of displaced fluid. F = r g V Physics 101: Lecture 17, Pg 13

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