Physics 101: Lecture 17 Fluids


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1 Exam III Physics 101: Lecture 17 Fluids Exam 2 is Mon Nov. 4, 7pm Extra office hours on Fri. (see webpage!) Physics 101: Lecture 17, Pg 1
2 Homework 9 Help A block of mass M 1 = 3 kg rests on a table with which it has a coefficient of friction µ = A string attached to the block passes over a pulley to a block of mass M 3 = 5 kg. The pulley is a uniform disk of mass M 2 = 0.7 kg and radius 15 cm. As the mass M 3 falls, the string does not slip on the pulley. Newtons 2 nd Law 1) T 1 f = M 1 a 1 2) T 1 R T 3 R = I a 2 /R M 1 N M 1 g 3) M 3 g  T 3 = M 3 a 3 x f T 1 T 1 y M 2 T 3 M 3 T 3 M 3 g Notes: 1) f = m M 1 g 2) a 2 = a 1 / R 3) I = ½ M R 2 4) a 1 = a 3 Rewrite (a = a 1 ) 1) T 1 = M 1 a+mm 1 g 2) T 3 T 1 = ½ M 2 a 3) T 3 = M 3 g M 3 a Physics 101: Lecture 17, Pg 2
3 Overview F Net = m a F Net Δx = DK F Net Δt = Dp t Net = I a = DL/Dt Today: apply these ideas to molecules in fluids! Physics 101: Lecture 17, Pg 3
4 Fluids States of Matter Solid Hold Volume Hold Shape Liquid Hold Volume Adapt Shape Gas Adapt Volume Adapt Shape Physics 101: Lecture 17, Pg 4
5 Qualitative Demonstration of Pressure Force due to molecules of fluid colliding with container. Impulse F av Dt = Dp Average Pressure = F / A y average vertical force = f y = Dp y Dt = D mv y Dt Physics 101: Lecture 17, Pg 5
6 Atmospheric Pressure Basically weight of atmosphere! Air molecules are colliding with you right now! Pressure = 1x10 5 N/m 2 = 14.7 lbs/in 2! Example: Sphere with r = 0.1 m Magdeburg Spheres demo A = 4 p r 2 =.125 m 2 F = 12,000 Newtons (over 2,500 lbs)! Can demo Physics 101: Lecture 17, Pg 6
7 Pascal s Principle A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid. Hydraulic Lift DP 1 = DP 2 F 1 /A 1 = F 2 / A 2 F 1 = F 2 (A 1 /A 2 ) lift demo Compare the work done by F 1 with the work done by F 2 A) W 1 > W 2 B) W 1 = W 2 C) W 1 < W 2 W = F d cos q W 1 = F 1 d 1 = F 2 (A 1 / A 2 ) d 1 but: A 1 d 1 = V 1 = V 2 = A 2 d 2 = F 2 V 2 / A 2 = F 2 d 2 = W 2 Physics 101: Lecture 17, Pg 7
8 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 2 A) P 1 > P 2 B) P 1 = P 2 C) P 1 < P 2 P = F/A = mg / A Cup 1 has greater mass, but same area Under water P = P atmosphere + r g h Physics 101: Lecture 17, Pg 8
9 Pascal s Principle (Restated) 1. Without gravity: Pressure of a confined fluid is everywhere the same. 2. With gravity: P = P atm + 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 9
10 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) F A > F B B) F A =F B C) F A < F B F = P A, and pressure is rgh. Same pressure, same area, same force even though more water in B! Physics 101: Lecture 17, Pg 10
11 Pressure and Depth Barometer: a way to measure atmospheric pressure For nonmoving fluids, pressure depends only on depth. p 2 = p 1 + rgh P atm  0 = rgh Measure h, determine p atm examplemercury p 1 =0 p 2 =p atm h r = 13,600 kg/m 3 p atm = 1.05 x 10 5 Pa h = m = 757 mm = (for 1 atm) Physics 101: Lecture 17, Pg 11
12 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 p=0 p a h P a = r gh h = Pa rg The atmospheric pressure applied to the liquid will only be able to lift the liquid to a limited height no matter how hard you suck on the straw. Certainly trees have developed some mechanism to transport water from ground level, through their vascular tissues, and to their leaves (which can be higher that 50 feet) [demo] Physics 101: Lecture 17, Pg 12
13 Archimedes Principle Determine force of fluid on immersed cube Draw FBD» F B = F 2 F 1» = P 2 A P 1 A» = (P 2 P 1 )A» = r g d A» = r g V» = (M fluid /V) g V» = M fluid g Buoyant force is weight of displaced fluid! Physics 101: Lecture 17, Pg 13
14 Archimedes Example A cube of plastic 4.0 cm on a side with density = 0.8 g/cm 3 is floating in the water. When a 9 gram coin is placed on the block, how much does it sink below the water surface? Mg F b mg F Net = m a = 0 F b Mg mg = 0 h r g V disp = (M+m) g V disp = (M+m) / r h A = (M+m) / r h = (M + m)/ (r A) = (51.2+9)/(1 x 4 x 4) = 3.76 cm [coke demo] M = r plastic V cube = 4x4x4x0.8 = 51.2 g Physics 101: Lecture 17, Pg 14
15 Summary Pressure is force exerted by molecules bouncing off container P = F/A Gravity/weight affects pressure P = P 0 + rgd Buoyant force is weight of displaced fluid. F = r g V Physics 101: Lecture 17, Pg 15
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