# Physics 101: Lecture 17, Pg 1 Physics 101: Lecture 17 Fluids Exam III.

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Physics 101: Lecture 17, Pg 1 Physics 101: Lecture 17 Fluids Exam III

Physics 101: Lecture 17, Pg 2 Homework Help A block of mass M1 = 3 kg rests on a table with which it has a coefficient of friction µ = 0.73. A string attached to the block passes over a pulley to a block of mass M3 = 5 kg. The pulley is a uniform disk of mass m2 = 0.7 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley. M3 T3T3 m3gm3g m2 T3T3 T1T1 M1 N m1gm1g T1T1 f Newtons 2 nd Law 1) T 1 – f = M 1 a 1 2)T 1 R – T 3 R = I   3)T 3 – M 3 g = M 3 a 3 x y Notes: 1)f =  M 1 g   = -a 1 / R 3)I = ½ M R 2 4)a 1 = -a 3 Rewrite (a = a 1 ) 1)T 1 = M 1 (a+  g) 2)T 3 – T 1 = ½ M 2 a 3)T 3 = M 3 (g – a) 10

Physics 101: Lecture 17, Pg 3Overview  F = m a  Fx = Change in Kinetic Energy  Ft = Change in momentum  = I   oday look at application of these ideas to molecules in fluids 10

Physics 101: Lecture 17, Pg 4 States of Matter l Solid è Hold Volume è Hold Shape l Liquid è Hold Volume è Adapt Shape l Gas è Adapt Volume è Adapt Shape Fluids 11

Physics 101: Lecture 17, Pg 5 Qualitative Demonstration of Pressure y 16 l Force due to molecules of fluid colliding with container.  Impulse =  p l Average Pressure = F / A

Physics 101: Lecture 17, Pg 6 Atmospheric Pressure l Basically weight of atmosphere! l Air molecules are colliding with you right now! l Pressure = 1x10 5 N/m 2 = 14.7 lbs/in 2 ! l Example: Sphere w/ r = 0.1 m è Demo A = 4  r 2 =.125 m 2 F = 12,000 Newtons (over 2,500 lbs)! 21

Physics 101: Lecture 17, Pg 7 Pascal’s Principle l A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid. l Hydraulic Lift  P 1 =  P 2 F 1 /A 1 = F 2 / A 2 F 1 = F 2 (A 1 /A 2 ) l 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 26 W = F d cos  W 1 = F 1 d 1 = F 2 (A 1 / A 2 ) d 1 = F 2 V 1 / A 2 = F 2 d 2 = W 2

Physics 101: Lecture 17, Pg 8 Gravity and Pressure l 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. A) P 1 > P 2 B) P 1 = P 2 C) P 1 < P 2 Under water P = P atmosphere +  g h 30 1 2 P = F/A = mg / A Cup 1 has greater mass, but same area

Physics 101: Lecture 17, Pg 9 Dam ACT 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  gh. Same pressure, same area same force even though more water in B! 33 B A A

Physics 101: Lecture 17, Pg 10 p 2 = p 1 +  gh P atm - 0 =  gh Measure h, determine p atm example--Mercury  = 13,600 kg/m 3 p atm = 1.05 x 10 5 Pa  h = 0.757 m = 757 mm = 29.80” (for 1 atm) h p 2 =p atm p 1 =0 35 Pressure and Depth Barometer: a way to measure atmospheric pressure For non-moving fluids, pressure depends only on depth.

Physics 101: Lecture 17, Pg 11 Preflight 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? 60% 1. No 40% 2. Yes 40 h p a p=0 What a cool situation-it must be possible. I mean we don't know how really tall trees get water all the way up to their highest leaves, but they obviously do, so this must also be possible. Even if a person could remove all the air from the straw, the height to which the outside air pressure moves the water up the straw would not be high enough for the person to drink the water.

Physics 101: Lecture 17, Pg 12 Archimedes’ Principle l Determine force of fluid on immersed cube è Draw FBD »F B = F 2 – F 1 » = P 2 A – P 1 A » = (P 2 – P 1 )A » =  g d A » =  g V l Buoyant force is weight of displaced fluid! 43

Physics 101: Lecture 17, Pg 13 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? 49 mg FbFb Mg  F = m a F b – Mg – mg = 0  g V disp = (M+m) g V disp = (M+m) /  h A = (M+m) /  h = (M + m)/ (  A) = (51.2+9)/(1 x 4 x 4) = 3.76 cm M =  plastic V cube = 4x4x4x0.8 = 51.2 g h

Physics 101: Lecture 17, Pg 14Summary l Pressure is force exerted by molecules “bouncing” off container P = F/A l Gravity/weight affects pressure  P = P 0 +  gd Buoyant force is “weight” of displaced fluid. F =  g V

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