Pressure VS Depth For a fluid in an open container pressure same at a given depth independent of the container p(y Fluid level is the same everywhere in a connected container, assuming no surface fo rces Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium Imagine a tube that would connect two regions at the same depth If the pressures were different, fluid would flow in the tube However, if fluid did flow, then the system was Not in equilibrium since no equilibrium system will spontaneously leave equilibrium Physics 121: Lecture 20, Pg 6
Physics 121: Lecture 20, Pg 6 If the pressures were different, fluid would flow in the tube! However, if fluid did flow, then the system was NOT in equilibrium since no equilibrium system will spontaneously leave equilibrium. Pressure vs. Depth For a fluid in an open container pressure same at a given depth independent of the container p(y) y Fluid level is the same everywhere in a connected container, assuming no surface forces Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium? Imagine a tube that would connect two regions at the same depth
Lecture 20. ACT 1 Pressure What happens with two fluids? Consider a U tube containing liquids of density p, and p, as shown Compare the densities of the liquids A)P1< p2 B)P1=p2 C)Pl Physics 121: Lecture 20, Pg 7
Physics 121: Lecture 20, Pg 7 Lecture 20, ACT 1 Pressure What happens with two fluids?? Consider a U tube containing liquids of density 1 and 2 as shown: Compare the densities of the liquids: A) 1 < 2 B) 1 = 2 C) 1 > 2 1 2 dI
Pascals Principle So far we have discovered (using Newtons Laws) Pressure depends on depth:△p=pg△y Pascal's Principle addresses how a change in pressure is transmitted through a fluid Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel Pascal's Principle explains the working of hydraulic lifts e. the application of a small force at one place can result in the creation of a large force in another Does this"hydraulic lever"violate conservation of energy? ) Certainly hope not. Let's calculate Physics 121: Lecture 20, Pg 8
Physics 121: Lecture 20, Pg 8 Pascal’s Principle So far we have discovered (using Newton’s Laws): Pressure depends on depth: p = gy Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. Pascal’s Principle explains the working of hydraulic lifts i.e. the application of a small force at one place can result in the creation of a large force in another. Does this “hydraulic lever” violate conservation of energy? »Certainly hope not.. Let’s calculate
Pascal's Principle Consider the system shown a downward force F1 is applied to the piston of area A1 This force is transmitted through d he liquid to create an upward force f Pascals principle says that increased pressure from F Anout A A2 F,Al is transmitted through the liquid F A A F2>F1: Have we violated conservation of energy?? Physics 121: Lecture 20, Pg 9
Physics 121: Lecture 20, Pg 9 Pascal’s Principle Consider the system shown: A downward force F1 is applied to the piston of area A1 . This force is transmitted through the liquid to create an upward force F2 . Pascal’s Principle says that increased pressure from F1 (F1 /A1 ) is transmitted throughout the liquid. F F 1 2 d 2 d 1 A1 A2 2 2 1 1 A F A F = 1 2 2 1 A A F = F F2 > F1 : Have we violated conservation of energy??
Pascals Principle F Consider F1 moving through a distance d1 How large is the volume of the ↑d liquid displaced? △V=dA This volume determines the displacement of the large piston A △V2=△Vm A A 2=F2d2= Therefore the work done by F1 equals the work done by We have NoT obtained"something for nothing Physics 121: Lecture 20, Pg 10
Physics 121: Lecture 20, Pg 10 Pascal’s Principle Consider F1 moving through a distance d1 . How large is the volume of the liquid displaced? F F 1 2 d 2 d 1 A1 A2 V1 = d1A1 V2 = V1 2 1 2 1 A A d = d Therefore the work done by F1 equals the work done by F2 We have NOT obtained “something for nothing”. 1 2 1 1 1 2 2 2 2 1 W A A d A A W = F d = F = This volume determines the displacement of the large piston