This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: F P A P gh liq disp B V g m V Av const. 2 1 2 P v const. 130  density = M kg V m 3 Pressure: P = = F force A area units: = Pa N m 2 or or lb in 2 lb ft 2 Atmospheric pressure 1 Atm = 14.7 1 Atm = 1.01 x 10 5 Pa lb in 2 131 Fluid  Statics 131a sample P Area, A 2 F 2 P A 2 F 1 PA 1 Pressure is transmitted through fluid without loss F 1 (Total) = F 1 P A 1 F 2 (Total) = F 2 P A 2 Piston at rest : 0 = F 1 PA 1 0 = F 2 P A 2 F 1 F 2 A 1 A 2 = P = F 2 = F 1 A 2 A 1 Pascal’s Principle 132 F s F L A s A L = P = F s = ( ) F s A L A s This factor can be made to be 1,000 or more.  Hydraulic Press  Hydraulic Jack Applications of Pascal’s Principle 133 P ATM h A Pressure Force per unit Area Any arbitrary cylinder V=Ah Pressure in a Liquid (or Fluid) vacuum F t =0 mg V=Ah mg= (Ah) g F h = P h A V F(total) = F tmg+ F h top 0 = 0 Ahg + P h A P h = hg vac fluid at rest Pressure due to weight of water on top If P o added at top (instead of vacuum) it is transmitted to bottom. P h = P o + hg 134 135 Note: h=0 at surface and all h’s are + #’s 135 Torricelli Barometer 136 A barometer compares the pressure due to the atmosphere to the pressure due to a column of fluid, typically mercury. The mercury column has a vacuum above it, so the only pressure is due to the mercury itself....
View
Full
Document
This note was uploaded on 04/03/2012 for the course PHY 1020 taught by Professor Kodera during the Spring '11 term at University of Florida.
 Spring '11
 KODERA
 Force

Click to edit the document details