Submitted by: Submitted by bmcarter92
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Category: Science and Technology
Date Submitted: 10/08/2015 08:44 PM
Fluid Mechanics
Density (p):
p= M/V
SI: kg/m^3
V=AΔX
Specific Gravity: ratio density to water at 4 degrees Celsius
pwater= 1000 kg/m^3
sp. gr. = psubstance/ pwater
Pressure (P):
P= F/A
SI: Pascals or Pa (1N/1m2)
Newton’s Third Law connection = distributing the force over a greater surface
Water Bed’s have the density of water! (Disregard other materials)
Po = atmospheric pressure, @ sea level = 101,325 Pa
1 atm = 101,325 Pa
1 atm = 14.7 lb/in2
1 atm = 760 mmHG or torr
1 torr or mmHG = 133.32 Pa
1 bar = 1,000,000 Pa
The force exerted by a fluid on an object is always perpendicular no matter which angle
Fg = mg= pVg = p(lwh)g (from density equation)
Pliquid = F/A = F(g of liquid) / A = p(lwh)g/lw = pgh
Pliquid = pgh
Static Equilibrium:
Fluid at Rest (all points at the same depth must be at the same pressure)
When an object is suspended in a liquid there is an upward force from the liquid at the bottom and a downward force that is equal and opposite from BOTH the force of gravity of the substance and the liquid on the top
Σ F = F2 – (F1 + Fg ) = 0
P2A – (P1A + mg) = 0
*M= pV=pA(Y1 – Y2)*
P2 = P1 + pgΔY
* water level = 0, meaning that
the depth below that is going to
be negative. (-Y1 – (- Y2 )) or
simply (-Y1 +Y2 )
Hydrostatic Pressure:
Ptotal = Patm+ Pliquid
Pliquid = pgh
Patm = Po
Ptotal = Po + pgh
Pguage= Ptotal – Po
Pguage = (Po + pgh) – Po
Pguage = pgh
*when there is a lid Po=0 Pa, atmospheric pressure is negligible*
Pascal’s Principle:
A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container
P1 = P2
F1/ A1= F2/ A2 (F1< F2 but A1> A2)
F1= (A1/ A2)( F2)
F1/ F2= A1/ A2
Application: Hydraulic Press/Brakes ~ greater Δx over a smaller A1 from a small F1 means less Δx over a greater A2 with a larger F2
Archimedes’s Principle:
Any object completely or partially submerged in a fluid is buoyed...