Pressure and Density
Fluid Pressure
Archimedes Principle
Buoyant Force
Pascal’s Principle
Table 4-4, p. 145
Pressure in Fluids
Fluid Pressure
Consider a column of a fluid. What is the
pressure at the bottom of the column?
Fluid: a matter which can flow (gases or liquids).
Pressure is the same on every
direction in a fluid at a given
depth.
Lab 5. Pressure & density.
1. The weight of a fluid:
Examples of fluids: air, water, alcohol,
mercury.
weight = FG = mg
m = DV
weight = FG = DVg
h
m - mass, D - density, V - volume, g - acceleration due to gravity
1. Fluids may have different density.
2. Fluids exerts pressure in all directions.
2. The pressure created by a fluid at the bottom
of the column:
P=
3. The force due to fluid pressure always
acts perpendicular to any surface it is
in contact with.
FG DVg D (abh)g
=
=
= D hg
A
A
ab
V = abh
P = Pgauge = D hg
P - pressure, D - density, h - height, g - acceleration due to gravity
Note that the fluid is not in motion!
b
a
Gauge
Pgauge is so-called “gauge” pressure.
How we can calculate pressure producing by any fluid?
Gauge pressure depends on density of fluid and depth!
3
Fig. 4-26, p. 149
4
Table 4-3, p. 139
Measurement of Atmospheric Pressure: Mercury Barometer
Atmospheric Pressure
The glass tube is completely filled with
mercury and then inverted into the bowl of
mercury.
A column of mercury 76 cm = 760 mm exerts
the same pressure as the atmosphere:
P= Dgh
Air
pressure
kg
Density of Hg : D = 13 600 3
m
Why use such a dense liquid?
Atmospheric pressure is defined as:
1 atm = 76 cm Hg = 1.01 105 Pa
How high would water rise in a closed
inverted column, at 1 atm?
P
h=
Dg
P = Dg h = (13.6 !103 kg / m3 )(9.8 m / s 2 )(0.76 m) = 1.013 !105 N / m 2 = 1 atm
h = 10.34 m water = 1 atm
The pressure at the base of the water column,
!gh, is 1 atm:
1.01 105 Pa = (103kg/m3)(9.8m/s2)(h);
thus h=10.3 m
8
7
A balloon that was partially inflated near
the sea level expanded as the experimenters climbed the mountain.
9
Fig. 4-27, p. 150
Gauge Pressure
We have a narrow pipe with water of 20 m height and water tower with 20 m of
water height (see below). Pressure gauge placed at the bottom shows:
water
1) Larger pressure for the water tower
2) The same pressure
3) Larger pressure for the pipe with water
4) Depends on the radius of the pipe
pipe
20 m
water
tower
Gauge pressure:
Pgauge = !hg
Pressure gauge
Fig. 4-28, p. 151
11
Archimedes Principle and Buoyancy
Archimides’ principle and Buoyant force.
Some objects placed in fluid can float on the surface or at some depth.
(FB=!Vg, Buoyant force=Weight of fluid displaced)
The reason: buoyant force
FB acts on them.
Fig. 9.15
Static equilibrium:
FB = weight = FG = mO g
P1
FB=(P2-P1)A = !fg(h2-h1)A = !fgV
P2
!f is the density of the fluid
The pressure acting on a bottom of the suspended metal block
is greater than that on the top due to increase of pressure with depth 12
1. The buoyant force on the object floating in a fluid is equal to the
weight of the object.
2. The buoyant force on the object immersed in a fluid is equal to the
weight of the submerged fluid.
13
Buoyant force = Weight of water
Buoyant Force
A
!A
Fbouyant
Fbouyant
B
!B
Wobject
! A < ! B < !C
C
Wobject
Fbouyant
!C
Water
Wobject
Fbouyant = Wobject
Wobject = mobject g = ( !objectVobject )g
Fbouyant = Wdisplaced
water
= mdisplaced
water
Fbouyant < Wobject
g=
= ( !waterVdisplaced water )g = ( !waterVsubmerged
object
)g
14
Fig. 4-35, p. 155
BUOYANT FORCE, ARCHIMEDES PRINCIPLE
Fig. 4-32, p. 153
Buoyant Force
Will an ice cube float higher in water or in alcohol?
Problem 1:
Density of water is approx 1g/ml
Density of ice is approx 0.93g/ml
Density of 100% pure ethanol is 0.79g/ml
Weight of ice = W = mg = !Vg
A 1.5-kg block of wood is floating in water. What is the magnitude of the
buoyant force acting on the block?
FB = weight of the object = mO g = ( 1.5 kg)( 9.8
Problem 2:
W = (0.93g / cm 3 )(1cm 3 )10m / s 2 ) = 9.3N
A floating boat displaces 3 m3 of water (Dw = 1000 kg/m3).
FBouyant = !FluidVg
Water :
1. What is the mass of the water displaced by the boat?
FBouyant = !FluidVg = (1g / cm )(1cm )(10m / s ) = 10N
3
3
2
FBouyant > W
Dw =
Ice cubes floats in water
Alcohol :
FBouyant = !FluidVg = (0.79g / cm )(1cm )(10m / s ) = 7.9N
3
m
) = 14.7 N
ss
3
2
mw
Vw
mw = DwVw = ( 1000
kg
)( 3 m 3 ) = 3000 kg
m3
2. What is the buoyant force acting on the boat?
FB = weight of the displaced water = mw g = ( 3000 kg)( 9.8
FBouyant < W
3. What is the weight of the boat?
Thus ice cubes would not float at all in 100% pure ethanol.
m
) = 29 400 N
ss
weight = 29 400 N
15
Fig. 4-40, p. 160
Fig. 4-38a, p. 158
Zeppelin Hindenburg: Buoyant Force in Air
Pascal’s Principle
Launched in 1936, crashed
in 1937 in Lakehurst, NJ
The zeppelin LZ-129 Hindenburg was one
of the largest aircraft ever built: 245 m
long, 41 m in diameter, 211 890 m3 of gas.
Designed to use helium, forced to use
hydrogen due to US military embargo.
P=
Pressure in fluids:
F
A
(pressure =
force
)
area
In a fluid, the applied force creates a pressure that is transmitted everywhere
throughout the fluid.
Pascal’s principle:
pressure applied to an enclosed fluid is transmitted undiminished to all
parts of the fluid and to the walls of the container.
LZ-129
force
Boeing 747
piston
fluid
The buoyant force occurs due to difference in density of air and helium or hydrogen
Air : ! air = 1.2 kg/m Helium : ! He = 0.18 kg/m
3
3
Hydrogen : ! H 2 = 0.089 kg/m
3
cylinder
Pascal’s principle is widely
applied in hydraulic systems.
Can carry : ( " air ! " H )Vg = [(1.2 ! 0.089) kg/m ](211 890 m 3 )(9.8 m / s 2 ) = 2 307 015 N
2
16
17
Use Pascal’s Principle in Hydraulic Systems
Hydraulic systems consists of two or more pistons:
Small piston
Large piston
Fs
Al
As
Ps =
Fl
Al = !r 2 l
Fs
As
Pl =
Fl =
Ps = Pl
A = ! r2
Fs Fl
=
As Al
Fs Al
As
Fs =
Fout =
Fl
Al
Fl As
Al
Fin Aout
Ain
18