Fluids at Rest: Pressure
• Pressure is the Force per unit Area:
ΔF
P = lim
Δ → 0 ΔA
F
P=
A
(definition of pressure)
(uniform force on flat area)
1 Pa = 1 N/m2
• Standard Atmospheric Pressure:
1 atm = 1.013× 105 N/m2 = 1.013×105 Pa = 14.7 lb/in2
• Gauge Pressure: Equal to the absolute pressure minus 1 atm.
• Pressure in a Fluid at Rest (density ρ):
∑ Fy = PA − ( P + ΔP) A + ρgAΔy = 0
P0
Area A
PA
ΔP = ρgΔy
Δy
P( y ) − P0 = ρgy
P( y ) = P0 + ρgy (pressure at depth y)
R. Field 10/24/2013
University of Florida
y=0
PHY 2053
ρ
(P+ΔP)A
Mg
y-axis
Page 1
Pascal’s and Archimedes’ Principles
P0= F/A
• Pascal’s Principle:
piston
Pressure applied to an enclosed fluid
is transmitted undiminished to all
parts of the fluid and to the walls of
the container.
y=0
P ( y ) = P0 + ρgy
ρ
y-axis
• Archimedes’ Principle:
FB = Mf g
Any body, wholly or partially
submerged in a fluid, is buoyed up by
a force equal to the weight, Mfg, of the
displaced fluid.
ρ
R. Field 10/24/2013
University of Florida
PHY 2053
Mf g
ρ
Page 2
Fluids at Rest: Example
• A hollow spherical iron shell floats almost completely submerged
in water (i.e. just touching the surface). If the density of iron is a
factor f greater than the density of water (i.e. ρiron = f × ρwater) and if
the outer radius of the iron shell is R2, what is the inner radius R1?
FB − M iron g = 0
M water g − 43 π ( R23 − R13 ) ρ iron g = 0
3
3
3
4
4
R
g
R
R
π
ρ
−
π
(
−
2 water
2
1 ) ρ iron g = 0
3
3
FB
R13 = R23 ( ρ iron − ρ water ) / ρ iron
R13 = R23 ( f − 1) / f
(
R. Field 10/24/2013
University of Florida
1
f
)R
R1
ρwater
R23 ( ρ water − ρ iron ) + R13 ρ iron = 0
R1 = 1 −
R2
R2
R1
y-axis
1
3
ρwater
2
PHY 2053
Mg
Page 3
Fluids at Rest: Examples
• The U-tube shown in the figure contains a liguid of
density ρ. What is the pressure difference ΔP = PA - PB?
B
h
A
PA = PB + ρgh
PA = PB + ρgh
ΔP = PA − PB = ρgh
Manometer
• The U-tube shown in the figure contains two liquids in
d
static equilibrium: Water of density ρW is in the right arm,
and an oil of unknown density ρX is in the left arm. What is
the density of the unknown oil in terms of d and L?
PB = P0 + ρ x g ( L + d )
ρX =
R. Field 10/24/2013
University of Florida
PB = P0 + ρW gL
L
B
B
L
ρW
L+d
PHY 2053
Page 4
Exam 2 Fall 2011: Problem 36
• Two small balls are simultaneously released from rest in a deep pool
of water (with density ρwater). The first ball is reseased from rest at the
surface of the pool and has a density three times the density of water
(i.e. ρ1 = 3ρwater). A second ball of unknown density is released from
rest at the bottom of the pool. If it takes the second ball the same
amount of time to reach the surface of the pool as it takes for the first
ball to reach the bottom of the pool, what is the density of the second
F
ball? Neglect hydrodynamic drag forces.
d = 12 a y t 2
ρ V g − ρ waterVball g = ρ downVball adown
Mg
Answer: 3ρwater/5 down ball
2d
⎛ ρ
⎞
− ρ water
ρ
% Right: 22%
d water
adown = down
g = ⎜⎜1 − water ⎟⎟ g
t=
y-axis
ρ down
ay
⎝ ρ down ⎠
⎛ ρ water ⎞
⎛ ρ
⎞
− 1⎟ g = adown = ⎜⎜1 − water ⎟⎟ g
aup = ⎜
⎜ ρ
⎟
water
⎝ ρ down ⎠
⎝ up
⎠
ρ waterVball g − ρ upVball g = ρ upVball aup d
y-axis
ρ water
ρ water 3
F
ρ up =
=
= 5 ρ water
⎛ ρ water
⎞
ρ water − ρ up
1
2 − ρ water / ρ down 2 − 3
− 1⎟ g
aup =
g =⎜
⎜
⎟
ρ up
Mg
⎝ ρ up
⎠
buoyancy
buoyancy
R. Field 10/24/2013
University of Florida
PHY 2053
Page 5
Exam 2 Spring 12: Problem 46
• What is the minimum radius (in m) that a
spherical helium balloon must have in order to
lift a total mass of m = 10-kg (including the mass
of the empty balloon) off the ground? The
density of helium and the air are, ρHE = 0.18 kg/m3
and ρair = 1.2 kg/m3, respectively.
Answer: 1.33
% Right: 37%
r
y-axis
m
Fbuoyancy − ( M HE + m) g = 0
(MHE +m)g
ρ airVballon g − ( ρ HEVballon + m) g = 0
Vballon = 43 πr 3 =
m
ρ air − ρ HE
1
3
1
3
⎛
⎞ ⎛
⎞
3m
3(10kg )
3
⎟⎟ = ⎜⎜
⎟
r = ⎜⎜
=
2
.
34
m
3 ⎟
⎝ 4π ( ρ air − ρ HE ) ⎠ ⎝ 4π (1.02kg / m ) ⎠
R. Field 10/24/2013
University of Florida
Fbuoyancy
PHY 2053
(
)
1
3
≈ 1.33m
Page 6