Thur Nov 5
• Assign 11 Friday
• Exam 2
– Mon Nov 16 - 7:15PM
– Morton 235
– Cumulative with emphasis on
new material (Chapter 5 on)
– One page 8.5x11 note
– Contact about class conflicts
Pressure
N/m2
or Pascals (Pa)
1 atm = 14.7 lb/in2 = 1.01x105 Pa
Absolute pressure – not < zero
Gauge pressure – relative to
atmospheric pressure
Fluids
• Pressure – columns of fluid
• Buoyancy
• Fluids in Motion
• Continuity
• Bernoulli's Equation
M
ρ=
V
P = F/A
P = Po + ρ gh
ρ1 A1 v1 = ρ2 A2 v2
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
Measuring Pressure Using a Column of Fluid
Height of column and density of fluid
determine pressure difference.
Manometer
Po
P
h
• Absolute Pressure, P
• Gauge Pressure = P – Po
Barometer
P=0
• Po = Atmospheric Pressure
= 1.01 x
105
Pa
h
P = Po + ρ gh
Pressure sometimes given in inches of Mercury (Hg)
30.2”
Po
Barometer
! Sect 11.4 – First paragraph
• P2 = P1 + ρgh
• Choose point 1 inside barometer (higher) and
point 2 at lower surface.
• PATM = 0 + ρgh
• Historically used Mercury (Hg) – high
density (13600 kg/m3)
• If know height of mercury and density,
then know atmospheric pressure
• PAtm 76cm or 29.9 in of Hg
• Use ρHg ghHg
As you travel up a mountain, your ears pop. The figure below is a
pitiful attempt to show the ear drum (in red). As you travel up a
mountain, what will the ear drum do between 'pops'?
(1) Bow towards the middle ear
(2) Hold its shape
(3) Bow towards the outside
Pressure outside decreases.
Any time an object is submerged either partially or completely in a
fluid, there is a buoyant force. Suppose I hang an object which has a
weight of 7N submerged in a fluid and find that the force I need to
support it is 5N. What is the buoyant force?
(1) 1 N downward
(4) 2 N upward
(7) 12 N downward
(2) 1 N upward
(5) 5 N downward
(8) 12 N upward
(3) 2 N downward
(6) 5 N upward
5N
Just another force
Σ FY = maY
5N – 7N + FB = 0
FB = 7N – 5N
FB
7N
Given three cylinders, rank the buoyant force when completely
submerged in water.
1.
2.
3.
4.
5.
6.
7.
BA < BB < BC
BA = BB < BC
BC < BB < BA
BA < BC < BB
BC < BB = BA
BA < BB = BC
BC = BB < BA
Weight
A
2N
Volume
VA
B
7N
VB =VA
C
7N
VC > VA
Buoyant force depends on weight of
fluid displaced. The greater the volume
of the submerged object, the greater the
buoyant force.
The weight of the object does not affect
the buoyant force.
Archimede's Principle - Buoyancy
• Difference in pressure
• Any time in fluid, buoyant force
• upwards
• What determines force?
• FB = weight of fluid displaced
= ρFLUID VFLUID DISPLACED g
Buoyant force is just another force that we can add
to our collection.
Floating versus Completely Submerged
FB
Floating:
FB = FG
VDISPLACED< VOBJECT
ρOBJ < ρFLUID
FG
FB = FG
ρFLUID VDISPLACED g = ρOBJ VOBJECT g
Completely Submerged
VDISPLACED=VOBJECT
FB = ρ VOBJECT g
ρOBJ > ρFLUID
Example: If ice has a density of 920 kg/m3, what percentage of an
iceberg sticks out above the waterline?
Percentage by volume. Can find volume sticking out of water and
divide by total volume. Actually easier to find fraction submerged
and subtract from 1.
FBUOYANT = Weight
weight of water displaced = weight of iceberg
ρw VDISPLACED g= ρICE VICE g
ρw VSUBMERGED g= ρICE VICE g
VSUB / VOBJ = ρICE / ρw
VSUB / VOBJ = 0.92
So fraction not submerged is 1-0.92 or 0.08, 8%
Object A is submerged in water. It would like to float at the surface,
but is tied to the bottom by a string. Object B has the same mass as
Object A, but has twice the volume. In which case is the tension in the
string the greater?
(1) A
(2) B
B
A
(3) Both same
FB
T
Fg
ΣF = 0
FB – T – Fg = 0
T = FB – F g
Same weight but B displaces larger
volume, therefore larger buoyant
force and larger tension
A beaker is filled with water up to the level of the spigot. A block of
wood weighing 2N is placed in the beaker and floats in the fluid. The
weight of the water which spills out the spigot is
(1) less than 2N
(2) 2N
(3) greater than 2N
Floating in equilibrium: Weight of object = Buoyant Force
Buoyant Force = Weight of Fluid Displaced
So Weight of Object = Weight of Fluid Displace
Things to Consider Memorizing
• Prefixes: mega, kilo, centi, milli,
micro
velocity
acceleration
sin, cos, tan
F=ma
Fg = mg (at surface of Earth)
FFRICTION = µ FN
aC = v2/r
KE = (1/2)mv2
PE = mgh
Total Mech Energy = KE + PE
p = mv
impulse = FΔt
work = F s cosθ
density = M/V
pressure = F/A
Fluids in Motion
• Ideal Fluid:
– incompressible: density constant
– nonviscous: no friction between layers
Flow:
• Steady: v doesn't change at point
• Unsteady: v changes magnitude
• Turbulent: v erratic
http://www.ceb.cam.ac.uk/data/images/
groups/CREST/Teaching/hydro/
iain1.gif
Continuity
m = ρV
m = ρ (A ·length)
m = ρ (A vΔt)
m/Δt = ρ A v = mass flow rate
ρ1 A1 v1 = ρ2 A2 v2
Same amount of stuff
flows past each point per
second
If incompressible: (ρ1= ρ2)
Q = A1 v1 = A2 v2
Q: Volume flow rate
If size of pipe changes, continuity determines the velocity
Continuity
If split, can still use continuity arguments
Mass flow past A is same as mass flow past combination of
B, C and D.
Example: Arteries and Capillaries
B
A
C
D
An incompressible fluid is flowing through a pipe. At which point is
the fluid traveling the fastest?
(6) All points have the same speed
Smallest cross-sectional area, highest speed
An incompressible fluid is flowing through a pipe. At which point is
the fluid traveling the slowest?
(6) All points have the same speed
Largest cross-sectional area, slowest speed
which one of these water tap photos is not real?
1
2
3
as water falls down, its speed is increasing,
the cross-sectional area is decreasing!
4
Fountain
water speed is decreasing, the cross-sectional area is increasing!
Bernoulli's Equation – Conservation of Energy
One way to look at it:
WNC =
ΔKE
F*d
Divide by volume: ΔPressure
(1/2)mv2
(1/2)ρv2
+
ΔPE
mgh
ρgh
Careful! In many cases continuity determines speed
For two points in incompressible, nonviscous fluid with steady flow:
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
If height not changing much, higher speed leads to lower
pressure
An incompressible fluid flows through a pipe. Compare the speed of
the fluid at points 1 and 2.
(1) Greater at 1
(2) Greater at 2
(3) Both the same
(4) Not enough information
Area is smaller at 2, therefore fluid must travel faster in order
to conserve the volume flow rate.
An incompressible fluid flows through a pipe. Compare the pressure
at points 1 and 2.
(1) Greater at 1
(2) Greater at 2
(3) Both the same
(4) Not enough information
P + (1/2) ρv2 + ρgh = constant
(1/2) ρv2 is larger (higher v due to smaller A)
ρgh is larger (higher)
Therefore pressure must decrease at point 2
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
An incompressible fluid flows through a pipe. At two different points,
small tubes are attached above the pipe to allow for the observation of
the pressure in the pipe.
(1) h2 < h1
?
?
(2) h2 = h1
h2
h1
(3) h > h
2
1
reference dot line: Height not changing
continuity: Larger area, lower speed (v2 < v1)
Bernoulli's Equation: lower speed, greater pressure (P2 > P1)
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
Greater pressure supports greater column of fluid (as measured from
the middle of the pipe)
Example: Lift force of an airplane wing (Video)
Calculate the lift force of an airplane wing with a surface area of
12.0m2. Assume the air above the wing is traveling at 70.0 m/s and
the air below the wing is traveling at 60.0m/s. Assume the density
of the air to be constant at 1.29 kg/m3.
FLIFT = FUP - FDOWN
= PBELOW A – PABOVE A
= (ΔP) A
From Bernoulli, find (ΔP) assuming Δy negligible.
PA + (1/2)ρvA2 + ρgyA = PB + (1/2) ρvB2 + ρgyB
(ΔP) = PB – PA = ((1/2)ρvA2 – (1/2)ρvB2) + (ρgyA – ρgyB)
(assume ΔPE negligible – if yA-yB = 10cm, only 1.26 Pa)
ΔP = (1/2) (1.29)(702 – 602 ) = 838.5 Pa
FLIFT = (838.5Pa)(12m2) = 1.01 x104 N
I will attempt to levitate a beach ball using an air blower. Under
which scenarios will the beach ball levitate in a stable state (it may
bounce around a little, but it won't fall).
(1) A
(6) A and B
(2) B
(7) B and C
(3) C
(8) A and C
(4) None
(5) All
Gravity has to bring the ball back closer to the nozzle to keep the
speed of the air high enough.
Consider a small, horizontal artery in which there is a constriction due
to placque. This constriction reduces the cross sectional area of the
artery. The pressure in the constricted region is _____ the pressure in
the unconstricted region.
1. greater than
2. less than
3. the same as
Continuity: greater speed in constriction
Bernoulli: if change in height negligible, greater speed, lower
pressure
Vascular flutter: If pressure too low, can close, then open,
close, then open, ….
Aneurism: weak walls, area expands, higher pressure, expands,
higher pressure, expands, higher pressure, expands, ….
The Water Tank
Water is leaving a tank at a speed of 3.0 m/s.
The tank is open to air on the top. What is
the height of the water level above the
spigot?
Assume the area of the tank is much larger
than the area of the spigot.
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
P0 + (1/2)ρv12 + 0 = P0 + (1/2)ρv22 + ρgh
Continuity: Since A2 >> A1, v2 << v1
(1/2)ρv12 = (essentially 0) + ρgh
h = 0.459m
In an action movie, the heroine is supposed
to jump with a motorcycle from the roof of
one skyscraper to another one. The roof of
the second skyscraper is h=8 meters lower,
and the gap between the buildings is d=15
meters wide. Neglecting air resistance, what
minimum initial speed would she needs to
make the jump?
Initial: Leave roof
Final: Just before hit roof
x = 15m
vx0 = v0 (find)
ax = 0
m/s2
t=?
y= 8m
vy0 = 0 m/s
vy = ?
ay = 9.8 m/s2 down
= - 9.8 m/s2
1 2
x = v x 0t + a x t
2
1 2
y = v y 0t + a y t
2
v0
+y
+x
Curve Ball