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Physics 203 – College Physics I
Department of Physics – The Citadel
Physics 203
College Physics I
Fall 2012
S. A. Yost
Chapter 10
Part 2
Pascal’s Principle, Archimedes’
Principle, Buoyancy
Physics 203 – College Physics I
Department of Physics – The Citadel
Announcements
Thursday: exam on chapters 7 – 9 (only the sections
covered in the homework)
Today: Problem set 10A was due.
(You remembered, right?)
Problem set 10B is scheduled for the Tuesday after
Thanksgiving, but this depends on what we
discuss in today’s class. Watch the due date for
possible changes.
Physics 203 – College Physics I
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Atmospheric Pressure
The weight of the air above us produces
atmospheric pressure at sea level equal to
1 atm = 1.013 × 105 N/m2.
Pressure is also measured in Pascals:
1 Pa = 1 N/m2.
Physics 203 – College Physics I
Department of Physics – The Citadel
Gauge Pressure
Pressure gauges are normally set to zero
when only atmospheric pressure is present.
Gauge pressure is then the additional
pressure beyond that due to the
atmosphere.
The total pressure including atmospheric
pressure is the absolute pressure.
Physics 203 – College Physics I
Department of Physics – The Citadel
Suction
Pump
A negative gauge pressure
corresponds to suction. If we
produce a negative gauge
pressure on a straw, water
will be “sucked” up the straw.
P < Patm
h
Why does the water go up
straw?
the
Physics 203 – College Physics I
Department of Physics – The Citadel
Suction
Pump
What is the highest a suction
pump can draw water up a
tube?
The absolute pressure can’t be
less than zero.
Assume the pressure in the
tube is reduced to P = 0.
P < Patm
h
Physics 203 – College Physics I
Department of Physics – The Citadel
Suction
The water must be supported by
atmospheric pressure.
Pump
P=0
rgh = 1 atm = 1.013 × 105 N/m2
h=
1.013 × 105 N/m2
(1000 kg/m3)(9.8 m/s2)
= 10.3 m.
h
P = rgh
Physics 203 – College Physics I
Department of Physics – The Citadel
Hydraulic Lift
A hydraulic lift is a simple machine which
uses the fact that any fluid pushed into a
cylinder on one end
must push out the
same volume into a
cylinder at the other
end.
V
L1
V
L2
Physics 203 – College Physics I
Department of Physics – The Citadel
Hydraulic Lift
Since the volumes are the same, the
distances the cylinders move is related to
their areas:
A1 L1 = V = A2 L2.
→
L2/L1 = A1/A2.
A1
V
L1
A2
V
L2
Physics 203 – College Physics I
Department of Physics – The Citadel
Hydraulic Lift
The work done by a force on
one cylinder will equal the
work done by the other
cylinder.
F2
V
F1 L1 = W = F2 L2.
V
F1
L1
L2
Physics 203 – College Physics I
Department of Physics – The Citadel
Hydraulic Lift
Mechanical advantage
F2
F2 / F1 = L1 / L2
= A2 / A1.
A1
V
F1
L1
A2
V
L2
Physics 203 – College Physics I
Department of Physics – The Citadel
Pascal’s Principle
Know: F2 / F1 = A2 / A1
F2
Rewrite:
A2
F2/A2 = F1/A1
→
V
DP2 = DP1.
A1
F1
V
Physics 203 – College Physics I
Department of Physics – The Citadel
Pascal’s Principle
When an force is applied to
a closed vessel, the
pressure increases by the
same amount throughout A2
the vessel.
F2
DP
A1
F1
DP = F1/A1 = F2/A2
Physics 203 – College Physics I
Department of Physics – The Citadel
Pascal’s Demonstration
d
Pascal demonstrated his principle
by inserting a thin 12 m long tube
of diameter 6 mm into a wine
barrel of diameter 40 cm.
h
Filling the tube with water caused
the barrel to burst!
(a) What was the weight of water in
the tube?
(b) What was the force on the lid?
D
Physics 203 – College Physics I
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Pascal’s Demonstration
d = 0.006 m
What was the weight of the
water in the tube?
h = 12 m
m = r (pr2) h
r = 0.003 m
= 1000 kg/m3
× (2.8 × 10-5 m2) × 12 m
= 340 g.
mg = 0.340 kg × 9.8 m/s2
= 3.33 N
(0.75 lb)
D
Physics 203 – College Physics I
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Pascal’s Demonstration
d
h
What was the force on the lid of
the barrel?
The force on the bottom of the
tube is
D
F1 = mg = 0.75 lb on area
A1 = pd2/4
(= 2.8 x 10-5 m2)
Physics 203 – College Physics I
Pascal’s Demonstration
Department of Physics – The Citadel
d
h
The force on the lid of the
barrel is
F2 = mg (A2/A1) with
A2 = pD2/4
Then F2 = mg (D/d)2
Area is proportional to the length squared.
D
Physics 203 – College Physics I
Department of Physics – The Citadel
Pascal’s Demonstration
d = 0.006 m
h
The force on the lid of the
barrel is
F2 = mg (D/d)2
= 0.75 lb (0.40 m / 0.006 m)2
= 0.75 lb × 4400
= 3300 lb
big number!
D = 0.4 m
Physics 203 – College Physics I
Department of Physics – The Citadel
Pressure in water
Balancing the net force of
pressure and the weight for an
imaginary box of water gives
F2 – F1 = mw g
where F1 = P1 A and F2 = P2 A.
F1
V
F2
submerged box
Physics 203 – College Physics I
Department of Physics – The Citadel
Buoyancy
Suppose I replace the imaginary
box of water by an actual box.
What is the net force due to
pressure on the box?
F2 – F1 = mw g , just as before!
This is the buoyant force.
Archimedes Principle:
The buoyant force equals the
weight of the water displaced.
F1
V
F2
submerged box
Physics 203 – College Physics I
Floating Example
A plastic block with specific
gravity 0.3 floats in water.
What fraction of the block’s
volume is under water?
The weight of the block equals
the buoyant force.
Department of Physics – The Citadel
Physics 203 – College Physics I
Department of Physics – The Citadel
Floating Example
V
If the volume of the block is V and
the volume under water is Vw,
we need to find Vw/V.
Vw
Use FB = rw Vwg
= mg = rVg.
Then rw Vw = rV, and
Vw/V = r/rw = SG = 0.3
30% of the
block is
under water.
Physics 203 – College Physics I
Department of Physics – The Citadel
Buoyancy Puzzle
Two identical cups have
water in them, to the
same level. But one also A:
has a plastic block
floating in it.
Which is heavier?
A
B
C they are the same
B:
Physics 203 – College Physics I
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Buoyancy Puzzle
A:
V
The total mass in case A is
MA = rV
What is the mass in case B?
Don’t guess – use physics.
B:
v1
v2
Physics 203 – College Physics I
Department of Physics – The Citadel
Buoyancy Puzzle
Remember Archimedes’
Principle:
A:
The weight of the plastic
block equals the weight
of the water that would
B:
have occupied the
submerged volume of the
block.
Physics 203 – College Physics I
Department of Physics – The Citadel
Buoyancy Puzzle
This means the mass of the A:
whole block equals the
mass of water that would
occupy volume v2.
Therefore the two cups
have the same total
mass.
B:
V
v1
v2
Physics 203 – College Physics I
Department of Physics – The Citadel
Buoyancy Puzzle
Another way to look at it:
The force on the bottom of
the cup is the weight of
the contents.
The force on the bottom of
the cup is also the
pressure times the area.
Both have the same depth,
so they have the same
pressure.
A:
V
B:
v1
v2
Physics 203 – College Physics I
Department of Physics – The Citadel
Volume Rate of Flow
The volume rate of flow (also called flux) is
defined to be the volume of fluid that flows
past a point in a pipe per unit time.
Q = DV / Dt = Av
where A is the area, v the flow velocity.
DV A
v
Physics 203 – College Physics I
Department of Physics – The Citadel
Volume Rate of Flow
The volume rate of flow is measured in m3/s.
Another common unit is liters/second (L/s)
1 liter = 1000 mL = 1000 cm3
1 m3 = 1000 L.
DV
Physics 203 – College Physics I
Department of Physics – The Citadel
Volume Rate of Flow
The volume rate of flow of an incompressible
fluid is the same throughout a pipe.
Q
DV
Physics 203 – College Physics I
Department of Physics – The Citadel
Volume Rate of Flow
Q = Av is fixed: the fluid’s velocity is inversely
proportional to the
crosssectional area.
v3
v1
v2
DV
DV
A
DV
Physics 203 – College Physics I
Department of Physics – The Citadel
P2
v2
Bernoulli Principle
Suppose a volume V of incompressible
fluid is pushed into a pipe with a
pressure difference and height
difference between the ends.
h
P1
v1
The same volume of fluid will be pushed
out the other end.
Physics 203 – College Physics I
Department of Physics – The Citadel
Bernoulli Principle
Pushing a volume V of fluid into a pipe under
pressure P does work on it:
W = F L = F V/A = PV.
F
L
P
V
A
Physics 203 – College Physics I
Department of Physics – The Citadel
Bernoulli’s Principle is an expression of
energy conservation:
P2
v2
Bernoulli Principle
PV + ½ m v2 + mgh = constant.
h
constant.
P1
+ potential energy =
v1
Work + kinetic energy
Physics 203 – College Physics I
Department of Physics – The Citadel
P2
v2
Bernoulli Principle
Dividing by volume,
P + ½ r v2 + rgh = constant.
h
Pressure + kinetic energy density
P1
constant.
v1
+ potential energy density =
Physics 203 – College Physics I
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Question
1. A fluid flows through the pipe shown. In which
section is the flow velocity the greatest?
Selections: A
A
B
B
C
D The same
C
Physics 203 – College Physics I
Department of Physics – The Citadel
Answer
The volume rate of flow Q = vA is constant for
an incompressible fluid. The fluid moves
fastest where the pipe is narrowest, section B.
(It moves slowest in section C.)
A
B
C
Physics 203 – College Physics I
Department of Physics – The Citadel
Question
2. In which section is the pressure of the fluid the
greatest?
Selections: A
A
B
B
C
D The same
C
Physics 203 – College Physics I
Department of Physics – The Citadel
Answer
This is Bernoulli’s principle: The pressure in a
fluid decreases when the flow velocity
increases. The fluid moves most slowly at C, so
the pressure is highest there (and lowest at B).
A
B
C
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
A water tower feeds a
fountain, which shoots
water straight up in the air.
h
How fast does the water
leave the fountain?
Assume the top of the water
is a height h = 55 m above
the fountain.
v
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
1
We’ll assume the tank is
big, so the top of the
water stays fixed:
h
h1 = 55 m,
v
v1 = 0,
P1 = 0
gauge
pressure
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
1
At the fountain,
h2 = 0
v2 = v
unknown
h
What is P2?
v
2
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
1
Careful!
This is not hydrostatics.
If the fountain were turned off,
the pressure would be
P2 = rgh = 1000 kg/m3
x 9.8 m/s2 x 55 m
= 5.4 x 105 N/m2.
h
no flow!
2
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
1
When the fountain is
flowing, this changes!
The pressure just outside
the pipe is P2 = 0,
normal atmospheric
pressure.
h
P2
v
2
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
1
The velocity is given by
Bernoulli’s equation with
P1 = P2 = 0
h1 = h, h2 = 0
v2 = 0, v1 = v
h
P2
v
2
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
1
The only terms remaining are
½ r v2 = rgh
h
The result is the same as if
the water had fallen from
the top of the tower:
v = √ 2gh = 33 m/s.
v
2
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
1
What is the volume rate of
flow if the pipe has
diameter 1 cm?
h
Q = Av
A = p (0.5 cm)2
= 0.785
cm2
v
2
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
1
Q = Av
A = 0.785 cm2
h
v = 33 m/s = 3300 cm/s
Q = 2600 cm3 /s
= 2600 mL /s = 2.6 L/s
v
2
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
How high does the water rise
from the fountain?
1
2
Bernoulli’s equation between
points 1 and 2:
P1 = P2 = 0,
v1 = v2 = 0
implies rgh1 = rgh2.
h
v
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
The water rises to the height
of the tower.
1
2
This assumes energy
conservation:
No friction (viscosity or air
resistance) or turbulence is
considered.
h
v
Physics 203 – College Physics I
Department of Physics – The Citadel
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
What is the pressure inside the pipe
feeding the fountain?
If there is no nozzle constricting the
flow, then v is the same in both
places, so the pressure must be
zero both inside and outside the
pipe.
v
2
1
v
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
Putting a nozzle on the hose would
change this.
Then v is bigger outside the hose,
and the pressure is higher inside
the hose. If the height is about the
same, then
v2
2
P1 + ½ rv12 = P2 + ½ rv22.
1
v1
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
The gauge pressure in the hose is
P1 - P2 = ½ r(v22 – v12).
v2
Continuity: A1 v1 = A2 v2.
P1 – P2 = ½ rv2
2 (1
-
2
A2
A22/A12)
A1
1
v1
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
What is the gauge pressure in the
pipe, assuming it’s diameter is still
1 cm at point 1, but the nozzle has
diameter 1/8 cm at point 2.
v2
2
P1 – P2 = ½ rv22 (1 - A22/A12)
A2/A1 =
(1/8)2
= 1/64
A2
A1
1
v1
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
P1 – P2 = ½ rv22 (1 - A22/A12)
= 0.5 (1000 kg/m3)(33 m/s)2 (1 – 1/4096)
= 5.4 x 105 Pa
= 0.9998
Note that P1 – P2 ≈ rgh = 5.4 x 105 Pa
The pressure inside the pipe is nearly the
same as when the pipe is shut off
entirely!
v2
2
1
v1
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
If v2 is the same with or
without the nozzle, the
water shoots as high
either way.
What good is the nozzle?
h
v2
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
This was an idealized problem:
we assumed the water loses
no energy in the pipe, not
matter how fast it goes.
In reality, there is viscosity, a
fluid friction that increases
with speed. And 33 m/s is
really fast! Bernoulli’s
equation would be a poor
approximation without the
nozzle.
h
v2
v1
Physics 203 – College Physics I
Department of Physics – The Citadel
Water Tower and Fountain
The nozzle allows the water
to move more slowly
through the pipe:
h
v1 = A2v2/A1
= v2/64 = 0.52 m/s
just inside the nozzle,
decreasing the energy
lost to viscosity.
v2
v1