Transcript Physics 2053C – Fall 2001

Outline
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Review Pressure and Density.
Begin
Buoyant Forces.
 Continuity Equation.
 Bernoulli’s Principle.
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CAPA and Examples.
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Pressure and Density
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Density = Mass/Volume
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A property of the material.
Pressure = Force/Area
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Depends on the height of the fluid.
Same in all directions.
Units are:
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Force/Area = N/m2.
Pascals  1 Pa = 1 N/m2.
Atmosphere  1 atm = 1.013 X 105 N/m2.
2
Pressure
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Pressure = Force/Area.
P = F/A.
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Depends on the height of the fluid.
Same in all directions.
Units are:
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Force/Area = N/m2.
Pascals  1 Pa = 1 N/m2.
Atmosphere  1 atm = 1.013 X 105 N/m2.
3
Pressure
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Pressure = Force/Area.
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P = F/A.
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Depends on the height of the fluid.
F = Mg = (V)g = (  (A*h) )g
F/A = (  (A*h) )g / A = gh
P = gh
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Buoyancy
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Fluid or gas exerts a force on materials
it touches
For an immersed object, the sum of
forces from surrounding material is the
buoyancy
If the net force on an object is upward,
then it “rises”, if it is downward then it
“sinks”, if zero then it “floats”
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Buoyant Forces
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Force exeted by a displaced liquid.
Ft-Fb = B
gAht - gAhb = B
B = gA(ht – hb) = Wt - Wb = B
B = A(ht – hb) * g
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Ice and Water
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Why does ice float?
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Solids are generally denser than liquids
Liquids are generally denser than liquids
Ice should sink
In fact, “really” cold water is denser
o
than ice (I.e. water at 34 F is denser
o
than ice at 32 F)
If this wasn’t true, oceans and lakes
would freeze from the bottom up
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Equation of Continuity
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The flow of a liquid is constant
throughout a system (if no liquid is
added or subtracted)
The amount of water going into a
pipe is the same as coming out
Amount of liquid flow is
Flow = Av
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Equation of Continuity
Flow1 = Flow2
1A1v1 = 2A2v2
assuming 1 = 2 (same liquid)
A1v1 = A2v2
A1
so v2 =
x v1
A2
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Bernoulli’s Principle
where the velocity of a fluid is high
the pressure is low and where the
velocity is low, the pressure is high
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This is why baseballs spin, aspirators
work, a roof can blow off a house in high
winds and sailboats can sail into the wind
Let’s see it work, take out a piece of
paper…
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Bernoulli’s Equation
Energy in a moving fluid:
Pressure: W = Fd F = PA (pressure does work)
Gravity: fluid flowing up or down changes its
potential energy
Kinetic: fluid speeding up or slowing down
changes its kinetic energy
Since energy is conserved, we can relate the
pressure, velocity and height between two
points in a flowing system
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Bernoulli’s Equation
P1 + _21 v12 + gy1 = P2 + _21 v22 + gy2
P = Pressure
v = velocity
 = density of fluid
1
y = height
g = acceleration due to gravity
2
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Example 10-12
Water heater in the basement,
faucet on the 2nd floor
Basement: dpipe = 4.0cm v = 0.50m/s P = 3 atm
2nd floor: dpipe = 2.6cm height = 5m
What is the flow speed and pressure on the 2nd floor?
First, find the flow rate (continuity):
v1A1 = v2A2
v2 = A1/A2 x v1
v2 = r12/r22 x v1
v2 = 2.02  /1.32 x 0.50
v2 = 1.2 m/s
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Example 10-12(cont)
Basement: dpipe = 4.0cm v = 0.50m/s P = 3 atm
2nd floor: dpipe = 2.6cm height = 5m
v2 = 1.2 m/s
What is the flow speed and pressure on the 2nd floor?
Next, find the pressure (Bernoulli):
P1 + _21 v12 + gy1 = P2 + _21 v22 + gy2
1
P2 = P1 + _2 v12 - _21 v22 + gy1 - gy2
P2 = P1 + _21 (v12 - v22) + g(y1 - y2)
P2 = (3.0 N/m2) + (1.0x103 kg/m3)(9.8 m/s2)(-5.0 m)
.
+ 0.5(1.0x103 kg/m3)[(0.5 m/s)2 – (1.2 m/s)2]
P2 = 2.5 x 105 N/m2
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CAPA #1
What is the absolute pressure on the bottom of a
swimming pool 20.0 m by 11.60 m whose uniform
depth is 1.92 m?
Pw = gh = (1.0x103 kg/m3)(9.8 m/s2)(1.92m)
= 1.89x104 N/m2
But, we need absolute pressure…
P = Pw + Patm
P = 1.89x104 N/m2 + 1.013x105 N/m2
= 1.20x105 N/m2
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CAPA #2-3
2. What is the total force on the bottom of that
swimming pool?
Area = 20.0 m x 11.60 m
F = P x A = (1.20x105 N/m2)(20.0 m)(11.60 m)
= 2.79x107 N
3. What will be the pressure against the side of the
pool near the bottom?
The pressure near the bottom is the same as on the
bottom
P = 1.20x105 N/m2
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CAPA #4
4. A dentist’s chair of mass 236.0 kg is supported by a hydraulic
lift having a large piston of cross-sectional area 1434.0 cm2.
The dentist has a foot pedal attached to a small piston of
cross-sectional area 76.0 cm2. What force must be applied to
the small piston to raise the chair?
Fchair
F?
Fchair = mg
= (236.0 kg)(9.80 m/s2)
= 2312.8 N
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Fchair
F?
CAPA #4(cont)
4. A dentist’s chair of mass 236.0 kg is supported by a hydraulic
lift having a large piston of cross-sectional area 1434.0 cm2.
The dentist has a foot pedal attached to a small piston of
cross-sectional area 76.0 cm2. What force must be applied to
the small piston to raise the chair?
P1 = P2
F1 F
2
___
= ___
A1 A2
P = F/A
F2 = (A2/A1)F1 = ((76.0 cm2)/(1434.0 cm2)) x (2312.8 N)
F2 = 123 N
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Next Time
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Larry Dennis will be back Monday.
Continue Chapter 10.
CAPA Problems.
Prepare for Quiz 7.
Please see me with any questions or
comments.
See you Monday.
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