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Warm-up For light of a given frequency, ice has an
index of refraction of 1.31 and water has an index of refraction of 1.33. Find the critical angle θc for the ray of light at the ice/water interface.
ice
waterθc
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Answer 80º
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Problems to skip/add to the HW Sets
Problem Set Skip Don't skip
63 11, 12, 13
64 9, 12, 16
65 11, 13, 14, 17 15, 18
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Fluid Mechanics
Buoyancy and Pressure
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Fluids and Buoyant Force Fluids
Have no definite shape Gases do not have definite volume They flow
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Density Density = mass/volume (kg/m3) ρ = m/v
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Common Densities Water = 1000kg/m3
Air = 1.29kg/m3
Objects will float if their density is less than these
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Pressure Force/Area
P=F/A N/m2 or Pa 1 N/m2 = 1Pa
Pair = 101,300 Pa
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Relationship between ρ, Fg, and P 1m3 of water has a weight of 9800N (1000kg/m3 x 1m3 x
9.8) Use 9.8 for all pressure problems
1.0m
1.0m
1.0m9800N
The pressure on the bottom equals the weight of the cubeP = F/A = 9800N/1.0m2 = 9800Pa
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Relationship between ρ, Fg, and P If we stack three cubes atop one another we get the
following pressure
P = 9800x 3/1.0m2 = 29,400Pa.
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Relationship between ρ, Fg, and P Thus the pressure at a depth in fluid in Pa is numerically
equivalent to the weight of a column of fluid whose cross sectional area is 1m2.
To find the pressure at any depth in a column of fluid of any given cross sectional area, the weight per unit volume is multiplied by the depth!
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Relationship between ρ, Fg, and P Works for gases also!
The average sea level density of air at 25°C and 1 atm is 1.29kg/m3. The density decreases rapidly with altitude. The total mass of a column of air extending from the top of the atmosphere down to seal level whose cross sectional area is 1m2 is 10, 330kg and its weight is 101300N. Thus the atmospheric pressure is 101,300 Pa or 101.3kPa.
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Fluid Pressure P = ρgh
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Pressure Varies with Depth Remember that air pressure is 1.01x105Pa https://www.youtube.com/watch?v
=cEnJITnn_4U
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Pressure Increasing with Depth The column of water above you builds as
you go further and further down. PT = P + Po
PT is the absolute pressure
Po is the atmospheric or original pressure
P is the fluid pressure (pressure from depth)
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Gauge Pressure Gauge Pressure
Pg = PT - Po
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Example I On a particular day, the atmospheric
pressure was 101,000 Pa. A) What was the pressure at a depth of 94.0m below the surface of the sea? B) What was the gauge pressure at this depth? The density of seawater is 1025kg/m3
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Answer 1.05x106Pa, 9.49x105Pa
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Example II The fluid in a u-tube is glycerin, whose
density is 1260kg/m3. Atmospheric pressure is 1.01x105Pa. A) What is the pressure P in the container? B) What is the gauge pressure in the container?
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gas
Pgas
Px Px
P14 14cm taller
Po
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Answer 1.03x105Pa, 2.0x103Pa
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Problems with Pressure Diving and ascending too fast
The Bends
https://www.youtube.com/watch?v=LfCOnGHheok
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Pascal’s Principle
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Pascal’s Principle P1 = F1/A1 and P2 = F2/A2 , but P2 = P1
Therefore Pincrease = F1/A1 = F2/A2
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Example IV The pressure produced by the force F1 on a
frictionless piston with a cross sectional area A1 is transmitted through the perfect
fluid. Find the resulting upward force F2
on the piston with area A2. (F1 = 75N and
A1 = 10cm2, A2 = 80cm2)
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Answer 600N
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Off Topic but Cool https://www.youtube.com/watch?v=
wXCDeqjWuDg
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Buoyancy Keeps objects afloat Archimedes’s Principle
The fluid that is displaced by an object that is submerged is equal to the volume of the object.
Any object completely or partially submerged in a fluid experiences an upward buoyant force equal in magnitude of the weight of the fluid displaced by the object.
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What????? This means that
FB = ρVfluidg (density of fluid)(volume of fluid)(9.8)
Buoyant force = weight of the displaced fluid
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Buoyancy Extended No matter what depth the cube is
submerged, the difference in the total force pressing down on the cube and the total force pushing up on the cube will always equal the weight of the volume of water displaced by the cube!!!!
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Further Extensions Archimedes’ Principle works no matter the
shape of the object. It also works for hollow objects compared to full objects (hollow sphere of lead compared to a ball of lead with the same volume). Both volumes of lead would have the same buoyant force acting upon them
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Example V A hollow sphere whose mass is 250kg has
a radius of 1.0m. If it is submerged in salt water at a depth of 10m by a cable. What is the tension in the cable? The density of salt water is 1030kg/m3
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Answer 40,000N
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Example VI The rectangular block below is placed in
water and 2.0cm of the block remains above the water. What is the density of the block?
10.0cm
40.0cm
10.0cm
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Fluids in Motion Fluid Statics – the study of the
relationships between forces and fluids at rest.
Fluid Dynamics – the study of the relationships between moving fluids and the forces that cause the motion
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Fluid Dynamics When a fluid is in motion in a pipe there
are friction forces at play between the individual particles of the fluid and between the fluid and pipe.
If the viscosity is not too great, all particles follow similar streamlines.
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Fluid Dynamics Streamlines form flow tubes which depict
fluid motion. Particles in a streamline do not cross paths
with other streamlines in laminar (smooth) motion
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In the figure below we see that the streamlines get closer together at the middle due to the decrease in area. The fluid is moving faster at this point.
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The streamlines above an airplane wing are closer indicating
greater speed.
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Fluids Dynamics Fluid Flow
Laminar – Smooth and repeating motion Turbulent – flow becomes irregular
Eddy currents
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Ideal Fluids We will only deal with “ideal fluids”
Nonviscous The density of the fluid is constant. (cannot
be compressed) The fluid motion is steady (P, ρ, and v at any
given point does not change). The flow is streamline and laminar.
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So What????? http://www.youtube.com/watch?v
=p08_KlTKP50
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Flow Rate Volume of fluid/unit time
Rate = Area x speed of fluid R = Av
A fluid is pumped into a pipe at a flow rate of 80cm3/s. If the diameter of the pipe is 1.4cm, what is the average speed of the fluid at this point?
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Making Fluids Flow Faster Pump it faster (no duh!) Pinch the hose or make the pipe smaller
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Continuity Equation
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Continuity Equation A1v1 = A2v2
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Example VII The radius of a fluid-carrying pipe at a
certain point is 2.0cm and the average speed of the fluid is 14cm/s. What is the average speed of the fluid at a point where the radius is only 1.3cm?
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Answer 33cm/s
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Warm-Up 4/15/14 Get out your notes and a calculator
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Bernoulli’s Principle The pressure in a fluid decreases as the
velocity increases This is how lift on an airplane wing works https://www.youtube.com/watch?v
=IDhH91mfCf8
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Bernoulli’s Equation
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Bernoulli’s Equation Based upon conservation of Energy Flow rate changes when a pipe changes
area (thickness) and elevation. Pressure + PE/volume + KE/volume =
constant Sum of pressures before = sum of pressures
after
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Bernoulli’s Equation P1 + ½ 1v1
2 + gh1 = P2 + ½ 2v2
2 + gh2
Uses Pipes moving water throughout a city Water flow into your house
Works for all fluids (liquids and gases) moving below the speed of sound
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Example VIII Water whose density is 1000kg/m3flows
through a pipe with a flow rate of 0.80m3/s. The cross sectional area of the pipe is initially 0.25m2 and the pressure is 5.2Pa. The cross sectional area of the pipe becomes 0.40m2, what is the new pressure of the fluid.
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Answer 3100Pa
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Example IX Ethyl alcohol (ρ = 810kg/m3) has a flow
rate of 0.24m3/s through a pipe whose cross sectional area is 0.060m2 and in which the pressure is 9.5x104Pa. The centerline of the pipes rises 4.0m and the area is reduced to 0.020m2. What is the fluid pressure in the elevated pipe?
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Answer 11000Pa