Chapter 9-10 - Eastchester High School

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AP Physics – Mechanics – Chapter 9-10– Fluid Mechanics Text Ch. 9 - Select topics - 9.4-9.7 - Reading pp. 255-263 – textbook HW -#15,17,18,19,20,22,29,32,34,35,36,39,40
Ch 10 - Select topics - 10.2-10.3- Reading pp. 278-286 - textbook HW -#1,4,5,6,7,13,15,16,20
Ch 9 – Fluid Statics
9.4 Density and Pressure
9.4 A. Density – The density of a substance of uniform
composition is defined as its mass per unit volume (or
mass:volume ratio)
Note: assume uniform density for all fluids in this chapter
Units of Density: mks: kg/m3, cgs: g/cm3, chemistry: g/mL
**For conversions remember the density of water**
Density of water = 1 g/cm3= 1 g/mL = 1 kg/L = 1000 kg/m3
Note: it’s no coincidence that the density of water is 1 g/cm 3. Here is original SI definition of the gram: “1 gram
= the absolute weight of a volume of pure water equal to the cube of the hundredth part of a metre (1 cm 3),
and at the temperature of melting ice"
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ρwater= 1000 kg/m3 (at 4⁰C, 1 atm)
ρice= 917 kg/m3
ρair= 1.29 kg/m3
ρhelium= .179 kg/m3
9.4 B. Specific gravity (S)
Specific gravity (S) = the ratio of an object’s density to the
density of water.
 water
IF ρobject > ρwater then S > 1 and the object sinks in water.
IF ρobject < ρwater then S < 1 and the object floats in water.
IF ρobject < ρair the object floats in air
IF ρobject > ρair the object sinks in air
#1) An object has density of 3 x 10 kg /m . Find its specific gravity. [3]
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Ex 2) Here’s one to ponder: “A pint’s a pound the world around.” Is it
Well that depends…
Some approximate English Metric Conversions:
1 L= 2.11 US pints
1 L = 1.76 Imperial pint
1 US pint = 16 fl. oz
1 Imperial pint = 20 oz
1USP int x
1kg 2.2lb
 1.04lb
2.11P int 1L 1kg
1P int x
1kg 2.2lb
 1.25lb
1.76 P int 1L 1kg
A better saying might be: “a liter’s a kilogram in every land.” Done.
#3) What is the weight of a gallon of milk? [8.3 lb]
4Quart .94 L 1kg 2.2lb
 8.3lb
1USGal 1quart 1L 1kg
9.4 C. Pressure
Pressure is defined as the ratio of the force to area
Units: the N/m2 , a.k.a. the Pa (Pascal), thus 1 N/m2 = 1 Pa
#4) Find the pressure exerted by 1 leg of your chair on the floor.
[5 x 107 Pa or 70 psi, assuming r=1 cm and m=60 kg]
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How can this device be used to measure fluid pressure?
Note: The force of the fluid at any point on a submerged
object is perpendicular to the surface of the object.
Note: we do not say the pressure is perpendicular, as
pressure is a SCALAR.
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#5) Hmmm. Water beds. Does anyone still have one?
a) Find the weight (in Newtons) of a king-sized water bed that is 2 m x 2 m x 0.3 m :
[12000 N]
b) Find the pressure the floor exerts on the bed.
[3000 Pa]
c) If bed is tilted on its side, find the pressure the floor exerts on the bed. [20000 Pa]
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9.5 A) Variation of Pressure with Depth
Consider this:
A) If a fluid is at rest in a container, all portions of the fluid
must be in equilibrium. Therefore:
B) All points at the same depth must be at
the same pressure.
Box of water
C) As depth increases, pressure increases.
ΣFy = Fdepth - Fair - mg = 0 N
Fdepth = Fair + mg
and since F=PA:
Box of
PAdepth = PAair + mg and since m = ρV=ρAh:
PAdepth = PAair + (ρAh)g
canceling like terms:
Pabsolute = Pair + ρgh
#6) An ehrlenmeyer flask and a beaker, both at sea level, are filled to the same
level with water. Point P is at the same depth in both containers. Compare the
absolute pressure at point P in each container.
Pressure is only affected by height of the liquid above a
point, and is not affected by the shape of the container.
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#7) a) Calculate the absolute pressure at a pool depth of 4 m.
Remember: ρwater = 1000 kg/m3 patmospheric = 15psi or 101.3 kPa) (1.413 x 105 Pa or 141.3 kPa or 21 psi)
b) What would a Pressure Gauge read at this depth?
(4.0x104 Pa
or 6 psi)
What is the relationship between gauge pressure and
absolute pressure?
Think: what does an air gauge read on a bike tire with “NO
AIR” in it? 0 psi or 0 Pa
A Pressure gauge always reads gauge pressure, which is:
Gauge Pressure = Absolute pressure – Air Pressure
And since: Pabsolute = Pair + ρgh
ρgh = Pabsolute - Pair Thus: Gauge Pressure = ρgh
Note: at 4m, Pabs= 21 psi. What if the depth was the deepest SCUBA Dive on record of 318.25m P=3.28 x 106
Pa or 486 psi. Why don’t you get crushed under this pressure? Because your body is made of fluids and exert
an equal force outward. The air spaces however, shrink in size because air is compressible, making it
eventually impossible to breath at very low depths, thus limiting how far one can SCUBA. The better question
is what would happen if you REMOVED the air pressure around the body? (Total Recall Mars Scene)
Now go home and do Homework - #15, 17, 18, 19, 20, 22 (ρethanol = 806 kg/m3)
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9.6 Pressure Measurements with pressure gauges
A. The open-tube manometer (Mahna, Mahna)
Remember: Gauge Pressure = ρgh
Pabs > Pair
Pabs = Pair
Pabs < Pair
Pabs = Pair + ρgh
Pabs = Pair - ρgh
Note: AP B problems deal with + gauge pressure, only
B. The Barometer (Draw one be sure to point out P = 0, thus P = P )
Air pressure of 1.013x10 Pa can support a
column of mercury of what height?
ΣF = 0 N = Fatm - FHg or
Fatm = FHg
Fatm = mHgg
and since F=PA
PatmA = ρHgVg
and since V = Ah
PatmA = ρHgAhg
Patm = ρHggh
since ρHg = 13.595 x 103 kg/m3
h = (1.013x105 Pa)/( 13.595 x 103 kg/m3)(9.80665m/s2) = .7598 mHg
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9.5 B) Pascal’s Principle
(Blaise Pascal 1623-1662)
Pressure applied to an enclosed fluid is transmitted
undiminished to every point of the fluid and to the walls of
the containing vessels (hydraulic press).
Derive equation:
P1 = P2
F1 = F2 or F2 = A2 F1
A1 A2
Greater area = greater force
Note: Vdisplaced1 = Vdisplaced2
A1h1 = A2h2 or h2 = A1 h1
F1h1 = F2h2
Greater area = less height
(See Video Pascal’s Principle- Hydraulics)
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#8) What is the minimum force exerted on the smaller piston to hold a 1500-kg car
if the diameters of the pistons have a 4:1 ratio? (938 N)
Now go home and do HW -- #29
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9.7 Buoyant Forces - Archimedes Principle
Why is it easy to lift someone in the pool?
When using the ladder to climb out of the pool, why do you feel so much heavier (and it’s not because your
Speedo® is wet)
Archimedes principle- any body that completely or partially
submerged in a fluid is buoyed up by a force equal to the
weight of the fluid displaced by the body.
Question: Why is there an upward force on an object
immersed in a liquid?
Since gauge pressure INCREASES with depth, the deeper
surface of the object experiences a greater force.
For a box of water:
ΣFy= 0 N = F2 – F1– mg or
F2 – F1 = mg
FB is defined as F2 – F1
FB = mg
Box of water
since FB=∆PA
and ∆P = ρg∆h
FB = (ρg∆h)A = ρVg = mg
In Summary:
FB = ρfVg = mfg
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What if the box of water is replaced with a steel object of
same volume?
FB is still the weight of water displaced, But the steel object
will NOT be in equilibrium.
ΣFy= FB – msteelg or ΣFy= ρfVg – ρoVg
Box of steel
Case I: Sinking object (ρo > ρf) Vf displaced = Vo
#9) A 1 cm x 1 cm x 1 cm cube of silver is placed in the middle of a large cup of
water. (ρsilver =10.5 x 103 kg/m )
a)Find the buoyant force on the silver block. FB = ρfVg = mfg (0.01 N)
b) Does it float or sink comparing forces acting on block? (Hint: FBD)
w=mog = ρoVg (0.105 N)
c) Describe the motion of the block initially.
Find the block's acceleration.
It accelerates.
(9.05 m/s2)
d) If the block is now replaced by a gold block of the same dimension find the
buoyant force on the block. FB = ρfVg = mfg (0.01 N)
e) What happens when it reaches the bottom of the cup? (buoyant force?)
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Case II: Floating object (ρo < ρf)
When an object is floating in a fluid, part of it is submerged.
In this case: Vf displaced < Vo
ΣFy= 0 N = FB - mog
FB = m o g
#10) A wooden raft with density ρraft = 600 kg/m3 is floating on the water.
The dimensions of the raft are 2.5m x 2.5m x 0.1m
a) Find buoyant force. Since floating: FB = mog = ρoVog (3750 N)
b) Find depth that the raft is below the water level.
FB = ρfVfg = ρf(Ah)g (0.06m)
See any ratio relationship?
FB = mog = ρoVog when floating and FB= ρfVfg always,
ρoVog= ρfVfg or ρoVo= ρfVf (6/10) (1) = (1) (6/10)
The density of the raft is 6/10x the density of water, thus it
displaces 6/10x its volume in water.
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#11) a)A person pushes a hollow metal sphere (radius=5 cm) filled with air below
the surface of water. Determine force required to hold the sphere below the
surface. (for simplicity, assume skin of sphere has negligible mass) (5.24 N)
Holding it means ΣF = 0 N, Fext= FB = ρfVfg
b) As the sphere is pushed deeper, ideally, the buoyant force_________
Although Pabs increases with depth, FB = ρfVfg = (ρfg∆h)A. The ∆h is the same!
c) In reality, if the incompressible hollow metal sphere is replaced with a balloon,
and the balloon is pushed deeper and deeper in the water, the buoyant force will
__________. (Dec, Inc, or RS?)
Think: when submerged: FB = ρfVfg As depth increases, Pressure increases, thus DECREASING the volume of
the balloon. Since this displaces less fluid, FB decreases.
#12) Estimate a person’s buoyant force in air. (0.903 N)
Assume m = 70 kg and ρo= 1000 kg/m3
Thus: Vo = m/ρo 70/1000 = 0.070 m3
FB = ρfVfg
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#13) And now the legend behind Archimedes…
Archimedes(287 - 212 B.C.), The “staff scientist” for King Hiero II of Syracuse, uncovered
a fraud in the manufacture of a golden crown commissioned by the King, like the one to
the right. Suspecting that the goldsmith might have replaced some of the gold given to
him by an equal weight of silver, Hiero asked Archimedes to determine whether the
wreath was pure gold. Because the wreath was a holy object dedicated to the gods,
Archimedes could not disturb the wreath in any way. The story goes that Archimedes happened to go to the
bath, and on getting into a tub observed that the more his body sank into it the more water ran out over the
tub. As this pointed out the way to explain the case in question, he jumped out of the tub and rushed home
naked, crying with a loud voice that he had found what he was seeking; he shouted repeatedly in Greek,
"Eureka, eureka." meaning "I have found (it), I have found (it)."
How did Archimedes supposedly solve this problem?
Submerge crown of mass m in water, collect displaced water.
Submerge equal mass of pure gold in water, collect displaced water.
If Volume displaced is the same, then VC = VAu, and mC=mAu , then ρC=ρAu
Note: VC > VAu and mC=mAu , so ρC<ρAu and Goldsmith was killed.
Galileo in 1586 felt that this method would not have been precise enough to show such a small difference in
density. He offered this technique as more precise:
Given: Spring Scale Reading in Air = 7.84 N
Scale Reading in the water = 6.86 N
Determine the density of the crown. (m = .784 kg, FB= 0.98 N, V=9.8 x 10-5 m3, ρ=7960 kg/m3
Now go home and do HW -- #32, 34, 35, 36, 39, 40
LAB: Density - irregularly shaped object
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AP Physics – Mechanics – Chapter 10– Fluid Dynamics Ch 10 - Select topics - 10.2-10.3- Reading pp. 278-286 - textbook HW -#1,4,5,6,7,13,15,16,20
Moving Fluids are really cool. Does a curve ball curve? Does a rising fastball actually rise? Why are knuckle
balls so hard to hit? Don’t like baseball? How does an airplane fly? During a hurricane, should you open
windows a crack to equalize pressure?
To understand fluid dynamics, we need to assume something: That we are dealing with an IDEAL FLUID.
10.2 A. Ideal Fluid assumptions –
1) Fluid is non viscous – no internal friction between
adjacent fluid levels
2) Fluid is incompressible -- density is constant
3) Fluid motion is steady – velocity, density and pressure at
each point in fluid does not change in time
4) Fluid flow is streamlined or laminar. This means it moves
without turbulence – each element of the fluid has zero
angular velocity about its center (no eddy currents)
Note: liquids are nearly incompressible, and gases are
compressible, but when not confined can be treated as
incompressible (such as wind)
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10.2 B. Equation of Continuity – for an ideal fluid moving
through a pipe of changing diameter:
the FLOW RATE (Volume/Time) must be constant (since
fluid incompressible). Therefore the volume flowing
through cross-section A1 must be the same as the volume of
fluid flowing through cross-section A2 in the same amount
of time.
What does this mean about the velocity at A1 and A2?
Velocity1(through narrower pipe) > Velocity2(through wider pipe).
Derive the equation of continuity:
Vol1 Vol 2
The equation of continuity is:
A1x1 A2 x2
A1v1  A2v2
And since:
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#1) A pipe under the sink with a 2-cm diameter is reduced to
1-cm diameter to bring water to the dishwasher. If water enters the larger section
at 100 cm/s, what is the speed as it exits the smaller hose? Demo: Fun with the garden hose
(400 cm/s)
#2) A garden hose with a 2-cm diameter is used to fill a 20-liter bucket. If it takes
60 s to fill the bucket, what is the speed at which water leaves the hose?
(1 ml = 1 cm3) (use A1v1 = Vol2/t) (V=20,000 cm3, 318 cm/s)
Now go home and do HW - # 1, 4 , 5
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10.2 C. Bernoulli’s Equation - The sum of the pressure, the
kinetic energy per unit volume, and the potential energy per
unit volume , has the same value at all points along a
Bernoulli’s Equation derived:
W1 = +F∆x1 = (P1A1)∆x1 = P1V
W2 = -F∆x2 = -(P2A2)∆x2 = -P2V
Note: F1 is + due to water flow from the left. F2 is – due to work done by
gravity being negative.
Wnet = ∆KE + ∆PE
P1V -P2V = (½mv22 -½mv12)+ (mgh2 - mgh1) Divide all terms by V:
P1V -P2V = (½mv22 - ½mv12)+ (mgh2 - mgh1) Since ρ = m/V:
P1 -P2 = (½ρv22 - ½ρv12)+ (ρgh2 - ρgh1)
P1 + ½ρv12 + ρgh1 = P2 +½ρv22 + ρgh2
Or............. P + ½ρv2 + ρgh = Constant
Moving terms for 1 & 2 to same side:
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Bernoulli’s Equation states that:
As a fluid moves through a pipe of varying cross sections
and elevations, the Pressure and velocity will change along
the pipe, but the SUM of the pressure, KE per unit volume,
and the PE per unit volume will remain constant.
P1 + ½ρv12 + ρgh1 = P2 + ½ρv22 + ρgh2
Things to consider:
1) When using Bernoulli's equation, it is important to choose 2
points on your diagram and label them points A and B.
2) If the fluid speed increases (assume same height), then the
pressure must decrease.
Therefore, swiftly moving fluids exert LESS pressure than slowly
moving fluids.
Practical applications of Bernoulli’s Equation:
1. The Venturi Tube- This leads to # 2-3 (Do demo 1-3 first)
DEMO 1: Blow under and over a piece of paper. Which gives it lift?
DEMO 2: Air over 8.5" x 11" paper
DEMO 3: Air between two balloons, or leaf blower and two bowling balls
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2) Water aspirators in the AP Chem room that produce a partial vacuum using the
kinetic energy from the faucet water pressure
3) Household plumbing sink trap and roof ventilation system
4) The Magnus Force – Which is the fastball? The Curve ball?
How is a Slider thrown so that it curves horizontally? How is a knuckleball thrown?
DEMO 4: Big Air Bag DEMO 5: Styrofoam Ball in air stream DEMO 6: Defying Gravity—Ping Pong Ball upsidedown in funnel THEN DISCUSS AIRPLANE WING
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#3) A large open tank filled with liquid of density ρ leaves the bottom of the vessel
through a small hole. (Assume A2 >> A1, then fluid level drops very slowly and v2=0m/s)
a) Find the speed that the liquid emerges from the hole. (Toricelli's Law)
b) Find horizontal distance the water travels
DEMO: (Toricelli’s Law)
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More Problems with Torricelli’s Law:
#4) A college student opens a beverage can and notices that, for some reason,
there is a small pinhole on the side of the can and the beverage shoots out as
shown. Find h. and determine if the can was full when he opened it. (6.96 cm, yes)
#5) If wind blows at 30 m/s over the roof of a house, what is the pressure
difference between air inside and outside? (Assume the roof is VERY thin) (∆P=581Pa
or .09 psi)
So, During a hurricane, should you open windows a crack to equalize pressure?
Now go home and do HW -- #6, 7, 13, 15, 16, 20
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Remember: Continuity equation and Bernoulli’s equation are for when a
fluid is moving.
In the Bernoulli’s equation, what happens when both v1= v2 = 0 ?
The equation boils down to: ΔP = ρg∆h
LAB ACTIVITY: Toricelli’s law. Where does the water stream hit the ground?
Place cup on ground! (assume that the speed when it leaves the hole is about 90%
of the theoretical speed due to viscosity/surface tension/friction)
Ch 9-10 and Mechanics. Done.