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AP Physics 2:
Fluids….
web studyguide included from: http://alisoniguel.com/staffpages/rhjansen/
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Vocab
definition / examples/ units
Ideal Fluid
Flow rate
Viscosity
Gauge pressure
Hydrostatic
Specific gravity
Cross-sectional area
Atmospheric pressure
Density of water
Effective weight
Contact force
Bernoulli effect
Pascal’s principle
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Misconceptions
(all are false, can you correct them?)
1. Objects float in water because they are lighter than water.
2. Objects sink in water because they are heavier than water.
3. Mass/volume/weight/heaviness/size/density may be perceived as equivalent.
4. Wood floats and metal sinks.
5. All objects containing air float.
6. Liquids of high viscosity are also liquids with high density.
7. Adhesion is the same as cohesion
8. Heating air only makes it hotter.
9. Pressure and force are synonymous.
10. Pressure arises from moving fluids.
11. Moving fluids contain higher pressure.
12. Liquids rise in a straw because of "suction".
13. Fluid pressure only acts downward.
Just the math
1. Density:
2. Pressure:
 = m/V
P=F/A
3. Hydraulics(same height): F2/ A2 = F1/ A1
4. Changes in height:
5. Buoyancy:
Pbelow = Pabove + gh
FB = Wfluid = fVd g (Vd is displaced- volume of object in the fluid)
note….
Wobject= oVog
(Vo is entire volume of object)
(floats if weight of object is less than the weight of water displaced)
6. Moving fluids (assume density doesn’t change, smooth velocity not turbulent):
P 1 + gh1 + ½  v12 = P2 + gh2 + ½  v22 (conservation of energy)
A1 v1 = A2 v2
(conservation of mass)
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Cliff notes : dr.V’s super condensed soup
I. Fluids
A. No shape, includes gas & liquids
B. Ideal: no turbulence, no energy loss, can’t compress
C. Density = mass/volume = kg/m3
(1000 for water)
D. Specific gravity: density of unknown/ density of water
II. Pressure
A. Force/area = 1 Newton/m2 = 1 Pascal (Pa)
B. F*d/ A*d = Energy/volume = J/m3 (energy density)
C. 1 atm = 105 Pascal = 760 torr = 760 mm of mercury
D. Gauge Pressure = Absolute pressure/ atmosphere pressure
III.
Static Fluids Don’t flow
A. Pascal’s principle- pressure gets transmitted everywhere
and is the same at any given height
1. Hydraulics: F2/ A2 = F1/ A1 (little force≫>big force; same pressure)
2. Diving:
Pbelow = Pabove + gh
(more pressure below!)
B. Archimedes’ Principle: buoyant force = weight of fluid displaced
1. Buoyant force depends on volume of object Vo but density of fluid f
Wfluid= fVog (floats if smaller than buoyant force)
compared to Wobject= oVog (floats if smaller than buoyant force)
2. Sinking or rising objects: Fnet = mg-Fb = ma
IV.
Moving Fluids
A. Continuity: Can’t lose mass: mass/sec = constant
1. 1 A1 v1 = 2A2 v2
B. Bernoulli’s equation:
energy conserved (faster fluid=lower pressure)
1. P1 + gh1 + ½  v12 = P2 + gh2 + ½  v22 = = constant energy/cubic meter
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Demos: None of these suck, only push or pull
Picture:
math or explanation
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Demos: more Buoyancy, Pressure, & Bernoulli
Picture:
math or explanation
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Strategies: see the flow chart, then read this if still confused
For problems involving density and specific gravity:
1. Use information given in the problem to calculate the object’s mass and volume.
2. Determine object’s specific gravity by dividing the object’s density by the density of water.
For problems involving Pascal’s principle:
1. Complete a data table with the information given, identify which are input and which output.
2. Use the concept of pressure in a fluid and Pascal’s principle to solve the problem.
For problems involving Archimedes’ principle
1. Complete a data table with the information given.
2. Determine the volume of the object from its mass and density.
The volume of the object equals the volume of the fluid the object displaces.
3. Use Archimedes’ principle to determine the buoyant force.
4. For a floating object the buoyant force equals the object’s weight.
5. The weight of the fluid displaced equals the buoyant force.
For problems involving the rate flow equation and the equation of continuity:
1. Make a data table: cross-sectional area, velocity, and density of the fluid at the each point
2. Use the rate flow equation and equation of continuity to solve the problem.
For problems involving Bernoulli's equation:
1. Make a data table: pressure, velocity, and height of the fluid at and above a given point.
If there is only a closed pipe, then determine the cross-sectional area at each point in question.
2. First try mass conservation: mass= A*v= A*v , also called the equation of continuity
3. Next try the full blown Bernoulli’s equation to solve the problem.
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Flow chart for Fluids
No
Is it moving?
Yes
Is an object
floating or
sinking in the
fluid?
No
Try:
P2 = P1 + gh
Yes
Try Archimedes’ Principle:
FB = Wfluid
and force diagrams
Does a pipe
change sizes?
(cross sectional
areas)
No
Try: Bernoulli’s Equation:
Yes
P1 + ½ v12 + gy1 = P2 + ½  v22 + gy2
Try: Equation of Continuity:
A1v1 = A2v2
And also if 2
unknowns
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Sample Problems
1. A scuba student stands on the bottom of a deep swimming pool, and lifts a
bowling ball 1m above the bottom of the pool. If the bowling ball has a specific
gravity 5.0, how much time does it take for the bowling ball to hit the bottom?
(hint: find the ball’s acceleration first)
Key:
since the ball is entirely submerged. We can replace the mass of the ball by
.
Solving for a,
Now using kinematics (TNEOMS), the time of the drop becomes
=
0.5sec
using tneom 1/2at2=y= 1m
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2. Water pressure: Find the depth to reach the following pressures under water.
Note: 1 atm = 101kPa and the units for density should by kg/m3 if using pa!
a. 1atm
b. 2 atm
c. 3 atm
key:
Pbelow = 1 atm + gh where = 1000 kg/m3 is the density of water
P
= 101,000 + 1000 kg/m3 * 10 *h
a. h = 0
b. P = 2 atm = 2.02 kPa = 1.01Kpa + 1000 *10*h
h=10.1 m
c. P = 3 atm = 3.03 kPa = 1.01Kpa + 1000 *10*h
h=20.2 m
d. P = 10 atm = 10.1 kPa = 1.01Kpa + 1000 *10*h
h=90.9 m
d. 10 atm (why not? As long as you don’t come up too fast & get the bends!)
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3. a toy boat has an empty mass of 210 g and a volume of 280 cm3.
D
How much water must you add to make it sink?
Set the density of the sub equal to density of water, 1 g/cm3.
To make this submarine sink, its mass, equal to
, should be
3
3
(1 g/cm )(280 cm ) = 280 g. Its empty mass is 210 g, so add 70 g (= 70
cm3, or 70 mL) of water.
4. Prove only about 10% of a floating ice berg is visible.
The density of ice is 0.9167 g/cm³ at 0°C, water has a density of 0.9998 g/cm³
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5. a 200 g mass has a volume of 50 cm3, and hangs from
a spring. What is the effective weight (that is what will
the spring scale read) when the mass (none of the
spring) is completely submerged in water?
The mass is in equilibrium, so
. ( is the tension in
the rope, which is also the force probe reading.)
now substitute for the buoyant force
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6. A hydraulic lift consists of a cylindrical piston of radius 2cm, which when pushed upon
with a force of only 100N is able to lift a 2000kg car sitting on a much larger cylindrical
piston. What is the area of the larger piston ?
Pressure felt same everywhere if at same height & speed
P = F/A = F/A
Fbig /F = Abig /A
Flarge/Fsmall = wt of car/ pushing force = mg/100 = 2000*10 /100 = 200
Alarge/Asmall = Area/ *(2)2 = 200 so A = .25 m2 or r = 28.3cm
7. Blood flows from your heart (.009m radius for the aorta) at about 0.33m/s. It then
travels through your capillary system at a much smaller speed of 0.00034 m/s. What is
the effective cross-sectional area of your capillary system? (assume no blood loss!)
Av = Av since mass is not lost and density is not changed
Aorta * v = *r2 *v = *(.009)2(0.33)
Acap *v = A* v
=*(3.4e -4)2so A = 0.25m2
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8. How fast will the water come out of a 2 liter bottle with a hole 5m from the ground and
the water filled to 25cm? (assume top of water and side of water at atmospheric pressure)
Water level
hole
Pascal principle: Pressure at top same as at bottom since exposed to air
P1 = P2 = 1 atm or 105 Pa
Initial velocity at top is at rest so v1 = 0
Bournoulli:
P1
+ gh1 + ½  v12 = P2 + gh2 + ½  v22
+ gh1 + ½  (0)2 = P1 + g()) + ½  v22
gh1 = ½  v22
v2 = square root (2gh1) =√2*10*0.20m = 2m/s
P1
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AP Problems
50m to top of well
35m deep well
9. A pump, submerged at the bottom of a well that is 35 m deep, is used to pump water
uphill to a house that is 50 m above the top of the well. The density of water is 1,000
kg/m3. All pressures are gauge pressures. Neglect friction, turbulence, and viscosity.
a. Residents of the house use 0.35 m3 of water per day. The day's pumping is completed in 2 hours during the day.
i. Calculate the minimum work required to pump the water used per day (300,000 J)
Work = PE = total wt * height= 1000kg/m3 * .35m3 * 10m/s2 * (50 + 35 m) = 300,000 J
ii.
b.
Calculate the minimum power rating of the pump. (41 Watts)
P = work/time (sec) = 300,000 J/ (2hrs * 3600 sec) = 41 watts
The average pressure the pump actually produces is 9.20 x 10 5 N/m2. Within the well the water flows at 0.50
m/s and the pipe has a diameter of 3.0 cm. At the house the pipe diameter is 1.25 cm.
i.
Calculate the flow velocity when a faucet in the house is open. (2.88 m/s)
radius = diameter /2 used in area of circle =  (radius)
Area * velocity at house = area * velocity at pump
*(.0125m/2) 2 * v = *(.03m/2)2 * 0.5m/s
v =2.88 m/s
ii.
Explain how you would calculate the minimum pressure at the faucet.
+ gh1 + ½  v12 = P2 + gh2 + ½  v22
9.2x105 + 1000*10*85 + ½*1000* (0.5)2 = P + 0 + ½ 1000 (2.88)2
solve for P since given all info (height = 0 at top of house for fluids)
P1
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10.
10.
a.
An unoccupied, 10 kg wooden raft made of driftwood with specific gravity 0.60 floats on the surface
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of a calm pond. [The density of water is 1,000 kg/m .]
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What is the volume of the raft? (0.0017 m )
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density=mass/volume so v= mass/density = 10kg/ (0.6*1000 kg/m ) = .017m
b.
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Determine the buoyant force acting on the raft while it floats. (100 N)
2
Since floats, no net force so buoyant force = wt = mg = 10kg *10m/sec = 100N
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Now a student under the water hangs brass (density = 8,000 kg/m ) masses on a light string attached to the raft, as
shown in the diagram above. The student adds brass a little at a time until the moment when the raft becomes
completely submerged.
c.
Determine the new buoyant force on the raft. (170N)
buoyant force = wt water displaced = density water * volume of raft and hangers (negligible) * g
3
3
2
= 1000 kg/m * .017m * 10m/sec = 170N
d.
Calculate the minimum mass of brass that the student must hang from the raft to allow the raft to become
completely submerged. (70 plug 8 = 78kg total)
If neglect volume of brass adding to the buoyant force, then
total weight (100N raft + brass) ≥ buoyant force = (170N )
so brass can weigh= 70N = m*10 and mass of brass = 7kg (but this is an underestimate)
extra buoyant force = water wt displaced by brass = volume of brass*water density *10m/s 2
100 + wt. brass = 170N + extra buoyant force
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Vbrass*8000kg/m *10 = 70 + Vbrass * 1000kg/m3 *10
3
Vbrass (7000 *10) = 70 so Vbrass =.001m
3
3
and mass of brass = density * volume = 8000 kg/m * .001m = 8kg
e.
If instead the 10 kg raft were made of freshly cut wood with specific gravity 0.80, how would the answer to
part (d) change? Check one box and justify your answer briefly.
X Less brass would be necessary since denser wood takes less space, there’s less buoyant force to hold up the
same weight of wood (same mass, so same weight). Need less extra weight to sink it !
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11.
Derive an equation for the velocity of the stream using the variables in the picture:
Conserve PE and KE and divide by volume to convert mass to density:
2
top: 1/2mv + mgh but h = y2 -y1 and v= v2 (or zero if assume water at rest before pull plug
2
bottom: 1/2mv + mgh but h= 0 and v = v1
2
2
1/2mv2 + mg(y2 -y1) = 1/2m*v1 + 0
Divide by volume so all mass is density but cancels out everywhere anyway
2
2
V1 = v2 + 2g(y2 -y1)
***** same as TNEOM Equation!
12.
What is the lifting force of an airplane wing having an area of 24m2 if the air flows 251
m/s above the wing, and 225 m/s below the wing? Assume densty of air is 1.29kg/m3.
balance energy using Bernoulli’s
above: P + pgh + 1/2p(v)2 but v = 251
below: P + pgh + 1/2p(225)2 but v= 225
see equal and solve for difference in pressure:
Ptop-Pbelow = 1/2p(v12- v22) where h=0 for both
= ½* 1.29 * (2512 – 2252) = 7982.52
Then force = pressure difference * area
= 7982.52 N/m2* 24 m2 = 191580N
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13.
A venture pump is basically a pipe that starts wide and ends narrow. The pressure is
lower at the narrow end and can be used to pull a vacuum to “suck” out air or liquids.
Suppose the wider end has an area of .07 m2 and a pressure of 200 Pa, and the narrow
end (middle section of picture) has an area of .05 m2 and a pressure of 80 Pa. The fluid
has a density of1.3 kg/m3. .
a. What is the pressure difference between the two ends? (120 Pa)
P = 200- 80 = 120 Pa
b. What is the ratio of speeds between the two ends, ie. what is V1/V2? (0.71)
mass is constant so Area*speed* density is constant or A*v = A*v (density cancels)
.07m2 * V1 = .05m2 * V2 , then rearrange so V1/V2 = .05/.07 = 0.71
c. Use Bernoulli’s equation to solve for V1 and V2. (14m/s, 19.4 m/s for V2)
use P + 1/2pv2 = P + 1/2pv2 and v1 = .71*V2
so 200 + ½*1.3*(.712* V2)2 = 80 + ½*1.3*V22
solve and get v1 = 13.7 m/s and v2 = 19.3 m/s
d. What is the volume flow rate Q2 in m3/sec at the initial larger end? (0.21m3/sec)
Q= A*v = .07 * 13.7 = 0.96 m3/sec
e. What is the volume flow rate Q2 in m3/sec at the narrow section? (trick question!)
Q= A*v = .57 * 19.3 = 0.96 m3/sec (about the same as should be, no lost mass)
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Buoyant force is just weight of water displaced. Fb= density* g* volume
All volumes same so only difference is density of lquid. A=B > C > D=F > E
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Buoyant force is just weight of water displaced. Fb= density* g* volume
Depth not important, only volume: B=D=E > A=C=F
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Buoyant force = total weight of blocks since floating. Just rank by total mass of blocks:
D=G > F > A=B=C > E
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Tension= weight – buoyant foce. So look for biggest difference between mass and volume
C > B=D > A > D > E
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Pressure depends on just height of water above hole: P= atm + pgh
H > C=F > A > B=E >G
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P= atm + pgh so must find largest multiple of height times density of liquid:
C > E > F > A=B > D
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