UNIT 1 FLUID STATICS AND DYNAMICS CHAPTER …...10/16/2018 1 AP PHYSICS 2 UNIT 1 FLUID STATICS AND...

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AP PHYSICS 2

UNIT 1

FLUID STATICS

AND DYNAMICS

CHAPTER 10

FLUIDS AT REST

MASS vs. VOLUME

0

10

20

30

40

50

60

70

80

90

100

0 10 20 30 40 50 60 70 80 90 100

MA

SS

[gra

ms]

VOLUME [cm3]

𝑚 = 𝜌 ∙ 𝑉

slope of a diagonal straight

line is constant.

Density () = slope.

𝒚 = 𝟏 ∙ 𝒙

DENSITY OF WATER

𝜌𝑤𝑎𝑡𝑒𝑟 = 1000𝑘𝑔

𝑚3

𝜌𝑤𝑎𝑡𝑒𝑟 = 1𝑔

𝑐𝑚3

DENSITY

𝑚 = 𝜌 ∙ 𝑉

𝜌 =𝑚

𝑉

𝑘𝑔

𝑚3

Density:

• Slope of mass vs. volume

of graph.

• The ratio of the mass m of a

substance divided by the

volume V of that substance.

• If volume increases, then density: ___________________

• If mass increases, then density: ___________________

stays the same

stays the same

Density

Mass describe solid objects that have real

boundaries.

For fluids, a much more useful physical quantity

than mass of the individual particles is the mass

of one unit of volume—density.

Tip

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DENSITY OF AN IRREGULAR SHAPED

OBJECT

AMAZING 9 LAYER DENSITY TOWER

DENSITY

DENSITY OF WATER (4C)

1000 kg/m3

SPECIFIC GRAVITY

The specific gravity of a substance is defined as

the ratio of the density of that substance to the

density of water at 4C (1,000 kg/m3). Specific

gravity is just a ratio (no units).

What is the specific gravity of:

• Platinum 𝜌 =21,450 𝑘𝑔

𝑚3 : ___________________

• Acetone 𝜌 =791 𝑘𝑔

𝑚3 : ___________________

• Hydrogen 𝜌 =0.090 𝑘𝑔

𝑚3 : ___________________

21.45

0.791

0.00009

WHITEBOARD – DENSITY 1

Substance A has a density of 6 g/cm3 and

substance B has a density of 9 g/cm3. In order to

obtain equal masses of these two substances,

what must be the ratio of the volume of A to the

volume of B? (Note: ratios are expressed as

fractions)

𝑉𝐴𝑉𝐵=3

2

𝜌𝐴 = 𝑚𝐴 ∙ 𝑉𝐴

𝜌𝐵 = 𝑚𝐵 ∙ 𝑉𝐵

𝑚𝐵 = 𝑚𝐵

𝜌𝐴𝑉𝐴=𝜌𝐵𝑉𝐵

𝑉𝐴𝑉𝐵=𝜌𝐴𝜌𝐵


A hollow sphere of negligible mass and radius R is

completely filled with a liquid so that its density is

ρ. You now enlarge the sphere so its radius is 2R

and completely fill it with the same liquid.

a. What is the density of the enlarged sphere?

b. what must be the ratio of the mass of the first

sphere to the mass of the second sphere?

𝑚1𝑚2

=1

8𝜌1 = 𝜌2

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A styrofoam sphere of radius R has a density ρ.

You now carefully compress the sphere so its

radius is R/2. What is the density of the

compressed sphere?

𝜌2 = 8𝜌1 m = 80 kg

A = 0.45 m x 0.30 m

h = 1.8 m


Estimate the density of a person.

1.Simplify and diagram.

2.Represent mathematically.

𝜌 =329.2 𝑘𝑔

𝑚3

PRESSURE

𝑃 =𝐹

𝐴

𝑁

𝑚2

Pressure:

• The ratio of the force F that

a fluid exerts perpendicular

to a surface area.𝑃𝑎

The SI unit of pressure is

the Pascal (Pa), where

1 Pa = 1 N/m2.

• Pressure is caused by fluid

particles colliding elastically

with objects in contact with

the fluid.

PRESSURE

𝑁

𝑚2

Pressure math model:

Operational definition (calculations)

Does not explain what causes pressure

𝑃𝑎

Macroscopic approach.

WHITEBOARD – PRESSURE 1

When a heavy metal block is supported by a

cylindrical vertical post of radius R, it exerts a

force F on the post. If the radius of the post is

increased to 2R:

A) what force does the block now exert on the

post?

B) what is the new average pressure?

𝐹1 = 𝐹2 𝑃2 =𝑃14


When a box rests on a round sheet of wood on

the ground, it exerts an average pressure P on

the wood. If the wood is replaced by a sheet that

has half the diameter of the original piece, what

is the new average pressure?

𝑃2 = 4𝑃1

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ATMOSPHERIC PRESSURE

Earth exerts a force on air molecules. The weight ofthese air molecules all around us (area) is called theatmospheric pressure.

Measurements show that the pressure of theatmospheric air at sea level is on average:

𝑃𝑎𝑡𝑚 =101,000 N

𝑚2

𝑃𝑎𝑡𝑚 =101,000 𝑃𝑎

Atmosphere: 1.0 atm = 𝑃𝑎𝑡𝑚 =101,000 𝑃𝑎

Atmospheric

pressure at sea level

is on average 105

N/m2,

or 105 Pa.


Estimate the total force that air exerts on

the front side of your body, assuming that

the pressure of the atmosphere is

constant.

𝐹𝐴𝑜𝑛𝑃 = 𝑃 ∙ 𝐴A = 1.8 m x 0.45 m P = 101,000 Pa

𝐹𝐴𝑜𝑛𝑃 = 81,810 𝑁

Pressure exerted by a fluid

(ask Mr. Largo for demonstration)

Pressure exerted by a fluid

Take a water bottle and poke four holes at the same height

along its perimeter.

• Parabolic-shaped streams of

water shoot out of the holes.

• The water inside must push out

perpendicular to the wall of the

bottle.

• Because the four streams are

identically shaped, the pressure

at all points at the same depth in

the fluid is the same.

Pascal's first law

An increase in the pressure of a static, enclosed

fluid at one place in the fluid, causes a uniform

increase in pressure throughout the fluid.

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Pascal's first law at a microscopic level

Particles inside a container move randomly in all

directions.

When we push harder on one of the surfaces of

the container, the fluid becomes compressed.

The molecules near that surface collide more

frequently with their neighbors, which in turn

collide more frequently with their neighbors.

The extra pressure exerted at one surface

quickly spreads, such that soon there is

increased pressure throughout the fluid.

Glaucoma

A person with glaucoma has closed drainage

canals. The buildup of fluid causes increased

pressure throughout the eye, including at the retina

and optic nerve, which can lead to blindness.

PASCAL’S FIRST LAW APPLIED TO THE

HYDRAULIC LIFT

Pressure changes uniformly throughout the liquid,

so the pressure under piston 2 is the same as the

pressure under piston 1 if they are at the same

elevation.

𝑃1 = 𝑃2𝐹1𝑜𝑛𝐿𝐴1

=𝐹𝐿𝑜𝑛2𝐴2

Force piston 1 on

liquid

Force liquid on

piston 2

HYDRAULIC LIFT: Examples

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𝑊1 = 𝑊2𝑃1 = 𝑃2

𝑃 =𝐹

𝐴𝑊 = 𝐹 ∙ ∆𝑑

Conservation of Energy

WHITEBOARD PROBLEM

A hydraulic lift has a small piston with surface

area 0.0020 m2 and a larger piston with surface

area 0.20 m2. Piston 2 and the car placed on

piston 2 have a combined mass of 1800 kg.

What is the minimal force that piston 1 needs to

exert on the fluid to slowly lift the car? (write an

expression for force first)

How far down will you need to push the piston in

order to lift the car 20 cm? (write an expression

for distance first)

WHITEBOARD SOLUTION

𝑊1 = 𝑊2𝑃1 = 𝑃2

𝐹1𝐴1=𝐹2𝐴2

𝐹1 ∙ ∆𝑑1 = 𝐹2 ∙ ∆𝑑2

𝐹1𝐴1=𝑚 ∙ 𝑔

𝐴2

𝐹1 =𝑚 ∙ 𝑔 ∙ 𝐴1

𝐴2

𝑭𝟏 = 𝟏𝟕𝟔.𝟒 𝑵

∆𝑑1=𝐹2 ∙ ∆𝑑2𝐹1

∆𝑑1=𝑚 ∙ 𝑔 ∙ ∆𝑑2𝑚 ∙ 𝑔 ∙ 𝐴1

𝐴2

∆𝑑1=𝐴2 ∙ ∆𝑑2𝐴1∆𝑑1= 𝟐0m m

WHITEBOARD . . . . . QUICKLY !

Mr. Largo has a soda bottle filled with

water.

The bottle has been punctured with 3

tacks.

The bottle is open.

Predict (draw) what will happen

when Mr. Largo removes all tacks

at the same time.

Bottle is

open

OUTCOME

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OUTCOME : Experiments .. . . .. Yay !

Experiment #1

• Two tacks on each side of the plastic bottle, one above

the other.

Pascal's first law fails to explain

this pressure variation at

different depths below the

surface.

Experiments .. . . .. Yay !

Experiment #2

• Fill the bottle with water to the same distance above the

bottom tack as it was filled above the top tack in

experiment 1.

We also see that the pressure at

a given depth is the same in all

directions.

Experiments .. . . .. Yay !

Experiment #3

• Repeat experiment 1 with a thinner or wider bottle with

the water level initially same distance above the top

tack in experiment 1.

The pressure of the liquid at the

hole depends only on the

height of the liquid above the

hole, and not on the mass of

the liquid above.

WHITEBOARD . . . . . QUICKLY !

Mr. Largo has a soda bottle

filled with water.

The bottle has been punctured

with 2 tacks

The bottle is closed.

Predict what will happen when

Mr. Largo removes the bottom

tack.

Predict what will happen when

Mr. Largo then removes the top

tack.

Testing our model of pressure in a liquid

Patm + Pdepth

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Testing our model of pressure in a liquid Why does pressure vary at different levels?

Why does pressure vary at different levels? Why does pressure vary at different levels?

WHITEBOARD

Draw a force diagram

(system is book 5)

Mind the length of your

arrows. FEonB5

FBabove on B5

FBbelow on B5

Quantifying pressure change with depth

WHITEBOARD

Draw a force diagram

(system is cylinder of fluid C)

Mind the length of your

arrows.


FEonB5FFabove on C

FFbelow on C

FLUID AT REST

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FEonB5FFabove on C

FFbelow on C

FLUID AT REST

Object is in equilibrium

Fy = 0

Fx = 0

a = 0 m/s2

v = 0 m/s

v = constant


WHITEBOARD

FEonB5FFabove on C

FFbelow on CWhat PHYSICS concept do

we use when we have a force

diagram?

Newton’s second law !

a = 𝐹

𝑚

Apply Newton’s second law to this force diagram.

NEWTON’S SECOND LAW

a = 𝐹

𝑚

m = V

ΣF = 0

FFBonC – FFAonC – FEonC = 0

FFBonC – FFAonC – gm = 0

FFBonC – FFAonC – gFLUIDV = 0

FFBonC – FFAonC – gFLUIDAh = 0

a = 0 m/s2 (fluid is at rest)

V = Ah

P = F/A

FEonC = gm

P1 – P2 – gFLUIDh = 0

P1 = P2 + gFLUIDh

Variation of

pressure with

depth

PASCAL’S SECOND LAW: VARIATION OF

PRESSURE WITH DEPTH

The pressure P in a static fluid at position y can

be determined in terms of the pressure P0 at

position y0:

= fluid = density of the fluid

g = 9.8 N/kg (positive)

Identify P, P0

P = Patm + fluidg(h0 – h)

P0

WHITEBOARD:

P = P0 + Pdepth

P

PASCAL’S PARADOX

When water is poured into the apparatus

below, the water level was the same in all

parts despite the shapes and the mass of

water. Why do you think this happens?

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Pascal’s 2nd law: All points

inside the fluid experience

the same pressure at the

same depth.

Measuring pressure

SIMPLE FLUID BAROMETER

Fluid height

Pascal’s 2nd law: All points

inside the fluid experience

the same pressure at the

same depth.

Measuring atmospheric pressure

Its been known since Galileo's

time that a pump consisting of

a piston in a long cylinder that

pulls up water.

Suction pumps where used to

pump waters out of wells and

to remove water from flooded

mines.

Water cannot not be pumped

more than 10.3 m high.

WHITEBOARD – Finding Patm

Use Pascal’s second law and

find the pressure at point 2? (Assumption: You do not “know” the

magnitude of the atmospheric pressure).

Note: in the picture, point 2 is

at the bottom and point 3 at

the top.

P2 = P3 + fluidgh

P2 = 0 + 1000 9.8 (10.3)

P2 = 100,940 Pa

P2 = P1 = Patmosphere

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WHITEBOARD

Use Pascal’s second law and

find the maximum height that

mercury can be pumped.

Write an expression for the

height before finding its

magnitude.

Assume: mercury = 13600 kg/m3

𝑃1 = 𝑃2 + 𝜌𝑓𝑙𝑢𝑖𝑑 ∙ 𝑔 ∙ ℎ

ℎ =𝑃1 − 𝑃2𝜌𝑓𝑙𝑢𝑖𝑑 ∙ 𝑔

ℎ =100,000 − 0

13600 ∙ 9.8 ℎ = 0.750 𝑚

Pressure is often

measured and reported in

mm Hg and the magnitude

of atmospheric pressure

is 760 mm Hg.

The atmospheric pressure

(101,000 N/m2) can push

mercury of density (13,560

kg/m3) 760 mm up a

column, and it can push

water of density (1,000

kg/m3) 10,300 mm up a

column.

PHYSICS UNDERSTANDING THE BAROMETER

This is a mercury barometer,

developed by Evangelista

Torricelli to measure atmospheric

pressure of mercury is such that

the pressure in the tube at the

surface level is 1 atm.

Therefore. The height of the

column, pressure is often quoted

in millimeters (or inches) of

mercury.

WHITEBOARD

Find the pressure at the top of each column of

mercury . Assume: mercury = 13560 kg/m3 and

Patm = 101,000 Pa

P0 = 5 Pa P0 = 3194 Pa P0 = 72429 Pa

WHITEBOARD SOLUTION

BAROMETER - b

P0 = ?

P = PATM

• Same height

• Same pressure

P = P0 + FLUIDgh

P0 = P - FLUIDgh

P0 = 101,000 –13560 9.8 0.736

P0 = 101,000 – 97805

P0 = 3194 Pa

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WB When you drink

liquid through a

straw, what causes

the liquid to rise?

Explain your answer

using a diagram.

WHITEBOARD

The pressure at the

top of the water in

Madison’s gravity fed

reservoir is

atmospheric pressure.

Determine the

pressure at the faucet

of a home 30 m below

the reservoir. P = 394,000 Pa

SUCTION CUPS

When a suction cup ispushed onto a glossy, nontextured, non poroussurface, the air underneathis pushed out and a

vacuum is created.

As there is now no air pressure under thesuction cup, only atmospheric pressure isbeing exerted on the outside of the cup, sopushing it down onto the surface. Frictionkeeps the cup from sliding

SUCTION CUP

A square suction cup is held against the ceiling,and the air is pressed out from inside it.

A 5 kg kitten hangs by a rope from the suctioncup (maximum mass the suction cup canwithstand)

1. Draw a Force diagram (suction cup, cat, ropeis the system).

2. Write a mathematical expression.

3. Now, write an expression for the side L of thesuction cup.

4. Calculate the side L of the suction cup.

5. If the side L is tripled, how many kitties canhang from the suction cup?

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SUCTION CUP

Suction cup + Cat + string

Fatm on SC

FE on SC

𝐹𝑎𝑡𝑚 𝑜𝑛 𝑆𝐶 − 𝐹𝐸 𝑜𝑛 𝑆𝐶 = 0

SUCTION CUP

𝑳 =𝒈 ∙ 𝒎

𝑷𝒂𝒕𝒎

L = 0.022 m 9 kitties

𝐹𝑎𝑡𝑚 𝑜𝑛 𝑆𝐶 − 𝐹𝐸 𝑜𝑛 𝑆𝐶 = 0

𝑃𝑎𝑡𝑚 ∙ 𝐴 − 𝑔 ∙ 𝑚 = 0

𝑃𝑎𝑡𝑚 ∙ 𝐿2 − 𝑔 ∙ 𝑚 = 0

MANOMETER WHITEBOARD

See the classroom set up

and find the pressure at

the top of the column of

water (point P).

Sorry I don’t have the

answer, I change the

balloon every time !

P

balloon

WHITEBOARD

What is the pressure of

the gas at “point P” if the

water raises 40 cm?

P0

• Same height

• Same pressure

• P = Patm

P = 96080 Pa

WHITEBOARD

1. Divide your whiteboard in 4 sections.

2. Mr. Largo is about to show you four experiments.

3. You must draw a FD for each experiment.

4. Include the magnitude of each force in each FD.

5. Let’s go . . . Everybody to the front of the room !

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OBSERVATIONAL EXPERIMENTS

1. Mr. Largo hangs a mass at the bottom of aspring scale.

2. Mr. Largo hangs a mass at the bottom of aspring scale. Then he dips half of the mass inwater.

3. Mr. Largo hangs a mass at the bottom of aspring scale. Then he dips the mass in water(water barely covers the top of the mass).

4. Mr. Largo hangs a mass at the bottom of aspring scale. Then he dips the mass completely inwater.

9.2 N 9.0 N

9.0 N

a = 𝐹

𝑚

FSonB + FWonB – FEonB = 0

FWonB = FEonB – FSonB

a = 0 m/s2 (fluid is at rest)

9.0 N

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THE MAGNITUDE OF THE FORCE THE

FLUID EXERTS ON A SUBMERGED OBJECT

The force exerted by the fluid pushing up on the

bottom is greater than the force exerted by the

fluid pushing down on the top of the block.

To calculate the magnitude of the upwardbuoyant force exerted by the fluid on theblock, we start from Pascal’s second law todetermine the upward pressure of the fluidon the bottom surface of the block,

compared to the downward pressure of thefluid on the top surface of the block.

PA = P0 A + fluid g h A

F = F0 + fluid g V

P = F/A

F = PA

BUOYANT FORCE (FWonO)

FWonO = FLUID Vdisp g

V = volume of fluid displaced

= density of the fluid

ARCHIMEDES' PRINCIPLE:

THE BUOYANT FORCE

A stationary fluid exerts an upwardbuoyant force (FWonO) on an objectthat is totally or partially submergedin the fluid.

The magnitude of this force is theproduct of the fluid density FLUID,the volume of the fluid VDISP that isdisplaced by the object, and thegravitational constant g

NOTES:

If the object is completely submerged in the fluid,

the volume V in the equation is just the whole

volume of the object.

If the object floats, the volume V in the equation

is the volume submerged in the fluid.

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Buoyant Force

density pressure

Pascal’s 2nd Law

WHITEBOARD

You hold on your hand a 50 kg rock ofdensity 2200 kg/m3 that is completelysubmerged in water of density 1000kg/m3 .

A) Sketch

B) Force Diagram

C) Find the magnitudes of:

• FEonR

• FWonR

• FHonR

490 N

222.7 N

267.3 N

WHITEBOARD - SOLUTION

𝐹𝑊𝑜𝑛𝑅 + 𝐹𝐻𝑜𝑛𝑅 = 𝐹𝐸𝑜𝑛𝑅

𝐹𝑊𝑜𝑛𝑅 = 𝐹𝐸𝑜𝑛𝑅 − 𝐹𝐻𝑜𝑛𝑅

𝐹𝐻𝑜𝑛𝑅 = 𝑔 ∙ 𝑚𝑅 − 𝜌𝑊 ∙ 𝑉𝑅 ∙ 𝑔

FEonR

FWonR

FHonR

𝐹𝐻𝑜𝑛𝑅 = 9.8 ∙ 50 − 1000 ∙50

2200∙ 9.8

𝐹𝐻𝑜𝑛𝑅 = 𝑔 ∙ 𝑚𝑅 − 𝜌𝑊 ∙𝑚𝑅

𝜌𝑊∙ 𝑔

𝐹𝐻𝑜𝑛𝑅 = 490 𝑁 − 222.73 𝑁

𝐹𝐻𝑜𝑛𝑅 = 267.27 𝑁

WHITEBOARD

Duckie is the system

1. Draw a force diagram

2. Use Newton’s secondLaw and find anexpression for theratio of densities:𝜌𝑑𝑢𝑐𝑘𝑖𝑒

𝜌𝑤𝑎𝑡𝑒𝑟


𝐹𝑊𝑜𝑛𝐷 = 𝐹𝐸𝑜𝑛𝐷

𝜌𝑊 ∙ 𝑉𝑆 ∙ 𝑔 = 𝑚𝐷 ∙ 𝑔

𝜌𝑊 ∙ 𝑉𝑆 = 𝑚𝐷

𝜌𝑊 ∙ 𝑉𝑆 = 𝜌𝐷 ∙ 𝑉𝐷

𝜌𝑑𝑢𝑐𝑘𝑖𝑒𝜌𝑤𝑎𝑡𝑒𝑟

=𝑉𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑

𝑉𝑑𝑢𝑐𝑘𝑖𝑒FEonD

FWonD

BUOYANT FORCE (FwonO)

If OBJECT < FLUID, then the object floats partiallysubmerged.

If OBJECT = FLUID, then the object remains whereverit is placed totally submerged at any depth in thefluid.

If OBJECT > FLUID, then the object sinks atincreasing speed until it reaches the bottom of thecontainer.

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WHITEBOARD

You have four objects at rest, each of the samevolume. Object A is partially submerged, and objectsB,C, and D are totally submerged in the samecontainer of liquid.

A) Determine if the object floats, sinks or remainsin place (inside the water).

B) Determine if water is >, <, or = to the cube

WHITEBOARD

Cube A

Cube B

Cube C

Cube D

𝜌𝐹𝜌𝐶=𝑉𝑐𝑉𝐹𝐷

𝜌𝐹 > 𝜌𝑐


𝜌𝐹 = 𝜌𝑐


𝜌𝐹 = 𝜌𝑐


𝜌𝐹 < 𝜌𝑐

Cube floats above water

Cube floats submerged

Cube floats submerged

Cube sinks

WHITEBOARD

Find the minimum mass required for a uniform

sphere of radius 10 cm to sink (have nothing

above the surface) in fresh water.

m = 4.19 kg

FWonS = FEonS

𝜌𝑤 ∙ 𝑔 ∙ 𝑉 = 𝑔 ∙ 𝑚

𝜌𝑤 ∙ 𝑉 = 𝑚

𝑚 = 𝜌𝑤 ∙4

3𝜋𝑟3

𝑚 = 𝜌𝑤 ∙4

3𝜋𝑟3

WHITEBOARD

You stand on a bathroom scale. Suppose your

mass is 70.0 kg and your density is 970 kg/m3.

Determine the reduction of the scale reading

(FSonP) due to air.

Assume density of air 1.29 kg/m3

A) Force diagram.

B) Find the magnitudes of:

• FEonYOU

• FAIRonYOU (Buoyant Force)

• FSonYOU

FEonY = 686 N

FFonY = 0.912 N

FFonY = 685.088 N

WHITEBOARD

Pretend for a moment that you areArchimedes, who needs to determinewhether a crown made for the king is puregold or some less valuable metal.

You find that the forcethat a string attached toa spring scale exerts onthe crown is 25.0 N whenthe crown hangs in air

and 22.6 N when thecrown hangs completelysubmerged in water.

WHITEBOARD

FD + N 2nd

What is the buoyant force exerted on thecrown?

What is the mass of the crown?

What is the volume of the crown?

What is the density of the crown?

Compare the crown to the density of gold(GOLD = 19,300 kg/m3). Is the crown made ofgold?

m = 2.55 kg

FFonC = 2.4 N

V = 0.000245 m3

= 10,416 kg/m3

Crown is not made of gold, what a rip off !!!

10/16/2018

18

WHITEBOARD

When analyzing a sample of ore, ageologist finds that it weighs 2.00 Nin air and 1.13 N when immersed inwater.

Assume the ore sinks in water.

A) Sketch

B) FD + N 2nd

D) Volume of water displaced

E) Density of ore

WHITEBOARD - SOLUTION𝐹𝑊𝑜𝑛𝑂 +𝐹𝑆𝑜𝑛𝑂 = 𝐹𝐸𝑜𝑛𝑂

FEonO

FWonO

FSonO

𝐹𝑊𝑜𝑛𝑂 = 𝐹𝐸𝑜𝑛𝑂 − 𝐹𝑆𝑜𝑛𝑂

𝐹𝑊𝑜𝑛𝑂 = 2 𝑁 − 1.13 𝑁

𝐹𝑊𝑜𝑛𝑂 = 0.87 𝑁

𝐹𝑊𝑜𝑛𝑂 = 𝜌𝑊 ∙ 𝑉𝑆 ∙ 𝑔

𝑉𝑆 =𝐹𝑊𝑜𝑛𝑂𝜌𝑊 ∙ 𝑔

𝑉𝑆 =0.87

1000 ∙ 9.8

𝑉𝑆 = 8.88𝑥10−5𝑚3

𝜌𝑂 =𝑚𝑂

𝑉

𝜌𝑂 =

𝐹𝐸𝑜𝑛𝑂𝑔

𝑉

𝜌𝑂 =𝐹𝐸𝑜𝑛𝑂𝑔 ∙ 𝑉

𝜌𝑂 =2

9.8 ∙ 8.88𝑥10−5

𝜌𝑂 = 2298.85𝑘𝑔

𝑚3

WHITEBOARD

A typical human being has a density of 900

kg/m3. as this typical human floats in sea water

(1250 kg/m3)

A) FD + N 2nd

B) what percent of his/her body is above water?

a) 10%

b) 28%

c) 72%

d) 90%


𝐹𝑊𝑜𝑛𝑃 = 𝐹𝐸𝑜𝑛𝐻

𝜌𝑊 ∙ 𝑉𝑆 ∙ 𝑔 = 𝑚𝐻 ∙ 𝑔

𝜌𝑊 ∙ 𝑉𝑆 = 𝑚𝐻

𝜌𝑊 ∙ 𝑉𝑆 = 𝜌𝐻 ∙ 𝑉𝐻

𝜌ℎ𝑢𝑚𝑎𝑛𝜌𝑤𝑎𝑡𝑒𝑟

=𝑉𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑𝑉ℎ𝑢𝑚𝑎𝑛

FEonP

FWonP

900

1250=𝑉𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑𝑉ℎ𝑢𝑚𝑎𝑛

0.72 =𝑉𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑𝑉ℎ𝑢𝑚𝑎𝑛

0.28 =𝑽𝒐𝒖𝒕𝒔𝒊𝒅𝒆𝑉ℎ𝑢𝑚𝑎𝑛

WHITEBOARD

A 3.6 kg goose floats on a lake with 40% of its

body below the 1000 kg/m3 water level..

A) FD + N 2nd

B) Find the density the goose.


𝐹𝑊𝑜𝑛𝐺 = 𝐹𝐸𝑜𝑛𝐺

𝜌𝑊 ∙ 𝑉𝑆 ∙ 𝑔 = 𝑚𝐺 ∙ 𝑔

𝜌𝑊 ∙ 𝑉𝑆 = 𝑚𝐺

𝜌𝑊 ∙ 𝑉𝑆 = 𝜌𝐺 ∙ 𝑉𝐺

𝜌𝐺𝑜𝑜𝑠𝑒𝜌𝑤𝑎𝑡𝑒𝑟

=𝑉𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑𝑉𝐺𝑜𝑜𝑠𝑒

FEonG

FWonG

𝜌𝐺𝑜𝑜𝑠𝑒1000

= 0.4

𝜌𝐺𝑜𝑜𝑠𝑒 = 400𝑘𝑔

𝑚3

10/16/2018

19

WHITEBOARD

A 3.0m x 1.0m rectangular plastic container that

is 1.0 m high, has a mass of 1500 kg. The

container floats in fresh water (density 1000

kg/m3), partially submerged.

A) FD + N 2nd

B) Write an expression for the depth of the

container submerged in water.

C) Find the depth of the container submerged in

water.


𝐹𝑊𝑜𝑛𝑃𝐶 = 𝐹𝐸𝑜𝑛𝑃𝐶

𝜌𝑊 ∙ 𝑉𝑆 ∙ 𝑔 = 𝑚𝑃𝐶 ∙ 𝑔

𝜌𝑊 ∙ 𝑉𝑆 = 𝑚𝑃𝐶

𝜌𝑊 ∙ 𝐴 ∙ ℎ𝑆 = 𝑚𝑃𝐶

FEonPC

FWonPC

ℎ𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 =𝑚𝑃𝐶

𝜌𝑊 ∙ 𝐴

h = 0.5 m

WHITEBOARD

An empty 400 kg life raft of cross-sectional area

2.0 m x 3.0 m has its top edge 0.36 m above the

waterline. How many 75-kg passengers can the

raft hold before water starts to flow over its

edges? The raft is in seawater of density 1025

kg/m3.


𝐹𝑊𝑜𝑛𝑅 = 𝐹𝐸𝑜𝑛𝑅 + 𝐹𝑃𝑜𝑛𝑅

𝜌𝑊 ∙ 𝑉𝑆 ∙ 𝑔 = 𝑚𝑅 ∙ 𝑔 + 𝑁 ∙ 𝑚𝑃 ∙ 𝑔

𝜌𝑊 ∙ 𝑉𝑆 = 𝑚𝑅 +𝑁 ∙ 𝑚𝑃

FEonR

FWonR

FPonR

𝑁 =𝜌𝑊 ∙ 𝑉𝑆 −𝑚𝑅

𝑚𝑃

𝑁 =1025 ∙ 2 ∙ 3 ∙ 0.36 − 400

75

𝑁 = 24 𝑝𝑒𝑜𝑝𝑙𝑒

WHITEBOARD

At a pool party, you pull an inflated beach ball

under water by tying a light string around it. The

ball has a mass m and an overall specific gravity

of 0.25. While you hold the string, and therefore

the ball, under water motionless, what is the

tension in the string?

a) FT = gm

b) FT = 2gm

c) FT = 3gm

d) FT = 4gm


𝐹𝑊𝑜𝑛𝐵 = 𝐹𝐸𝑜𝑛𝐵 +𝐹𝑇

𝐹𝑇 = 𝐹𝑊𝑜𝑛𝐵 − 𝐹𝐸𝑜𝑛𝐵

𝐹𝑇 = 𝜌𝑊 ∙ 𝑉𝑊𝐷 ∙ 𝑔 − 𝑚𝐵 ∙ 𝑔

𝐹𝑇 = 𝑔 ∙ 𝜌𝑊 ∙ 𝑉𝑊𝐷 −𝑚𝐵

𝐹𝑇 = 𝑔 ∙ 𝜌𝑊 ∙ 𝑉𝐵 −𝑚𝐵

𝐹𝑇 = 𝑔 ∙ 𝜌𝑊 ∙𝑚𝐵

𝜌𝐵−𝑚𝐵

𝐹𝑇 = 𝑔 ∙ 𝑚𝐵

𝜌𝑊𝜌𝐵− 1

𝑉𝑊𝐷 = 𝑉𝐵

𝐹𝑇 = 𝑔 ∙ 𝑚𝐵

1000

250− 1

𝐹𝑇 = 𝑔 ∙ 𝑚𝐵 4 − 1

𝐹𝑇 = 3 ∙ 𝑔 ∙ 𝑚𝐵

10/16/2018

20

The average balloon has a

volume of 0.01 m3. The density

of helium is 0.164 kg/m3. The

density of air is 1.3 kg/m3. How

many balloons would you need

to attach to an average house

(50,000 kg) in order for it to

float?

Hint:

Start with a Force Diagram

System = balloons


𝐹𝐴𝑜𝑛𝐵 = 𝐹𝐸𝑜𝑛𝐵 + 𝐹𝐻𝑜𝑛𝐵

𝜌𝑎𝑖𝑟𝑉𝐵𝑔 = 𝑛𝑔𝑚𝐵 + 𝑔𝑚𝐻

𝜌𝑎𝑖𝑟𝑛𝑉𝐵 = 𝑛𝑚𝐵 +𝑚𝐻

𝜌𝑎𝑖𝑟𝑛𝑉𝐵 = 𝑛𝑉𝐵𝜌𝐻𝑒𝑙𝑖𝑢𝑚 +𝑚𝐻

𝑛 = 4.4𝑀

𝜌𝑎𝑖𝑟𝑛𝑉𝐵 − 𝑛𝑉𝐵𝜌𝐻𝑒𝑙𝑖𝑢𝑚 = 𝑚𝐻

𝑛𝑉𝐵 𝜌𝑎𝑖𝑟 − 𝜌𝐻𝑒𝑙𝑖𝑢𝑚 = 𝑚𝐻

𝑛 =𝑚𝐻

𝑉𝐵 𝜌𝑎𝑖𝑟 − 𝜌𝐻𝑒𝑙𝑖𝑢𝑚

balloons

FAir on B

FE on B

FH on B

HOW DO SUBMARINES MANAGE TO SINK

AND THEN RISE IN THE WATER?

A submarine's density increases when water fills

its compartments.

With enough water in the compartments, the

submarine's density is greater than that of the

water outside, and it sinks.

When the water is pumped out, the submarine's

density decreases.

With enough air in the compartments, its

density is less than that of the outside water,

and the submarine rises toward the surface.

STABILITY OF SHIPS

A challenge to building

watercraft is to maintain

stable equilibrium for the

ship, allowing it to right itself

if it tilts to one side due to

wind or rough seas.

WHITEBOARD

CUP + WATER + INDEX CARD

Fill the cup with some water (half empty/half full).

Cover the cup with the index card.

Turn the cup around (remove hand).

ON WHITEBOARD:

Describe all your observations.

Use all possible representations to explain

the outcome.

Groups will present in front of the class.

UNIT 1 FLUID STATICS AND DYNAMICS CHAPTER …...10/16/2018 1 AP PHYSICS 2 UNIT 1 FLUID STATICS AND...

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