Liquids and Gases

Pressure, Pascal’s principle

How do they lift your car for service?

Atmospheric pressure

We’re submerged!

How can you drink a Coke?

Archimedes ! Eureka!!!

Balloons of all sizes

Bubbles of all sizes

3/1/2017 Physics 214 Spring 2017 1

8th week Lectures Feb. 27. – March 03. 2017.

3/1/2017 Physics 214 Spring 2017 2

Liquids and Gases

As we know liquids and gases act very differently than solids.

Liquids and gases have mass but their constituent atoms are not bound so that each part of the liquid or gas can move.

The atoms of a liquid are more tightly bound so a liquid can be kept in an open container whereas gas usually requires a closed container.

Liquids, like solids are not very compressible, that is, it is difficult to change the volume.

A volume of gas can have it’s volume changed fairly easily. Both have the property of being able to flow, for example water and gas lines in a house.

The unit of volume is

the meter cubed , m3,

which is a very large

volume. Very often we use

cm3 = cc. Other everyday

units are gallons, quarts,

pints

3/1/2017 Physics 214 Spring 2017 3

Pressure

A volume of liquid or gas has mass and F = ma is still a good law except a force at a point on the surface of water only moves the water near that point

So generally we work with liquids and gases in containers

and exert forces over a surface.

We define pressure (P) as: P = F/A

that is the force (F) divided by the area (A) over

which the force acts.

Any change in pressure of an enclosed fluid at

rest is transmitted uniformly throughout the liquid.

This is Pascal’s principle.

The dimension of pressure is N/m2 and its unit is 1N/m2 = 1Pascal

3/1/2017 Physics 214 Spring 2017 4

Hydraulic Jack

If the liquid levels in each cylinder

on each side are at the same height

then the pressure F/A just below

each piston is the same and

F1/A1 = F2/A2 so F2 = F1A2/A1

If we make the ratio of the surfaces

A2/A1 = 100 then an F1 force of 50lbs

can generate a lift force .

Of course if we push F1 down a distance h1 then the F2

Side will only rise h1 A1/A2 because h1A1 = h2A2, that is the displaced volumes must be the same.

Work done = F1h1 = F2h2

2F = 5000Lbs

3/1/2017 Physics 214 Spring 2017 5

Atmospheric pressure

If I stack a pile of bricks each brick has to support all

the bricks above it. So as we go down the stack the

force increases as does the pressure. Denote the

mass of a brick by “m” and its surface “A”

In the example shown the gravitational force on the face

of the bottom brick would be and the

pressure

At the earths surface we are supporting

a column of air which exerts a force and

because it is a gas it exerts an equal pressure

In all directions.

g

6F mg

6F mg

6mgP

A

3/1/2017 Physics 214 Spring 2017 6

2A-01 Suction Cups

How does a

suction cup

work ?

Remember atmospheric pressure can support 32 feet of water so the force on 1 square foot is ~ 2000 lbs!

How does a suction cup ‘hold on’ to objects?

Holding the suction cup by

itself I only have to support the

weight since the force due to

atmospheric pressure acts on

the top and bottom of the cup.

If I place it on a surface and

exclude all the air the cup is

held to the surface by a force

due to atmospheric pressure

of 1.013x105 Pascals per

square meter F = PAA

PA

F F

If two cups are

pushed together,

does this make

it twice as

difficult to

separate ?

PA

3/1/2017 Physics 214 Spring 2017 7

Otto von Guericke 1602 - 1686 was the “inventor” of

the “nothing” we now call a vacuum.

Von Guericke created a

vacuum by attaching two

hemispheres and then

evacuating the air from

the resultant sphere. Von

Guericke demonstrated

the force of the vacuum

before the German

emperor Ferdinand III by

having two teams of

eight horses attempt to

disengage the

hemispheres.

He was the mayor of the German city of Magdeburg. He built the “Hemispheres

of Magdeburg” and constructed the first vacuum pump.

3/1/2017 Physics 214 Spring 2017 8

2D-01 Hydraulic Press

THE INPUT FORCE IS GREATLY ENHANCED BY THE HYDRAULIC FLUID AND BY THE LEVERAGE GAINED USING THE LONG

HANDLE.

Within the Hydraulic fluid the pressure is uniform:

F1/A1 = F2/A2 F1/F2 = A1/A2

Pump piston diameter = 0.5 in

Lift piston diameter = 1.25 in

F1= 6.25*F2

This is not enough mechanical advantage to

crush the wood. How is it done?

Is it possible to

crush a 2x4 only

with the force

exerted by one

hand ?

Remember, the lever arm also gives a mechanical

advantage. We have: F1≈ (16)*6.25*F2 = 100*F2

Using the uniformity of pressure within a

liquid as a mechanical advantage

3/1/2017 Physics 214 Spring 2017 9

2D-03 Pascal's Principle (Wine Bag)

BY CREATING A PRESSURE WITHIN A VESSEL, A PRESSURE IS EXERTED ON ALL THE WALLS OF THE VESSEL. THIS MEANS A

FORCE = PA WHICH IS LARGE ENOUGH TO LIFT A PERSON

Creating a Pressure to exert a Force

Is it possible to

lift a heavy

object using

only your

breath ?

P = F/A F = PA

A pressure applied over an area

can generate a large force.

3/1/2017 Physics 214 Spring 2017 10

2A-05 Pressure Force Milk Jug

Liquid takes up a much smaller volume than steam. As the

steam cools and condenses a partial vacuum forms and the

pressure drops. Atmospheric pressure then crushes it.

Effect of atmospheric pressure on a partially evacuated milk bottle

The hot liquid heats the air

which expands and some

leaves the container. The

hot liquid also causes

vapor to form in the jug

which then cools and

condenses.

What will happen

to the container

if it is sealed

after putting in a

small amount of

very hot water?

3/1/2017 Physics 214 Spring 2017 11

Density

If one takes two objects of exactly the same volume made of

different materials they have different weights. So we define

a useful quantity called density and denoted by .

Density ρ = mass/unit volume, kg/m3 or grams/cc so the

mass of an object is ρV and the weight ρVg

If an object is put into a container of liquid

it will float if ρobject is less than ρliquid

It will sink if ρobject is greater than ρliquid

3/1/2017 Physics 214 Spring 2017 12

Density and pressure

If I take a container of liquid the pressure at any depth is the weight of the water above that depth divided by the area.

It is very useful to define

Mass/unit volume ρ = kg/m3.

Very often we use grams/cubic centimeter

1 gram/cm3 = 1000kg/m3

Water is 1 gram/cc

W = mg = ρAhg

Pressure P = W/A = ρgh and the pressure is the same at all depths increasing with depth. This is the pressure increase from the surface

which is also subject to atmospheric pressure

g

3/1/2017 Physics 214 Spring 2017 13

Can you guess what happens when

Shaving Cream is placed in vacuum ?

Do the balloons

burst in vacuum

differently then they

normally burst ?

2A-03 Vacuum Demos

Effects of Vacuum on objects made largely of air or air pockets.

Why do the

balloons burst

in the vacuum ?

What will happen when

the marshmallows are

returned to normal

pressure ?

Why do the

marshmallows

get bigger in

vacuum ?

AN AIR-POCKET/BALLOON WILL EXPAND WHEN THE PRESSURE IS REDUCED AND IT WILL DEFLATE WHEN THE PRESSURE IS INCREASED. SO BALLOONS WILL EXPAND AS THEY RISE IN THE ATMOSHPERE AND THE EXPANSION OF A PARTIALLY EVACUATED CAN IS USED IN BAROMETERS.

3/1/2017 Physics 214 Spring 2017 14

2B-04 Liquid Pressure

AT ANY GIVEN POINT IN A STATIONARY LIQUID, THE PRESSURE IS THE SAME IN ALL DIRECTIONS.

PRESSURE IS NOT A VECTOR. IT ACTS

EQUALLY IN ALL DIRECTIONS

Investigating Pressure in different directions within a liquid in equilibrium.

IF A LIQUID IS IN EQUILIBRIUM, THE

FORCES ACTING AT A POINT CANNOT

HAVE A PREFERENTIAL DIRECTION OR

THE LIQUID WOULD MOVE.

The increase in pressure ρgh is

measured by the difference in

height of the liquid in the U tube.

A B B A

h g

What will happen

to the reading on

the manometer as

the sensor is

rotated ?

3/1/2017 Physics 214 Spring 2017 15

2B-05 Pressure Forces in Liquids

THE LIQUID PRESSURE DEPENDS ONLY ON DEPTH P = ρgh.

THE UPWARD FORCE DEPENDS ON THE AREA F = PA

What happens as

the submerged

cylinder filled

with air is filled

with water ?

Air

Water

An open ended cylinder kept shut by liquid pressure

There are two forces acting

on the plate. It’s weight

down and PA up. When PA

exceeds the weight the

cylinder stays intact

In this situation the plate

has to now support the

weight of the water and

when the weight of the

water plus plate exceeds PA

the cylinder opens

PA

PA

3/1/2017 Physics 214 Spring 2017 16

If the different

vessels were filled

to the same height,

how does the

pressure differ at

each vessel base ?

2B-02 Pascal's Vases

THE PRESSURE IS DEPENDENT ONLY ON THE HEIGHT h AND NOT THE VOLUME OR SHAPE OF THE VESSEL.

P = ρgh

Examining the Pressure at the bottom of

differently shaped vessels filled with fluid

3/1/2017 Physics 214 Spring 2017 17

2B-03 Water Seeks Own Level

LIQUID PRESSURE DEPENDS ONLY ON VERTICAL HEIGHT (MEASURED STRAIGHT DOWN THAT IS PARALLEL TO g).

P = ρgh

The slanted cylinder

and twisted cylinder

hold a longer “total

length” of water. But in

each case the vertical

height is the same.

Liquid pressure

depends on the

“height” of the

liquid column.

But how is this

height measured ?

Investigating the Dependence of Pressure on Height

3/1/2017 Physics 214 Spring 2017 18

In our everyday lives we do not feel we are supporting a very large weight because it is the environment in which we have developed.

Galileo’s disciple Torricelli invented the Barometer. He used a meter long glass tube, sealed one end and left open the other.

He filled the tube with mercury and hold his finger over the open end. Inverted the tube and placed the open end into a container of mercury. So that all the air bubbles were removed he raised the tube to a vertical position.

Some of the mercury flowed from the glass tube into the mercury container until an equilibrium was reached. A 760mm high column of mercury was supported by the atmospheric pressure.

No Air

Vacuum

g

P is the same

Determination of Atmospheric Pressure

Using a Barometer

3/1/2017 Physics 214 Spring 2017 19

Knowing the density of mercury and the height h of the column of mercury

We can calculate the mass of a 0.76m high and base cube of mercury

313600 /

0.76

mercury kg m

h m

21m

2 3 2

2

2 2 2 2 5

2 2

2 2

2 5 2

/ 13600 / 0.76 10336 /

Next we calculate the weigh/m = pressure

Weight/m / 10336 / 9.8 / 1.013 10

Since / and thus / (See page 4)

/ 1.1013 10 / 1.

Mass m kg m m kg m

mkgF m kg m m s

s m

mkg s Newton F m Pascal

F m p N m

5

5

1013 10 ( )

1Atmosphere 1.013 10

1Atmospere 14.696 / in

This is equal to for the Mercury

Pa Pascal

Pa

pounds sq

gh

3/1/2017 Physics 214 Spring 2017 20

Everyday examples

Suction cups – remove the air and the atmosphere holds it in place.

Drinking through a straw – create a partial vacuum in your mouth and the atmospheric pressure pushes the fluid up the straw

Pressure is lower at higher altitudes – water boils at a lower temperature

Pressure is higher the deeper you go in the ocean – leads to more nitrogen being absorbed and the bends.

3/1/2017 Physics 214 Spring 2017 21

Suction cup

Question: What is the minimum area of a

suction cup that can be used to lift 100kg.

The suction cup will stay attached to the block

providing PA is greater than F/A. In practice

there would only be a partial vacuum under the

cup so the cup would break away from the

block at a lower force and pressure.

1.013 x 105 x A = 100 x 9.8 A = 9.67/1000m2

Which would be a circle of radius just over 2

inches

mg

F PA

3/1/2017 Physics 214 Spring 2017 22

Archimedes Principle If an object is lowered into a liquid

the volume it occupies was being supported by an

upward force that exactly balanced the weight of

the same volume of liquid so the object will feel the

same upward buoyant force.

buoyant force = weight of liquid displaced

this is true for objects that are immersed and for

objects that float. So a boat made of steel can float

because it can displace a volume of liquid greater

than it’s own weight. A large volume of the boat is

air so the average density is less than that of water.

T + FB = W = mg

g

T

mg

Fb

3/1/2017 Physics 214 Spring 2017 23

Floating and buoyant force If a floating object is submerged to a depth h then the pressure on the bottom is ρgh and the upward force is ρghA but hA is the volume of liquid displaced, V, and ρgV is the weight of liquid displaced so the

Fb = weight of liquid displaced = weight of object

ρliquid x Vdisplaced x g = ρobject x Vobject x g

If an object has a density larger than the liquid it will sink. Suppose the top is at a depth of dt and the bottom at db. Then there is downward pressure on the top ρgdt and an upward pressure on the bottom ρgdb so the net buoyant force

Fb = ρg(db – dt)A

= ρgV = weight of liquid displaced.

T + Fb = weight of the object = mg

g

Fb

g

Fb

T

3/1/2017 Physics 214 Spring 2017 24

Which object

experiences the

greater buoyant

force, the

heavier one or

lighter one ?

2B-08 Buoyant Force

BUOYANT FORCE DOES NOT DEPEND ON THE MATERIAL OF

THE OBJECT DISPLACING THE FLUID. THE BUOYANT FORCE

DEPENDS ONLY ON THE VOLUME OF FLUID DISPLACED.

T

Mg

T

Mg

FB

Scale reads

tension in cord:

T = Mg – FB

Mg = FB + T

We find Fb to be exactly equal

for both masses

Compare the Buoyant Force between two

cylinders of equal volume and different mass.

3/1/2017 Physics 214 Spring 2017 25

What happens to

the reading on

the upper scale

when the block is

lowered into the

beaker of water ?

2B-09 Archimedes I

EVEN THOUGH THE BLOCK DOES NOT ‘TOUCH’ THE LOWER SCALE, THE FORCE ON THE

FLUID DUE TO THE BLOCK IS TRANSMITTED TO THE SCALE. THE REDUCTION IN READING ON

THE UPPER SCALE IS EXACTLY EQUAL TO THE INCREASE IN READING ON THE LOWER SCALE.

IF THE CONTAINER WAS FULL SO THAT WHEN THE BLOCK WAS INSERTED THE VOLUME THE

BLOCK DISPLACED SPILLS OUT OF THE CONTAINER THEN THE BOTTOM SCALE WOULD NOT

CHANGE.

The fluid exerts a

buoyant force on the

block, which reduces

the tension on the

cord. The reading on

the scale is lowered.

T

Mg

FB

T = Mg – FB

Should

anything

happen to the

reading on the

lower scale ?

Since the fluid exerts a force on the

block, the block exerts an EQUAL

and OPPOSITE force on the fluid.

3/1/2017 Physics 214 Spring 2017 26

2B-10 Archimedes II

A. The block is not immersed T1 = Mg

B. The block is immersed but the liquid runs out T2 = Mg – FB

C. The displaced liquid is poured into the can T3 = Mg – FB + WDF

T1 is found to equal T3 which means that the bouyant force FB is

equal to WDF the Weight of the displaced Fluid

THE BUOYANT FORCE IS EQUAL TO THE WEIGHT OF THE FLUID DISPLACED.

Use a scale to establish the relationship between the Buoyant

Force on an object and the Weight of Fluid Displaced by the object

T1

Mg

A

T2

Mg

FB

B

T3

Mg

FB

WDF

C

3/1/2017 Physics 214 Spring 2017 27

Does the air

exert a buoyant

force ?

2A-08 Buoyancy of Air

IF SENSITIVE WEIGHING OF AN OBJECT IS REQUIRED,

UNEQUAL BUOYANT FORCES COULD AFFECT THE RESULTS.

mbg

mag

ρairgVb ρairgVa

Setting the sum of torques on equal-arm balance

about pivot equal to zero, we have in the presence

of air:

mag – ρairgVa = mbg - ρairgVb

Vb > Va implies mb > ma which is demonstrated in

vacuum

Investigating the Buoyant force resulting from Air

3/1/2017 Physics 214 Spring 2017 28

Pressure and Volume of a Gas

In the apparatus shown the pressure at

point A is the same as at point B so the

pressure exerted by the gas is equal to

ρgh.

If we add more mercury we can

measure both pressure and volume

and if the temperature does not change

we find

PV = constant

A B

B A

h

g

3/1/2017 Physics 214 Spring 2017 29

Summary of Chapter 9

P = F/A Pascals

F1/A1 = F2/A2

Work done = F1h1 = F2h2

1 Atmosphere = 1.013 105 Pa

and will support

76cm of mercury

32 feet of water

No Air

g

3/1/2017 Physics 214 Spring 2017 30

Liquids

P = W/A = ρgh

Water is 1 gram/cc

1 gram/cm3 = 1000kg/m3

buoyant force = the weight of liquid displaced

T + FB = W = mg

For a floating object T = 0

g

T

mg

Fb

3/1/2017 Physics 214 Spring 2017 31

Balloons

Any object in the atmosphere is subject

to a buoyant force and Archimedes law

applies so if the buoyant force is greater

than the weight of an object it will rise.

So since the material of a balloon has a

density greater than air then the balloon

must be filled with a gas having a

density less than air. In practice balloons

either use Helium or hot air.

As the balloon rises the buoyant force

decreases because the density of the air

decreases and the balloon will float at

constant altitude when the buoyant force

is equal to the weight.

mg

FB

3/1/2017 Physics 214 Spring 2017 32

Surface tension

The molecules and atoms in a liquid are

continually in motion so that a molecule at

the surface can escape and this is

evaporation. However a molecule at the

surface feels an attractive force pulling it

back into the liquid and this is surface

tension. This is the reason that one can

form bubbles and water drops

3/1/2017 Physics 214 Spring 2017 33

Questions Chapter 9

Q1 Is it possible for a 100-lb woman to exert a greater pressure on

the ground than a 250-lb man? Explain.

Q3 The same force is applied to two cylinders that contain air.

One has a piston with a large area, and the other has a piston with

a small area. In which cylinder will the pressure be greater?

Yes. The pressure will be mg/A so if A is small e.g. small

heels the pressure will be very large

The pressure is F/A so the one with the smallest A

3/1/2017 Physics 214 Spring 2017 34

Q4 A penny and a quarter are embedded in the concrete bottom of

a swimming pool filled with water. Which of these coins

experiences the greater downward force due to water pressure

acting on it?

Q5 Why are bicycle tires often inflated to a higher pressure

than automobile tires, even though the automobile tires must

support a much larger weight?

Each coin has to support the weight of water in a vertical column

so the quarter has the biggest force F =PA

Once again the upward force has to support the weight so

F = mg but F = PA where A is the area of the tire on the road so

M/m = (PA)car/(Pa)bike so Ma/mA = Pcar /Pbike the ratio of the areas is

smaller than the ratio of the weights so

Pbike is higher.

3/1/2017 Physics 214 Spring 2017 35

Q6 The fluid in a hydraulic system pushes against two pistons, one

with a large area and the other with a small area.

A. Which piston experiences the greater force due to fluid pressure

acting on it?

B. When the smaller piston moves, does the larger piston move

through the same distance, a greater distance, or a smaller distance

than the smaller piston?

Q8 When a mercury barometer is used to measure atmospheric

pressure, does the closed end of the tube above the mercury

column usually contain air?

A. The pressure is the same and F = PA so the larger piston has

the larger force.

B. The work done is the same so the small piston moves the most

No it needs to be a vacuum

3/1/2017 Physics 214 Spring 2017 36

Q9 Could we use water instead of mercury to make a barometer?

What advantages and disadvantages would be associated with the

use of water?

Q10 If you climbed a mountain carrying a mercury barometer,

would the level of the mercury column in the glass tube of the

barometer increase or decrease (compared to the mercury

reservoir) as you climb the mountain?

The height of the liquid depends on the density. So one can use

water but the column would be 32 feet high

The pressure decreases because you have a smaller column of

air to support so the height would decrease

3/1/2017 Physics 214 Spring 2017 37

Q11 If you filled an airtight balloon at the top of a mountain,

would the balloon expand or contract as you descend the

mountain?

Q12 When you go over a mountain pass in an automobile,

your ears often “pop” both on the way up and on the way

down. How can you explain this effect?

It would contract because the atmospheric pressure would

increase and the pressure inside the balloon would increase to

balance this change

As you go up or down the atmospheric pressure changes and the

popping is the inner ear adjusting to the pressure change

3/1/2017 Physics 214 Spring 2017 38

Q15 Is it possible for a solid metal ball to float in mercury?

Q16 A rectangular metal block is suspended by a string in a

breaker of water so that the block is completely surrounded by

water. Is the water pressure at the bottom of the block equal to,

greater than, or less than the water pressure at the top of the

block?

The upward force is the weight of liquid displaced and the

downward force is the weight of the ball. If the density of the

liquid is greater than that of the ball it will float.

The pressure is ρgh so the pressure is higher at the bottom. The

difference in pressure provides the upward force on the block

3/1/2017 Physics 214 Spring 2017 39

Q19 A large bird lands on a rowboat that is floating in a swimming

pool. Will the water level in the pool increase, decrease, or remain

the same when the bird lands on the boat?

Q20 A rowboat is floating in a swimming pool when the anchor

is dropped over the side. When the anchor is dropped, will the

water level in the swimming pool increase, decrease, or remain

the same?

The buoyant force is the weight of liquid displaced so to

support a larger weight more liquid is displaced and the level

rises

When the anchor is in the boat it’s whole weight is supported and the

amount of water displaced balances that weight. When it is thrown

overboard it sinks and only displaces it’s volume so the water level

falls

3/1/2017 Physics 214 Spring 2017 40

Q22 If an object has the same density as water, will the object float

to the top, sink to the bottom, or take neither course?

Providing the object and water are incompressible the object

will stay at whatever depth it is placed. It will not sink or rise.

3/1/2017 Physics 214 Spring 2017 41

Ch 9 E 4

Pressure of gas in piston = 300 N/m2.

Area of Piston = 0.2m2.

What is force exerted by piston on gas?

P = F/A, F = PA = 300 N/m2 (0.2 m2) = 60 N

A = 0.2m2

p = 300 N/m2

3/1/2017 Physics 214 Spring 2017 42

Ch 9 E 6

Hydraulic system: A2 = 50 A1

F2 = 6000 N. What is F1?

Pressure is the same just underneath each piston

F1/A1 = F2/A2

F2/F1 = A2/A1 = 50A1/A1 = 50

F1 = F2/50 = 6000/50 = 120N

A1 A2

F2 = 6000N.

3/1/2017 Physics 214 Spring 2017 43

Ch 9 E 8 T = constant, P1 = 10 kPa, V1 = 0.6 m3.

P2 = 90 kPa, V2 = ?

P1V1 = P2V2

(10 kPa)(0.6 m3)/(90 kPa) = V2 = 1/15 = 0.067 m3

P1 = 10 kPa

V1 = 0.6 m3 P2 = 90 kPa

V2 = ?

3/1/2017 Physics 214 Spring 2017 44

Ch 9 E 12

Boat displaces 2.5 m3 of water.

Density of water H2O = 1000 kg/m3.

a) What is the mass of water displaced?

b) What is the buoyant force?

a) Mass of fluid displaced

(mFD) = volume x density of fluid.

MFD = VFDH2O = (2.5 m3)(1000 kg/m3) = 2500 kg

b) Buoyant force equals weight of fluid displaced.

Fb = WFD = mFD g = (2500 kg)(9.8 m/s2) = 24500 N

Fb = WFD

3/1/2017 Physics 214 Spring 2017 45

Ch 9 E 14

Stream moves at v1 = 0.5 m/s in cross sectional area A1.

Stream reaches point where A2 = ¼ A1.

What is v2?

a) v1A1 = v2A2

(0.5 m/s)(A1) = v2(¼A1)

v2 = 2 m/s

V1 V2

A1 A2

3/1/2017 Physics 214 Spring 2017 46

Ch 9 E 16

Wing has a cross sectional area A = 10 m2. Wing

experiences Lift = 60000 N.

What is the difference in air pressure b/w top and

bottom of wing?

a) P = F/A , Pb – PA = F/A = 60000N/10m2 = 6000Pa

A = 10m2

Pb

Pt

60000 N

3/1/2017 Physics 214 Spring 2017 47

Ch 9 CP 2 Water density = H2O = 1000 kg/m3. Depth of swimming pool = 3m.

a) What is the volume of a column of water 3m deep and cross

sectional area 0.5 m2?

b) What is its mass?

c) What is its weight?

d) What is the excess pressure exerted on the pool bottom?

e) Compare to atmospheric pressure.

a) V = Ad = (0.5 m2)(3m) = 1.5 m3

b) M = V = (1.5 m3)(1000 kg/m3) = 1500 kg

c) W = Mg = (1500 kg)(9.8 m/s2) = 14700 N

d) P = F/A = 14700N/0.5m2 = 29400 Pa

e) Atmospheric Pressure is about 100 kPa

P is about 30 kPa

P/Atm = (29400 Pa)/(1.013 x 105 Pa) = 0.29

3m

0.5m2

3/1/2017 Physics 214 Spring 2017 48

Ch 9 CP 4 Wooden boat: 3m x 1.5m x 1m that carries five people.

Total mass of boat and people equals 1200 kg.

a) What is total weight?

b) What is buoyant force required to float?

c) What volume of water must be displaced to float?

d) How much of the boat underwater?

a) W = Mg = 1200 kg (9.8 m/s2)

W = 11760 N

W Fb

3m

1m

b) Fnet = Fb – W = 0

Fb = 11760 N

c) Fb = H2O Vg (see Ch 9 E 12)

Fb/H2Og = 11760N/(1000 kg/m3)(9.8 m/s2) = V = 1.2 m3

d) V = LWh = (3m)(1.5m)h = 1.2 m3 h = 0.27 m