This Week

Liquids and GasesPressure

How do they lift your car for service?Atmospheric pressure

We’re submerged!How can you drink a Coke?

Archimedes ! Eureka!!!Balloons of all sizesBubbles of all sizes

7/20/2017 Physics 214 Fall 2017 1

http://wimp.com/slinkyanswer/

7/20/2017 Physics 214 Fall 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

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PressureA 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 containersand exert forces over a surface.

We define pressure as P = F/A

that is the force divided by the area over which the force acts.

Any change in pressure is transmitted uniformly throughout the liquid.Units are N/m2 and 1 N/m2 = 1Pascal

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Hydraulic JackIf the liquid levels oneach side are the same heightthen the pressure just beloweach piston is the same andF1/A1 = F2/A2 so F2 = F1A2/A1

If we make A2/A1 = 100 then anF1 force of 50lbs can lift 5000lbs.

Of course if we push F1 down a distance h1 then the F2Side will only rise h1 A1/A2 because h1A1 = h2A2, that is the displaced volumes must be the same.Work done = F1h1 = F2h2

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Atmospheric pressureIf I stack a pile of bricks each brick has to support allthe bricks above it. So as we go down the stack theforce increases as does the pressure.

In the example shown the force on the faceof the bottom brick would be 6mg and the pressure = 6mg/A.

At the earths surface we are supportinga column of air which exerts a force andbecause it is a gas it exerts an equal pressureIn all directions.

g

F

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2A-01 Suction CupsHow 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

FF

If two cups are pushed together, does this make

it twice as difficult to separate ?

PA

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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 horses attempt to disengage the hemispheres.

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2A-02 Magdeburg Hemispheres

This more modern version of the famous Magdeburg Hemispheres uses two disks instead of hemispheres, but the principle is the same. Two disks are placed together face to face. The inner portions of the disks are hollowed out so that when they are placed together, a hollow chamber is formed. If the chamber is evacuated, the disks are almost impossible to separate because of the large pressure forces holding the disks together.

Otto von Guericke, Burgomeister von Magdeburg, used 16 horses, eight on each side.What would be different if only eight horses were used and the other half of the sphere attached to a solid support such as a tree?

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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 inLift piston diameter = 1.25 inF1= 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

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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 = PAA pressure applied over an area

can generate a large force.

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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?

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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

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

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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/ccW = 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

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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 DemosEffects 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.

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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 CANNOTHAVE 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 BBA

hg

What will happen to the reading on the manometer as

the sensor is rotated ?

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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

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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 hAND NOT THE VOLUME OR SHAPE OF THE VESSEL.

P = ρgh

Examining the Pressure at the bottom of differently shaped vessels filled with fluid

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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

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2B-07 Pressure in a liquid

The speed with which the liquid exits the holeis given by

v = sqrt(2gh)

where h is the depth below the surfaceof the liquid. The time for the water to fall to the surface is

t = sqrt(2(H-h)/g)which gives for the range

R = vt =2sqrt(h(H-h))

This shows that there are two positions (interchange h and H - h) that give the same range and also that the range is maximized when h = H - h, or h = H/2.

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Effects of Atmospheric PressureIn our everyday lives we do not feel we are supporting a very large weightbecause it is the environment in which we developed.

If we immerse a tube in a liquidSo that all the air bubbles are removed and then we raise the tube to a vertical position we find that atmospheric pressure supports 76cm of mercury or 32 feet of water.

Knowing the weight of the Mercury and the area of the tube we find that 1 Atmosphere = 1.013 105 Pa1 Atmosphere = 14.696 pounds/sq inThis is equal to ρgh for the Mercury

No Air

g

P is the same

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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.

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Suction cupQuestion: 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

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Archimedes PrincipleIf 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 displacedthis 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

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Floating and buoyant forceIf 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 objectρliquid 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

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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

FBScale reads tension in cord: T = Mg – FBMg = 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.

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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.

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2B-10 Archimedes II

A. The block is not immersed T1 = Mg B. The block is immersed but the liquid runs out T2 = Mg – FBC. The displaced liquid is poured into the can T3 = Mg – FB + WDFT1 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

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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.

mbgmag

ρairgVbρairgVa

Setting the sum of torques on equal-arm balance about pivot equal to zero, we have in the presence of air:

mag – ρairgVa = mbg - ρairgVbVb > Va implies mb > ma which is demonstrated in

vacuum

Investigating the Buoyant force resulting from Air

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2B-11 Buoyancy ParadoxWhen the hanging weight is submerged, a buoyant force F pushes up on the weight. At the same time a reaction force equal to Fpushes down on the right hand plate. Thus a weight equal to 2F must be added to the left side to restore balance. For this apparatus, the hanging weight has a volume of 100 cm3, so the “weight” on the left pan is reduced by 100 g when the weight is completely submerged. At the same time the “weight” on the right hand side is increased by 100 g (see discussion above regarding the reaction force). Therefore, to restore equilibrium 200 g must be added to the left pan.

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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 BBA

h

g

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Summary of Chapter 9

P = F/A Pascals

F1/A1 = F2/A2

Work done = F1h1 = F2h2

1 Atmosphere = 1.013 105 Paand will support

76cm of mercury 32 feet of water

No Air

g

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Liquids

P = W/A = ρghWater is 1 gram/cc1 gram/cm3 = 1000kg/m3

buoyant force = the weight of liquid displacedT + FB = W = mgFor a floating object T = 0

g

T

mg

Fb

Our World

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BalloonsAny 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

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Surface tensionThe 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

Worked Questions and Problems

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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

7/20/2017 Physics 214 Fall 2017 38

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.

7/20/2017 Physics 214 Fall 2017 39

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

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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

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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

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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

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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

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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.

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Ch 9 E 4Pressure 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

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Ch 9 E 6Hydraulic system: A2 = 50 A1

F2 = 6000 N. What is F1?

Pressure is the same just underneath each pistonF1/A1 = F2/A2

F2/F1 = A2/A1 = 50A1/A1 = 50F1 = F2/50 = 6000/50 = 120N

A1 A2

F2 = 6000N.

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Ch 9 E 8T = 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 = ?

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Ch 9 E 12Boat 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

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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

A1A2

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Ch 9 E 16Wing 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

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Ch 9 CP 2Water 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 kgc) 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 kPaP is about 30 kPaP/Atm = (29400 Pa)/(1.013 x 105 Pa) = 0.29

3m

0.5m2

7/20/2017 Physics 214 Fall 2017 52

Ch 9 CP 4Wooden 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 = 0Fb = 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