Fluids Statics Mp

7/22/2019 Fluids Statics Mp

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Fluids Statics

Due: 8:00pm on Wednesday, September 21, 2011

Note: You will receive no credit for late submissions.To learn more, read your instructor'sGrading Policy

[Switch to Standard Assignment View]

Fluid Pressure in a U-Tube

A U-tube is filled with water, and the two arms are capped.

The tube is cylindrical, and the right arm

has twice the radius of the left arm. The caps have negligible mass, are watertight, and canfreely slide up and down the tube.

Part A

A one-inch depth of sand is poured onto the cap on each arm.

After the caps have moved (if necessary)
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to reestablish equilibrium, is the right cap higher, lower, or the same height as the left cap?

Hint A.1 Pressure at the surface

H in t not displayed

Hint A.2 Evaluate the weight of the sand


Hint A.3 Evaluate the pressure applied by the sand


ANSWER: higher

lower

the same height

Correct

Although one inch of sand on the right cap is much heavier than one inch of sand on the leftcap, the pressures exerted by the sand are the same on both caps. Since the pressures exertedby the sand are equal, the pressures at the base of each arm due to the water must be equal.This requires equal heights of water in the two arms.

Part B

The sand is removed and a 1.0- -mass block is placed on each cap.

After the caps have moved (if necessary)

to reestablish equilibrium, is the right cap higher, lower, or the same height as the left cap?Hint B.1 Evaluate the pressure applied by the 1.0- blocks


ANSWER: higher

lower


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the same height

Correct

Although the masses of the blocks are equal, the pressures exerted by them on the caps are

not equal. There is a greater pressure on the left cap, which results in a greater pressure at thebase of the left arm. To compensate for this increased pressure, the height of the water

column in the right arm will have to be greater than in the left arm.

Part C

If a 1.0- -mass block is on the left cap, how much total mass must be placed on the right

cap so that the caps equilibrate at equal height?

Hint C.1 Meaning of equal water levels


Express your answer in kilograms.

ANSWER: 4.0Correct

Part D

The locations of the two caps at equilibrium are now as given in this figure.

The dashed line represents the level of thewater in the left arm.

What is the mass of the water located in the right arm between the dashed line and the rightcap?

Hint D.1 Pressure at the dashed line


Express your answer in kilograms.

ANSWER: 3.0


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Correct

Pressure in the Ocean

The pressure at 10 below the surface of the ocean is about 2.00105 .

Part A

Which of the following statements is true?

Hint A.1 How to approach the problem


ANSWER:The weight of a column of seawater 1 in cross section and 10 high

is about 2.00105 .

The weight of a column of seawater 1 in cross section and 10 highplus the weight of a column of air with the same cross section extending

up to the top of the atmosphere is about 2.00105 .

The weight of 1 of seawater at 10 below the surface of the ocean is

about 2.00105 .

The density of seawater is about 2.0010 times the density of air at sea

level.

Correct

The pressure at a given level in a fluid is caused by the weight of the overlying fluidplusany

external pressure applied to the fluid. In this case, the external pressure is the atmosphericpressure. Tropical storms are formed by low-pressure systems in the atmosphere. When such

a storm forms over the ocean, both the atmospheric pressure and the fluid pressure in the

ocean decrease.

Part B

Now consider the pressure 20 below the surface of the ocean. Which of the followingstatements is true?

Hint B.1 Variation of pressure with height


ANSWER: The pressure is twice that at a depth of 10 .

The pressure is the same as that at a depth of 10 .

The pressure is equal to that at a depth of 10 plus the weight per 1cross sectional area of a column of seawater 10 high.

The pressure is equal to the weight per 1 cross sectional area of a


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column of seawater 20 high.

Correct

Pressure on a Dam

As the reservoir behind a dam is filled with water, the pressure that the water exerts on the

dam increases. Eventually, the force on the dam becomes substantial, and it could cause the

dam to collapse. There are two significant issues to be considered: First, the base of the dam

should be able to withstand the pressure , where is the density of the water behind the

dam, is its depth, and is the magnitude of the acceleration due to gravity. This means thatthe material of which the dam is made needs to be strong enough so that it doesn't crack

(compressive strength).

The second issue has to do with the strength of the foundation of the dam. The water pressureexerts a clockwise torque on the dam, as shown in the figure. The foundation of the damshould be strong enough so that the dam does not topple. The material has to be strong

enough that the dam does not snap (shear strength).

To study this phenomenon, consider the simple model of a dam depicted in the diagram.

A reservoir of water (density ) behind the

dam is filled to a height . Assume that the width of the dam (the dimension pointing into the

screen) is .

Part A

Consider a horizontal layer of the dam wall of thickness located a distance above the

reservoir floor. What is the magnitude of the force on this layer due to the water in the

reservoir?


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Hint A.1 Find the pressure as a function of depth

What is the pressure of the water at a vertical height above the floor of the reservoir?

Express your answer in terms of and any quantities from the problem introduction.

ANSWER:= Answer not displayed

Express your answer in terms of , , the magnitude of the acceleration due to gravity ,and any quantities from the problem introduction.

ANSWER:=

All attempts used; correct answer displayed

Part B

The force of the water produces a torque on the dam. In a simple model, if the torque due to

the water were enough to cause the dam to break free from its foundation, the dam would

pivot about its base (pointP). What is the magnitude of the torque about the pointPdue tothe water in the reservoir?

Hint B.1 How to approach the problem


Hint B.2 Definition of torque


Hint B.3 Find the torque on a horizontal layer of the dam


Hint B.4 Some helpful integrals


Express your answer in terms of quantities given in the problem introduction and the

magnitude of the acceleration due to gravity .

ANSWER:

=

Correct

Thus the average moment arm is . Therefore, since the water exerts a greaterforce at the bottom of the dam than at the top, the net effect is as if the force were acting at a

height of from the bottom.

A Water Tank on Mars

You are assigned the design of a cylindrical, pressurized water tank for a future colony on

Mars, where the acceleration due to gravity is 3.71 meters per second per second. The


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pressure at the surface of the water will be 100 , and the depth of the water will be 14.3

. The pressure of the air in the building outside the tank will be 95.0 .

Part A

Find the net downward force on the tank's flat bottom, of area 2.20 , exerted by the waterand air inside the tank and the air outside the tank.

Hint A.1 The net force


Hint A.2 What is (a Pascal)?


Hint A.3 Density of water


Hint A.4 Find the force exerted on the tank's bottom by the air outside the tank


Hint A.5 Find the force exerted on the tank's bottom by the water


Express your answer numerically in Newtons, to three significant figures.

ANSWER: 1.2810

Correct

Exercise 14.4: Gold BrickYou win the lottery and decide to impress your friends by exhibiting a million-dollar cube of

gold. At the time, gold is selling for 426.60 per troy ounce, and 1.0000 troy ounce equals31.1035 .

Part A

How tall would your million-dollar cube be?

ANSWER:=

15.6

Correct

Exercise 14.8: Black Smokers

Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them

teem with exotic creatures, and some biologists think that life on earth may have begunaround such vents. The vents range in depth from about 1500 to 3200 below the surface.

Part A

What is the gauge pressure at a 3200- deep vent, assuming that the density of water does


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not vary? Express your answer in pascals.

ANSWER:=

3.2310

Correct

Part B

Express your answer in atmospheres.

ANSWER:=

319Correct

Exercise 14.16

The liquid in the open-tube manometer in Figure 14.9a in the textbook is mercury, = 3.00

, and = 7.00 . Atmospheric pressure is 980 millibars.

Part A

What is the absolute pressure at the bottom of the U-shaped tube?ANSWER:

=1.0710Correct

Part B

What is the absolute pressure in the open tube at a depth of 4.00 below the free surface?

ANSWER:=

1.0310

Correct

Part C

What is the absolute pressure of the gas in the tank?ANSWER:

=1.0310

Correct

Part D

What is the gauge pressure of the gas in pascals?

ANSWER:=

5330Correct

Exercise 14.25: Hydraulic Lift IIThe piston of a hydraulic automobile lift is 0.30 in diameter.

Part A

What gauge pressure, in pascals, is required to lift a car with a mass of 1200 ?

Express your answer using two significant figures.


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ANSWER:=

1.710Correct

Part B

Express the result of part A in atmospheres.Express your answer using two significant figures.

ANSWER:=

1.6Correct

Test Your Understanding 14.1: Average Density

Part A

Rank the following objects in order of their mass, from largest value to smallest value.

ANSWER:

View

Correct

The average density of an object is its mass divided by its volume. Hence an object's mass

equals its average density multiplied by its volume. Being careful with units, we find thefollowing masses for the five objects:

Cube: mass = ( )(3.00 ) =

Sphere: mass = ( )( )(1 ) =

Rectangular solid: mass = ( )(1.50 ) =

Cylinder: mass = ( )(3.20 ) =

Cone: mass = ( )( )(1 ) =

Test Your Understanding 14.2: Pressure in a Fluid

Two identical vertical tubes, each 1.00 in diameter and 50.0 tall, are each filled with a

liquid. The liquid in tube #1 has density 1500 , whereas the liquid in tube #2 has density2000 . The top of each tube is exposed to the air.

Part A

The pressure in the air above the two tubes is now increased from 1.00 atm to 2.00 atm. In

which tube does this cause the greatest pressure increaseat the bottom of the tube?
http://openappletwindow%28%27850x500%27%2C%27assignmentproblemid%3D12493715%26partid%3D1%26expressionid%3D%26printview%3D1%26hidedetails%3D1%27%2C%27applet_id5849062%27%29/http://openappletwindow%28%27850x500%27%2C%27assignmentproblemid%3D12493715%26partid%3D1%26expressionid%3D%26printview%3D1%26hidedetails%3D1%27%2C%27applet_id5849062%27%29/http://openappletwindow%28%27850x500%27%2C%27assignmentproblemid%3D12493715%26partid%3D1%26expressionid%3D%26printview%3D1%26hidedetails%3D1%27%2C%27applet_id5849062%27%29/


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ANSWER: tube #1

tube #2

the pressure increase is the same in both tubes

not enough information is given to decide

Correct

Pascal's law states that an additional pressure applied at one point in a fluid is transmitted

undiminished to all other points in the fluid. In this situation an additional pressure of 1.00

atm is applied to the top of each fluid-filled tube, so this causes an additional pressure of1.00 atm at the bottom of each tube. It's true that the value of the totalpressure at the bottom

of tube #1 is different from that at the bottom of tube #2 (because the liquids have different

densities), but that's not what the question is about.

Problem 14.83

A U-shaped tube with a horizontal portion of length (the figure

) contains a liquid.

Part A

What is the difference in height between the liquid columns in the vertical arms if the tube

has an acceleration toward the right?

Express your answer in terms of the variables , and appropriate constants.

ANSWER:=

Correct

Part B

What is the difference in height between the liquid columns in the vertical arms if the tube is

mounted on a horizontal turntable rotating with an angular speed with one of the vertical


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arms on the axis of rotation?

Express your answer in terms of the variables , and appropriate constants.

ANSWER:

=

Correct

Part C

Explain why the difference in height does not depend on the density of the liquid or on thecross-sectional area of the tube.

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer:The density is identical and constant for both ends of thetube. As expressed in poth part b and a, delta y isindependent of rho due to cancellation in the derivation ofdelta y expressions.

Part D

Would answer be the same if the vertical tubes did not have equal cross-sectional areas?

ANSWER: Yes

No

Correct

Part E

Would answer be the same if the horizontal portion were tapered from one end to the other?

ANSWER: Yes

No

Correct

Part F

Explain.Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer:Similar to considering the effect of density, cross sectional

area does not appear within the expression for rotational orlinear motion.

Problem 14.85

An incompressible fluid with density is in a horizontal test tube of inner cross-sectional area

. The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed


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Gravitational forces are negligible. Consider a volume element of the fluid of area and

thickness a distance from the rotation axis. The pressure on its inner surface is and on

its outer surface is .

Part A

Apply Newton's second law to the volume element to show thatEssay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer:Newton's second law States that F=ma cp. In our situation,acceleration is substituted with acp= v^2/r, with angularvelocity v=wr replacing v.We now have m(w^2r) m=Arho*dr.Arho(w^2*r)=F, and F=Ap, A cancels and we have p=rho(w^2*r)dr

Part B

If the surface of the fluid is at a radius where the pressure is show that the pressure at a

distance is .Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer:Integrating the given formula from part A yields results inthe formula from part B, where rho amd omega are constants, ris the variable and p and p_0 are the endpoints ofintegration.

Part C

An object of volume and density has its center of mass at a distance from the axis.

Show that the net horizontal force on the object is , where is the distance fromthe axis to the center of mass of the displaced fluid.Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer:I'm sorry Professor, but I didn't have time for these two. Iwill complete them in writing and submit them to you directly.

Part D

Explain why the object will move inward if and outward if .Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer:I'm sorry i ran out of time for these last two questions

Part E

For small objects of uniform density, . What happens to a mixture of smallobjects of this kind with different densities in an ultracentrifuge?

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).


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ANSWER: My Answer:Objects with larger densities will move away from the centerof mass and vice versa for objects with less density

Exercise 14.12

A barrel contains a 0.130-m layer of oil floating on water that is 0.235 deep. The density of

the oil is 610

Part A

What is the gauge pressure at the oil-water interface?

ANSWER:=

777

Correct

Part B

What is the gauge pressure at the bottom of the barrel?

ANSWER:=

3080Correct

Exercise 14.14

You are designing a diving bell to withstand the pressure of seawater at a depth of 280 .

Part A

What is the gauge pressure at this depth? (You can ignore changes in the density of the water

with depth.)ANSWER:

=2.8310

Correct

Part B

At this depth, what is the net force due to the water outside and the air inside the bell on a

circular glass window 31.0 in diameter if the pressure inside the diving bell equals thepressure at the surface of the water? (You can ignore the small variation of pressure over the

surface of the window.)

ANSWER:=

2.1410Correct

Exercise 14.18

A tall cylinder with a cross-sectional area 12.0 is partially filled with mercury; thesurface of the mercury is 6.00 above the bottom of the cylinder. Water is slowly poured in

on top of the mercury, and the two fluids don't mix.


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Part A

What volume of water must be added to double the gauge pressure at the bottom of thecylinder?

ANSWER:=

979

Correct

Exercise 14.21

An electrical short cuts off all power to a submersible diving vehicle when it is a distance of

26 below the surface of the ocean. The crew must push out a hatch of area 0.750 and

weight 200 to escape.

Part A

If the pressure inside is , what downward force must they exert on the hatch to open

it?Express your answer using two significant figures.

ANSWER:=

2.010

Correct

Archimedes' Principle

Learning Goal: To understand the applications of Archimedes' principle.Archimedes' principleis a powerful tool for solving many problems involving equilibrium in

fluids. It states the following:

When a body is partially or completely submerged in a fluid (either a liquid or a gas), the fluidexerts an upward force on the body equal to the weight of the fluid displaced by the body.

As a result of the upward Archimedes force (often called the buoyant force), some objects

may float in a fluid, and all of them appear to weigh less. This is the familiar phenomenon ofbuoyancy.

Quantitatively, the buoyant force can be found as

,

where is the force, is the density of the fluid, is the magnitude of the accelerationdue to gravity, and is the volume of the displaced fluid.

In this problem, you will be asked several qualitative questions that should help you develop afeel for Archimedes' principle.

An object is placed in a fluid and then released. Assume that the object either floats to the

surface (settling so that the object is partly above and partly below the fluid surface) or sinks


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to the bottom. (Note that for Parts A through D, you should assume that the object has settled

in equilibrium.)

Part A

Consider the following statement:

The magnitude of the buoyant force is equal to the weight of fluid displaced by the object.

Under what circumstances is this statement true?

Hint A.1 Archimedes' principle


ANSWER: for every object submerged partially or completely in a fluid

only for an object that floats

only for an object that sinks

for no object submerged in a fluid

Correct

Use Archimedes' principle to answer the rest of the questions in this problem.

Part B


The magnitude of the buoyant force is equal to the weight of the amount of fluid that has thesame total volume as the object.


Hint B.1 Consider Archimedes' principle


ANSWER: for an object that is partially submerged in a fluid

only for an object that floats

for an object completely submerged in a fluid

for no object partially or completely submerged in a fluid

Correct

Part C


The magnitude of the buoyant force equals the weight of the object.



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Hint C.1 Forces and equilibrium



for an object that floats

only for an object that sinks


Correct

Part D


The magnitude of the buoyant force is less than the weight of the object.Under what circumstances is this statement true?


for an object that floats

for an object that sinks


Correct

Now apply what you know to some more complicated situations.

Part E

An object is floating in equilibrium on the surface of a liquid. The object is then removedand placed in another container, filled with a denser liquid. What would you observe?

Hint E.1 Density and equilibrium


ANSWER: The object would sink all the way to the bottom.

The object would float submerged more deeply than in the firstcontainer.

The object would float submerged less deeply than in the first container.

More than one of these outcomes is possible.


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Correct

Part F

An object is floating in equilibrium on the surface of a liquid. The object is then removed

and placed in another container, filled with a less dense liquid. What would you observe?Hint F.1 Density and equilibrium


ANSWER: The object would sink all the way to the bottom.

The object would float submerged more deeply than in the first

container.

The object would float submerged less deeply than in the first container.

More than one of these outcomes is possible.

Correct

If the fluid in the second container is less dense than the object, then the object will sink all

the way to the bottom. If the fluid in the second container is denser than the object (though

less dense than the fluid in the original container), the object will still float, but its depth willbe greater than it was in the original container.

Part G

Two objects, T and B, have identical size and shape and have uniform density. They are

carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of

object T is submerged than of object B, which floats, fully submerged, closer to the bottomof the container. Which of the following statements is true?

ANSWER: Object T has a greater density than object B.

Object B has a greater density than object T.

Both objects have the same density.

Correct

Since both objects float, the buoyant force in each case is equal to the object's weight. Block

B displaces more fluid, so it must be heavier than block T. Given that the two objects havethe same volume, block B must also be denser. In fact, since the weight equals the buoyant

force, and B is fully submerged, , where all the symbols have their usualmeaning. From this equation, one can see that the density of B must equal the density of the

fluid.

18-Karat Gold vs. 14-Karat Gold


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Gold in its pure form is too soft to be used for most jewelry. Therefore, the gold is mixed with

other metals to produce an alloy. The composition of gold alloys are always calculated by

mass, using the karat (kt) as a unit of measure. A karat represents a proportion by mass of one

part in twenty-four. The higher the karat value, the higher the proportion of gold in relation tothe total metal content. Pure gold is therefore 24 karat, while an 18-karat gold alloy contains

(at least) 18 parts (by mass) of gold out of 24 parts total.Part A

In a sample of 18-karat gold, 75 percent of the total mass is pure gold, while the rest istypically 16 percent silver and 9 percent copper. If the density of pure gold is

, while the densitites of silver and copper are respectively

and , what is the overall density of this alloy of 18-karat gold?



Hint A.2 Find the volume of pure gold in 1 gram of alloy


Express your answer in grams per cubic centimeter to three significant figures.

ANSWER:=

15.6

Correct

Part B

The percentage composition by mass of the less expensive 14-karat gold is typically 58.5

percent gold, 4.0 percent silver, 31.2 percent copper, and 6.3 percent zinc.

Find the percentage composition by volume of 14-karat gold. The density of zinc is

.



Hint B.2 Find the volume of pure gold in 1 gram of alloy


Enter four answers as percentages, separated by commas. Round each to the nearest integer.

ANSWER:=

39,5,45,11Correct

In a sample of 14-karat gold, less than 50 percent of the volume is pure gold!

A Submerged Ball


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A ball of mass and volume is lowered on a string into a fluid of density .

Assume that the object would sink to thebottom if it were not supported by the string.

Part A

What is the tension in the string when the ball is fully submerged but not touching thebottom, as shown in the figure?

Hint A.1 Equilibrium condition


Hint A.2 Find the magnitude of the buoyant force


Express your answer in terms of any or all of the given quantities and , the magnitude of

the acceleration due to gravity.ANSWER:

= Correct

Buoyant Ping-Pong Ball

A Ping-Pong ball is held submerged in a bucket of water by a string attached to the bucket's

bottom.

Part A

Salt is now added to the water in the bucket, increasing the density of the liquid. Whathappens to the tension in the string ?



Hint A.2 Determine the effects of a density change



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ANSWER: The tension does not change.

The tension increases.

The tension decreases.

Correct

Part B

What happens to the tension in the string if the Ping-Pong ball is replaced by a smaller

spherical object of equal weight?



Hint B.2 Determine the effect of a volume change


ANSWER: The tension does not change.

The tension increases.

The tension decreases.

Correct

Force on a Goldfish Vector Drawing

A fishbowl contains a single goldfish and is filled with water to the level indicated.

Part A

At each of the designated points, rotate the given vector to indicate the direction of the forceexerted by the water on either the inside of the fishbowl (for pointsAandB) or the outside of

the goldfish (for points C,D, andE).

Hint A.1 Direction of the force due to the water


The orientation of your vectors will be graded.

ANSWER:

View

Correct

Submerged Block

A beaker contains a thick layer of oil (shown in green) of density floating on water (shown
http://openappletwindow%28%270%27%2C%27assignmentproblemid%3D12531364%26partid%3D1%26expressionspos%3D1%26printview%3D1%26hidedetails%3D1%27%2C%27applet_id5818275%27%29/http://openappletwindow%28%270%27%2C%27assignmentproblemid%3D12531364%26partid%3D1%26expressionspos%3D1%26printview%3D1%26hidedetails%3D1%27%2C%27applet_id5818275%27%29/http://openappletwindow%28%270%27%2C%27assignmentproblemid%3D12531364%26partid%3D1%26expressionspos%3D1%26printview%3D1%26hidedetails%3D1%27%2C%27applet_id5818275%27%29/


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in blue), which has density . A cubical block of wood of density with side length isgently lowered into the beaker, so as not to disturb the layers of liquid, until it floats

peacefully between the layers, as shown.

Part A

What is the distance between the top of the wood cube (after it has come to rest) and theinterface between the oil and water?



Hint A.2 Find the buoyant force


Hint A.3 Find the gravitational force


Express your answer in terms of quantities given in the problem introduction.

ANSWER:

=Correct

In solving this problem, there is an implicit assumption that . In examining the

solution just obtained, one sees that since by assumption (the oil is less dense than

water), we must have (the wood block is less dense than water) and ,

or (the oil is less dense than the wood block). The only way that the situationdescribed in the problem introduction (with the block of wood floating "peacefully") can

exist is if . What would happen if this were not the case?

Submerged Sphere in a Beaker

A cylindrical beaker of height 0.100 and negligible weight is filled to the brim with a fluid


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of density = 890 . When the beaker is placed on a scale, its weight is measured to be

1.00 .

A ball of density = 5000 and volume = 60.0 is then submerged in the fluid,so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiffrod of negligible volume and weight. Throughout the problem, assume the acceleration due to

gravity is = 9.81 .

Part A

What is the weight of the ball?

Hint A.1 Find the mass of the ball


Hint A.2 Converting between cubic meters and cubic centimeters


Express your answer numerically in newtons.

ANSWER:=

2.94

Correct

Part B

What is the reading of the scale when the ball is held in this submerged position? Assumethat none of the water that spills over stays on the scale.



Hint B.2 Find the pressure at the bottom of the beaker



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Hint B.3 Find the pressure before the ball is submerged


Hint B.4 The force exerted by the ball on the water

H in t not displayedCalculate your answer from the quantities given in the problem and express it numerically in

newtons.

ANSWER:=

1.00

Correct

The "extra force" that the ball exerts on the water is equal to the force that the water exerts

on the ball, that is, the weight of the displaced water. Therefore, the reading does not change.

Part C

What is the force applied to the ball by the rod? Take upward forces to be positive (e.g., if

the force on the ball is downward, your answer should be negative).Hint C.1 How to approach the problem


Hint C.2 Find the buoyant force


Hint C.3 Find the expression for the net force



ANSWER:=

2.42Correct

The force does act upward as one would expect because the ball is denser than the fluid andwould have sunk if it weren't for the rod.

The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled

with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and


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submerged ball is placed on the scale.

Part D

What weight does the scale now show?

Hint D.1 How to approach the problem


Hint D.2 Find the weight of the fluid in the beaker



ANSWER:=

3.42Correct

Note that in this case it is not necessary to know the buoyant force acting on the ball in toanswer the question.

Pressure in a Centrifuge

A test tube filled with liquid of uniform density , as shown in the figure, is spun in a


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centrifuge with angular frequency . The

test tube lies perpendicular to the axis of rotation of the centrifuge. The pressure in the fluid at

the distance from the axis of rotation is . You may ignore the variation in pressure with

depth; asume that it is much smaller than the variation in pressure with radius.

Part A

What is the pressure of the fluid in the test tube at an arbitrary distance from the axis of

rotation?


The key to solving the problem is to find a relationship between pressure and centripetalacceleration. The pressure (or more appropriately, the pressure gradient) is what provides

the centripetal acceleration that keeps the fluid moving in a circle around the axis of

rotation.

Hint A.2 More on how to approach the problem

Using the hints if necessary, derive a differential equation for the change in pressure across

a "slice" of water between and . Integrate this equation to obtain the pressure as a

function of radius up to an arbitrary constant. Determine this constant of integration from

the condition .

Hint A.3 Centripetal force

In order for the "slice" of water between and to move in a circle around the axis of

the centrifuge, there must be a centripetal force acting on it. What is the magnitude of the

centripetal force needed to keep the slice of water between and moving in a circle of

radius with angular velocity ?

Hint A.3.1 Find the mass of a slice of water


Express your answer in terms of , , , , and , the cross-sectional area of the testtube.


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ANSWER:= Answer not displayed

Hint A.4 Find the change in pressure


Hint A.5 Rewrite the change in pressure


Hint A.6 Solving differential equations


Hint A.7 A helpful integral


Express your answer in terms of , , , , and .

ANSWER:

=Correct

Problem 14.64

A hydrometer consists of a spherical bulb and a cylindrical stem with a cross-section area of

0.400 . The total volume of bulb and stem is 13.2 . When immersed in water, thehydrometer floats with 8.00 of the stem above the water surface. When immersed in anorganic fluid, 3.20 of the stem is above the surface. (Note:This illustrates the precision of

such a hydrometer. Relatively small density differences give rise to relatively largedifferences in hydrometer readings.)

Part A

Find the density of the organic fluid.

ANSWER:=

839Correct

Problem 14.66: SHM of a Floating Object

An object with height , mass , and a uniform cross-sectional area floats upright in aliquid with density .

Part A

Calculate the vertical distance from the surface of the liquid to the bottom of the floatingobject at equilibrium.

Express your answer in terms of the given quantities.


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ANSWER:

=Correct

Part B

A downward force with magnitude is applied to the top of the object. At the newequilibrium position, how much farther below the surface of the liquid is the bottom of the

object than it was in part A? (Assume that some of the object remains above the surface ofthe liquid.)


ANSWER:

=

Correct

Part C

Your result in part B shows that if the force is suddenly removed, the object will oscillate upand down in SHM. Calculate the period of this motion in terms of the density of the liquid,

the mass , and cross-sectional area of the object. Neglect the damping due to fluidfriction.


ANSWER:

=

Correct

Exercise 14.28

You are preparing some apparatus for a visit to a newly discovered planet Caasi having

oceans of glycerine and a surface acceleration due to gravity of 4.15 .

Part A

If your apparatus floats in the oceans on earth with 27.0 of its volume submerged, whatpercentage will be submerged in the glycerine oceans of Caasi?

ANSWER:

=

22.1

Correct

Problem 14.60

A hot-air balloon has a volume of 2200 . The balloon fabric (the envelope) weighs 860 .

The basket with gear and full propane tanks weighs 1500 .


28/28

Part A

If the balloon can barely lift an additional 3200 of passengers, breakfast, and champagne

when the outside air density is , what is the average density of the heated gases in

the envelope?

ANSWER:=

0.972

Correct

Score Summary:

Your score on this assignment is 94.3%.

You received 218.77 out of a possible total of 232 points.[Print]

Fluids Statics Mp

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