Section 12.1­2

Due: 10:00am on Tuesday, August 4, 2015

You will receive no credit for items you complete after the assignment is due. GradingPolicy

An Iron Rod

Part A

On a part­time job, you are asked to bring a cylindrical iron rod of density7800 , length 92.6 and diameter 3.00 from a storage roomto a machinist. Calculate the weight of the rod, . The acceleration due togravity, = 9.81 .

Express your answer numerically in newtons.

Hint 1. General approach

To calculate the weight of a body you need to know its mass. If themass is unknown, and instead the problem gives you information onthe volume of the body and the density of the material of which it ismade, you can calculate the mass of that body by applying thedefinition of density.

Hint 2. Weight of a body

A body of mass has weight with magnitude given by

,

where = 9.81 is the acceleration due to gravity.

Hint 3. Find the mass of the rod

From the length and the cross­sectional area of the rod and thedensity of iron, find the mass of the rod.

Express your answer numerically in kilograms.

Hint 1. Finding mass from volume and density

Consider a mass of homogeneous material, whosevolume is . The density of that material is defined as

.

Hint 2. Volume of a cylinder

The volume of a cylinder of length and cross­sectionalarea is given by

If the cross section is a circle with radius , then

ANSWER:

ANSWER:

Correct

Part B

Now that you know the weight of the rod, do you think that you will be ableto carry the rod without a cart?

Hint 1. Weight and mass

= 5.11

= 50.1

A weight measured in newtons may not be very familiar to you;therefore it is more convenient to determine the correspondingmass measured in kilograms, or in pounds.

Hint 2. Unit conversion factors

The unit convertion factors for newtons, kilograms, and pounds are

,

.

ANSWER:

Correct

Pressure on a Dam

As the reservoir behind a dam is filled with water, the pressure that the waterexerts on the dam increases. Eventually, the force on the dam becomessubstantial, and it could cause the dam to collapse. There are two significantissues to be considered: First, the base of the dam should be able to withstandthe pressure , where is the density of the water behind the dam, is itsdepth, and is the magnitude of the acceleration due to gravity. This meansthat the material of which the dam is made needs to be strong enough so that itdoesn't crack (compressive strength).

The second issue has to do with the strength of the foundation of the dam. Thewater pressure exerts a clockwise torque on the dam, as shown in the figure.The foundation of the dam should be strong enough so that the dam does nottopple. 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

yes

no

diagram. A reservoirof water (density )behind the dam isfilled to a height .Assume that thewidth of the dam (thedimension pointinginto the screen) is .

Part A

Consider a horizontal layer of the dam wall of thickness located adistance above the reservoir floor. What is the magnitude of the forceon this layer due to the water in the reservoir?

Express your answer in terms of , , the magnitude of theacceleration due to gravity , and any quantities from the problemintroduction.

Hint 1. Find the pressure as a function of depth

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

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

ANSWER:

=

ANSWER:

Correct

Part B

The force of the water produces a torque on the dam. In a simple model, ifthe torque due to the water were enough to cause the dam to break freefrom its foundation, the dam would pivot about its base (point P). What isthe magnitude of the torque about the point P due to the water in thereservoir?

Express your answer in terms of quantities given in the problemintroduction and the magnitude of the acceleration due to gravity .

Hint 1. How to approach the problem

Using the answer to Part A, find the torque on a horizontal layer ofthe dam. Then integrate over the entire height of the dam to find thetotal torque.

Hint 2. Definition of torque

The torque due to a force about a point is the product of themagnitude of the force and the moment arm, which is theperpendicular distance from the line of action of the force to thepivot.

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

What is the torque due to the water acting on a horizontal layerof the dam of thickness at position above the bottom of thereservoir?

Express your answer in terms of , , the gravitationalaccleration , and any quantities needed from the problemintroduction.

=

Hint 1. Definition of torque

The torque due to a force about a point is the product of themagnitude of the force and the moment arm, which is theperpendicular distance from the line of action of the force tothe pivot.

ANSWER:

Hint 4. Some helpful integrals

and

.

ANSWER:

Correct

Thus the average moment arm is . Therefore, sincethe water exerts a greater force at the bottom of the dam than at thetop, the net effect is as if the force were acting at a height of fromthe bottom.

A Water Tank on Mars

=

=

You are assigned the design of a cylindrical, pressurized water tank for a futurecolony on Mars, where the acceleration due to gravity is 3.71 meters persecond per second. The pressure at the surface of the water will be 115 ,and the depth of the water will be 14.0 . The pressure of the air in thebuilding outside the tank will be 86.0 .

Part A

Find the net downward force on the tank's flat bottom, of area 2.10 ,exerted by the water and air inside the tank and the air outside the tank.

Express your answer numerically in Newtons, to three significantfigures.

Hint 1. The net force

The net force on the tank's flat bottom is the sum of the (downward)force exerted on the bottom of the tank by the water and the(upward) force exerted on the bottom of the tank by the air outsidethe tank.

Hint 2. What is (a Pascal)?

The SI unit of pressure, a Pascal, is a force per unit area: 1 = 1 .

Hint 3. Density of water

The density of water is 1 .

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

outside the tank

Write an expression for the force exerted on the tank's bottom bythe air outside the tank.

Express your answer numerically to three significant figures.

Hint 1. Pressure and force

The pressure in a gas is defined as the normal force

exerted by the gas on a surface in contact with it. If the forceis the same at all points of a finite plane surface with area ,then the pressure is uniform and given by

.

ANSWER:

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

water

Write an expression for the force exerted on the tank's bottom bythe water in the tank. Keep in mind that the water tank is located onMars, so weights depend on the acceleration due to gravity on thatparticular planet.

Express your answer numerically in Newtons, to threesignificant figures.

Hint 1. Find the pressure exerted by the water on the

bottom of the tank

In a fluid of uniform density , the pressure at a depth isgiven by the pressure at the surface plus a term due tothe weight of the fluid above that point. In other words,

where is the magnitude of the acceleration due to gravity.Given this information, find the pressure exerted by the wateron the bottom of the tank.

Express your answer numerically in pascals, to threesignificant figures.

ANSWER:

1.81×105

Hint 2. Pressure and force

The pressure in a gas is defined as the normal force exerted by the gas on a surface in contact with it. If the forceis the same at all points of a finite plane surface with area ,then the pressure is uniform and given by

.

ANSWER:

ANSWER:

Correct

Breaking a Glass Soda Bottle

A glass soda bottle is emptied of soda and filled to the very top with water. Acork is carefully fitted into the top of the bottle, leaving no air between the corkand the water. The top of the bottle has a diameter of = 2.00 and thebottom of the bottle has a diameter of = 6.50 . The glass breaks whenit is exposed to = 70.0 of pressure.

A student hits the cork sharply with her fist and the bottom of the bottle breaks.The student's fist has a mass of = 0.480 and moves downward at aspeed of = 5.00 . It collides elastically with the cork and rebounds with

1.67×105

3.51×105

1.70×105

the same speed. Thecollision lasts for =1.20×10−4 . In thisproblem, the positivedirection is upward.

Part A

What is the force that her fist exerts on the top of the bottle?

Express your answer in newtons.

Hint 1. How to approach the problem

This question cannot be solved using the pressure required to breakthe glass. Knowing the glass broke means the student's fist createdmore than the minimum pressure required to break the glass. Weneed to find out how much force was actually applied.

There are two methods for approaching this problem:

You could use Newton’s 2nd law to relate the mass andacceleration of the fist to the force exerted on the fist by the bottle.From Newton’s 3rd law, you know that the force that the bottleexerts on the fist is equal in magnitude to the force the fist exerts onthe bottle.

Alternatively, you could use the relationship between force andimpulse to find the force exerted on the fist by the bottle.

Hint 2. Find the acceleration of the fist during the collision

What is the acceleration of the fist during its collision with thebottle?

Express your answer in meters per second squared to foursignificant figures.

Hint 1. Definition of acceleration

The acceleration is equal to the change in velocity of the fist,divided by the time during the collision:

.

Hint 2. Change in velocity

The change in velocity of the fist is equal to its final velocityminus its initial velocity:

.

ANSWER:

Hint 3. Find the force the bottle exerts on the fist

What is the force of the fist during its collision with the bottle?

Express your answer in newtons to four significant figures.

Hint 1. Calculating force

Newton’s 2nd law states that force is equal to the mass timesthe acceleration:

.

ANSWER:

= 8.33×104

Hint 4. Newton's 3rd law

Newton’s 3rd law states that the force the fist exerts on the bottle isequal in magnitude and opposite in direction to the force the bottleexerts on the fist.

ANSWER:

Correct

Part B

What is the magnitude of the force exerted on the bottom of the bottle?

Express your answer in newtons to four significant figures.

Hint 1. How to approach this problem

You can find the pressure of the water at the top of the bottle.Pascal’s principle provides you with a way to relate the pressure atthe top of the bottle to that at the bottom . You can then use thedefinition of pressure to find the force exerted on the bottom of thebottle.

Hint 2. Find the area of the top of the bottle

Pressure is defined as force divided by area. You have solved forthe force. Now, you need to determine the area it is applied to. Whatis the area of the top of the bottle?

Express your answer in meters squared.

= 4.00×104

−4.00×104

Hint 1. Area of a circle

The area of a circle is , where is the radius ofthe circle.

ANSWER:

Hint 3. Find the change in pressure at the top of the bottle

What is the change in pressure in the water at the top of the bottle?

Express your answer in pascals.

Hint 1. How to treat pressure

The definition of pressure is the magnitude of the forcedivided by the area over which it is exerted. Pressure is ascalar quantity, unlike force, which is a vector.

ANSWER:

Hint 4. Find the increase in pressure at the bottom of the

bottle

What is the increase in pressure at the bottom of the bottle?

Express your answer in pascals.

Hint 1. Pascal’s principle

Pascal’s principle states that if there is an increase inpressure anywhere in a confined, incompressible fluid, thereis an equal increase in pressure throughout the fluid. The

3.14×10−4

1.27×108

pressure of the student’s fist striking the top of the bottle istransmitted equally throughout the bottle. The increase inpressure at the bottom is equal to the increase in pressure atthe top.

ANSWER:

Hint 5. Find the area of the bottom of the bottle

What is the area of the bottom of the bottle?

Express your answer in meters squared.

ANSWER:

Hint 6. Relationship between pressure and force

The definition of pressure is the magnitude of the force dividedby the area over which it is exerted:

.

ANSWER:

Correct

Part C

1.27×108

3.32×10−3

4.23×105

Which of the following explains why the bottle breaks?

Choose the correct explanation.

Hint 1. How to solve this problem

The text of the original problem states that “The glass breaks whenit is exposed to 70.0 of pressure." You calculated thepressure in Part B of this problem.

ANSWER:

Correct

Exercise 12.4

You win the lottery and decide to impress your friends by exhibiting a million­dollar cube of gold. At the time, gold is selling for 426.60 per troy ounce, and1.0000 troy ounce equals 31.1035 .

Part A

How tall would your million­dollar cube be?

The pressure that the water exerts on the walls of the glass bottleis greater than the pressure required to break the glass.

The force that the water exerts on the bottom of the bottle isgreater than the failure pressure of the glass.

The force that the water exerts on the bottom of the bottle isgreater than the force required to break the glass.

When the hand strikes the bottle, it sends a shock wave throughthe bottle, causing the water molecules to get very close together.When the mass of extra water molecules strike the bottom of thebottle, it breaks.

ANSWER:

Correct

Exercise 12.10

Part A

Calculate the difference in blood pressure between the feet and top of thehead for a person who is 1.80 tall.

ANSWER:

Correct

Part B

Consider a cylindrical segment of a blood vessel 2.80 long and 3.00 in diameter. What additional outward force would such a vessel need

to withstand in the person's feet compared to a similar vessel in her head?

ANSWER:

Correct

= 15.6

= 1.87×104

= 4.93

Exercise 12.19

An electrical short cuts off all power to a submersible diving vehicle when it is adistance of 35 below the surface of the ocean. The crew must push out ahatch of area 0.750 and weight 340 to escape.

Part A

If the pressure inside is , what downward force must they exert onthe hatch to open it?

Express your answer using two significant figures.

ANSWER:

Correct

Exercise 12.23

= 2.6×105

Part A

For the hydraulic lift shown in the figure , what must be the ratio of thediameter of the vessel at the car to the diameter of the vessel where theforce is applied so that a 1550 car can be lifted with a force ofjust 135 ?

ANSWER:

Correct

Score Summary:

Your score on this assignment is 100%.You received 8 out of a possible total of 8 points.

= 10.6