dynamics_solns_ch01_2009_11_02

Solutions ManualEngineering Mechanics: Dynamics

1st Edition

Gary L. GrayThe Pennsylvania State University

Francesco CostanzoThe Pennsylvania State University

Michael E. PleshaUniversity of Wisconsin–Madison

With the assistance of:Chris PunshonAndrew J. MillerJustin HighChris O’BrienChandan KumarJoseph Wyne

Version: November 2, 2009

The McGraw-Hill Companies, Inc.

Copyright © 2002–2010Gary L. Gray, Francesco Costanzo, and Michael E. Plesha

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. Itmay be used and/or possessed only by permission of McGraw-Hill, and must be surrendered uponrequest of McGraw-Hill. Any duplication or distribution, either in print or electronic form, withoutthe permission of McGraw-Hill, is prohibited.

Dynamics 1e 3

Important Information aboutthis Solutions Manual

Even though this solutions manual is nearly complete, we encourage you to visit

http://www.mhhe.com/pgc

often to obtain the most up-to-date version. In particular, as of September 11, 2009, please note the following:

_ The solutions for Chapters 1, 2, 5, 8, and 9 have been accuracy checked and are in their final form.

_ The solutions for Chapters 3, 4, and 7 have been accuracy checked and should be error free. We will beadding some additional detail to these solutions in the coming weeks.

_ The solutions for Chapter 6 have been accuracy checked through Section 6.3 and most of 6.4. Thischapter should be complete by November 15.

_ The solutions for Chapter 10 are a work in progress. The solutions for the first 29 problems in thechapter are complete.

Contact the Authors

If you find any errors and/or have questions concerning a solution, please do not hesitate to contact theauthors and editors via email at:

[email protected]

We welcome your input.

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permissionof McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without thepermission of McGraw-Hill, is prohibited.

November 2, 2009

4 Solutions Manual

Accuracy of Numbers in CalculationsThroughout this solutions manual, we will generally assume that the data given for problems is accurate to3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations inthis solutions manual using the rounded intermediate numerical results that are reported, you should obtainthe final answers that are reported to 3 significant digits.


November 2, 2009

Dynamics 1e 5

Chapter 1 SolutionsProblem 1.1

Determine .rB=A/x and .rB=A/y , the x and y components of the vector ErB=A, so as to be able to writeErB=A D .rB=A/x O{ C .rB=A/y O| .

Solution

Using the component system formed by the unit vectors O{ and O| , the vector ErB=A can be written as ErB=A D

.xB � xA/ O{ C .yB � yA/ O| , where .xA; yA/ and .xB ; yB/ are the Cartesian coordinates of points A and Brelative to the xy frame, respectively. By inspection we have .xA; yA/ D .1:000; 2:000/ ft and .xB ; yB/ D

.5:000; 1:000/ ft. Using these relations one obtains

ErB=A D .4:00 O{ � 1:00 O|/ ft:


November 2, 2009

6 Solutions Manual

Problem 1.2

If the positive direction of line ` is from D to C , find the component of the vector ErB=A along `.

Solution



.5:000; 1:000/ ft. Therefore, we have

ErB=A D .4:000 O{ � 1:000 O|/ ft: (1)

Next we find the unit vector parallel to ` and pointing from D to C . This is accomplished by finding ErC=D

and scaling it by its own magnitude.

ErC=D D .xC � xD/ O{ C .yC � yD/ O| D .3:000 O{ � 2:000 O|/ ft: (2)

Next, we derive the magnitude of ErC=D:

jErC=Dj D

q.3:000 ft/2 C .2:000 ft/2 D 3:606 ft: (3)

Hence we have

Ou` DErC=D

jErC=DjD

�3:000 ft3:606 ft

O{ �2:000 ft3:606 ft

O|

�D 0:8319 O{ � 0:5546 O| : (4)

The component of ErC=D along ` is then obtained by taking the dot product of ErB=A and Ou`, i.e.,

.rB=A/` D Œ.4:000 O{ � 1:000 O|/ ft� � .0:8319 O{ � 0:5546 O|/ D 3:882 ft; (5)

which, when expressed to three significant figures gives

.rC=D/` D 3:88 ft:


November 2, 2009

Dynamics 1e 7

Problem 1.3

Find the components of ErB=A along the p and q axes.

Solution



.5:000; 1:000/ ft. Therefore, we have

ErB=A D .4:000 O{ � 1:000 O|/ ft: (1)

Now we find the unit vectors Oup and Ouq , which are the unit vectors orienting the p and q axes, respectively,

Oup D cos �p O{ C sin �p O|; (2)

Ouq D cos �q O{ C sin �q O|; (3)

where �p and �q are the angles formed by the p and q axes with the positive x axis, respectively, and thereforeare given by �p D � D 22:5ı, �q D .� C 90

ı/ D 112:5ı. Consequently, Eqs. (2) and (3) become

Oup D 0:9239 O{ C 0:3827 O| and Ouq D �0:3827 O{ C 0:9239 O| : (4)

Then desired components are obtained by computing the dot product of the vector ErB=A with the unit vectorsin Eqs. (4), i.e.,

.rB=A/p D Oup � ErB=A D�.4:000 O{ � 1:000 O|/ ft

�� .0:9239 O{ C 0:3827 O|/ D 3:313 ft; (5)

.rB=A/q D Ouq � ErB=A D�.4:000 O{ � 1:000 O|/ ft

�� .�0:3827 O{ C 0:9239 O|/ D �2:455 ft; (6)

which, when expressed to three significant figures, give

.rB=A/p D 3:31 ft and .rB=A/q D �2:46 ft:


November 2, 2009

8 Solutions Manual

Problem 1.4

Determine expressions for the vector ErB=A using both the xy and the pq coordinate systems. Next,determine jErB=Aj, the magnitude of ErB=A, using both the xy and the pq representations and establishwhether or not the two values for jErB=Aj are equal to each other.

Solution


.xB � xA/ O{ C .yB � yA/ O| , where .xA; yA/ and .xB ; yB/ are the Cartesian coordinates of points A and Brelative to the xy coordinate system, respectively. By inspection we have .xA; yA/ D .1:000; 2:000/ ft and.xB ; yB/ D .5:000; 1:000/ ft. Therefore, we have

ErB=A D .4:000 O{ � 1:000 O|/ ft; (1)

which, when expressed to three significant figures, gives

ErB=A D .4:00 O{ � 1:00 O|/ ft:

Now we find the unit vectors Oup and Ouq , which are the unit vectors orienting the p and q axes, respectively,

Oup D cos �p O{ C sin �p O| and Ouq D cos �q O{ C sin �q O|; (2)

where �p and �q are the angles formed by the p and q axes with the positive x axis, respectively, and thereforeare given by �p D � D 22:5ı, �q D .� C 90

ı/ D 112:5ı. Consequently, Eqs. (2) and (3) become

Oup D 0:9239 O{ C 0:3827 O| and Ouq D �0:3827 O{ C 0:9239 O| : (3)

Then, using Eqs. (1) and (3), the expression for ErB=A in the pq component system is

ErB=A D . Oup � ErB=A/ Oup C . Ouq � ErB=A/ Ouq

D˚�.4:000 O{ � 1:000 O|/ ft

�� .0:9239 O{ C 0:3827 O|/

Oup

C˚�.4:000 O{ � 1:000 O|/ ft

�� .�0:3827 O{ C 0:9239 O|/

Ouq

D .3:313 Oup � 2:455 Ouq/ ft; (4)

which, when expressed to three significant figures gives

ErB=A D .3:31 Oup � 2:46 Ouq/ ft:


November 2, 2009

Dynamics 1e 9

Using Eq. (1), the magnitude of ErB=A expressed in the xy component system is

ˇ̌ErB=A

ˇ̌D

q.4:000 ft/2 C .�1:000 ft/2 D 4:123 ft: (5)

Using the result in Eq. (4), the magnitude of ErB=A expressed in the pq component system is

ˇ̌ErB=A

ˇ̌D

q.3:313 ft/2 C .�2:455 ft/2 D 4:123 ft: (6)

Expressing the results in the last two equations to three significant figures we have

ˇ̌ErB=A

ˇ̌xy system D

ˇ̌ErB=A

ˇ̌pq system D 4:12 ft:


November 2, 2009

10 Solutions Manual

Problem 1.5

Suppose that you were to compute the quantitiesˇ̌ErB=A

ˇ̌xy

andˇ̌ErB=A

ˇ̌pq

, that is, the magnitude of thevector ErB=A computed using the xy and pq frames, respectively. Do you expect these two scalar values tobe the same or different? Why?

Solution

The magnitude of a vector is a fundamental property of the vector in question and it must be independent ofhow the vector is represented. In other words, a choice of frame only affects the values of the components ofa vector and not its magnitude or direction. For this reason, we expect the value of

ˇ̌ErB=A

ˇ̌xy

to be same asthat of

ˇ̌ErB=A

ˇ̌pq

.


November 2, 2009

Dynamics 1e 11

Problem 1.6

The measure of angles in radians is defined according to the followingrelation: r� D sAB , where r is the radius of the circle and sAB denotes thelength of the circular arc. Determine the dimensions of the angle � .

Solution

Use the given definition of an angle in radians:

Œr�� D Œr� Œ�� D ŒsAB �: (1)

The radius r and the arc length s must have dimensions of length, we have that

ŒL� Œ�� D ŒL�: (2)

A simplification of the above relation yields that

Œ� � D 1; (3)

so that we conclude that

The angle � is dimensionless (or nondimensional):

Observe that since angles can be expressed in a variety of units, such as radian or degree, this problemillustrates the idea that some nondimensional quantities still require proper use of units.


November 2, 2009

12 Solutions Manual

Problem 1.7

A simple oscillator consists of a linear spring fixed at one end and a mass attached at the otherend, which is free to move. Suppose that the periodic motion of a simple oscillator is describedby the relation y D Y0 sin .2�!0t /, where y has units of length and denotes the verticalposition of the oscillator, Y0 is called the oscillation amplitude, !0 is the oscillation frequency,and t is time. Recalling that the argument of a trigonometric function is an angle, determine thedimensions of Y0 and !0, as well as their units in both the SI and the U.S. Customary systems.

Solution

Because the value of a trigonometric function must be nondimensional and because y has dimensions oflength, Y0 must also have dimensions of length:

ŒY0� D ŒL�:

A such, the units to express the quantity Y0 in the SI and the U.S. Customary systems are

Units of Y0 in the SI system: m.

Units of Y0 in the US Customary system: ft:

The quantity .2�!0t / is the argument of a trigonometric function and as such it must be nondimensional.Therefore we have

Œ2�!0t � D 1 ) Œ!0� Œt � D 1;

Œ!0� D1

ŒT �:

If we where to infer the units of !0 directly from its dimensions, we would conclude that the units of !0

are seconds to the power negative one. However, in this particular problem, this would lead to an erroneousconclusion. The reason for this is that, as indicated in the problem statement, the term !0 appears as partof the argument 2�!0t . When this happens, !0 is typically seen as the “number of 2� rad per unit time,”that is as a quantity that measures the number of oscillation cycles per unit time (for additional details, seethe discussion about Eqs. (9.4) and (9.8) on p. 674 of th textbook). In both the SI and the U.S. Customarysystems, the number of cycles per unit time are measured in hertz, which is a unit corresponding to 1 cycleper second. Hence, we have

In both the SI and the U.S. Customary systems, !0 has units of Hz (hertz), i.e., cycles per second:


November 2, 2009

Dynamics 1e 13

Problem 1.8

To study the motion of a space station, the station can be modeled as a rigidbody and the equations describing its motion can be chosen to be Euler’sequations, which read

Mx D Ixx ˛x ��Iyy � I´´

�!y !´;

My D Iyy ˛y � .I´´ � Ixx/ !x !´;

M´ D I´´ ˛´ ��Ixx � Iyy

�!x !y :

In the previous equations, Mx , My , and M´ denote the x, y, and ´ com-ponents of the moment applied to the body; !x , !y , and !´ denote thecorresponding components of the angular velocity of the body, where angu-lar velocity is defined as the time rate of change of an angle; ˛x , ˛y , and˛´ denote the corresponding components of the angular acceleration of thebody, where angular acceleration is defined as the time rate of change of anangular velocity. The quantities Ixx , Iyy , and I´´ are called the principalmass moments of inertia of the body. Determine the dimensions of Ixx , Iyy ,and I´´ and determine their units in SI as well as in the U.S. Customarysystem.

y

z��

M��

y

�

Solution

We consider only one of the three equations as their dimensions are identical:

ŒMx� D ŒIxx˛x � .Iyy � I´´/!y!´� D ŒIxx�Œ˛x� � ŒIyy � I´´�Œ!y �Œ!´� ) ŒMx� D ŒIxx�Œ˛x�; (1)

given that the additive terms in above equation must all the same dimensions.Because ˛x is the rate of change of an angular velocity, its units are:

Œ˛x� D1

ŒT �2: (2)

The term Mx is the component of a moment and therefore it must have the dimensions of force times length:

ŒMx� D ŒL� ŒF � D ŒM � ŒL�21

ŒT �2: (3)

Substituting Eqs. (2) and (3) into the last of Eqs. (1), we have

ŒM � ŒL�21

ŒT �2D ŒIxx�

1

ŒT �2:

Therefore, we conclude that the dimensions of the quantities Ixx , Iyy , and I´´ are

ŒIxx� D ŒIyy � D ŒI´´� D ŒM �ŒL�2:

Because the quantities Ixx , Iyy , and I´´ have units of mass time length squared, their units in the SI and U.S.Customary systems are as follows:

Units of Ixx , Iyy , and I´´ in the SI system: kg�m2;

Units of Ixx , Iyy , and I´´ in the U.S. Customary system: slug�ft2 D lb�s2�ft.


November 2, 2009

14 Solutions Manual

Problem 1.9

The lift force FL generated by the airflow moving over a wing is often expressed as follows:

FL D12�v2CL.�/A; (1)

where �, v, and A denote the mass density of air, the airspeed (relative to the wing), and the wing’s nominalsurface area, respectively. The quantity CL is called the lift coefficient, and it is a function of the wing’sangle of attack � . Find the dimensions of CL and determine its units in the SI system.

Solution

The dimensions of each of the quantities in the given equation are

ŒFL� D

�M�LT2

�; Œ�� D

�ML3

�; ŒA� D

�L2�; Œv� D

�LT

�: (2)

When the lift force .FL/ equation is written in terms of its dimensions, it becomes�M�LT2

�D

�ML3

��LT

�2�CL.�/

��L2�: (3)

Simplifying and solving for the dimensions of CL.�/ yields�M�LT2

�D ŒCL.�/�

�MT2

�ŒL� )

�CL.�/

�D 1: (4)

Therefore we conclude that

The lift coefficient is dimensionless and has no units associated with it.


November 2, 2009

Dynamics 1e 15

Problem 1.10

Are the words units and dimensions synonyms?

Solution

No, units and dimensions are not synonyms. “Dimensions” refer to a physical quantity which is independentof the specific system of units used to measure it.


November 2, 2009

dynamics_solns_ch01_2009_11_02

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