Engineering Mechanics: Dynamics -...

Solutions ManualEngineering Mechanics: Dynamics

2nd Edition

Gary L. GrayThe Pennsylvania State University

Francesco CostanzoThe Pennsylvania State University

Michael E. PleshaUniversity of Wisconsin–Madison

Version: May 12, 2012

The McGraw-Hill Companies, Inc.

Copyright © 2002–2012Gary 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 2e 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 December 16, 2009, please note the following:

_ The solutions for Chapters 1, 2, and 4–9 have been accuracy checked and are in their final form.

_ The solutions for Chapter 3 have been accuracy checked and should be error free. We will be addingsome additional detail to these solutions in the coming weeks.

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

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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.

May 12, 2012

4 Solutions Manual

Accuracy of Numbers in CalculationsThroughout this book, we will generally assume that the data given for problems is accurate to three significantdigits. When calculations are performed, intermediate results are stored in the memory of a calculator orcomputer using the full precision these machines offer. However, when these intermediate results are reportedin this solutions manual, they are rounded to four significant digits. Final results are also rounded to foursignificant digits.


May 12, 2012

Dynamics 2e 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. Recall that,

ErB=A D .xB � xA/ O{ C .yB � yA/ O| : (1)

Since .xA; yA/ D .1; 2/ ft and .xB ; yB/ D .5; 1/ ft, we then have

ErB=A D .5 � 1/ O{ C .2 � 1/ O| ) ErB=A D 4 O{ C 1 O|; (2)

which, expressed to four significant figures, is

ErB=A D .4:000 O{ � 1:000 O|/ ft:


May 12, 2012

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

Using the .O{; O|/ component system, the vector ErB=A can be written as follows:

ErB=A D .xB � xA/ O{ C .yB � yA/ O| D .4:000 O{ � 1:000 O|/ ft; (1)

where we have used the values xA D 1 ft, yA D 2 ft, xB D 5 ft, and yB D 1 ft, and where we have expressedthe result to four significant figures. 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 magnitude:

Ou` DErC=D

jErC=Dj) Ou` D

.xC � xD/ O{ C .yC � yD/ O|p.xC � xD/2 C .yC � yD/2

D 0:8321 O{ � 0:5547 O|; (2)

where we have used the values xC D 4 ft, yC D 3 ft, xD D 1 ft, and yD D 5 ft, and again expressed theresult to four significant figures.

The component of ErB=A along `, which we denote by .rB=A/`, is obtained by taking the dot product ofErB=A and Ou`:

.rB=A/` D ErB=A � Ou` ) .rB=A/` D 3:883 ft. (3)


May 12, 2012

Dynamics 2e 7

Problem 1.3

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

Solution

Using the .O{; O|/ component system, the vector ErB=A can be written as

ErB=A D .xB � xA/ O{ C .yB � yA/ O| : (1)

Next, we find the expressions of the unit vectors Oup and Ouq in terms of the unit vectors O{ and O| . Usingtrigonometry, we have

Oup D cos � O{ C sin � O| and Ouq D � sin � O{ C cos � O| : (2)

Denoting the components of ErB=A in the p and q directions by .rB=A/p and .rB=A/q , respectively, thesecomponents can be found by computing the dot product of ErB=A with the unit vectors Oup and Ouq , respectively:

.rB=A/p D ErB=A � Oup D .xB � xA/ cos � C .yB � yA/ sin �; (3)

.rB=A/q D ErB=A � Ouq D �.xB � xA/ sin � C .yB � yA/ cos �: (4)

Observing that we have xA D 1 ft, yA D 2 ft, xB D 5 ft, yB D 1 ft, and � D 22:5ı, we can evaluate theexpressions in Eqs. (3) and (4) to obtain

.rB=A/p D 3:313 ft and .rB=A/q D �2:455 ft:


May 12, 2012

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

In the xy coordinate system we can represent the vector ErB=A as follows:

ErB=Aˇ̌xy system D .xB � xA/ O{ C .yB � yA/ O| : (1)

Observing that we have xA D 1 ft, yA D 2 ft, xB D 5 ft, yB D 1 ft, we can evaluate the expression in Eq. (1)to obtain

ErB=Aˇ̌xy system D .4:000 O{ � 1:000 O|/ ft:

Next, we find the expressions of the unit vectors Oup and Ouq in terms of the unit vectors O{ and O| . Usingtrigonometry, we have

Oup D cos � O{ C sin � O| and Ouq D � sin � O{ C cos � O| : (2)

Denoting the components of ErB=A in the p and q directions by .rB=A/p and .rB=A/q , respectively, thesecomponents can be found by computing the dot product of ErB=A as expressed in Eq. (1) with the unit vectorsOup and Ouq , respectively:

.rB=A/p D ErB=Aˇ̌xy system � Oup D .xB � xA/ cos � C .yB � yA/ sin �; (3)

.rB=A/q D ErB=Aˇ̌xy system � Ouq D �.xB � xA/ sin � C .yB � yA/ cos �: (4)

Therefore, in the pq coordinate system, we have

ErB=Aˇ̌pq system D

�.xB � xA/ cos � C .yB � yA/ sin �

�OupC

��.xB � xA/ sin � C .yB � yA/ cos �

�Ouq: (5)

Since xA D 1 ft, yA D 2 ft, xB D 5 ft, yB D 1 ft, and � D 22:5ı, we can evaluate the expression in Eq. (5)to obtain

ErB=Aˇ̌pq system D .3:313 Oup � 2:455 Ouq/ ft:

Using Eq. (1), the magnitude of the vector ErB=A in the xy coordinate system isˇ̌ErB=A

ˇ̌xy system D

q.xB � xA/2 C .yB � yA/2: (6)


May 12, 2012

Dynamics 2e 9

Using Eq. (5), the magnitude of the vector ErB=A in the pq coordinate system is

ˇ̌ErB=A

ˇ̌pq system D

q�.xB � xA/ cos � C .yB � yA/ sin �

�2C��.xB � xA/ sin � C .yB � yA/ cos �

�2:

(7)Since xA D 1 ft, yA D 2 ft, xB D 5 ft, yB D 1 ft, and � D 22:5ı, we can evaluate Eqs. (6) and (7) to obtainˇ̌

ErB=Aˇ̌xy system D 4:123 ft and

ˇ̌ErB=A

ˇ̌pq system D 4:123 ft: (8)

Hence, we conclude that ˇ̌ErB=A

ˇ̌xy system D

ˇ̌ErB=A

ˇ̌pq system D 4:123 ft:


May 12, 2012

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 and it must be independent of how thevector is represented. The choice of frame only affects the values of the components of a vector and not itsmagnitude or direction. For this reason, we expect the value of

ˇ̌ErB=A

ˇ̌xy

to be same as that ofˇ̌ErB=A

ˇ̌pq

.


May 12, 2012

Dynamics 2e 11

Problem 1.6

The magnitude of the velocity vector of the car is jEvj D 80 ft=s. If the vector Ev forms an angle � D 0:09 radwith the horizontal direction, determine the Cartesian representation of Ev relative to the .O{; O|/ componentsystem.

Solution

We can represent the vector Ev as was done for the vector Er in Eq. (1.22) of p. 8 of the textbook (Er D jEr j Our ),that is as

Ev D jEvj Ouv; (1)

where Ouv is the unit vector describing the orientation of Ev. The expression for Ouv is

Ouv D cos � O{ C sin � O| : (2)

Letting v D jEvj and using Eq. (2), we can then re-write Eq. (1) as

Ev D v cos � O{ C v sin � O| : (3)

Recalling that v D 80 ft=s and that � D 0:09 rad, we can evaluate Eq. (3) to obtain

Ev D .79:68 O{ C 7:190 O|/ ft=s:


May 12, 2012

12 Solutions Manual

Problem 1.7

The velocity of the car has the following representation: Ev D .8:30 O{ C 0:726 O|/m=s. Determine themagnitude of Ev. In addition, knowing that the angle � describes the orientation of Ev, determine � andexpress its value in degrees.

Solution

The vector Ev is given the following form

Ev D vx O{ C vy O|; (1)

where vx and vy are the components of Ev in the x and y directions, respectively. Using Eq. (1.20) on p. 8 ofthe textbook, the magnitude of Ev is

jEvj Dqv2x C v

2y : (2)

Recalling that vx D 8:30m=s and vy D 0:726m=s, we can evaluate jEvj in Eq. (2) to obtain

jEvj D 8:332m=s:

To determine the angle � we use Eq. (1.21) on p. 8 of the textbook. This gives

� D tan�1

�vy

vx

�˙ �n; n D 0; 1; 2; : : : (3)

Since both vx and vy are positive, the parameter n can be taken to be equal to zero. Sincevx D 8:30m=s andvy D 0:726m=s, the value of � in degrees is

� D 4:999ı:


May 12, 2012

Dynamics 2e 13

Problem 1.8

The acceleration of the car has the following representation: Ea D �.3:53 O{C 0:309 O|/m=s2. Knowing thatEa is parallel to the incline, determine the angles � and � and express their value in radians.

Solution

The vector Ea can be represented asEa D ax O{ C ay O|; (1)

where ax and ay are the components of Ea in the x and y directions, respectively. Using Eq. (1.21) on p. 8,for � we can write

� D tan�1

�ay

ax

�˙ �n; n D 0; 1; 2; : : : (2)

Recalling that ax D �3:53m=s2 and ay D �0:309m=s2, and observing that both ax and ay are negative,the parameter n is equal to 1. Hence, the value of � in radians is

� D 3:229 rad:

Observing that � D � C � rad and that therefore � D � � � rad, we have

� D 0:08731 rad:


May 12, 2012

14 Solutions Manual

Problem 1.9

A jaguar A leaps from O with a velocity Ev0 to try and intercept a panther B . The unit vectors Oup andOuq are parallel and perpendicular to the incline, respectively. The unit vectors O{ and O| are horizontal andvertical, respectively. While airborne, the jaguar is subject to a constant acceleration with magnitude g anddirection opposite to O| . Denoting the magnitude of Ev0 by v0 and denoting the (vector) acceleration of thejaguar by EaA, provide the expression of Ev0 in the .O{; O|/ component system and the expression of EaA in the. Oup; Ouq/ component system. Treat the angles ˇ and � as known.

Solution

We observe that the vector Ev0 forms an angle equal to ˇ � � with respect to the horizontal direction. Hence,in the .O{; O|/ component system, we can write the vector Ev0 as

Ev0 D v0 cos.ˇ � �/ O{ C v0 sin.ˇ � �/ O| :

Using the given information, we can write the vector EaA as

EaA D �g O| : (1)

Following the strategy demonstrated in Eq. (1.26) on p. 9 of the textbook, we can express EaA in the . Oup; Ouq/component system as follows:

EaA D .EaA � Oup/ Oup C .EaA � Ouq/ Ouq ) EaA D��g O| � Oup

�Oup C

��g O| � Ouq

�Ouq; (2)

where, in writing the second of Eqs. (2), we have used Eq. (1). Using trigonometry, we now observe that

O| � Oup D � sin � and O| � Ouq D cos �: (3)

Substituting Eqs. (3) into the last of Eqs. (2) we then have

EaA D g sin � Oup � g cos � Ouq:


May 12, 2012

Dynamics 2e 15

Problem 1.10

The velocity vector of the airplane is Ev D v0 O{, with v0 D 420mph. Determine the components of thevector Ev in the Our and Ou� directions for � D 35ı. Express the result in feet per second.

Solution

We begin by expressing the unit vectors Our and Ou� in terms of the unit vectors O{ and O| :

Our D cos � O{ C sin � O| and Ou� D � sin � O{ C cos � O| : (1)

We denote the components of Ev along the Our and Ou� directions by vr and v� , respectively. Then we have

vr D Ev � Our and v� D Ev � Ou� : (2)

Recalling that Ev D v0 O{, and using Eqs. (1) and (2), we have

vr D v0 cos � and v� D �v0 sin �: (3)

Substituting the given values v0 D 420mph D .420mph/.5280 ft=mi/.3600 s=h/�1 D 616:0 ft=s and� D 35ı in Eqs. (3), we obtain

vr D 504:6 ft=s and v� D �353:3 ft=s:


May 12, 2012

16 Solutions Manual

Problem 1.11

The motion of the telescopic arm is such that the velocity and acceleration vectors of the gear B areEv D �v0 O| and Ea D �a0 O| , respectively, with v0 D 8 ft=s and a0 D 0:5 ft=s2. Determine the componentsof Ev and Ea in the direction of the unit vectors Our and Ou� for � D 32ı.

Solution

We begin by expressing the unit vectors Our and Ou� in terms of the unit vectors O{ and O| :

Our D cos � O{ C sin � O| and Ou� D � sin � O{ C cos � O| : (1)

We denote the components of Ev along the Our and Ou� directions by vr and v� , respectively. Similarly, thecomponents of Ea along the Our and Ou� directions will be denoted by ar and a� , respectively. We now observethat

vr D Ev � Our ; v� D Ev � Ou� ; ar D Ea � Our ; and a� D Ea � Ou� : (2)

Recalling that Ev D �v0 O| and Ea D �a0 O| , and using Eqs. (1) into Eqs. (2), we have

vr D �v0 sin �; v� D �v0 cos �; ar D �a0 sin �; and a� D �a0 cos �: (3)

Substituting the given values v0 D 8 ft=s, a0 D 0:5 ft=s2, and � D 32ı in Eqs. (3), we obtain

vr D �4:239 ft=s; v� D �6:784 ft=s; ar D �0:2650 ft=s2; and a� D �0:4240 ft=s2:


May 12, 2012

Dynamics 2e 17

Problem 1.12

At the instant shown, the velocity and acceleration vectors of the airplane have the following expressions:

Ev D .215 O{ C 332 O|/ ft=s and Ea D .�190 O{ C 76:0 O|/ ft=s2:

Use Eq. (1.17) on p. 8 to determine the angle �, the smaller of the two angles formed by Ev and Ea. Expressthe result in degrees.

Solution

We begin by expressing Ev and Ea as follows:

Ev D vx O{ C vy O| and Ea D ax O{ C ay O| : (1)

Applying Eq. (1.17) on p. 8 of the textbook and referring to the problem’s figure, we have

Ev � Ea D jEvjjEaj cos� ) cos� DEv � Ea

jEvjjEaj: (2)

As far as the term Ev � Ea is concerned, using Eqs. (1), we have

Ev � Ea D vxax C vyay : (3)

As far as the terms jEvj and jEaj are concerned, we can write

jEvj Dqv2x C v

2y and jEaj D

qa2x C a

2y : (4)

Substituting Eqs. (3) and (4) into the last of Eqs. (2), we can solve for � to obtain

� D cos�1

0B@ vxax C vyayqv2x C v

2y

qa2x C a

2y

1CA˙ 360ın; n D 0; 1; 2; : : : (5)

Recalling that � is the smallest of the angles formed by the vectors Ev and Ea, we select n D 0. Then, recallingthat vx D 215 ft=s, vy D 332 ft=s, ax D �190 ft=s2, and ay D 76 ft=s2, we can evaluate Eq. (5) to obtain

� D 101:1ı:


May 12, 2012

18 Solutions Manual

Problem 1.13

At the instant shown, when expressed via the . Out ; Oun/ component system, theairplane’s velocity and acceleration are

Ev D 135 Out m=s and Ea D .�7:25 Out C 182 Oun/m=s2:

Determine the angle � between the velocity and acceleration vectors. In addi-tion, treating the . Out ; Oun/ and .O{; O|/ component systems as stationary relativeto one another, express the airplane’s velocity and acceleration in the .O{; O|/component system.

Solution

We begin by expressing Ev and Ea as follows:

Ev D vt Out and Ea D at Out C an Oun: (1)

Applying the definition of dot product and referring to the problem’s figure, we have

Ev � Ea D jEvjjEaj cos� ) cos� DEv � Ea

jEvjjEaj: (2)

As far as the term Ev � Ea is concerned, using Eqs. (1), we have

Ev � Ea D vtat : (3)

As far as the terms jEvj and jEaj are concerned, we can write

jEvj D jvt j and jEaj D

qa2t C a

2n: (4)

Substituting Eqs. (3) and (4) into the last of Eqs. (2), we can solve for � to obtain

� D cos�1

0B@ vtat

jvt j

qa2t C a

2n

1CA˙ 360ın; n D 0; 1; 2; : : : (5)

Recalling that � is the smallest of the angles formed by the vectors Ev and Ea, we select n D 0. Then, recallingthat vt D 135m=s, at D �7:25m=s2, and an D 182m=s2, we can evaluate Eq. (5) to obtain

� D 92:28ı:

To determine the expressions of the vectors Ev and Ea in the .O{; O|/ component systems, letting � D 57ı, wefirst express the unit vectors Out and Oun in terms of the unit vectors O{ and O| :

Out D cos � O{ C sin � O| and Oun D � sin � O{ C cos � O| : (6)

Substituting Eqs. (6) into Eqs. (1) and simplifying, we then have

Ev D vt cos � O{ C vt sin � O| and Ea D .at cos � � an sin �/ O{ C .at sin � C an cos �/ O| : (7)

Then, recalling that vt D 135m=s, at D �7:25m=s2, an D 182m=s2, and � D 57ı, we can evaluateEqs. (7) to obtain

Ev D .73:53 O{ C 113:2 O|/m=s and Ea D .�156:6 O{ C 93:04 O|/m=s2:


May 12, 2012

Dynamics 2e 19

Problem 1.14

The components of the position vector Er of point P relative to the .O{1; O|1/ component system are rx1 D 2 ftand ry1 D 5 ft. If � D 30ı, determine coordinates of P relative to the .x2; y2/ coordinate system.

Solution

The coordinates of P in the .x2; y2/ coordinate system coincide with the components of the vector Er in thecomponent system .O{2; O|2/. These components can be determined via the dot product. Specifically, we have

xP2 D Er � O{2 and yP2 D Er � O|2: (1)

We now observe that the vector Er can be represented as follows:

Er D rx1 O{1 C ry1 O|1; (2)

where the components rx1 and rx2 are given in the problem statement. Next, we observe that the unit vectorsO{2 and O|2 can be written in terms of the unit vectors O{1 and O|1:

O{2 D cos � O{1 � sin � O|1 and O|2 D sin � O{1 C cos � O|1: (3)

Therefore, substituting Eqs. (2) and (3) into Eqs. (1), we have

xP2 D rx1 cos � � ry1 sin � and yP2 D rx1 sin � C ry1 cos �: (4)

Since rx1 D 2 ft, ry1 D 5 ft, and � D 30ı, we can evaluate the expressions in Eqs. (4) to obtain

xP2 D �0:7679 ft and yP2 D 5:330 ft:


May 12, 2012

20 Solutions Manual

Problem 1.15

The velocity of point P relative to frame A is EvP=A D .�14:9 O{A C 19:4 O|A/ ft=s, and the acceleration ofP relative to frame B is EaP=B D .3:97 O{B C 4:79 O|B/ ft=s2. Frames A and B do not move relative to oneanother. Determine the expressions for the velocity of P in frame B and the acceleration of P in frame A.

Solution

Since the frames do not move relative to one another, EvP=A D EvP=B and EaP=A D EaP=B , and we can solvethe problem using the approach demonstrated in Eq. (1.26) on p. 9. That is, we can write

EvP=B D�EvP=A � O{B

�O{B C

�EvP=A � O|B

�O|B ; (1)

and

EaP=A D�EaP=B � O{A

�O{A C

�EaP=B � O|A

�O|A: (2)

Using the given information, we can re-write Eqs. (1) and (2) as

EvP=B D�.�14:9 O{A � O{B C 19:4 O|A � O{B/ O{B C .�14:9 O{A � O|B C 19:4 O|A � O|B/ O|B

�ft=s (3)

and

EaP=A D�.3:97 O{B � O{A C 4:79 O|B � O{A/ O{A C .3:97 O{B � O|A C 4:79 O|B � O|A/ O|A

�ft=s2: (4)

We now observe that

O{A � O{B D cos 23ı; O{A � O|B D � sin 23ı; O|A � O{B D sin 23ı; and O|A � O|B D cos 23ı: (5)

Substituting Eqs. (5) into Eqs. (3) and (4), we have

EvP=B D .�6:135 O{B C 23:68 O|B/ ft=s and EaP=A D .1:783 O{A C 5:960 O|A/ ft=s2:


May 12, 2012

Dynamics 2e 21

Problem 1.16

Two Coast Guard patrol boats P1 and P2 are stationary while monitoring the motion of a surface vessel A.The velocity of A with respect to P1 is expressed by

EvA D .�23 O{1 � 6 O|1/ ft=s;

whereas the acceleration of A, expressed relative to P2, is given by

EaA D .�2 O{2 � 4 O|2/ ft=s2:

Determine the velocity and the acceleration of A expressed with respect to the land-based componentsystem .O{; O|/.

Solution

We begin by expressing the unit vectors O{1, O|1, O{2, and O|2 in terms of the unit vectors O{ and O| :

O{1 D cos 18ıO{ C sin 18ı

O|; O|1 D � sin 18ıO{ C cos 18ı

O|; (1)

O{2 D � cos 45ı.O{ C O|/; O|2 D sin 45ı.O{ � O|/: (2)

Substituting Eqs. (1) and (2) into the given expressions for EvA and EaA and simplifying, we have

EvA D �.20:02 O{ C 12:81 O|/ ft=s and EaA D .�1:414 O{ C 4:243 O|/ ft=s2:


May 12, 2012

22 Solutions Manual

Problem 1.17

The measure of angles in radians is defined according to the following relation: r� D sAB , where r is theradius of the circle and sAB denotes the length of the circular arc. Determine the dimensions of the angle� .

Solution

Let L denote length. Using the definition of angle measure in radians, we have

Œr�� D Œr� Œ�� D ŒsAB �: (1)

The radius r and the arc length s have dimensions of length. Hence, we have

LŒ�� D L: (2)

Simplifying Eq. (2) yieldsŒ� � D 1; (3)

so that we conclude that

The angle � is dimensionless (or nondimensional):


May 12, 2012

Dynamics 2e 23

Problem 1.18

Letting C denote the circumference of a circle, a 1ı angle is, by definition, an angle that subtends an arc oflength ` such that C=` D 360. Apply the definition of degree and determine the radius of the circle shownknowing that the length s of the arc subtended by the 4ı angle in the figure is 1:84mm.

Solution

The circumference of a circle is related to the radius of the circle as follows: C D 2�r . Therefore, therelationship between s and r is governed by the following proportionality relation:

2�r

sD360

4: (1)

Solving Eq. (1) for r , we have

r D45

�s: (2)

Recalling that s D 1:84mm, we can evaluate the result in Eq. (2) to obtain

r D 26:36mm:


May 12, 2012

24 Solutions Manual

Problem 1.19

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 the oscillation amplitude, !0 is the oscillation frequency, andt 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

Let L and T denote length and time, respectively. Because the value of a trigonometric function must benondimensional and because y has dimensions of length, Y0 must also have dimensions of length:

ŒY0� D L:

As 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

T:

If we inferred the units of !0 directly from its dimensions, we would conclude that the units of !0 are secondsto the power negative one. However, in this particular problem, !0 appears as part of the argument 2�!0t .When this happens, !0 is seen as the “number of 2� rad per unit time.” Since 2� rad denotes a full oscillationcycle, then !0 measures the number of cycles per unit time (for additional details, see the discussion aboutEqs. (9.4) and (9.8) on p. 664 of the textbook). Hence, in both the SI and the U.S. Customary systems, thenumber of cycles per unit time are measured in hertz, which is a unit corresponding to 1 cycle per 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:


May 12, 2012

Dynamics 2e 25

Problem 1.20

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

Let L, M , and T denote length, mass, and time, respectively. We consider only one of the three equations astheir 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 terms Ixx˛x and .Iyy � I´´/!y!´ must have the same dimensions. Because ˛x is the rate ofchange of an angular velocity, its dimensions are:

Œ˛x� D1

T 2: (2)

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

ŒMx� DML

T 2L: (3)

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

ML2

T 2D ŒIxx�

1

T 2:

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

ŒIxx� D ŒIyy � D ŒI´´� DML2:

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.


May 12, 2012

26 Solutions Manual

Problem 1.21

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

Let L, M , and T denote length, mass, and time, respectively. The dimensions of each of the quantities in thegiven equation are

ŒFL� DML

T 2; Œ�� D

M

L3; Œv� D

L

T; and ŒA� D L2: (2)

Hence, referring to the equation given in the problem statement, we have

ML

T 2DM

L3L2

T 2

�CL.�/

�L2: (3)

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

�D 1: (4)

Therefore, we conclude that

The lift coefficient is dimensionless and, in principle, it does not need the use of specific units.


May 12, 2012

Dynamics 2e 27

Problem 1.22

Are the words units and dimensions synonyms?

Solution

No, units and dimensions are not synonyms. Units are associated with dimensions. “Dimensions” refer to aphysical quantity (e.g., length, mass, time, etc.) which is independent of the specific system of units used tomeasure it. For example, the dimension of length can have units of meters or feet.


May 12, 2012

28 Solutions Manual

Problem 1.23

A rock is released from rest into water. The magnitude Fd of the drag force acting on the rock due to itsmotion through water can be modeled as Fd D Cdv, where v is the speed of the rock and Cd is a constantdrag coefficient. Determine the units used to measure Cd in the U.S. Customary system.

Solution

Let L, M , and T denote length, mass, and time, respectively. Since Fd is a force, then its dimensions are asfollows:

ŒFd � DML

T 2: (1)

Since v is a speed, its dimensions are

Œv� DL

T: (2)

Therefore, considering the given expression for Fd , we have

ŒFd � D ŒCd �Œv� ) ML

T 2D ŒCd �

L

T) ŒCd � D

M

T: (3)

Since in the U.S. Customary system the units of mass are lb�s2=ft and the unit of time is s, then

the units of Cd in the U.S. Customary system are lb�s=ft.


May 12, 2012

Dynamics 2e 29

Problem 1.24

In elementary beam theory, for a uniform beam supported as shown, the relation between the force Papplied at the end of a beam and the corresponding end deflection ı is P D .3EI=L3/ı, where E is aconstant called the modulus of elasticity, I is a constant called the centroidal area moment of inertia, andL is the length of the beam. If the dimensions of I are length to the power four, determine the SI unitsused to measure the constant E.

Solution

Let `, m, and t denote length, mass, and time, respectively. Since P is a force, its dimensions are

ŒP � D m`

t2: (1)

As stated in the problem, the dimensions of I are as follows:

ŒI � D `4: (2)

In addition, since ı represents a displacement and L is the length of the beam, we have

Œı� D ` and ŒL� D ` (3)

Applying dimensional analysis to the given expression for P , and using Eqs. (1)–(3), we have

ŒP � DŒE�ŒI �

ŒL3�Œı� ) m

`

t2D ŒE�

`4

`3` ) ŒE� D

m

`t2: (4)

Since in the SI system m, `, and t have units of kilogram, meter, and second, respectively, then we have that

the units of E are kg=�m�s2�.

It is important to note that, in practice, E is expressed with the same unit used to express pressure. This is aderived unit called pascal, abbreviated Pa where 1Pa D 1 kg=

�m�s2

�.


May 12, 2012

Engineering Mechanics: Dynamics -...

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