Solutions Manual
Engineering Mechanics: Dynamics 1st Edition
Gary L. Gray The Pennsylvania State University
Francesco Costanzo The Pennsylvania State University
Michael E. Plesha University of Wisconsin–Madison With the assistance of: Chris Punshon Andrew J. Miller Justin High Chris O’Brien Chandan Kumar Joseph Wyne
Version: August 10, 2009
The McGraw-Hill Companies, Inc.
Copyright © 2002–2010 Gary L. Gray, Francesco Costanzo, and Michael E. Plesha
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 permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.
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Dynamics 1e
Important Information about this 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 July 30, 2009, please note the following: _ The solutions for Chapters 1 and 2 have been accuracy checked and have been edited by us. They are in their final form. _ The solutions for Chapters 4 and 7 have been accuracy checked and should be error free. We will be adding some additional detail to these solutions in the coming weeks. _ The solutions for Chapters 3, 6, 8, and 9 are being accuracy checked and the accuracy checked versions should be available by the end of August 2009. We will be adding some additional detail to these solutions in the coming weeks. _ The solutions for Chapter 10 should be available in their entirety by the end of August 2009. All of the figures in Chapters 6–10 are in color. Color will be added to the figures in Chapters 1–5 over the coming weeks.
Contact the Authors If you find any errors and/or have questions concerning a solution, please do not hesitate to contact the authors and editors via email at:
[email protected] We welcome your input.
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Solutions Manual
Accuracy of Numbers in Calculations Throughout this solutions manual, we will generally assume that the data given for problems is accurate to 3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4 significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported, you should obtain the final answers that are reported to 3 significant digits.
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Dynamics 1e
Chapter 1 Solutions Problem 1.1 Determine .rB=A /x and .rB=A /y , the x and y components of the vector rEB=A , so as to be able to write rEB=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 rEB=A can be written as rEB=A D .xB xA / {O C .yB yA / |O, where .xA ; yA / and .xB ; yB / are the Cartesian coordinates of points A and B relative 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 rEB=A D .4:00 {O
1:00 |O/ ft:
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Problem 1.2 If the positive direction of line ` is from D to C , find the component of the vector rEB=A along `.
Solution Using the component system formed by the unit vectors {O and |O, the vector rEB=A can be written as rEB=A D .xB xA / {O C .yB yA / |O, where .xA ; yA / and .xB ; yB / are the Cartesian coordinates of points A and B relative 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. Therefore, we have rEB=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 rEC =D and scaling it by its own magnitude. rEC =D D .xC
xD / {O C .yC
yD / |O D .3:000 {O
2:000 |O/ ft:
(2)
Next, we derive the magnitude of rEC =D : jErC =D j D
q
.3:000 ft/2 C .2:000 ft/2 D 3:606 ft:
(3)
Hence we have uO ` D
rEC =D D jErC =D j
3:000 ft {O 3:606 ft
2:000 ft |O D 0:8319 {O 3:606 ft
0:5546 |O:
(4)
The component of rEC =D along ` is then obtained by taking the dot product of rEB=A and uO ` , 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:
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Dynamics 1e
Problem 1.3 Find the components of rEB=A along the p and q axes.
Solution Using the component system formed by the unit vectors {O and |O, the vector rEB=A can be written as rEB=A D .xB xA / {O C .yB yA / |O, where .xA ; yA / and .xB ; yB / are the Cartesian coordinates of points A and B relative 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. Therefore, we have rEB=A D .4:000 {O
1:000 |O/ ft:
(1)
Now we find the unit vectors uOp and uO q , which are the unit vectors orienting the p and q axes, respectively, uOp D cos p {O C sin p |O;
(2)
uO q 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 therefore are given by p D D 22:5ı , q D . C 90ı / D 112:5ı . Consequently, Eqs. (2) and (3) become uOp D 0:9239 {O C 0:3827 |O
and
uO q D
0:3827 {O C 0:9239 |O:
(4)
Then desired components are obtained by computing the dot product of the vector rEB=A with the unit vectors in Eqs. (4), i.e., .rB=A /p D uOp rEB=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 uO q rEB=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:
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Solutions Manual
Problem 1.4 Determine expressions for the vector rEB=A using both the xy and the pq coordinate systems. Next, determine jErB=A j, the magnitude of rEB=A , using both the xy and the pq representations and establish whether or not the two values for jErB=A j are equal to each other.
Solution Using the component system formed by the unit vectors {O and |O, the vector rEB=A can be written as rEB=A D .xB xA / {O C .yB yA / |O, where .xA ; yA / and .xB ; yB / are the Cartesian coordinates of points A and B relative 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 rEB=A D .4:000 {O
1:000 |O/ ft;
(1)
which, when expressed to three significant figures, gives rEB=A D .4:00 {O
1:00 |O/ ft:
Now we find the unit vectors uOp and uO q , which are the unit vectors orienting the p and q axes, respectively, uOp D cos p {O C sin p |O
and uO q 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 therefore are given by p D D 22:5ı , q D . C 90ı / D 112:5ı . Consequently, Eqs. (2) and (3) become uOp D 0:9239 {O C 0:3827 |O
and
uO q D
0:3827 {O C 0:9239 |O:
(3)
Then, using Eqs. (1) and (3), the expression for rEB=A in the pq component system is rEB=A D .uOp rEB=A / uOp C .uO q rEB=A / uO q ˚ D .4:000 {O 1:000 |O/ ft .0:9239 {O C 0:3827 |O/ uOp ˚ C .4:000 {O 1:000 |O/ ft . 0:3827 {O C 0:9239 |O/ uO q D .3:313 uOp
2:455 uO q / ft;
(4)
which, when expressed to three significant figures gives rEB=A D .3:31 uOp
2:46 uO q / ft: August 10, 2009
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Dynamics 1e Using Eq. (1), the magnitude of rEB=A expressed in the xy component system is ˇ ˇ q ˇrEB=A ˇ D .4:000 ft/2 C . 1:000 ft/2 D 4:123 ft:
(5)
Using the result in Eq. (4), the magnitude of rEB=A expressed in the pq component system is ˇ ˇ q ˇrEB=A ˇ D .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 ˇ ˇ ˇ ˇ ˇrEB=A ˇ D ˇrEB=A ˇpq system D 4:12 ft: xy system
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Solutions Manual
Problem 1.5 ˇ ˇ ˇ ˇ Suppose that you were to compute the quantities ˇrEB=A ˇxy and ˇrEB=A ˇpq , that is, the magnitude of the vector rEB=A computed using the xy and pq frames, respectively. Do you expect these two scalar values to be 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 of how the vector is represented. In other words, a choice of frame only affects the values ˇ of the ˇ components of ˇ a vector and not its magnitude or direction. For this reason, we expect the value of rEB=A ˇxy to be same as ˇ ˇ that of ˇrEB=A ˇpq .
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Dynamics 1e
Problem 1.6 The measure of angles in radians is defined according to the following relation: r D sAB , where r is the radius of the circle and sAB denotes the length 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 ΠD 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 problem illustrates the idea that some nondimensional quantities still require proper use of units.
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Solutions Manual
Problem 1.7 A simple oscillator consists of a linear spring fixed at one end and a mass attached at the other end, which is free to move. Suppose that the periodic motion of a simple oscillator is described by the relation y D Y0 sin .2!0 t /, where y has units of length and denotes the vertical position 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 the dimensions 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 of length, 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!0 t / is the argument of a trigonometric function and as such it must be nondimensional. Therefore we have Œ2!0 t D 1 ) Œ!0 Œt D 1; Œ!0 D
1 : 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 erroneous conclusion. The reason for this is that, as indicated in the problem statement, the term !0 appears as part of the argument 2!0 t . 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, see the discussion about Eqs. (9.4) and (9.8) on p. 674 of th textbook). In both the SI and the U.S. Customary systems, the number 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:
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Dynamics 1e
Problem 1.8 To study the motion of a space station, the station can be modeled as a rigid body and the equations describing its motion can be chosen to be Euler’s equations, which read Mx D Ixx ˛x Iyy I´´ !y !´ ; My D Iyy ˛y
.I´´
M´ D I´´ ˛´
Ixx
ជ z ជ M
␣ជ
y
Ixx / !x !´ ; Iyy !x !y :
In the previous equations, Mx , My , and M´ denote the x, y, and ´ components of the moment applied to the body; !x , !y , and !´ denote the corresponding components of the angular velocity of the body, where angular velocity is defined as the time rate of change of an angle; ˛x , ˛y , and ˛´ denote the corresponding components of the angular acceleration of the body, where angular acceleration is defined as the time rate of change of an angular velocity. The quantities Ixx , Iyy , and I´´ are called the principal mass 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. Customary system.
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 D
1 : 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 ŒL2
1 : ŒT 2
(3)
Substituting Eqs. (2) and (3) into the last of Eqs. (1), we have ŒM ŒL2
1 1 D ŒI : xx ŒT 2 ŒT 2
Therefore, we conclude that the dimensions of the quantities Ixx , Iyy , and I´´ are ŒIxx D ŒIyy D ŒI´´ D ŒM ŒL2 : 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: kgm2 ; Units of Ixx , Iyy , and I´´ in the U.S. Customary system: slugft2 D lbs2 ft. August 10, 2009
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Solutions Manual
Problem 1.9 The lift force FL generated by the airflow moving over a wing is often expressed as follows: FL D 12 v 2 CL . /A;
(1)
where , v, and A denote the mass density of air, the airspeed (relative to the wing), and the wing’s nominal surface area, respectively. The quantity CL is called the lift coefficient, and it is a function of the wing’s angle 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 ML M ŒFL D ; Œ D ; ŒA D L2 ; 2 3 T L
L : Œv D T
(2)
When the lift force .FL / equation is written in terms of its dimensions, it becomes
2 ML M L CL . / L2 : D 2 3 T T L
Simplifying and solving for the dimensions of CL . / yields ML M D ŒCL . / 2 ŒL ) 2 T T
CL . / D 1:
(3)
(4)
Therefore we conclude that The lift coefficient is dimensionless and has no units associated with it.
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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 independent of the specific system of units used to measure it.
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