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Solution Manual of Engineering Mechanics Dynamics 1st edition by Gray pdf

https://gioumeh.com/product/engineering-mechanics-dynamics-solution/<br>-----------------------------------------------------------------------------------<br>Authors: Gary Gray, Francesco Costanzo, Michael Plesha<br> Published: McGraw<br> Edition: 1st<br> Pages: 1265<br> Type: pdf<br> Size: 124MB<br> Sample: solution sample file<br> Download After Payment

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Solution Manual of Engineering Mechanics Dynamics 1st edition by Gray pdf

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  1. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ cl i ck h ere t o d ow nl oad 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. @solutionmanual1

  2. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ cl i ck h ere t o d ow nl oad 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. @solutionmanual1

  3. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Dynamics 1e 3 cl i ck h ere t o d ow nl oad 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: dyn_solns@email.esm.psu.edu We welcome your input. August 10, 2009 @solutionmanual1

  4. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Solutions Manual 4 cl i ck h ere t o d ow nl oad 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. August 10, 2009 @solutionmanual1

  5. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Dynamics 1e Chapter 1 Solutions 5 cl i ck h ere t o d ow nl oad Problem 1.1 Determine .rB=A/xand .rB=A/y, the x and y components of the vector E rB=A, so as to be able to write E rB=AD .rB=A/xO { C .rB=A/yO |. Solution Using the component system formed by the unit vectors O { and O |, the vector E rB=Acan be written as E rB=AD .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 E rB=AD .4:00O { ? 1:00 O |/ft: August 10, 2009 @solutionmanual1

  6. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Solutions Manual 6 cl i ck h ere t o d ow nl oad Problem 1.2 If the positive direction of line ` is from D to C, find the component of the vector E rB=Aalong `. Solution Using the component system formed by the unit vectors O { and O |, the vector E rB=Acan be written as E rB=AD .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 E rB=AD .4:000O { ? 1:000 O |/ft: (1) Next we find the unit vector parallel to ` and pointing from D to C. This is accomplished by finding E rC=D and scaling it by its own magnitude. E rC=DD .xC? xD/O { C .yC? yD/ O | D .3:000O { ? 2:000 O |/ft: Next, we derive the magnitude of E rC=D: q Hence we have ?3:000ft The component of E rC=Dalong ` is then obtained by taking the dot product of E rB=Aand O u`, i.e., .rB=A/`D Œ.4:000O { ? 1:000 O |/ft? ? .0:8319O { ? 0:5546 O |/ D 3:882ft; which, when expressed to three significant figures gives (2) .3:000ft/2C .2:000ft/2D 3:606ft: (3) jE rC=Dj D ? 3:606ftO { ?2:000ft E rC=D jE rC=DjD (4) O u`D 3:606ftO | D 0:8319O { ? 0:5546 O |: (5) .rC=D/`D 3:88ft: August 10, 2009 @solutionmanual1

  7. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Dynamics 1e 7 cl i ck h ere t o d ow nl oad Problem 1.3 Find the components of E rB=Aalong the p and q axes. Solution Using the component system formed by the unit vectors O { and O |, the vector E rB=Acan be written as E rB=AD .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 E rB=AD .4:000O { ? 1:000 O |/ft: (1) Now we find the unit vectors O upand O uq, which are the unit vectors orienting the p and q axes, respectively, O upD cos?pO { C sin?p O |; O uqD cos?qO { C sin?q O |; where ?pand ?qare the angles formed by the p and q axes with the positive x axis, respectively, and therefore are given by ?pD ? D 22:5ı, ?qD .? C 90ı/ D 112:5ı. Consequently, Eqs. (2) and (3) become O upD 0:9239O { C 0:3827 O | Then desired components are obtained by computing the dot product of the vector E rB=Awith the unit vectors in Eqs. (4), i.e., .rB=A/pD O up? E rB=AD?.4:000O { ? 1:000 O |/ft?? .0:9239O { C 0:3827 O |/ D 3:313ft; (2) (3) and (4) O uqD ?0:3827O { C 0:9239 O |: (5) (6) .rB=A/qD O uq? E rB=AD?.4:000O { ? 1:000 O |/ft?? .?0:3827O { C 0:9239 O |/ D ?2:455ft; which, when expressed to three significant figures, give .rB=A/pD 3:31ft and .rB=A/qD ?2:46ft: August 10, 2009 @solutionmanual1

  8. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Solutions Manual 8 cl i ck h ere t o d ow nl oad Problem 1.4 Determine expressions for the vector E rB=Ausing both the xy and the pq coordinate systems. Next, determine jE rB=Aj, the magnitude of E rB=A, using both the xy and the pq representations and establish whether or not the two values for jE rB=Aj are equal to each other. Solution Using the component system formed by the unit vectors O { and O |, the vector E rB=Acan be written as E rB=AD .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 E rB=AD .4:000O { ? 1:000 O |/ft; which, when expressed to three significant figures, gives (1) E rB=AD .4:00O { ? 1:00 O |/ft: Now we find the unit vectors O upand O uq, which are the unit vectors orienting the p and q axes, respectively, O upD cos?pO { C sin?p O | where ?pand ?qare the angles formed by the p and q axes with the positive x axis, respectively, and therefore are given by ?pD ? D 22:5ı, ?qD .? C 90ı/ D 112:5ı. Consequently, Eqs. (2) and (3) become O upD 0:9239O { C 0:3827 O | Then, using Eqs. (1) and (3), the expression for E rB=Ain the pq component system is E rB=AD .O up? E rB=A/ O upC .O uq? E rB=A/ O uq D˚?.4:000O { ? 1:000 O |/ft?? .0:9239O { C 0:3827 O |/?O up D .3:313 O up? 2:455 O uq/ft; which, when expressed to three significant figures gives and O uqD cos?qO { C sin?q O |; (2) and (3) O uqD ?0:3827O { C 0:9239 O |: C˚?.4:000O { ? 1:000 O |/ft?? .?0:3827O { C 0:9239 O |/?O uq (4) E rB=AD .3:31 O up? 2:46 O uq/ft: August 10, 2009 @solutionmanual1

  9. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Dynamics 1e 9 cl i ck h ere t o d ow nl oad Using Eq. (1), the magnitude of E rB=Aexpressed in the xy component system is ˇˇE rB=A Using the result in Eq. (4), the magnitude of E rB=Aexpressed in the pq component system is ˇˇE rB=A Expressing the results in the last two equations to three significant figures we have q ˇˇD .4:000ft/2C .?1:000ft/2D 4:123ft: (5) q ˇˇD .3:313ft/2C .?2:455ft/2D 4:123ft: (6) ˇˇE rB=A ˇˇxy systemDˇˇE rB=A ˇˇpq systemD 4:12ft: August 10, 2009 @solutionmanual1

  10. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Solutions Manual 10 cl i ck h ere t o d ow nl oad Problem 1.5 ˇˇE rB=A ˇˇxyand ˇˇE rB=A ˇˇpq, that is, the magnitude of the Suppose that you were to compute the quantities vector E rB=Acomputed 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 of the components of a vector and not its magnitude or direction. For this reason, we expect the value ofˇˇE rB=A ˇˇxyto be same as that ofˇˇE rB=A ˇˇpq. August 10, 2009 @solutionmanual1

  11. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Dynamics 1e 11 cl i ck h ere t o d ow nl oad 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 sABdenotes the length of the circular arc. Determine the dimensions of the angle ?. Solution Use the given definition of an angle in radians: (1) Œr?? D Œr?Œ?? D ŒsAB?: The radius r and the arc length s must have dimensions of length, we have that (2) ŒL?Œ?? D ŒL?: A simplification of the above relation yields that (3) Œ?? D 1; 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. August 10, 2009 @solutionmanual1

  12. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Solutions Manual 12 cl i ck h ere t o d ow nl oad 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 Y0sin.2?!0t/, where y has units of length and denotes the vertical position of the oscillator, Y0is called the oscillation amplitude, !0is the oscillation frequency, and t is time. Recalling that the argument of a trigonometric function is an angle, determine the dimensions of Y0and !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, Y0must also have dimensions of length: ŒY0? D ŒL?: A such, the units to express the quantity Y0in the SI and the U.S. Customary systems are Units of Y0in the SI system: m. Units of Y0in 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; 1 Œ!0? D ŒT?: If we where to infer the units of !0directly 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 !0appears as part of the argument 2?!0t. When this happens, !0is 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, !0has units of Hz (hertz), i.e., cycles per second: August 10, 2009 @solutionmanual1

  13. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Dynamics 1e 13 cl i ck h ere t o d ow nl oad 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 MxD Ixx˛x??Iyy? I´´ MyD Iyy˛y? .I´´? Ixx/!x!´; M´D I´´˛´??Ixx? Iyy 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 corre- sponding 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. ? ? z M? ? ? ? ?!y!´; ?!x!y: y y Solution We consider only one of the three equations as their dimensions are identical: ŒMx? D ŒIxx?Œ˛x?; (1) ŒMx? D ŒIxx˛x? .Iyy? I´´/!y!´? D ŒIxx?Œ˛x? ? ŒIyy? I´´?Œ!y?Œ!´? given that the additive terms in above equation must all the same dimensions. Because ˛xis the rate of change of an angular velocity, its units are: ) 1 (2) Œ˛x? D ŒT?2: The term Mxis the component of a moment and therefore it must have the dimensions of force times length: 1 ŒMx? D ŒL?ŒF? D ŒM?ŒL?2 (3) ŒT?2: Substituting Eqs. (2) and (3) into the last of Eqs. (1), we have 1 1 ŒM?ŒL?2 ŒT?2D ŒIxx? 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?ft2D lb?s2?ft. August 10, 2009 @solutionmanual1

  14. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Solutions Manual 14 cl i ck h ere t o d ow nl oad Problem 1.9 The lift force FLgenerated by the airflow moving over a wing is often expressed as follows: 2?v2CL.?/A; FLD1 (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 CLis called the lift coefficient, and it is a function of the wing’s angle of attack ?. Find the dimensions of CLand determine its units in the SI system. Solution The dimensions of each of the quantities in the given equation are ?M?L When the lift force .FL/ equation is written in terms of its dimensions, it becomes ?M?L Simplifying and solving for the dimensions of CL.?/ yields ?M?L Therefore we conclude that ? ?M ? ?L ? ŒA? D?L2?; (2) ŒFL? D Œ?? D Œv? D ; ; : T T2 L3 ? ?M ??L ?2?CL.?/??L2?: (3) D T T2 L3 ? ?M ? ?CL.?/?D 1: ŒL? (4) D ŒCL.?/? ) T2 T2 The lift coefficient is dimensionless and has no units associated with it. August 10, 2009 @solutionmanual1

  15. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Dynamics 1e 15 cl i ck h ere t o d ow nl oad 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. August 10, 2009 @solutionmanual1

  16. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ cl i ck h ere t o d ow nl oad 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. @solutionmanual1

  17. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ cl i ck h ere t o d ow nl oad 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. @solutionmanual1

  18. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Dynamics 1e 3 cl i ck h ere t o d ow nl oad 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: dyn_solns@email.esm.psu.edu We welcome your input. August 10, 2009 @solutionmanual1

  19. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Solutions Manual 4 cl i ck h ere t o d ow nl oad 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. August 10, 2009 @solutionmanual1

  20. https://gioumeh.com/product/engineering-mechanics-dynamics-solution/https://gioumeh.com/product/engineering-mechanics-dynamics-solution/ Solutions Manual 16 Chapter 2 Solutions cl i ck h ere t o d ow nl oad Problem 2.1 The position of a car traveling between two stop signs along a straight city block is given by r DŒ9t ? .45=2/sin.2t=5/?m, pute the displacement of the car between 2:1 and 3:7s as well as between 11:1 and 12:7s. For each of these time intervals compute the average velocity. where t denotes time and 0s?t ?17:7s. Com- STOP STOP Solution We start with observing that the two time intervals considered have the same length. More importantly, we observe that the argument of the sine function in the definition of the function r.t/ is understood to be expressed in radians. The displacement of the car along the street between 2:1s to 3:7s is calculated from the function r.t/ as ?r1D r.3:7s/ ? r.2:1s/ D 8:75m: (1) Similarly, the displacement between 11:1s and 12:7s is ?r2D r.12:7s/ ? r.11:1s/ D 13:7m: (2) The average velocity at which the car traveled over the first interval is calculated to be 1Dr.3:7s/ ? r.2:1s/ 3:7s ? 2:1s ?vavg ? D 5:47m=s; (3) while the average velocity of the car over the second interval is 2Dr.12:7s/ ? r.11:1s/ 12:7s ? 11:1s ?vavg ? D 8:58m=s: (4) August 10, 2009 @solutionmanual1

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