Engineering mechanics dynamics 14th edition pdf free + solutions manual

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A Proven Approach to Conceptual Understanding and Problem-solving Skills
Engineering Mechanics: Dynamics excels in providing a clear and thorough presentation of the theory and application of engineering mechanics. Engineering Mechanics empowers students to succeed by drawing upon Professor Hibbeler’s everyday classroom experience and his knowledge of how students learn. This text is shaped by the comments and suggestions of hundreds of reviewers in the teaching profession, as well as many of the author’s students.

 The Fourteenth Edition includes new Preliminary Problems, which are intended to help students develop conceptual understanding and build problem-solving skills. The text features a large variety of problems from a broad range of engineering disciplines, stressing practical, realistic situations encountered in professional practice, and having varying levels of difficulty.

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• 1193

  Engingeering Mechanics Dynamics in SI Units 14th Edition Hibbeler

  SOLUTIONS MANUAL

  Full download at: https://testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14th-edition-hibbeler-solutions-manual/

  221.

  A spring is stretched 175 mm by an 8-kg block. If the block is

  displaced 100 mm downward from its equilibrium position

  and given a downward velocity of 1.50 m > s, determine the

  differential equation which describes the motion. Assume

  that positive displacement is downward. Also, determine the

  position of the block when t = 0.22 s.

  SOLUTION $

  + T Fy = may; mg - k(y + yst) = my where kyst = mg

  $ k y +

  m y = 0

  k Hence p =

  B m Where k =

  8(9.81)

  0.175 = 448.46 N > m

  448.46

  = B 8

  = 7.487

  $ $ 6 y + (7.487)2y = 0 y + 56.1y = 0 Ans.

  The solution of the above differential equation is of the form:

  y = A sin pt + B cos pt (1)

  # v = y = Ap cos pt - Bp sin pt (2)

  At t = 0, y = 0.1 m and v = v0 = 1.50 m > s

  From Eq. (1) 0 .1 = A sin 0 + B cos 0 B = 0.1 m

  v0 1.50

  From Eq. (2) v0 = Ap cos 0 - 0 A = p

  = 7.487

  = 0.2003 m

  Hence y = 0.2003 sin 7.487t + 0.1 cos 7.487t

  At t = 0.22 s, y = 0.2003 sin [7.487(0.22)] + 0.1 cos [7.487(0.22)]

  = 0.192 m Ans.

  https://testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14th-edition-hibbeler-solutions-manual/https://testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14th-edition-hibbeler-solutions-manual/

• 1194

  Ans:

  y + 56.1 y = 0

  y 0 t = 0.22 s = 0.192 m

• 1195

  222.

  A spring has a stiffness of 800 N > m. If a 2-kg block is

  attached to the spring, pushed 50 mm above its equilibrium

  position, and released from rest, determine the equation

  that describes the blocks motion. Assume that positive

  displacement is downward.

  SOLUTION

  k p =

  A m =

  A

  800

  2 = 20

  2-Kg

  x = A sin pt + B cos pt

  x = - 0.05 m when t = 0,

  - 0.05 = 0 + B; B = - 0.05

  v = Ap cos pt - Bp sin pt

  v = 0 when t = 0,

  0 = A(20) - 0; A = 0

  Thus,

  x = - 0.05 cos (20t) Ans.

  Ans:

  x = - 0.05 cos (20t)

• 1196

  223.

  A spring is stretched 200 mm by a 15-kg block. If the block

  is displaced 100 mm downward from its equilibrium

  position and given a downward velocity of 0.75 m>s,

  determine the equation which describes the motion. What is

  the phase angle? Assume that positive displacement is

  downward.

  SOLUTION

  F

  k = y

  =

  15(9.81)

  0.2 = 735.75 N>m

  k vn =

  A m =

  A

  735.75

  15 = 7.00

  y = A sin vn t + B cos vn t

  y = 0.1 m when t = 0,

  0.1 = 0 + B; B = 0.1

  v = A vn cos vn t - Bvn sin vn t

  v = 0.75 m>s when t = 0,

  0.75 = A(7.00)

  A = 0.107

  y = 0.107 sin (7.00t) + 0.100 cos (7.00t)

  f = tan - 1 a B

  b = tan - 1 a 0.100

  b = 43.0

  Ans.

  Ans.

  A 0.107

  Ans:

  y = 0.107 sin (7.00t) + 0.100 cos (7.00t)

  f = 43.0

• 1197

  *224

  When a 2-kg block is suspended from a spring, the spring is

  stretched a distance of 40 mm. Determine the frequency and

  the period of vibration for a 0.5-kg block attached to the same

  spring.

  SOLUTION

  F

  k = y

  =

  2(9.81)

  0.040 = 490.5 N>m

  k vn =

  A m =

  A

  490.5

  0.5 = 31.321

  vn f =

  2p =

  31.321

  2p = 4.985 = 4.98 Hz

  Ans.

  1 1t =

  f =

  4.985 = 0.201 s

  Ans.

  Ans:

  f = 4.98 Hz

  t = 0.201 s

• 1198

  225.

  When a 3-kg block is suspended from a spring, the spring is

  stretched a distance of 60 mm. Determine the natural

  frequency and the period of vibration for a 0.2-kg block

  attached to the same spring.

  SOLUTION F

  k = = x

  3(9.81)

  0.060

  = 490.5 N>m

  k vn =

  A m =

  A

  490.5

  0.2 = 49.52 = 49.5 rad >s

  Ans.

  vn f =

  2p =

  49.52

  2p = 7.88 Hz

  1 1t =

  f =

  7.88 = 0.127 s

  Ans.

  Ans:

  vn = 49.5 rad > s

  t = 0.127 s

• 1199

  Engingeering Mechanics Dynamics in SI Units 14th Edition Hibbeler

  SOLUTIONS MANUAL

  Full download at: https://testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14th-edition-hibbeler-solutions-manual/

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  https://testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14th-edition-hibbeler-solutions-manual/https://testbankreal.com/download/engingeering-mechanics-dynamics-si-units-14th-edition-hibbeler-solutions-manual/