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- Title: HW3_solutions_
- Type: Notes
- School: Michigan
- Course: MECHENG 240
- Term: Spring
Coursehero >> Michigan >> Michigan >> MECHENG 240
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: Michigan >> MECHENG >> 240 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
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Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
Path: UF >> EGM >> 3520 Spring, 2009
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Path: Cornell >> NTRES >> 2010 Spring, 2009
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Additional Note: problem (solving for T ) to 14.86 is 14.85 on 5th edition text book. [solution] Using polar coordinates, Fr = T + f = mar = m r r 2 F = NH = ma = m r + 2r Considering vertical direction (through the paper) for gravity (g) and the normal force (Ng ) Fvert = mg + Ng = mavert = 0 Hence, Ng = mg From given angular acceleration, =6 = 6t = 3t 2 for initial string length r0 = 0.4m, the and radius of the fixed cylinder R = 0.05m, similarly to Problem 14.86, r can be expressed as, r = r0 R = r0 3Rt 2 r = R = 6Rt r = R = 6R for t=1, plugging all numbers into the equation force is, F = ma to get Nh = 2.1 N, and the friction force due to the total normal 2 2 f = k Ntotal = k NH + Ng = 10.03 N Hence, answer for the tension (T) is, T t = 1 = f m r r = 11.306 N
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Path: Michigan >> MECHENG >> 240 Spring, 2009
Description: Answers in back of book: 15.12: 3.27 m/s 15.30: 15.36: 3.55 m/s 15.66: 5.77 m/s ...
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HW7_solutions_Path: Michigan >> MECHENG >> 240 Spring, 2009
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Description: [ME240 2008W] Homework #9 - Problem Set (Due : 3/21 Fri) Chap. 18 - 38, 40, 45, 48, 65, 125 [Answers for even number problems from the text book (5th edition)] 18.38 (a) 14.8 rad/s^2 clockwise (b) 0.227 18.40 Velocity = 3.81 ft/s, time=1.97s 18.48 ...
HW9_solutions_Path: Michigan >> MECHENG >> 240 Spring, 2009
Description: ...
HW10_questions_Path: Michigan >> MECHENG >> 240 Spring, 2009
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HW11_questions_Path: Michigan >> MECHENG >> 240 Spring, 2009
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HW11_solutions_Path: Michigan >> MECHENG >> 240 Spring, 2009
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HW12_questions_Path: Michigan >> MECHENG >> 240 Spring, 2009
Description: [ME240 2008W] Homework #9 - Problem Set (Due : 4/14 Mon) Chap. 21 - 39, 44, 51, 61, 63, 70 [Answers for even number problems from the text book (5th edition)] 21.44 (a) t_d=2.32s , f_d=0.431 Hz (b) 5.28s 21.70 16.5 in ...
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Description: ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ...
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Description: ...
14_15Path: UF >> EGM >> 3520 Spring, 2009
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14_3Path: UF >> EGM >> 3520 Spring, 2009
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18_12Path: UF >> EGM >> 3520 Spring, 2009
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18_19aPath: UF >> EGM >> 3520 Spring, 2009
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18_19bPath: UF >> EGM >> 3520 Spring, 2009
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18_29Path: UF >> EGM >> 3520 Spring, 2009
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Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 9. First note that, by inspection of joint H: FCH = 0 ! and FDH = FGH FCG = 0 ! and FBC = FCD FBG = 0 ! and FFG = FGH FBF = 0 ! and FAB = FBC then, by inspection of j...
6_10Path: UF >> EGM >> 3520 Spring, 2009
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 10. FBD Truss: Fx = 0: H x = 0 By symmetry: A y = H y = 4 kips by inspection of joints C and G : FAC = FCE and FBC = 0 FEG = FGH and FFG = 0 also, by symmetry FAB = FFH...
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6_86Path: UF >> EGM >> 3520 Spring, 2009
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 86. (a) FBD AC: Note: CE is a two-force member M A = 0: 1 1 - ( 8 in.) FCE - ( 2 in.) FCE + 192 lb in. = 0 2 2 FCE = 19.2 2 lb, E x = 19.20 lb E y = 19....
7_38Path: UF >> EGM >> 3520 Spring, 2009
Description: ...
7_65Path: UF >> EGM >> 3520 Spring, 2009
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 65. (a) Shear Diag: dV V is zero at A with constant slope = - w = -1 kip/ft decreasing to dx - 3.6 kips at C. V then jumps 9 kips to 5.4 kips and is constant t...
7_74Path: UF >> EGM >> 3520 Spring, 2009
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 74. (a) Fy = 0: Ay - ( 2 kips/ft )( 8 ft ) - 3 kips + 7 kips = 0 A y = 12 kips M A = 0: Shear Diag: V jumps to 12 kips at A, then decreases at 2 kips/ft to - 4 kips...
7_78Path: UF >> EGM >> 3520 Spring, 2009
Description: COSMOS: Complete Online Solutions Manual Organization System Chapter 7, Solution 78. (a) Note: The 2 kip force at E has been replaced by the equivalent force and couple at C. M A = 0: - ( 6 ft )(1 kip/ft )(12 ft ) + 8 kip ft - (12 ft )( 2 kips )...
EMLch5Path: UF >> EML >> 3100 Spring, 2009
Description: SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 5 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION Correspondence table Concept-Study Guide Problems Kinetic and potential ene...
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Description: Introduction to Renewable Natural Resources Plato: Human intellect ponders a physical reality (nature) to which the intellect is not connected. Aristotle: There exists a fundamental connection between thinker intellect and the physical reality th...
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Description: Freshwater Resources Principal Issues in Renewable Natural Resources 1. sustainability: management approaches 2. conservation vs. preservation: wise use and the rights of nature A major difficulty in allocating groundwater is that the future value ...
Lec_27_Forest Management_03_27_09Path: Cornell >> NTRES >> 2010 Spring, 2009
Description: If any resource can be shifted from one configuration to another, and it would bring greater satisfaction in the second, then it should be so shifted. Maxim of resource economics McEcoy, A.F. 1987. Toward an interactive theory of nature and culture:...
648520Path: Cornell >> DSOC >> 1101 Spring, 2009
Description: Sending Sickness: Sorcery, Politics, and Changing Concepts of AIDS in Rural Haiti Author(s): Paul Farmer Source: Medical Anthropology Quarterly, New Series, Vol. 4, No. 1, Culture and Behavior in the AIDS Epidemic (Mar., 1990), pp. 6-27 Published by:...
2771989Path: Cornell >> DSOC >> 1101 Spring, 2009
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Description: Warning Concerning Copyright Restrictions The copyright law of the United States (Title 17, United States Code) governs the making of photocopies or other reproductions of copyrighted material. Under certain conditions specified in the law, libraries...