Engineering Mechanics Statics - Examples Part 4 potx
... r BC 550 0 200− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ mm= F v F cos φ () sin θ () cos φ () cos θ () sin φ () − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F v 44 .5 34 53.073 40 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M B r BC F v ×= M B 10.615 13.093 29.19 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4- 4 6 The x-ray machine is used for medical diagnosis. ... from the publisher. Engineering Mechanics - Statics Chapter 4 Given x y z ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F× M...
Ngày tải lên: 11/08/2014, 02:22
... publisher. Engineering Mechanics - Statics Chapter 6 The four-member "A" frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. ... from the publisher. Engineering Mechanics - Statics Chapter 6 Given: P 10 lb= a 2in= b 2in= c 0.5 in= d 0.75 in= e 1.5 in= Solution: Define φ atan b d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 69 .44 4 de...
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... publisher. Engineering Mechanics - Statics Chapter 8 r f D 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin φ k () = r f 0. 049 94 in= φ asin r f r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 0.1788 deg= PWtan φ () = P 62 .4 lb= Problem 8-1 24 The trailer ... publisher. Engineering Mechanics - Statics Chapter 8 φ atan μ s () = φ 21.80 deg= θ atan r l πd ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 4. 55 deg= Since there are two blocks, M 2T d 2 tan θφ +...
Ngày tải lên: 11/08/2014, 03:20
Engineering Mechanics Statics - Examples Part 16 ppsx
... writing from the publisher. Engineering Mechanics - Statics Chapter 10 I max I x I y + 2 R+= I max 4. 92 10 6 × mm 4 = I min I x I y + 2 R−= I min 13 644 44. 44 mm 4 = Problem 1 0-8 8 Determine the directions ... circle. Given: I x 45 0 in 4 = I y 1730 in 4 = I xy 138 in 4 = Solution: RI x I x I y + 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 I xy 2 += R 6 54. 71 in 4 = I max I x...
Ngày tải lên: 11/08/2014, 02:21
Engineering Mechanics Statics - Examples Part 1 pdf
... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 1 b 0.0 045 7 kg()= c 34. 6 N()= Solution: l ab c = l 26. 945 μmkg⋅ N = Problem 1-8 If a car is traveling at speed v, ... writing from the publisher. Engineering Mechanics - Statics Chapter 1 Problem 1-1 6 Two particles have masses m 1 and m 2 , respectively. If they are a distance d a...
Ngày tải lên: 11/08/2014, 02:22
Engineering Mechanics Statics - Examples Part 2 pps
... from the publisher. Engineering Mechanics - Statics Chapter 2 Given: F 40 20 50− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= L 25 m= Solution: r L F F = r 14. 9 7.5 18.6− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= Problem 2-9 7 Express each of ... publisher. Engineering Mechanics - Statics Chapter 2 α R β R γ R ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F R F R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α R β R γ R ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 75 .4 90 165 .4 ⎛ ⎜ ⎜ ⎝ ⎞...
Ngày tải lên: 11/08/2014, 02:22
Engineering Mechanics Statics - Examples Part 5 docx
... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 y F 1 cos θ () bF 3 c()+ F Rx = y 4. 617 ft= (Below point B) Problem 4- 1 24 Replace the force system acting on the ... publisher. Engineering Mechanics - Statics Chapter 4 Units Used: kN 10 3 N= Given: F 1 20 kN= a 3m= F 2 50 kN= b 8m= F 3 20 kN= c 2m= F 4 50 kN= d 6m= e 4m= Solut...
Ngày tải lên: 11/08/2014, 02:22
Engineering Mechanics Statics - Examples Part 6 doc
... any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 5 Given: a 0.5 ft= b 3ft= c 4ft= d 4ft= Solution: Σ M B = 0; N A − ab+()Wb ccos θ () − () + ... form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 5 Problem 5 -4 5 The mobile crane has weight W 1 and center o...
Ngày tải lên: 11/08/2014, 02:22
Engineering Mechanics Statics - Examples Part 7 docx
... in writing from the publisher. Engineering Mechanics - Statics Chapter 5 N B aPb c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 − 0= N B P b 2 c 4 a = N B 40 0 N= Σ F x = 0; A x N B = A x 40 0 N= Σ F y = 0; A y Pb c 2 = ... publisher. Engineering Mechanics - Statics Chapter 6 F AB F AF F BC F BF F FC F FE F ED F EC F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 330.0− 79 .4...
Ngày tải lên: 11/08/2014, 02:22
Engineering Mechanics Statics - Examples Part 8 pptx
... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Guess E y 1lb= F GJ 1lb= Given F 2 − a() F 3 2a()− F 4 3a()− E y 4a()+ 0= F 4 − a() E y 2a()+ F GJ sin θ () 2a()+ ... publisher. Engineering Mechanics - Statics Chapter 6 F 2 10 kN= a 4m= F 3 5kN= b 4m= Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Inital Guesses F AB 1kN= F BG 1kN= F BF 1kN= Give...
Ngày tải lên: 11/08/2014, 02:22