Engineering Mechanics Statics - Examples Part 8 pptx

Engineering Mechanics Statics - Examples Part 8 pptx

Engineering Mechanics Statics - Examples Part 8 pptx

... Engineering Mechanics - Statics Chapter 6 pppj socket joint at C. Given: M 50 kg= g 9 .81 m s 2 = a 0.25 m= b 0.5 m= c 2m= Solution: hb 2 a 2 −= AB a− c− h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = ... (C) A y A z B x B z C y C z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 200− 667− 0 667 300− 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= F BA F BC F BD F AD F AC F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 167 2...

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Engineering Mechanics Statics - Examples Part 16 ppsx

Engineering Mechanics Statics - Examples Part 16 ppsx

... Engineering Mechanics - Statics Chapter 10 θ p1 1 2 asin I xy R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ p1 6. 08 deg= θ p2 θ p1 90 deg+= θ p2 96. 08 deg= Problem 1 0-9 0 The right circular cone ... the publisher. Engineering Mechanics - Statics Chapter 10 c 0.1 m= d 0.3 m= Solution: y c b ρ r b 2 π d 2 ρ s bd+()+ π c 2 ρ s bd+()− b ρ r π d 2 ρ s + π c 2 ρ s − ρ r 2a+ = y c 0 .88 8 m= I G 1 12...

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Engineering Mechanics Statics - Examples Part 1 pdf

Engineering Mechanics Statics - Examples Part 1 pdf

... from the publisher. Engineering Mechanics - Statics Chapter 1 Given: a 1 0.631 Mm= b 1 8. 60 kg= a 2 35 mm= b 2 48 kg= Solution: a() a 1 b 1 2 8. 532 km kg 2 = b() a 2 2 b 2 3 135. 48 kg 3 m 2 ⋅= 12 © ... means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 1 b() S 2 8. 3 68 kN= c() S 3 89 3g= Problem 1-6 Represent eac...

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Engineering Mechanics Statics - Examples Part 2 pps

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 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠...

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Engineering Mechanics Statics - Examples Part 4 potx

Engineering Mechanics Statics - Examples Part 4 potx

... writing from the publisher. Engineering Mechanics - Statics Chapter 4 FM A c 2 d 2 + da ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 35.2 N= Problem 4-2 2 Determine the clockwise direction θ 0deg θ≤ 180 deg≤ () of the force ... publisher. Engineering Mechanics - Statics Chapter 4 Units Used: kN 10 3 N= Given: a 16 mm= F 2.30 kN= θ 60− 80 ()= Solution: M A θ () Fa( ) cos θ deg () = 50 0 50 10...

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Engineering Mechanics Statics - Examples Part 5 docx

Engineering Mechanics Statics - Examples Part 5 docx

... publisher. Engineering Mechanics - Statics Chapter 4 d 6m= g 5m= Solution: F R F= M R M f− e dg+ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F×+= F R 8 6 8 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= M R 46− 66 56− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Problem 4-1 27 Replace ... sin θ () rcos θ () ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F R F 1v F 2v += M A r 1 F 1v × r 2 F 2v ×+= F R 0 28. 28 68. 28 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M A 0 20.49− 8. 49 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅=...

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Engineering Mechanics Statics - Examples Part 6 doc

Engineering Mechanics Statics - Examples Part 6 doc

... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 5 Given: F 1 380 lb= b 12 in= c 10 in= F 2 500 lb= d 5in= F 3 80 0 lb= e 12 in= a 8in= f 12 in= Solution: The initail ... means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 5 F C F B cos θ ( ) = F C 9.3 78 lb= Problem 5-2 2 The uniform door h...

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Engineering Mechanics Statics - Examples Part 7 docx

Engineering Mechanics Statics - Examples Part 7 docx

... publisher. Engineering Mechanics - Statics Chapter 6 F AB F AF F BC F BF F FC F FE F ED F EC F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 377.1− 189 .7 266.7− 266.7 188 .6 56.4 56.4 0.0 188 .6− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Positive ... means, without permission in writing from the publisher. Engineering Mechanics - Statics...

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Engineering Mechanics Statics - Examples Part 9 potx

Engineering Mechanics Statics - Examples Part 9 potx

... 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. ... b 2 c 2 + = D z B z = B z 283 N= D z 283 N= B y D y + F max b b 2 c 2 + − 0= D y B y = B y F max b 2 b 2 c 2 + = D y B y = B y 283 N= D y 283 N= B x D x = 0= Problem 6-1 26 The struc...

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Engineering Mechanics Statics - Examples Part 10 ppsx

Engineering Mechanics Statics - Examples Part 10 ppsx

... publisher. Engineering Mechanics - Statics Chapter 7 024 681 0 5000 0 5000 Distance (ft) Force (lb) Vx() x ft 024 681 0 1 . 10 4 0 1 . 10 4 Distance (ft) Moment (lb-ft) Mx() x ft Problem 7-4 7 The shaft ... the publisher. Engineering Mechanics - Statics Chapter 7 024 681 0 200 0 200 Distance in m Force in kN V 1 x 1 () V 2 x 2 () x 1 x 2 , 024 681 0 400 200 0 200 400 Dis...

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