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. ... form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Problem 6 -9 9 The truck rests on the scale, which consists o...

Ngày tải lên: 11/08/2014, 02:22

... form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 Problem 4 -9 Determine the magnitude and directional sense of the moment of the forces ... from the publisher. Engineering Mechanics - Statics Chapter 4 Given x y z ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F× M O = y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find yz,()= y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m= Proble...

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... the publisher. Engineering Mechanics - Statics Chapter 8 r f D 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin φ k () = r f 0.0 499 4 in= φ asin r f r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 0.1788 deg= PWtan φ () = P 62.4 lb= Problem 8-1 24 The trailer ... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 Σ M B = 0; F'− rFr+ M+ 0= MF'rFr−= M 177N m⋅= Problem 8 -9 3 Blocks A and...

Ngày tải lên: 11/08/2014, 03:20

... 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 ... writing from the publisher. Engineering Mechanics - Statics Chapter 10 θ p1 1− 2 asin I xy R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ p1 31. 39 deg= θ p2 θ p1 π 2 += θ p2 58.61 deg= Problem 1 0-8 9 The area of the cross...

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... publisher. Engineering Mechanics - Statics Chapter 2 Solution: α atan c d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F' F 1 2 F 2 2 + 2F 1 F 2 cos 90 deg θ + α − () −= F' 30 .9 N= F' sin 90 deg θ − α + ()() F 1 sin 90 ... 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...

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... 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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... from the publisher. Engineering Mechanics - Statics Chapter 4 d 9m= Solution: F R 1 2 w 1 w 2 + () d= F R 90 kN= M RO w 2 d d 2 1 2 w 1 w 2 − () d d 3 += M RO 338kN m⋅= Problem 4-1 45 Replace the ... means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 θ 1 atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 1 62 .99 6 deg= F Ry x() MF 2 dc+()− F...

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... permission in writing from the publisher. Engineering Mechanics - Statics Chapter 5 Units used: Mg 10 3 kg= Given: M 1.4 Mg= a 0.4 m= g 9. 81 m s 2 = Solution: θ 090 = N A θ () Mgsin θ deg () 10 3 = N B θ () Mgcos θ deg () 10 3 = 0 ... means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 5 F C F B cos θ ( ) = F C 9. 378 lb= Pr...

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... from the 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. 4 233.3− 233.3 47.1− 112.7 112.7 300.0 377.1− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Positive ... from the publisher. Engineering Mechanics - Statics Chapter 6 F AB F AF F BC F BF...

Ngày tải lên: 11/08/2014, 02:22

... R B , R C , R D , () = A x A y R B R C R D ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0.601 0. 399 0.564 0. 399 0. 399 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Now find the critical load and scale the problem ans R B R C R D ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ... writing from the publisher. Engineering Mechanics - Statics Chapter 6 Positive (T) Negative (C) F AB F AC F AD F AE F BC F BE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞...

Ngày tải lên: 11/08/2014, 02:22