Engineering Mechanics Statics - Examples Part 1 pdf

... of each particle. Units Used: G 66.73 10 12 − × m 3 kg s 2 ⋅ = nN 10 9− N= Given: m 1 8kg= m 2 12 kg= d 800 mm= Solution: F Gm 1 m 2 d 2 = F 10 .0 nN= W 1 m 1 g= W 1 78.5 N= W 1 F 7.85 10 9 ×= W 2 m 2 g= ... Engineering Mechanics - Statics Chapter 2 F 2 500 N= β 45 deg= γ 70 deg= Solution: F 1u sin γα − () F 1 sin 18 0 deg γ − () = F 1u F 1 sin γα − () sin...

Ngày tải lên: 11/08/2014, 02:22

... publisher. Engineering Mechanics - Statics Chapter 7 0 2 4 6 810 1 214 1 618 20 10 00 500 0 500 10 00 15 00 Distance (ft) Force (lb) V 1 x 1 () V 2 x 2 () V 3 x 3 () x 1 ft x 2 ft , x 3 ft , 0246 810 1 214 1 618 20 1 . 10 4 5000 0 5000 Distance ... BwcC−= B 316 .67lb= x 1 0 0. 01 a, a = V 1 x 1 () BF+() 1 lb = M 1 x 1 () F− ax 1 − () Ba b+ x 1 − () − ⎡...

Ngày tải lên: 11/08/2014, 03:20

... the publisher. Engineering Mechanics - Statics Chapter 10 I x 2 1 2 m a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 md e−() 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 12 M 2d() 2 e 2 + ⎡ ⎣ ⎤ ⎦ += I x 3.25 10 3− × kg m 2 ⋅= Problem 1 0 -1 10 Determine ... 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 d...

Ngày tải lên: 11/08/2014, 02:21

... publisher. Engineering Mechanics - Statics Chapter 2 Solution: F 1v F 1 c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ d 0 c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F 1v 64 0 48 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= α 1 β 1 γ 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F 1v F 1v ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 1 β 1 γ 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 36.9 90 53 .1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= F Rv F R cos φ () sin θ () cos φ () cos θ () sin φ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = ... means, witho...

Ngày tải lên: 11/08/2014, 02:22

... publisher. Engineering Mechanics - Statics Chapter 4 θ 1 30 deg= θ 2 45 deg= a 5in= b 13 in= c 3in= d 6in= e 3in= f 6in= Solution: M RO = Σ M O ; M RO F 1 cos θ 2 () eF 1 sin θ 2 () f− F 2 cos θ 1 () b 2 a 2 −− ... 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...

Ngày tải lên: 11/08/2014, 02:22

... publisher. Engineering Mechanics - Statics Chapter 4 w 2 20 kN m = a 3m= b 3m= c 4.5 m= Solution: F R 1 2 w 2 w 1 − () cw 1 c+ w 1 b+ 1 2 w 1 a+= F R 95.6 kN= M Ro 1 2 − w 2 w 1 − () c c 3 w 1 c c 2 − ... the publisher. Engineering Mechanics - Statics Chapter 4 kip 10 3 lb= Given: w 1 50 lb ft = w 2 300 lb ft = w 3 10 0 lb ft = a 12 ft= b 9ft= Solut...

Ngày tải lên: 11/08/2014, 02:22

... publisher. Engineering Mechanics - Statics Chapter 5 Units Used: kN 10 3 N= Given: F 500 N= a 0 .15 m= L 3m= Solution: The initial guesses w 1 1 kN m = w 2 1 kN m = Given + ↑ Σ F y = 0; 1 2 w 1 a 1 2 w 2 a− ... the publisher. Engineering Mechanics - Statics Chapter 5 Given: W 850 lb= a 0.5 ft= b 1ft= c 1. 5 ft= θ 1 10 deg= θ 2 30 deg= θ 3 10 deg= Solution:...

Ngày tải lên: 11/08/2014, 02:22

... the publisher. Engineering Mechanics - Statics Chapter 6 Initial Guesses: F AB 1kN= F AG 1kN= F CF 1kN= F BC 1kN= F BG 1kN= F DE 1kN= F CG 1kN= F FG 1kN= F EF 1kN= F CD 1kN= F DF 1kN= Given Joint ... publisher. Engineering Mechanics - Statics Chapter 6 a 10 ft= b 10 ft= Solution: θ atan b a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Initial Guesses: F AB 1lb= F AG 1lb= F BG 1lb= F BC 1lb= F DC 1lb=...

Ngày tải lên: 11/08/2014, 02:22

... publisher. Engineering Mechanics - Statics Chapter 6 CB b 2− c 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = BD 2− b 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses F BA 1N= F BC 1N= F CA 1N= F DA 1N= F BD 1N= F DC 1N= B y 1N= B z 1N= C z 1N= Given F BA AB AB F CA AC AC + ... EF 0.5− b 0 h− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses F AB 1lb= F AC 1lb= F AD 1lb= F AE 1lb= F AF 1lb= F BC 1lb= F BD 1lb= F CD 1lb= F CF 1lb= F DE 1lb= F DF 1lb= F...

Ngày tải lên: 11/08/2014, 02:22

... from the publisher. Engineering Mechanics - Statics Chapter 6 F W 1 W 2 + 2 = F 10 2lb= Man: + ↑ Σ F y = 0; N C W 1 − 2 F 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= N C W 1 F−= N C 72.5 lb= Problem 6 -1 15 The piston C moves ... publisher. Engineering Mechanics - Statics Chapter 6 Given: w 1 500 N m = w 2 400 N m = w 3 600 N m = a 3m= b 3m= Solution: Guesses A x 1N= A y 1N= C x 1N= C...

Ngày tải lên: 11/08/2014, 02:22