Engineering Mechanics - Statics Episode 3 Part 4 potx

... publisher. Engineering Mechanics - Statics Chapter 9 Solution: V 2 3 π a 3 π 3 a 2 a−= π 3 a 3 = z c 1 V 5a 8 2 3 π a 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 4 a π 3 a 3 ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = z c a 2 = Problem 9-7 6 Determine ... Engineering Mechanics - Statics Chapter 9 Solution: A 2a 3 2 a 2 2 π = A 3 π a 2 = V 2 1 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 2 a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 6 a2 π...

Ngày tải lên: 21/07/2014, 17:20

... writing from the publisher. Engineering Mechanics - Statics Chapter 11 V'' 3 2 cos θ () 1 4 sin θ () − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= amg V'' 1.521= amg Stable Problem 1 1 -4 1 The homogeneous cylinder ... c 2 ce ea− = y c d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 1 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 1 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 1 c− 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 1 3c+ 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − e 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 2 3 ⎛...

Ngày tải lên: 21/07/2014, 17:20

... density ρ . Solution: m ρ 4 π r 3 3 = ρ 3m 4 π r 3 = I x r− r x 1 2 3m 4 π r 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ π r 2 x 2 − () r 2 x 2 − () ⌠ ⎮ ⎮ ⌡ d= 2 5 mr 2 = I x 2 5 mr 2 = Problem 1 0-9 4 Determine the radius of ... from the publisher. Engineering Mechanics - Statics Chapter 10 Solution: Moment of inertia I x and I y : I x 1 12 ba 3 = I x 90 10 3 × mm 4 = I y 1 12 ab 3 =...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 10 I x 1192in 4 = I y 1 3 cd+()ab+() 3 1 36 cb 3 1 2 bc a 2b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − π r 4 4 π r 2 a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= I y 36 4. 84 in 4 = Problem ... publisher. Engineering Mechanics - Statics Chapter 10 Solution: I y ab 3 3 1 36 ac 3 + 1 2 ac b c 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1 36 db c+() 3...

Ngày tải lên: 21/07/2014, 17:20

... the publisher. Engineering Mechanics - Statics Chapter 9 Units Used: Mg 10 3 kg= kN 10 3 N= Given: L 8m= ρ w 1.0 Mg m 3 = a 3m= b 2m= g 9.81 m s 2 = Solution: F 3 ρ w gabL= F 3 47 0.88 kN= F 2 ρ w gabL= ... integrals, we have L 0 a x1 4b 2 x 2 a 4 + ⌠ ⎮ ⎮ ⎮ ⌡ d= L 23. 6 63 m= x c 1 L 0 a xx 1 4b 2 x 2 a 4 + ⌠ ⎮ ⎮ ⎮ ⌡ d= x c 9.178 m= A 2 π x c L= A 1 .36 5 10 3 × m...

Ngày tải lên: 21/07/2014, 17:20

... from the publisher. Engineering Mechanics - Statics Chapter 9 Problem 9 -3 6 Locate the centroid of the quarter-cone. Solution: r a h hz−()= z c z= x c y c = 4r 3 π = V 0 h z π 4 a h hz−() ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⌡ d= 1 12 ha 2 π = z c 12 ha 2 π 0 h zz π 4 a h hz−() ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⌡ d= 1 4 h= x c 12 ha 2 π 0 h z 4 3 π a h hz−() ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ π 4 a h h ... publisher. E...

Ngày tải lên: 21/07/2014, 17:20

... 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 ... tan φ s θ + () = A x 32 9 N= + → Σ F x = 0; A x 2Tcos β () − 0= T 1 2 A x cos β () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 232 .36 N= C y Tsin β () = B y C y = B y C y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 64 .3 1 64 .3 ⎛...

Ngày tải lên: 21/07/2014, 17:20

... the publisher. Engineering Mechanics - Statics Chapter 8 of the wedge and the thickness of the beam. Units Used: kN 10 3 N= Given: F 1 2kN= a 3m= F 2 4kN= b 2m= F 3 4kN= c 3m= F 4 2kN= θ 15 deg= μ s 0.25= Solution: Guesses N 1 1kN= ... 2m= μ B 0 .4= b 40 0 mm= μ C 0.2= c 30 0 mm= w 800 N m = d 3= g 9.81 m s 2 = e 4= Solution: Member AB: Σ M A = 0; 1 2 wa ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −...

Ngày tải lên: 21/07/2014, 17:20

... from the publisher. Engineering Mechanics - Statics Chapter 6 Positive (T) Negative (C) F AB F AC F AD F AE F BC F BE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 30 0− 5 83. 095 33 3 .33 3 666.667− 0 500 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= F BF F CD F CF F DE F DF F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 30 0− 30 0− 0 42 4.2 64 30 0− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠...

Ngày tải lên: 21/07/2014, 17:20

... publisher. Engineering Mechanics - Statics Chapter 6 F AB F AF F BC F BF F FC F FE F ED F EC F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 33 0.0− 79 .4 233 .3 233 .3 47 .1− 112.7 112.7 30 0.0 37 7.1− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Positive ... writing from the publisher. Engineering Mechanics - Statics Chapter 6 F CB F C...

Ngày tải lên: 21/07/2014, 17:20