engineering mechanics statics episode 2
Engineering Mechanics - Statics Episode 3 Part 8 pdf
... 3a x + 2b x + c = dx x1 = − 2b + 4b − 4( 3a)c 2( 3a) x1 = 0.59 ft x2 = − 2b − 4b − 4( 3a c) ( 3a) x2 = −0. 424 ft Stability: d2 V = V'' = 6a x + 2b dx At x = x1 V'' = 6a x1 + 2b V'' = 12. 2 lb ft ... cos ( θ ) δy2 = −b sin ( θ ) δθ y3 = y2 + a δy3 = δy2 δU = −2WLδy1 − Wb δy2 − 2kδ δy3 − 2Mδθ = b⎞ δU = ⎡2WL⎛ ⎟ sin ( θ ) + Wb b sin ( θ ) + 2kδ b sin ( θ ) − 2M⎤ δθ = ⎢ ⎜ ⎥ ⎣ M = ⎝ 2 ⎡⎛ WL + ... 0) At ( , 0) ∂ 2 ∂x ∂ 2 ∂y At ( , 0) At ( , 0) N m V = 2b 2b = V = 2a 2a = 12 ∂ ∂ ∂x ∂ y N m >0 >0 V =0 ⎤ ⎡⎛ ⎞⎜ ⎢ ∂ ∂ ⎞ ⎛ 2 ⎟ ⎛ 2 ⎞⎥ ⎜ ⎟ V V = − 4a b V⎟ − ⎢⎜ ∂x ∂ y ⎠ ⎜ x2 ⎟ ⎜ y2 ⎟⎥ ⎣⎝ ⎝ ∂ ⎠...
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Engineering Mechanics - Statics Episode 3 Part 7 pptx
... ⌠ ⎮ 2 m = ⎮ ρ π b ⎜1 − ⎜ ⎮ ⎝ ⌡0 a 2 ⎟ dx = a ρ π b2 a ⎠ x ρ = 2 ⌠ ⎮ ⎛ 3m ⎞ ⎛ Ix = ⎮ π b ⎜1 − ⎜ 2 ⎜ ⎮ ⎝ 2 a b ⎠ ⎝ ⌡0 2 2 ⎟ b2 ⎛1 − x ⎟ dx = m b2 ⎜ 2 ⎜ 2 a ⎠ ⎝ a ⎠ x 3m 2 a b Ix = 2 mb ... Given: b = in t = in Solution: ⎡1 I x = 2 ⎣ 12 Ix = 55.55 in ⎡1 I y = 2 ⎣ 12 ⎛ ⎝ a t + a t ⎜b − t⎞ ⎟ 2 2 ⎥ + t ( 2b − 2t) ⎦ 12 ⎛ a − t + t ⎞ ⎤ + 2b t3 ⎟⎥ ⎠ ⎦ 12 ⎝ t ( a − t) + t( a − t) ⎜ ⎡a − t ... ( a + b) c + 2b d + 2b d ⎜ ⎟ 12 ⎝ 2 Iy = 4 Ix = 5.515 × 10 in 1 3 [ 2( a + b) ] c + ( 2b) d 12 12 Iy = 1.419 × 10 in Ixy = in Iu = Ix + Iy Ix − Iy + cos ( 2 ) − Ixy sin ( 2 ) 2 Iu = 3.47 ×...
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Engineering Mechanics - Statics Episode 3 Part 6 pps
... = 25 mm b = 25 0 mm c = 50 mm d = 150 mm Solutuion: ⎛ b ⎞ b2a + ⎛ b + ⎜ ⎟ ⎜ ⎝ 2 ⎝ yc = c⎞ ⎟ 2d c 2 b2a + c2d yc = 20 7 mm b⎞ 3 ⎞ ⎛ ⎛ c Ix' = 2a b + 2a b ⎜ yc − ⎟ + 2d c + c2d ⎜ b + − yc⎟ 12 2⎠ ... the publisher Engineering Mechanics - Statics Chapter 10 Solution: 2a b yc = Ix = b ⎛ c⎞ + 2c b⎜ b + ⎟ ⎝ 2 yc = 12. 50 mm 2a b + 2c b 12 2a b + 2a b ⎛ yc − ⎜ ⎝ b⎞ ⎟ 2 ⎡1 + 2 ⎣ 12 b c + b c ⎛b ... publisher Engineering Mechanics - Statics Chapter 10 Given: a = 150 mm b = 20 0 mm t = 20 mm θ = 20 deg Solution: Moments of inertia Ix and Iy: Ix = 1 3 2a ( 2b) − ( 2a − t) ( 2b − 2t) 12 12 Ix =...
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Engineering Mechanics - Statics Episode 3 Part 5 pps
... from the publisher Engineering Mechanics - Statics w = 2m g = 9.81 Chapter m s Solution: w1 = ρ w g( b − 2a)w w1 = 59 kN m w2 = ρ w g2a w w2 = 59 kN m F1 = 2a w1 F = w2 2a F = 88 kN F = 177 ... publisher Engineering Mechanics - Statics Chapter Given: r = 600 mm a = 25 mm b = 50 mm c = 75 mm Solution: ⎡⎛ ⎣⎝ V = 2 ⎢⎜ r + c ⎞ ⎛ 1⎞ ⎛ c⎞ ⎤ ⎟ 2 ⎟ a c + ⎜ r + ⎟ b c⎥ 3⎠ ⎝ 2 ⎝ 2 ⎦ V = 22 .4 × ... − xc = ⎡ −a ⎢ A⎣ 2 a πa − 2 ⎤ ⎛ 2 a⎞ − π a ⎛a − 4a ⎞⎥ ⎜ ⎟ ⎜ ⎟ ⎝ 3π ⎠⎦ ⎝3 ⎠ ⎡ −a 2a π a yc = ⎢ − A⎣ A = 97.7 in 2 xc = −0 .26 2 in ⎛ 4a − a⎞⎤ ⎜ ⎟⎥ ⎝ 3π ⎠⎦ yc = 0 .26 2 in 987 © 20 07 R C Hibbeler...
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Engineering Mechanics - Statics Episode 3 Part 4 potx
... = 50 mm d2 = 35 mm h = 110 mm t = 15 mm Solution: 2 ⎛ d1 ⎞ d1 ⎛ d + h ⎞ + π ⎛ d2 ⎞ ⎛ d + h + d2 ⎞ π⎜ ⎟ + h t⎜ ⎜ ⎟ ⎜ ⎟ ⎟ 2 2 2 ⎝ 2 ⎝ yc = 2 ⎛ d1 ⎞ ⎛ d2 ⎞ π ⎜ ⎟ + ht + π ⎜ ⎟ 2 2 yc = 85.9 ... publisher Engineering Mechanics - Statics Chapter d = in e = in f = in Solution: zc = ⎛ a⎞ ⎟ + 2( d − e)a⎜ ⎟ ⎝ 2 ⎝ 3⎠ ( 2e f a) + 2e a⎛ ⎜ 2d b + a⎞ ( 2c b) + zc = 1. 625 in 2d( a + f) Problem 9- 82 Each ... publisher Engineering Mechanics - Statics Chapter Given: a = 300 mm b = 400 mm Solution: L = πa 2 +2 a +b xc = 1⎡ π a ⎛ 2a ⎞⎤ 2a ⎢ a + b + ⎜ ⎟⎥ L⎣ 2 ⎝ π ⎠⎦ yc = 1⎡ π a ⎛ 2a ⎞⎤ 2a ⎢ a + b + ⎜ ⎟⎥ L⎣ 2...
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Engineering Mechanics - Statics Episode 3 Part 3 docx
... z dy Solution: b ⌠ ⎮ 2 V = ⎮ π a ⎜1 − ⎜ ⎮ ⎝ ⌡0 z =a 2 ⎜ ⎜ ⎝ 2 y 1− b ⎟ 2 ⎠ 2 2 ⎟ d y = b b − b a2 π 2 b ⎠ b y b ⌠ ⎮ 2 ⎜ yc = ⎮ yπ a − ⎜ 2b a π ⎮ ⎝ ⌡0 2 ⎟ d y = b2 2 8b b ⎠ y By symmetry ... the publisher Engineering Mechanics - Statics Chapter Solution: a ⌠ A=⎮ ⌡0 ( 2) 2 a a y dy = a ⌠ ⎮ xc = 2 2a ⌡ 2a A= a xc = a yc = a y dy = a a 5 a yc = ( ) ⌠ ⎮ y a y dy = a 2 2a 5a Problem ... writing from the publisher Engineering Mechanics - Statics Chapter Solution: 2 x +y =a y= 2 a −x a A = π xc = A W = Aγ W = 62. 8 32 lb a ⌠ ⎮ x a2 − x2 dx ⌡0 ⌠ ⎮ yc = A⎮ ⌡ a ( ) 2 a − x dx xc = 1.698...
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Engineering Mechanics - Statics Episode 3 Part 2 pdf
... Engineering Mechanics - Statics ⌠ ⎮ P= ⎮ ⌡ 2 R2 ⌠ ⎮ ⎮ ⌡R Chapter ⎛ R2 ⎞ ⎟ r dr dθ ⎝ r ⎠ p0 ⎜ P = 2 p0 R2 ( R − R ) p0 = P 2 R ( R − R ) 2 ⌠ ⌠ M = ⎮ r dF = ⎮ ⌡ ⌡ A ⌠ ⎮ M= ⎮ ⌡ 2 ⌠ ⎮ ⎮ ⌡ 2 ... writing from the publisher Engineering Mechanics - Statics Chapter ⎛d − c ⎟ ⎞ M2 = μ s P' ⎜ ⎜ d2 − c2 ⎟ ⎝ ⎠ P' = P' b − P a = P = P' 3M2 ⎛ d2 − c2 ⎞ ⎜ ⎟ P' = 88.5 N 2 s ⎜ d3 − c3 ⎟ ⎝ ⎠ ⎛ b⎞ ⎜ ... M= ⎮ ⌡ 2 ⌠ ⎮ ⎮ ⌡ 2 R2 ⌠ ⎮ ⌡R rμ s p0 r dr dθ ( ⎛ R2 ⎞ 2 ⎟ r dr dθ = π μ s p0 R R − R r ⎠ ⎝ μ s p0 ⎜ ) ( P 2 ⎤ ⎡ M = π μ s⎢ ⎥ R2 R2 − R1 2 R2( R2 − R1 )⎦ ) M= μ s P(R2 + R1) Problem 8-118...
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Engineering Mechanics - Statics Episode 3 Part 1 doc
... θ ) ⎟ ⎜ ⎟ ⎜ ⎟ 2 2 2 sin ⎜ ⎛ θ ⎞ − sin ⎛ θ ⎞ cos ( θ ) ⎟ ⎜ ⎟ 2 2 ⎛θ⎞ sin ⎜ ⎟ sin ( θ ) 2 sin ( θ ) cos ⎜ μs = 2 ⎛ θ ⎞ 2 cos ⎛ θ ⎞ − sin ⎛ θ ⎞ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ 2 − ⎝ 2 ⎝ ⎠ ⎠ = cot ... kN N2 = kN P = kN Given (F1 − N1)( a + b + c) + F2( b + c) + F3 c = N2 cos ( θ ) − μ s N2 sin ( θ ) − N1 = μ s N1 + μ s N2 cos ( θ ) + N2 sin ( θ ) − P = ⎛ N1 ⎞ ⎜ ⎟ ⎜ N2 ⎟ = Find ( N1 , N2 , ... = cot ⎛ θ ⎞ − cos ⎛ θ ⎞ + tan ⎛ θ ⎞ μs = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 2 2 ⎛θ⎞ ⎛θ⎞ ⎛θ⎞ sin ⎜ ⎟ sin ⎜ ⎟ cos ⎜ ⎟ 2 2 2 cos ⎜ ⎛θ⎞ μ s = tan ⎜ ⎟ 2 θ = atan ( μ ) Problem 8-57 The carpenter slowly pushes...
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Engineering Mechanics - Statics Episode 2 Part 10 ppt
... kN⋅ m ⎝ M2 ( x2 ) = ⎡−M + C( a + b − x2 )⎤ ⎣ ⎦ kN kN⋅ m Force (kN) V1( x1) V2( x2) 10 10 10 x1 , x2 Distance (m) Moment (kN-m) 20 M1( x1) M2( x2) 20 40 60 x1 , x2 Distance (m) 7 62 © 20 07 R C ... publisher Engineering Mechanics - Statics Chapter ⎛ T ⎞ ⎛ 173 .21 ⎞ ⎜ ⎟ NA ⎟ ⎜ 21 5.31 ⎟ ⎜ ⎜ ⎟ ⎜ FA ⎟ = ⎜ 107.66 ⎟ N ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 21 5.31 ⎟ ⎜ Nground ⎟ ⎝ 538 .28 ⎠ ⎝ ⎠ mass = 54.9 kg Problem *8 -20 The ... = 28 0 .2 N ΣMO = 0; P = 140 N ⎛ e P ⎞⎛ b ⎞ = ⎟ − N1 x + ⎜ ⎜ ⎟ 2 ⎝ 2 d + e ⎠⎝ ⎠ ⎝ −μ s NC⎛ ⎜ x = a⎞ −1 μ s NC a Thus, the distance from A is NC 2 d +e −ePb x = 123 .51 mm d +e A = x+ b A = 523 .51...
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Engineering Mechanics - Statics Episode 2 Part 9 ppsx
... ⎥ TBC + W2 = TAB − ⎢ ⎜ a2 + xB2 ⎟ ⎢ b2 + ( xB − d) 2 ⎝ ⎠ ⎣ ⎦ a b ⎛ ⎡ ⎞ ⎤T = BC ⎜ ⎟ TAB − ⎢ 2 2⎥ a + xB ⎠ b + ( xB − d) ⎦ ⎝ ⎣ xB − d ⎡ ⎤ ⎢ ⎥ TBC − ⎛ d ⎞ TCD = ⎜ 2 ⎢ b2 + ( xB − d) 2 ⎝ c +d ... permission in writing from the publisher Engineering Mechanics - Statics Chapter 20 Moment (kip-ft) M1p( x1) M2p( x2) M3p( x3) 20 40 60 10 15 20 x1 x2 x3 , , ft ft ft Distance (ft) Problem 7-88 ... Engineering Mechanics - Statics TDE = lb Chapter yD = ft Given c ⎛ −b ⎞ T + ⎡ ⎤T = BC ⎜ ⎟ AB ⎢ 2 2⎥ b + yB ⎠ c + ( f − yB) ⎦ ⎝ ⎣ yB f − yB ⎛ ⎞ ⎤ ⎜ ⎟ TAB − ⎡ ⎢ ⎥ TBC − P2 = ⎜ b2 + yB2 ⎟ ⎢ c2...
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Engineering Mechanics - Statics Episode 2 Part 8 pdf
... kN⋅ m M2 ( x2 ) = B ( a + b − x2 ) kN kN⋅ m Force (kN) V1( x1) V2( x2) 0 .2 0.4 0.6 0.8 x1 , x2 Distance (m) Moment (kN-m) M1( x1) 0.5 M2( x2) 0.5 0 .2 0.4 0.6 0.8 x1 , x2 Distance (m) 708 © 20 07 ... from the publisher Engineering Mechanics - Statics Chapter Force (kN) V1( x1) V2( x2) 0.5 1.5 2. 5 2. 5 3.5 x1 , x2 Distance (m) Moment (kN-m) M1( x1) M2( x2) 0.5 1.5 3.5 x1 , x2 Distance (m) Problem ... publisher Engineering Mechanics - Statics x1 = , 0.01a a Chapter V ( x) = −F x2 = a , 1.01a a + b M2 ( x2 ) M1 ( x) = −F1 x lb V ( x2 ) = ⎡−B + F2 + w( a + b − x2 )⎤ ⎣ ⎦ lb⋅ in lb ⎡ (a + b − x2 )2 ⎥...
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Engineering Mechanics - Statics Episode 2 Part 7 pdf
... from the publisher Engineering Mechanics - Statics Chapter Force in kN 20 0 V1( x1) V2( x2) 20 0 10 x1 , x2 Distance in m Moment (kN-m) 400 M1( x1) 20 0 M2( x2) 20 0 400 10 x1 , x2 Distance (m) Problem ... ⎠ 2 ⎡ ⎢Ay x − w ( x − a) ⎥ M2 ( x) = ⎣ ⎦ kN⋅ m kN Force (kN) V1( x1) V2( x2) 0.5 1.5 2. 5 2. 5 3.5 x1 , x2 Distance (m) Moment (kN-m) M1( x1) M2( x2) 0.5 1.5 3.5 x1 , x2 Distance (m) 676 © 20 07 ... x) M2 = 150 lb⋅ ft M1 − M2 + w Ay = ⎛ L2 ⎞ ⎜ ⎟ 2 L Ay = 25 00 lb ⎡ ⎛ 2 ⎤ ⎢Ay x − w⎜ x ⎟ − M1⎥ M ( x) = ⎣ 2 ⎦ lb⋅ ft lb Force in lb 20 00 V( x ) 20 00 10 15 20 x ft Distance in ft 677 © 20 07...
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Engineering Mechanics - Statics Episode 2 Part 6 pot
... −P c + 2 b +c Fmax c = F max b P = 2 P = 28 2.843 N b +c c B z + Dz − Fmax b +c F max c Bz = 2 =0 Dz = B z b +c Dz = Bz B z = 28 3 N Dz = 28 3 N b B y + Dy − Fmax b +c F max b By = 2 =0 2 b +c Dy ... the publisher Engineering Mechanics - Statics Chapter MD = Mmax MD w1 = w1 = 100 w2 = 22 5 ab N m N m Assume that the maximum normal force in BC has been reached T = Pmax T2 d w2 = ( a + b) c ... from the publisher Engineering Mechanics - Statics Chapter b − Ay( b + c) + F1 ( a + b + c) + w b⎛ c + ⎞ + F2 c = ⎜ 2 ⎝ ⎠ F1 ( a + b + c) + w b⎛ c + ⎜ ⎝ Ay = b⎞ ⎟ + F2 c 2 b+c Ay = 4514 lb...
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Engineering Mechanics - Statics Episode 2 Part 5 ppsx
... Solution: L ⎛ θ ⎞ − 2L sin ⎛ θ ⎞ N = ⎟ ⎜ ⎟ A 2 2 cos ⎜ ΣMD = 0; 2W ΣΜC = 0; T L cos ⎜ ⎛ θ ⎞ − N L sin ⎛ θ ⎞ − W L cos ⎛ θ ⎞ = ⎟ ⎜ ⎟ ⎜ ⎟ A 2 2 2 W ⎛θ⎞ NA = cot ⎜ ⎟ 2 T=W Problem 6-113 ... from the publisher Engineering Mechanics - Statics F CG = F Chapter a F CG = 100 N b Sector gear : ⎞d = 2 ⎝ c +d ⎠ ΣMH = 0; F CG( d + e) − FAB⎛ ⎜ c ⎛ c2 + d2 ⎞ ⎟ F = 29 7. 62 N F AB = FCG ( d ... permission in writing from the publisher Engineering Mechanics - Statics F = Chapter W1 + W2 F = 1 02 lb Man: + ↑ Σ F y = 0; NC − W1 + ⎛F⎞ = ⎜ ⎟ ⎝ 2 NC = W1 − F NC = 72. 5 lb Problem 6-115 The piston C...
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Engineering Mechanics - Statics Episode 2 Part 4 pdf
... ⎟ ⎝ 2 uAD = ⎛ −a ⎟ ⎜ ⎞ ⎜ ⎟ 2 ⎟ ⎛ a⎞ ⎜ ⎟ h +⎜ ⎟ ⎝ 2 ⎝ h ⎠ uBD = ⎛a⎟ ⎜ ⎞ 2 2 ⎟ a⎞ h +⎛ ⎟ ⎜h⎟ ⎜ ⎝ 2 ⎝ ⎠ uAC = ⎛−a ⎟ ⎜ ⎞ ⎜ 2 2 b ⎟ a⎞ 2 h +b +⎛ ⎟ ⎜ h ⎟ ⎜ ⎝ 2 ⎝ ⎠ uBC = ⎛a⎟ ⎜ ⎞ 2 2 b ... publisher Engineering Mechanics - Statics b −F CD − −b + a +b 2 2 b +c −c F AD − c −F − F DE − + FBD a +b +c a +b +c Chapter =0 b 2 b +c F DF F DF = F AD Joint B a −F BC − 2 a +b FBD = b 2 a +b ... publisher Engineering Mechanics - Statics Chapter Given Ay − w1 a − MA − w1 a a − d b +d F CD = d d +b Ax − b b +d b F CD2 a = 2 b +d d FCD = FCD − Bx = c c B y c − w2 =0 c FCD − w2 + B y = 2 b +d...
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Engineering Mechanics - Statics Episode 2 Part 3 ppsx
... the publisher Engineering Mechanics - Statics Chapter Given: F = 30 kN F = 20 kN F = 20 kN F = 40 kN a = 4m b = 4m Solution: −F a − F 3( 2a) − F 4( 3a) + Gy( 4a) = Gy = F2 + 2F + 3F4 Gy = 45 ... publisher = Chapter Engineering Mechanics - Statics Solution: Guess E y = lb F GJ = lb Given −F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = −F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = ⎞ ⎛ Ey ⎟ ⎜ = Find ... Ax = (F1 − Ay)( 4a) + F2( 3a) + F3( 2a) + F4( a) = (F1 − Ay)( a) + Ax( c) − FBC ( c) = (F1 − Ay)( 2a) + F2( a) + ( A y − F 1) ( d) − F 2( a + d ) + a a +b F HG( c) + c 2 a +c b a +b FHC ( a +...
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Engineering Mechanics - Statics Episode 2 Part 2 potx
... publisher Engineering Mechanics - Statics Chapter ⎛ FAB ⎞ ⎜ ⎟ ⎛ 22 .361 ⎞ ⎜ FBF ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FBC ⎟ = ⎜ 20 ⎟ kN ⎜F ⎟ ⎜ ⎟ ⎜ ED ⎟ ⎜ ⎟ ⎝ 20 ⎠ ⎜ FAF ⎟ ⎝ ⎠ ⎛ FFC ⎞ ⎜ ⎟ ⎛ 28 .28 4 ⎞ ⎜ FEC ⎟ ⎜ 20 ⎟ ... ⎞ FBC ⎟ ⎜ 26 .25 ⎟ ⎜ ⎜ ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCG ⎟ ⎜ 6 .25 ⎟ ⎜ F ⎟ ⎝ −18.75 ⎠ ⎝ CD ⎠ ⎛ FAG ⎞ ⎜ ⎟ ⎛ 26 .25 ⎞ FBG ⎟ ⎜ 35 ⎟ ⎜ ⎜ ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FFG ⎟ ⎜ 22 .5 ⎟ ⎜ F ⎟ ⎝ 25 ⎠ ⎝ DF ⎠ ⎛ FCF ⎞ ⎛ −6 .25 ⎞ ⎜ ⎟ ⎜ ⎟ ... Joint D 2 ⎛ a⎞ + ⎛ b⎞ ⎥ = ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎝ 4⎠ ⎦ ⎡a − ρ g⎢ + 2 2 ( a) + b b ( F BD + FCD + FDA − FDE (FCD − FDA − FDE) ⎡b − ρ g⎢ + 2 ⎣4 ( a) + b ) b 2a 2 2⎤ ⎛ a⎞ + ⎛ b⎞ ⎥ = ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎝ 4⎠ ⎦...
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Engineering Mechanics - Statics Episode 2 Part 1 ppsx
... Az = 100 lb Given c ⎞ ⎞ ⎛ ⎟ − TBD a⎜ 2 ⎟ = 2 ⎝ a +b +c ⎠ ⎝ a +c ⎠ (F1 cos (θ ) + F2)c − TBC a⎛ ⎜ c ⎞=0 ⎝ a +b +c ⎠ F sin ( θ ) c − b TBC ⎛ ⎜ c 2 2⎟ 425 © 20 07 R C Hibbeler Published by Pearson ... permission in writing from the publisher Engineering Mechanics - Statics ⎛ Chapter ⎞ ⎟=0 2 a +b +c ⎠ ⎝ Ax + F sin ( θ ) − b⎜ TBC TBC ⎞ ⎞ + a⎛ ⎜ 2 2⎟ = ⎟ 2 ⎝ a +c ⎠ ⎝ a +b +c ⎠ ⎛ Ay − F cos ( θ ... publisher Engineering Mechanics - Statics NB a − P ⎛ b ⎜ Chapter c⎞ b ⎟ =0 ⎝ 2 2 NB = P b c NB = 400 N 4a ΣF x = 0; Ax = NB Ax = 400 N ΣF y = 0; Ay = P b FA = c Ay = 600 N 2 Ax + Ay F A = 721 N Problem...
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Engineering Mechanics - Statics Episode 1 Part 10 potx
... the publisher Engineering Mechanics - Statics Chapter a = ft b = ft x1 = 1.5 ft x2 = 7.5 ft Solution: The maximum occurs when x = x2 ΣMA = 0; −F x2 + B x a = Bx = F x2 a B x = 1.4 62 × 10 lb + → ... the publisher Engineering Mechanics - Statics Chapter d = e = f = 12 g = Solution: Initial Guesses: NA = 20 lb NB = 10 lb MA = 30 lb ft Given f g ⎞ ⎞ ⎛ ⎛ ⎜ 2 ⎟ ( a + b) − NB ⎜ 2 ⎟ c = ⎝ f +g ... publisher Engineering Mechanics - Statics Chapter Given: W1 = 800 lb W2 = 450 lb W3 = 400 lb a = 2ft b = ft c = ft Solution: + → Σ Fx = 0; + ↑Σ Fy = 0; Dx = Dy − ( W1 + W2 + W3 ) = Dy = W1 + W2 + W3...
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Engineering Mechanics - Statics Episode 1 Part 9 pdf
... 800 lb w2 = 500 lb ft ft a = 12 ft b = ft Solution: FR = a w1 + (w1 − w2)b + w2 b F R = 10.65 kip a 1 b b F R x = − a w1 + ( w1 − w2 ) b + w2 b 2 x = a b b − a w1 + ( w1 − w2 ) b + w2 b 3 2 FR x ... the publisher Engineering Mechanics - Statics ⎛α ⎞ ⎜ ⎟ ⎜ β ⎟ = acos ( −u) ⎜γ ⎟ ⎝ ⎠ Chapter ⎛ α ⎞ ⎛ 109 .22 6 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β ⎟ = ⎜ 140 .20 6 ⎟ deg ⎜ γ ⎟ ⎝ 123 .28 6 ⎠ ⎝ ⎠ Problem 4-1 62 Determine the ... 800 N w2 = 20 0 N m m a = 2m b = 3m Solution: F R = w2 b + w1 a + x F R = w1 a w1 a x = a + (w1 − w2)b F R = 3.10 kN b b (w1 − w2)b⎛a + ⎞ + w2 b⎛a + ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ a ⎛ b⎞ ⎛ b⎞ + ( w1 − w2 ) b⎜...
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Tìm thêm: xác định các mục tiêu của chương trình xác định các nguyên tắc biên soạn khảo sát các chuẩn giảng dạy tiếng nhật từ góc độ lí thuyết và thực tiễn khảo sát chương trình đào tạo của các đơn vị đào tạo tại nhật bản khảo sát chương trình đào tạo gắn với các giáo trình cụ thể tiến hành xây dựng chương trình đào tạo dành cho đối tượng không chuyên ngữ tại việt nam điều tra đối với đối tượng giảng viên và đối tượng quản lí điều tra với đối tượng sinh viên học tiếng nhật không chuyên ngữ1 khảo sát các chương trình đào tạo theo những bộ giáo trình tiêu biểu nội dung cụ thể cho từng kĩ năng ở từng cấp độ xác định mức độ đáp ứng về văn hoá và chuyên môn trong ct phát huy những thành tựu công nghệ mới nhất được áp dụng vào công tác dạy và học ngoại ngữ mở máy động cơ lồng sóc mở máy động cơ rôto dây quấn đặc tuyến mômen quay m fi p2 đặc tuyến tốc độ rôto n fi p2 đặc tuyến dòng điện stato i1 fi p2 thông tin liên lạc và các dịch vụ từ bảng 3 1 ta thấy ngoài hai thành phần chủ yếu và chiếm tỷ lệ cao nhất là tinh bột và cacbonhydrat trong hạt gạo tẻ còn chứa đường cellulose hemicellulose chỉ tiêu chất lượng 9 tr 25