engineering mechanics statics episode 2

... 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 ⎟⎥ ⎣⎝ ⎝ ∂ ⎠...

Ngày tải lên: 21/07/2014, 17:20

... ⌠ ⎮ 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 ×...

Ngày tải lên: 21/07/2014, 17:20

... = 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 =...

Ngày tải lên: 21/07/2014, 17:20

... 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...

Ngày tải lên: 21/07/2014, 17:20

... = 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...

Ngày tải lên: 21/07/2014, 17:20

... 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...

Ngày tải lên: 21/07/2014, 17:20

... 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...

Ngày tải lên: 21/07/2014, 17:20

... 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...

Ngày tải lên: 21/07/2014, 17:20

... 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...

Ngày tải lên: 21/07/2014, 17:20

... ⎥ 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...

Ngày tải lên: 21/07/2014, 17:20

... 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 ⎥...

Ngày tải lên: 21/07/2014, 17:20

... 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...

Ngày tải lên: 21/07/2014, 17:20

... −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...

Ngày tải lên: 21/07/2014, 17:20

... 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...

Ngày tải lên: 21/07/2014, 17:20

... ⎟ ⎝ 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...

Ngày tải lên: 21/07/2014, 17:20

... 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 +...

Ngày tải lên: 21/07/2014, 17:20

... 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⎠ ⎦...

Ngày tải lên: 21/07/2014, 17:20

... 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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... 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...

Ngày tải lên: 21/07/2014, 17:20

... 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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