# 7 ed j l meriam engineering mechanics statics

## Engineering Mechanics - Statics Episode 3 Part 7 pptx

... publisher Engineering Mechanics - Statics Chapter 10 a ⌠ 2 ⎮ ⎛ 3m ⎞ ⎛ b x ⎞ ⎛ b x + b⎞ dx = 93 m b2 Ix = ⎮ ⎟ ⎜ ⎟ π ⎜ + b⎟ ⎜ a 70 ⎠ ⎝ ⎠ ⎮ ⎝ 7 a b2 ⎠ ⎝ a ⌡0 Ix = 93 mb 70 Problem 1 0-1 16 Determine ... writing from the publisher Engineering Mechanics - Statics Chapter 10 m = ρa c − ρ πb c m = 4 .7 233 kg I0 = IG + m ( a sin ( 45 deg) ) I0 = 6. 23 kg⋅ m 2 Problem 1 0-1 01 Determine the moment of ... permission in writing from the publisher Engineering Mechanics - Statics Chapter 10 Solution: b ⌠ a Ixy = ⎮ ⎮ ⌡0 Ixy = 0.6 67 in y b ya y b dy 10 73 © 20 07 R C Hibbeler Published by Pearson Education,...
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## Engineering Mechanics - Statics Episode 2 Part 7 pdf

... 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 7- 5 3 Draw ... 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 R ... ⎠ 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...
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## Engineering Mechanics - Statics Episode 1 Part 7 pptx

... N r1 = 17 5 mm r2 = 17 5 mm Solution: ⎛ −F1 r1 ⎞ ⎛ −F2 2r2 cos ( θ ) ⎞ ⎜ ⎟ ⎜ ⎟ M = ⎜ ⎟ + ⎜ −F2 r2 sin ( θ ) ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 16 .63 ⎞ M = ⎜ 7. 58 ⎟ N⋅ m ⎜ ⎟ ⎝ ⎠ M = 18 .3 N⋅ m 280 © 20 07 R C Hibbeler ... writing from the publisher Engineering Mechanics - Statics Chapter Units Used: kN = 10 N Given: x = 1m ⎛6⎞ ⎜ ⎟ F = −2 kN ⎜ ⎟ 1 ⎛ ⎞ ⎜ ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ 14 ⎠ Solution: y = 1m Initial Guesses: Given ... Engineering Mechanics - Statics Chapter Given: M = 15 0 kg a = 1. 2 m b = 1. 5 m θ = 60 deg g = 9. 81 m s Solution: ⎡ −( b) cos ( θ ) ⎤ ⎛ ⎞ ⎢ ⎥×⎜ ⎟ MO = a ⎢ ⎥ ⎜ ⎟ ⎣ ( b)sin ( θ ) ⎦ ⎝ −M g ⎠ ⎛ 1. 77 ...
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## Engineering Mechanics Statics - Examples Part 7 docx

... permission in writing from the publisher Engineering Mechanics - Statics ⎛ FAB ⎞ ⎜ ⎟ ⎛ − 377 .1 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 189 .7 ⎟ ⎜ BC ⎟ ⎜ −266 .7 ⎟ ⎜ FBF ⎟ ⎜ 266 .7 ⎟ ⎟ ⎜ ⎟ ⎜ FFC ⎟ = ⎜ 188.6 ⎟ lb ⎜ ⎜ F ... publisher Engineering Mechanics - Statics Chapter Given Joint B (FBC − FAB) +P=0 ( −F BD − FAB + FBC Joint D (FCD − FAD) ) =0 =0 17 ( ) F BD − P − F AD + F CD Joint C −F BC − F CD =0 17 =0 17 ⎛ FAB ... publisher Engineering Mechanics - Statics Chapter ⎛ FAB ⎞ ⎜ ⎟ ⎛ −330.0 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 79 .4 ⎟ ⎜ BC ⎟ ⎜ −233.3 ⎟ ⎜ FBF ⎟ ⎜ 233.3 ⎟ ⎟ ⎜ ⎟ ⎜ FFC ⎟ = ⎜ − 47. 1 ⎟ lb ⎜ ⎜ F ⎟ ⎜ 112 .7 ⎟ ⎟ ⎜ FE...
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## Engineering Mechanics - Statics Episode 3 Part 8 pdf

... publisher Engineering Mechanics - Statics At x = x2 Chapter 11 V'' = 6a x2 + 2b V'' = −12.2 lb ft V'' < Unstable Problem 1 1-2 8 If the potential energy for a conservative one-degree-of-freedom ... Engineering Mechanics - Statics Chapter 11 M = 15 kg b = 150 mm c = 30 0 mm g = 9 .81 m s Solution: δU = 0; P a δθ + P a+ P = Mg ( −c δθ + bδθ ) = Mg ( b − c) = Mg ⎛ c − b⎞ ⎜ ⎟ ⎝ a ⎠ P = 13. 8 ... publisher Engineering Mechanics - Statics Chapter 11 θ = deg At V'' = −3W a cos ( θ ) + a ( k1 + 4k2 ) cos ( 2θ ) Since V'' = 1.62 × 10 lb ft > 0, then the vertical position is stable Problem 1 1 -3 8 If...
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## Engineering Mechanics - Statics Episode 3 Part 6 pps

... publisher Engineering Mechanics - Statics Ix = 1192 in Iy = Chapter 10 2b ⎞ ⎡1 ⎛ ( c + d) ( a + b) − ⎢ c b + b c ⎜ a + ⎟ 36 3 ⎣ Iy = 36 4.84 in 2⎤ ⎛ π r4 2⎞ ⎥−⎜ + πr a ⎟ ⎦ ⎝ ⎠ Problem 1 0 -3 5 Determine ... from the publisher Engineering Mechanics - Statics Chapter 10 Solution: Ix' = 36 bh ⎛ b − a ⎞ h ( b − a) a h a + ⎜a + ⎟ 3 ⎠2 ⎝ xc = = 1 h a + h ( b − a) 2 Iy' = Iy' = 1 36 36 + (2 ha⎛ ⎜ b+a ⎝ ... the publisher Engineering Mechanics - Statics Chapter 10 Solution: 2 ⎛b + c ⎞ + d ( b + c) + d( b + c) ⎡ 2( b + c) ⎤ Iy = + ac + ac⎜ ⎟ ⎢ ⎥ 36 3 36 ⎣ ⎦ ab Iy = 1971 in Problem 1 0-4 5 Locate the...
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## Engineering Mechanics - Statics Episode 3 Part 5 pps

... publisher Engineering Mechanics - Statics Chapter lb γ = 50 ft p = 0.2 Solution: b ⌠ A = ⎮ a ⎮ ⌡ y dy b A = 133 .3 ft ⌠ ⎮ xc = A⎮ ⌡ b 1⎛ ⎜a 2⎝ y⎞ ⎟ dy b⎠ xc = 3. 75 ft 3 V = 2π A xc V = 3. 142 × 10 ... publisher Engineering Mechanics - Statics Chapter Given: a = 1 .5 ft lb γ = 55 ft Solution: a ⌠ 2 F R = ⎮ γ a − y ( a − y) d y ⌡− a F R = 5 83 lb a ⌠ ⎮ yγ a2 − y2 ( a − y) d y d = a− FR ⌡− a d = 1.8 75 ... from the publisher Engineering Mechanics - Statics Chapter 10 Alternatively h ⌠ 2⎞ ⎮ 2⎛ ⎜ b − b y ⎟ d y = b h3 Ix = ⎮ y 2⎟ ⎜ 15 ⎮ h ⎠ ⎝ ⌡0 Ix = bh 15 Ix = ab 3( + 3n) Problem 1 0-7 Determine the...
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## Engineering Mechanics - Statics Episode 3 Part 4 potx

... + M3 d + M4 g xc = xc = 1.29 ft M1 + M2 + M3 + M4 M1 0ft + M2 b + M3 e + M4 h M1 + M2 + M3 + M4 yc = yc = 1.57 ft M1 0ft + M2 c + M3 f + M4 i M1 + M2 + M3 + M4 zc = 0 .42 9 ft zc = Problem 9 -4 6 ... the publisher Engineering Mechanics - Statics Chapter Solution: a ( ) ⌠ 2 V = ⎮ π a − z dz = a π 24 ⎮a ⌡ a ( ) 24 ⌠ ⎮ zπ a2 − z2 dz = 27 a zc = 40 5π a ⎮a ⌡ zc = 27 a 40 Problem 9 -4 3 Determine ... Engineering Mechanics - Statics Chapter Solution: a b⎛ ⎜ b⎞ ⎟ ⎝ 2⎠ xc = ab + zc = a c⎛ ⎜ xc = 1.1 43 in ac c⎞ ⎟ ⎝ 3 ab + zc = 0.857 in ac ⎛ xc ⎞ ⎟ ⎝ zc ⎠ θ = atan ⎜ θ = 53. 13 deg Problem 9-7 3...
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## Engineering Mechanics - Statics Episode 3 Part 3 docx

... from the publisher Engineering Mechanics - Statics Chapter Given: M = 1.2 Mg D = 30 mm ag = 0.4 mm as = 0.2 mm Solution: P = M g( ag + as) D⎞ 2⎛ ⎟ ⎜ ⎝2⎠ P = 235 N Problem 8-1 33 A machine of mass ... the publisher Engineering Mechanics - Statics Chapter Given: a = ft b = ft Solution: b ⌠ a z ⎮ V = ⎮ π dz b ⌡ V = 12.566 ft b ⌠ 1⎮ a z zc = zπ dz V⎮ b ⌡ zc = 1 .33 3 ft Problem 9 -3 5 Locate the ... writing from the publisher Engineering Mechanics - Statics Chapter Problem 9 -3 6 Locate the centroid of the quarter-cone Solution: r= a ( h − z) h ⌠ ⎮ V=⎮ ⌡ zc = z xc = yc = 4r 3 h π ⎡a ⎢ ⎣h ⎤ ⎦ (...
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## Engineering Mechanics - Statics Episode 3 Part 2 pdf

... 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⎞ ⎜ ⎟ ⎝ a⎠ ... 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 ... Engineering Mechanics - Statics Chapter a = 700 mm b = 25 mm c = 30 0 mm d = 20 0 mm e1 = 60 mm μ s = 0 .3 e = 2. 718 Solution: Initial guesses: T1 = N T2 = N M = kg Given Drum: ⎛ 3 ⎞ μ s⎜ ⎟ T2...
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## Engineering Mechanics - Statics Episode 3 Part 1 doc

... publisher Engineering Mechanics - Statics Chapter Given: M1 = 75 kg M2 = 10 0 kg μ A = 0 .3 μ B = 0.25 μ C = 0.4 θ = 30 deg r = 15 0 mm g = 9. 81 m s Solution: Initial guesses: NA = 10 0 N F A = 10 N NB ... publisher Engineering Mechanics - Statics Chapter Given: W1 = 30 0 lb W2 = 200 lb μ s = 0.2 μ's = 0 .35 a = 4.5 ft c = ft b = 3. 5 ft d = 4.5 ft Solution: ΣF y = 0; NC − W1 = NC = W1 ΣF x = 0; ... publisher Engineering Mechanics - Statics Chapter Given: mp = 50 kg P = 15 0 N N w = 800 m a = 2m b = 400 mm c = 30 0 mm d = e = Solution: Member AB: 2a − wa + NB a = NB = NB = 533 .33 N wa Post:...
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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 -2 0 The ... publisher Engineering Mechanics - Statics Chapter P y = W L − Nc P y = 110. 00 lb P = Px + Py P = 110 lb The length on the ground is supported by L = Nc = 50.00 lbthus Nc W L = 6 .25 ft Problem 8 -2 8 The...
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## Engineering Mechanics - Statics Episode 2 Part 9 ppsx

... 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 7-8 8 Draw the shear and ... publisher Engineering Mechanics - Statics Chapter wa d y = tan ( θ max) = 2a FH dx Tmax = (θ max) FH ⎛ w a2 ⎞ = atan ⎜ ⎟ ⎝ 2a FH ⎠ (θ max) = 57 .99 deg Tmax = 4. 42 kip cos ( θ max) Problem 7-1 02 The ... from the publisher Engineering Mechanics - Statics Chapter Force (kip) V( x ) 0 10 12 14 x ft Distance (ft) Moment (kip-ft) 40 M( x) 20 0 10 12 14 x ft Distance (ft) Problem 7-8 9 Determine the force...
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## Engineering Mechanics - Statics Episode 2 Part 8 pdf

... 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 7-7 0 Draw ... −w x wL V max = w2 = −w x w2 L w1 = 88 .9 lb ft ⎛ Mmax ⎞ ⎜ ⎟ ⎝ L ⎠ w2 = 22 .2 Now choose the critical case lb ft w = ( w1 , w2 ) w = 22 .22 lb ft Problem 7-5 8 The beam will fail when the maximum ... (kN) V1( x1) 40 V2( x2) 20 V3( x3) V4( x4) 20 40 8 x1 , x2 , x3 , x4 Distance (m) Moment (kN-m) M1( x1) 20 M2( x2) M3( x3) 40 M4( x4) 60 80 x1 , x2 , x3 , x4 Distance (m) 707 © 20 07 R C Hibbeler...
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