Assignment 3 Solution

Assignment 3 Solution - Problem3.27 p A H 2O L sin a h Hg h...

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Problem 3.27 A B θ L a h 22 2 3 2 (s i n ) or sin [( 1)* sin 62.4 / [(13.6 1)*0.5 2.5 ] 237 / 237 AH O H g H O B A B Hg HO f f pL a h h a p pp h h L SG h L lb ft ft ft lb ft psf γ θγ γγ γθ −+ + + + = −= = =− = == 2 237( / )( /144 ) 1.65 f lb ft ft in psi Problem 3.49 2 2 3 ˆ 2 2*(9810 / )*0.1 1960 1.96 ˆˆ ˆ 1960 (13.6*9810 / )*(0.1 ) 11.4 1.5 8.64 2.5 21.9 1.5 1.99 O D A Hg A H O Hg BD H g CD H g EC H g p hN m m P a k P a h p S G h Pa Pa m m kPa h k P a h k P a h k P a = = = = =+ = = = h 1.5h h 2 H O Hg Air Air A B C D E
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The Forms of dA Natural forms: dA wd ξ = θ R dA wRd = Other forms: n yc x = dA x dA y dA sin ( i ) cos ( j ) x y dA dA n dA dA dA n dA == x y x y dA cos x dA dA = sin y dA dA = x y dA sin ( i ) x dA dA n dA = =⋅ cos ( j ) y dA dA n dA = dA () x =
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Problem 3.50 a b Liquid Concrete y 22 00 23 3 ˆ Hydrostatic eqn: Force on port: 1 ˆ 3 Moment about point O: 1 ˆ 4 Point of action of force: ' c aa cc c A oc c c A O py b F p dA y wdy y dy ba a b M y pdA y wdy y dy ba a yF M γ γγ = == = = = = = ∫∫ or ' / 3 or ' 0.938 4 o yM F ya f t = From 0 at 0 we get =0 and from at we get / , or y wy wb ya ba b w a α β =+ = = O
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Problem 3.70 y Gate 2 zc x = O
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Assignment 3 Solution - Problem3.27 p A H 2O L sin a h Hg h...

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