2-4.Determine the height h of the mercury in the tubeif the level of water in the tube is h'= 0.3m and thedepths of the oil and water in the tank are 0.6and 0.5mrespectively.TakePo=900kg/m,pw=1000kg/m,andPHg=13550kg/m3131SOLUTIONReferringtoFig.a,hAB=0.6m,hBc=0.8m-0.3m=0.5mandhcD=h.Then themanometerrule givesPA+PoghAB+PughBC-PHghcD=PDHere,Pa=Pp=O,sincepointsAandDareexposedtotheatmosphere.0+(900kg/m)(g)(0.6m）+(1000kg/m2）g)(0.5m)-(13550kg/m)(g）(h）=0h=0.07675m=76.8mmAns.hAB=0.6mBDhBc=0.5mhcD=h.0=0.3m(a)17
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2-5.TheuniformrectangularreliefgateABhasa weightof 800Ibandawidthof2ft.Determinethecomponentsofreaction atthepinB and thenormal reaction atthe smoothsupport A.BWBBr8001b(6f)=3ft(Fp)1 = 6739.2 1b60°(Fp)2=1945.441b(6f)=4ft6ft60°INA(a)SOLUTIONHere,hg=9ft and hA=9ft+6ftsin 60°=14.20ft.Thus, the intensitiesof thedistributed loadatBandAareWg=whgb=(62.4 Ib/ft)(9ft)(2ft)=1123.21b/ftwA=whab=(62.41b/ft)(14.20ft)(2ft)=1771.681b/ftThus,(Fp);=(1123.2 Ib/ft)(6ft)=6739.2 b(Fp)2=(1771.68 lb/ft-1123.2 lb/ft)(6ft)=1945.44 lbWrite the momentequation of equilibrium aboutBbyreferring to theFBD of thegate, Fig.a.+Mg=0;（800lb)cos60(3ft）+（6739.2lb)(3ft)+（1945.44lb)(4ft)-Ncos60°(6ft)=0Ans,N=9733.12Ib=9.73kipUsing this resultto write the force equations ofequilibrium alongx and y axes,±≥F=0;Bx(6739.2 1b)sin 60°- (1945.44Ib)sin 60 = 0Ans.B,=7521.12lb=7.52kip+EF,=0;：9733.12lb-800lb-（6739.2lb）cos60°-（1945.44lb)cos60-B=0B,=4590.80 lb=4.59kipAns.18
18 2-5
2-6.Thetide gate opens automatically when thetidewater at B subsides,allowing themarsh at A todrain.Forthe water level h=4m,determinethe horizontal reactionat the smooth stop C.Thegate has a width of 2m.At whatheight h willthe gate be on the verge of opening?F, =156.96(10), = 120.17(10) NPw,=68.67(10) N/mw, = 78.48(10) N/mSOLUTIONSince the gate has a constant width of b =2m, the intensities of the distributedload on the left and rightsides ofthegate at Care(wc)=pwghBc(b)=(1000kg/m)(9.81m/s2)(4m)(2m)=78.48(103)N/m(wc)R = pwghAc(b) = (1000 kg/m)(9.81 m/s2)(3.5m)(2 m)=68.67(10)N/mTheresultant triangular distributed load on the left and right sides of the gate isshown on its free-body diagram,Fig.a,= (78.48(10) N/m)(4 m) = 156.96(10) NFi-2(wo)iLBc=(86())m)=207()These results can also be obtained as follows:F,=htA,= (1000kg/m)(9.81m/s2)(2m)[(4m)(2m)) =156.96(103)NFR=hrAR=(1000kg/m2)(9.81m/s)(1.75m)[3.5m(2m))=120.17(103)NReferring to thefree-bodydiagram ofthe gate in Fig.a,(+Mp=0;[156.96(10)N)2m+(4m)[120.17(10)N)2.5m+(3.5m) Fc(6 m) = 0Fc=25.27(103)N=25.3kNAns.Whenh=3.5m,the waterlevels areequal.Since Fc=O,the gate will openh=3.5mAns.19
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2-7.Theuniformplate,which ishinged at C,is used tocontrol the level of the water at A to maintain its constantdepthof6m.Iftheplatehasawidthof 1.5mandamassof30 Mg,determine the required minimum height h of thewater at B so that seepage will not occur at D.SOLUTIONReferring to the geometry shown in Fig.a,_hx=54The intensities of thedistributedload shown in theFBDof the gawi=pwghb=(1000kg/m)(9.81m/s)(2m)(1.5m)=29.4n(a)W2=pwghzb=(1000kg/m)(9.81m/s2)(6m)(1.5m）=88.29(10)Nw3=Pwghsb=(1000kg/m2)(9.81m/s2)(h)(1.5m）=[14.715(103)h)Then, theresultant forces of these distributed loads areFi=wilc=[29.43(103)N/mJ(5m)=147.15(103)N[88.29(10)N/m2943(103)N/m)(5m)=14F2(W2-Wi)/cD=wlD=[14.715(10)h() = [9.196875(10)n] NF3=and actat(5m）=2.5mdz=(5m）=3.3333md=15d,=5m-)= (5- 0.4167h)m214The seepage is on the verge of occurring when the gate is about to open.Thus, it isrequired that Np=O.Writethemoment equation of equilibrium about point Cbyreferring to Fig.a.(+Mc=0;[147.15(103)NJ(2.5m)+[147.15(103)NJ(3.3333m)-[30(103)(9.81) N)((2.5m）-[9.196875(103)h2](5-0.4167h)= 03.8320/h3-45.9844h2+416.925=0Solving numericallyAns.h=3.5987m=3.60m20
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2-8.The wall is in the form of a parabola.Determinethemagnitude and direction of the resultant force on thewall if it is 8 ft wide.2SOLUTIONThehorizontal loading on the wall is due to the pressure on the vertical projectedarea of the wall, Fig.a. Since the wall has a constant width of b =8 ft, theintensityofthehorizontal distributed load atthebaseofthewallisw=hb=(62.4lb/ft3)(12ft)(8ft)=5.9904(103)lb/ftThus,[5.9904(103)1b/ft)(12ft）=35.9424(103)IbFh=wh=PThevertical force acting on the wall is equal to theweightof thewater contained intheblock abovethewall (shownshaded inFig.a).Fromtheinsidebackcoverofthetext,thevolumeofthisblock(paraboliccross-section)isL_ahb=(12ft)(12ft)(8ft)=768ft33Thus,F=yw=(62.41b/ft)(768ft3)=47.9232(103)1bThen the magnitude of the resultant force isFR=VF+F=V[35.9424(103)1b]2+[47.9232(103)1b]=59.904(103)lb=59.9kipAns.Andits direction is「47.9232(103）1bHe=tantan53.13°=53.1×Ans.35.9424（103）1b21
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