notes - Pa ’-Pb ’ =0.1 for deltah=1ft Asin((Pa ’-Pb...

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Unformatted text preview: Pa ’-Pb ’ =0.1 for deltah=1ft Asin((Pa ’-Pb ’ )/(deltah*(gamma-gammaB))) p atm =gamma* h+p vapor I_xc=PI*R^4/4 y_R=I_xc/(y_c*A)+y_C distance below the shaft to the center of the pressure is y_R-y_C ∑ M_C=0=F_R*(y_R-Y_C) Example 2.8 A pressurized tank contains oil (SG = 0.90) and has a square, 0.6-m by 0.6-m plate bolted to its side. When the pressure gage on the top of the tank reads 50kPa, what is the magnitude and location of the resultant force on the attached plate? The outside of the tank is at atmospheric pressure. F_1=(P_s+gamma*h1)*A F_2=1/2*gamma(h2-h1)/A F_R=F_1+F_2 M_0=-F_R*y_0+F_1*.3 Example 2.9 The 6-ft-diameter drainage conduit is half full of water at rest. Determine the magnitude and line of action of the resultant force that the water exerts on a 1-ft length of the curved section BC of the conduit wall. F1=gamma*hc*A Weight=gamma*volume Fh=F1 Fv=Weight FR=sqrt(Fh^2+Fv^2) Fbouyant=gamma*Volume Buoyant force passes through the centroid of the displaced volume. The point through which the buoyant force acts is called the center of buoyancy. buoyant force acts is called the center of buoyancy....
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