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+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 3-Hydrostatic Forces on surfaces
+// Problem 3.20
+
+//Data given in the Problem
+b=2
+l=3
+A=b*l
+W=343350
+dens=1000
+g=9.81
+theta=45
+
+
+//Calculations
+h=poly(0,"h")
+H=horner(h,0)
+H=h-(3*sin(theta/180*%pi)-0.6*tan(theta/180*%pi))
+F=dens*g*A*H
+IG=b*l^3/12
+H0=IG*(sin (theta*%pi/180))^2/(A*H)+H
+//Taking moments about the hinge,
+AK=W*0.6*sin(theta/180*%pi)/F
+//but AK = H0-AC=H0-(CD-AD)
+//Therefore,
+CD=h
+AD=l*sin (theta/180*%pi)
+AC=CD-AD
+//Hence.AK=H-(CD-AD)
+ak=H0-AC
+//We know ak=AK
+//hence,solving AK-ak=0
+function [f]=F(h)
+    f=(b*l^3/12*(sin (theta*%pi/180))^2/(A*h-(3*sin(theta/180*%pi)-0.6*tan(theta/180*%pi)))+(h-(3*sin(theta/180*%pi)-0.6*tan(theta/180*%pi)))-(h-l*sin (theta/180*%pi)))-W*0.6*sin(theta/180*%pi)/(dens*g*A*(h-(3*sin(theta/180*%pi)-0.6*tan(theta/180*%pi))))
+endfunction
+h=10
+h=fsolve(h,F)
+mprintf("The  height of water that just causes the gate to open is %f m.\n",h)