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+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 3-Hydrostatic Forces on surfaces
+// Problem 3.19
+
+//Data given in the Problem
+theta=60
+AC=h/sin(theta*%pi/180)
+s=sin(theta/180*%pi)
+h=poly(0,"h")
+H=h/2
+b=1
+d=AC
+A=AC
+IG=b*d^3/12
+//COP=(IG/(A*H)+H)
+//COP=(h/sin(theta/180*%pi)^3/12/(h/sin(theta*%pi/180)/(h/2)+h/2
+//We know that COP is equal to (h-3),THAT IS ,the depth of centre of pressure
+//hence
+function f=F(h)
+f=((h/s)^3/12*s^2/(h/s*h/2))+(h/2)-(h-3);
+endfunction
+h=100;
+y=fsolve(h,F)
+mprintf("The height of wahter for tipping the gate is %f m",y)