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+clc;
+clear;
+dia=6;//ft
+l=1;//ft
+
+//horizontal force f1=sw*hc*A
+hc=dia/4;//ft
+sw=62.4;//lb/ft^3
+A=dia/2*l;//ft^2
+f1=sw*hc*A;//lb
+//this force f1 acts at a height of radius/3 ft above the bottom
+ht=(dia/2)/3;//ft
+//weight w = sw*volume
+w=sw*((dia/2)^2)*%pi/4*l;//lb
+//this force acts through centre of gravity which is 4*radius/(3*%pi) right of the centre of conduit
+dist=(4*dia/2)/(3*%pi);//ft
+//horizontal force that tank exerts on fluid = f1
+//vertical force that tank exerts on fluid = w
+//resultant force fres =((f1)^2+(w)^2)^0.5
+fres =((f1)^2+(w)^2)^0.5;//lb
+disp("lb",fres,"The resultant force exerted by the tank on the fluid=");
+disp("ft",dist,"above the bottom of the conduit and to the right of the axis of the conduit at a distance of","ft",ht,"The force acts at a distance of")
+