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+// Problem no 14.1,Page No.325
+
+clc;clear;
+close;
+b=2 //m //width 
+FOS=1.5 //Factor of safety
+//rho_mason=2.5*rho_w
+mu=0.5 //coeffeicient of friction
+
+//Calculations
+
+//Let L=1 m (length of dam)
+L=1
+//W=b*H*L*rho
+//After substituting values and Further simplifying we get
+//W=2*H*rho
+
+//Total Pressure
+//P=W*H**2*2**-1
+
+x_bar=b*2**-1 //Distance of Line of action of W from waterface
+
+//Distance of pt where resultant cuts the base measured from Line of action
+//x=P*W**-1*H*3**-1
+//After substituting values and Further simplifying we get
+//x=H**2*30**-1
+
+//x_bar+x=2*b*3**-1
+//After substituting values and Further simplifying we get
+//1+H**2*30**-1=2*b*3**-1
+H=(30*(2*b*3**-1-1))**0.5 //height of dam
+
+//Frictional Resistance offered at the base
+//F=mu*W
+//After substituting values and Further simplifying we get
+//F=3.16*rho
+
+//Total Lateral Pressure
+//P=W*H**2*2**-1
+//P=4.99*W
+
+//Factor of safety against sliding
+//FOS1=F*P**-1=3.16*4.99**-1*rho_mason*rho_w**-1
+FOS1=3.16*4.99**-1*2.5
+
+//FOS1>FOS
+
+//Result
+printf("Dam is safe against sliding  = %.2f m",FOS1)