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+// Problem no 14.7,Page No.333
+
+clc;clear;
+close;
+H=8 //m //height of dam
+h=7.5 //m //Height of water
+a=1 //m //top width
+mu=0.6 //Coeffeicient of friction
+rho_mason=22.4 //KN/m**3 //weight of mason
+rho_w=9.81 //KN/m**3 //density of water
+
+//Calculations
+
+//weight of dam
+//W=(a+b)*2**-1*L*H*rho2*10**-3
+//After substituting values and further simplifying we get
+//W=89600*(b+1)
+
+//Distance of Line of action from vertical base
+//x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+//After substituting values and further simplifying we get
+//x_bar=(1+b+b**2)*(3*(1+b))**-1
+
+//Lateral thrust
+P=rho_w*h**2*2**-1
+
+//Min width to avoid tension at base
+//Z=x_bar+P*W**-1*h*3**-1
+//Z=2*3**-1*b
+//After substituting values and further simplifying we get
+//b**2+b-24.09=0
+a=1
+b=1
+c=-24.09
+
+X=b**2-4*a*c
+
+b1=(-b+X**0.5)*(2*a)**-1
+b2=(-b-X**0.5)*(2*a)**-1
+
+//Thus width cannot be negative,b1 is considered
+
+//Min width to avoid sliding
+//mu*W>P
+//After substituting values and further simplifying we get
+b=P*10**3*(mu*89600)**-1-1
+
+//Result
+printf("The Min bottom width is %.2f",b);printf(" m")