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+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal

+// Chapter 4-Buoyancy and Floatation

+//// Problem 4.16

+

+//Given Data Set in the Problem

+dens=1000

+g=9.81

+D=1

+H=2

+w=7.848*10^3

+dens1=1030

+

+

+//calculations

+//1) to show that it cannot float vertically

+function [f]=F(h)

+    f=w-dens1*g*%pi/4*D^2*h

+endfunction

+h=1

+h=fsolve(h,F)

+//distance of the centre of gravity G,from A is AG 

+AB=h/2

+AG=H/2

+BG=AG-AB

+//now ,meta centric height is equal to 

+I=%pi/64*D^4

+Vol=%pi/4*D^2*h

+GM=I/Vol-BG

+mprintf("The meta centric height is at %f m \n",GM)

+if GM<0 then mprintf("Since M lies below G,Hence,The body cannot float vertically \n")

+    else mprintf("Since M lies above G,Hence,The body can float vertically \n")

+end

+//2)

+T=poly(0,"T")

+F_d=w+T

+//equating the total downward force to weight of awter displaced

+h0=(F_d)/(dens1*g*%pi/4*D^2)

+AB=h0/2

+//Combined CG due to weight of cylinder and the rension in the chain is

+AG=(w*H/2+T*0)/(w+T)

+BG=AG-AB

+//the metacentric height is GM

+I=%pi/64*D^4

+function [g]=G(T)

+    g=(%pi/64*D^4)/(%pi/4*D^2*(w+T)/(dens1*g*%pi/4*D^2))-((w*H/2+T*0)/(w+T))+((w+T)/2/(dens1*g*%pi/4*D^2))

+endfunction

+T=1

+T=fsolve(T,G)

+mprintf("The Force necessary in the chain to keep it vertical is minimum %f N \n",T)