1 files changed, 49 insertions, 0 deletions
diff --git a/3819/CH4/EX4.16/Ex4_16.sce b/3819/CH4/EX4.16/Ex4_16.sce
new file mode 100644
index 000000000..d0ad05d15
--- /dev/null
+++ b/3819/CH4/EX4.16/Ex4_16.sce
@@ -0,0 +1,49 @@
+// 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)