Question

Fluids static question:

A solid cylindrical pine (S = 0.50) spar buoy has a cylindrical lead (S = 11.3) weight attached, as shown in Fig. P2.144. Determine the equilibrium position of the spar buoy in seawater (i.e.. find d). Is this spar buoy stable or unstable? For seawater. S= 1.03.

Answer 1

>> Now, downward force exerting by system =

             $\rho$1*g*h + $\rho$2*g*y ,

where, $\rho$1 = Density of Cylindrical Pine = 0.50*$\rho$w,

$\rho$w = Density of Water

^,$\rho$2 = Density of Lead = 11.3*$\rho$w

h = height of Cyindrical Pine = 16 feet

y = Height of Lead = 0.5 ft

>> Upward force due to fluid acting = $\rho$s*g*d

, where, $\rho$s = Density of Sea Water = 1.03*$\rho$w

>> As, for equilibrium, Upward Force = Downward Force

=> $\rho$s*g*d = $\rho$1*g*h + $\rho$2*g*y

=> 1.03*d = 0.5*16 + 11.3*0.5

=> d = 13.25 feet   ...ANSWER....

>> Now, for body, let's find Center of Gravity,

>> Let, it is at y" distance from bottom

and, M = Mass of Cylindrical Pine, and it s COG = at 0.5 + 16/2, feet from bottom

, m = Mass of Lead, and it s COG = at 0.5/2, feet from bottom

=> (M + m)y" = M*(0.5+8) + m*0.25 ,        .(1)...

>> As, Mass = Volume*Density = Area*Length*Density

=> Equation 1 becomes:

(0.5*$\rho$2 + 16*$\rho$1)y" = 8.5*16*$\rho$1 + 0.25*0.5*$\rho$2

=> (0.5*11.3 + 16*0.5)y" = 8.5*16*0.5 + 0.25*0.5*11.3

=> y" = 5.085 feet ...Center of Gravity from bottom ...

>> And, Center of Buoyancy = at, d/2 = 13.25/2 = 6.625 feet from bottom

>> As, Center of Buoyancy is above than Center of Gravity

=> So, system, i.e. spar buoy is stable .....ANSWER......