Question

10. The weight of a cylindrical wooden block of 1 m diameter and 1 m length is 6.28 kN. Will it float in a stable manner in water with its axis vertical as shown in the given figure? (6 pts.) (Hint: Ymc = Ycb + MB and MB 1 m diameter 1 m yeg = 0.5 m ch.

Answer 1

.The weight of the block = weight of the water displaced by the block.

6280=m*g

6280=Density of water*Volume of water displaced *9.81

6280=1000*V*9.81

Volume of water displaced by block = V=0.640163 m³

V=0.640163 =Area*length of block in water = 0.785*1² * x.

Distance X = 0.8154 m.

Center of buoyancy = X/2 = 0.8154/2 = 0.40774 m.

Center of buoyancy from free surface = 1-0.40774 = 0.59226 m

Metacentric height = MB

Second moment of inertia = I = 3.14*1⁴ / 64 = 0.0490625

Volume of water displaced by block = V=0.640163 m³

Metacentric height = GM = I/V-BG

I/V= 0.0490625/0.640163 = 0.076640 m.

BG = Distance between center of buoyancy and center of gravity from free surface = 0.59226-0.50 = 0.09226 m.

Metacentric height = GM = 0.076640-0.09226 = -0.01562

Negative sign indicates it is in unstable condition.