## Q. 3.12

A uniform body 4 m long, 2 m wide and 1 m deep floats in water. What is the weight of the body if the depth of immersion is 0.6 m? Determine the metacentric height of the body .

## Verified Solution

Weight of the body = Weight of the water displaced

= (1000 × 9.81 × 4 × 2 × 0.6) N = 47088 N

From using Eq. (3.25), we have

Metacentric height MG = $\frac{I}{V}-BG=\frac{\frac{1}{12}\times 4\times 2^{3} }{4\times 2\times 1} – (\frac{1}{2}-\frac{0.6}{2} )$

MG = $\frac{{8}/{3}}{8}-\frac{0.4}{2}=0.1333 m$