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# Chapter 3-flotation Exercise 3

### 1. A rectangular log of wood 8m long, 2m wide and 2m high floats in FW at a draft of 1.6m with one face of horizontal. Find its mass and RD.

Solution :

Volume of rectangular log = (L x b x h) = (8m x 2m x 2m) = 32m3

Weight = (U/W volume ) x (Density of displaced water) Weight = (8 x 2 x1.6) x(1) = 25. 6t

RD = (mass / volume) = 25.6 / 32 = 0.8t/m3 .

### 2. A rectangular log of wood 5m x 1.6m x1.0m weight 6t and floats with it largest face horizontal. Find its draft in SW and its RD.

Solution :

Volume of rectangular log = (L x B X H ) = (5m x 1.6m x 1.0m)

Weight of log = 6t SW RD = 1.025

Weight = (u/w volume )x (Density of water displaced) 6 = (5 x 1.6 x D) x (1.025) D = (6 /( 5 x 1.6 x1.025) = 0.73m Hence draft = 0.73m Density = (mass /volume) = (6 / (5 x 1.6 x 1 ) =0.75 t/m3

### 3. A rectangular log 3m broad and 2m high floats with its breadth horizontal. If the density of the log is 0.7 tm-3, find its draft In water of RD 1.01.

Solution :

Area of rectangular log = ( B X H ) = 3m x 2m RD of log = 0.7 t/m3 Draft in water of RD 1.01 can be calculated as; Density= (mass/volume) 0.7 = mass/ volume Mass = volume x 0.7 Mass = (L x 3 x2) x ( 0.7) = 4.2L t

Now mass = (U/w volume) at depth of D m x (1.01) 4.2L =( L x 3 x D )x( 1.01) D = ( 4.2 / 3 x 1.01 ) = 1.386m.

Hence draft in water of RD 1.01 is 1.38