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.
Volume of rectangular log = (L x b x h)
= (8m x 2m x 2m)
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 .
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.
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)
Hence draft = 0.73m
As we know that
Density = (mass /volume)
= (6 / (5 x 1.6 x 1 )
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.
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;
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 )
Hence draft in water of RD 1.01 is 1.386m