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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 1.386m
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A cylinder 2m in diameter and 10 m log floats in FW, with its axis horizontal, at draft of 0.6m. Find its mass.
Solution :
Diameter of cylinder= D = 2m,
Radius = ( D / 2 )
= 1m
length = 10m, depth = 0.6m
Mass = (u/w volume) x (density of displaced water)
Weight = ( πr2h x density )
= (3.1416 x 1 x 1 x 0.6) x (1)
= 1.88t.
PROBLEM NUMBER 4 AND 5 ARE COMPLETELY WRONG, HOW CAN ONE USE FORMULA FOR A VOLUME OF A RECTANGLE TO DERIVE VOLUME OF A TRIANGLE
Question 4 and 5 are wrong.
Questions 4 and 5 are wrong.
Question 4&5 are wrong .
Really helpful…
Solution No.4 is incorrect. It says the log floats on its horizontal axis thus we cant use the normal cylinder formula. We have to use the other formula ‘Area of a circle segment given height and radius”. Google it. Pls resolve and update.