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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 .
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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
As we know that
Density = (mass /volume)
= (6 / (5 x 1.6 x 1 )
=0.75 t/m3
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.