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A rectangular barge of 10 m long and 5m wide, floating in SW at a draft of 3m, is being lifted out of the water by a heavy-lift crane. Find the load on the crane when the draft has reduced to 1m.
Solution:
Area of rectangular barge = ( L x B)
= 10m x 5m,
RD = 1.025
Mass when draft 3m = (u/w volume at 3m) x (density)
= (10 x 5 x 3) x (1.025)
= 153.75t.
Mass after being lifted by the heavy crane =(u/w volume at 1m depth )x (density)
= (10 x5 x 1) x (1.0250
= 51.25 t
Hence ,Load taken by the crane = (153.75 – 51.25)
= 102.5 t
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A rectangular box 2.4 m long, 1.2m wide and 0.8m high, floats in water of RD 1.012m at an even keel draft of 0.2m. find maximum mass of SW that can be poured in to it without sinking it.
Solution :
Given :
L = 2.4m, B = 1.2m, H = 0.8m
RD = 1.012 , Depth = 0.2m
Mass of rectangular box = (L x B x H )x (1.012)
= (2.4 x 1.2 x 0.8) x (1.012)
= 2.33t
Mass of box even keel when draft is 0.2m
= (2.4 x 1.2 x 0.2) x(1.012)
= 0.583
Hence, Salt water that can poured is (2.33 – 0.58)
= 1.75 t.
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.