A ship with vertical sides near the waterline weighs 4000 tons and draws 22 ft in salt water (γ = 64.0 lb/ft3). Discharge of 200 tons of water ballast decreases the draft to 21 ft. What would be the draft d of the ship in fresh water?

Question

A ship with vertical sides near the waterline weighs 4000 tons and draws 22 ft in salt water (γ = 64.0 lb/ft³). Discharge of 200 tons of water ballast decreases the draft to 21 ft. What would be the draft d of the ship in fresh water?

Accepted Answer

Because the shape of the underwater section of the ship is not known, it is best to solve the problem on the basis of volumes displaced.
A 1-ft decrease in draft was caused by a reduction in weight of 200 tons, or

200 x 2000 = γv = 64.0(A x 1)

where v represents the volume between drafts of 22 ft and 21 ft, and (A x 1) represents the waterline area x 1 ft, or the same volume v. Then

v = A x 1 = (200)(2000)/64.0 - 6250 ft³/ft depth

Buoyant force F_B = γx volume of displaced liquid. Then F_B/γ = volume of displaced liquid.
From Fig. 4-8, the vertically hatched volume is the difference in displaced fresh water and salt water.

This difference can be expressed as [latex]\left( \frac{3800 imes 2000}{62.4}-\frac{3800 imes 2000}{64.0} ight) [/latex] , and this volume is also equal to 6250_y. Equating these values, y = 0.49 ft.

The draft d = 21 + 0.49 = 21.49 ft.

Question 4.10: A ship with vertical sides near the waterline weighs 4000 to...

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