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# Buoyancy Question watch

1. A cuboid boat (Shown in a picture) has been made entirely out of steel. The bottom of the boat is a sheet of steel with dimensions 10.0m X 5.0m X 2.0cm.
The sides of the boat are 0.5cm thick steel.

Assuming that ρ(steel)=7900kgm-3, determine the minimum height, h, that the sides of the boat must have in order for it to be able to float on calm water? Give your answer in units of cm.

Ive also the method i tried, but i got a negative awnser.. not entirely sure what to try now aha

Any help would be appreciated!
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2. (Original post by seb5321)
A cuboid boat (Shown in a picture) has been made entirely out of steel. The bottom of the boat is a sheet of steel with dimensions 10.0m X 5.0m X 2.0cm.
The sides of the boat are 0.5cm thick steel.

Assuming that ρ(steel)=7900kgm-3, determine the minimum height, h, that the sides of the boat must have in order for it to be able to float on calm water? Give your answer in units of cm.

Ive also the method i tried, but i got a negative awnser.. not entirely sure what to try now aha

Any help would be appreciated!
You don't need to go anywhere near forces if you remember that the boat will float when the mass of water displaced = mass of steel.

water mass displaced = water_density x volume = 1000kgm-3 x width x length x height

1000 x 5 x 10 x h = (50x103 x h) kg

i.e. each 1m draught (boat height) will displace 50 metric tons of water.

So now do the same for the steel and equate the two. You should get two 'h' terms which can be solved by rearranging the expression to isolate.

Hint: be careful when finding an expression for the steel mass because the boat bottom is a different thickness to the boat walls, also, be careful with the dimensions for the walls - both sets because only one parallel pair of sides will be exactly 5m or 10m.

My expression in terms of boat height:

displaced mass = steel mass

50x103h = (1184h + 7876)

h = 16.1cm

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