What is the draft of the barrel?

[kyaranger]

Homework Statement

The oil drum floats in such a way that one circular end is always directly up. The total mass of drum is 200 kilograms. The drum has a cross-sectional area of 0.4 m2, and the North Sea is level.

What is the draft (the distance from the surface to the deepest point of the barrel at equilibrium) of the barrel?

Homework Equations

N/A

The Attempt at a Solution

I can't begin to solve it because I have no idea how to begin.


I could use a good initial starting position because I am completely clueless about this problem

 

[SteamKing]

A good starting point is to remember Archimedes principle: The barrel will sink until the weight of water displaced equals the weight of the barrel.

 

[kyaranger]

So, the weight of the barrel is Fw = ma = 200kg * 9.8 m/s^2

So the amount of water displaced is equal to that.

Can I replace Mass of water displaced with m = volume * density (density of water * 200kg of water displaced). Solve for volume, then for volume I can substitute w * l * h? solve for h?

 

[BvU]

" m = volume * density "
Yep, except w*l is no good for a cylinder.

"(density of water * 200kg of water displaced)" is no good: gets you kg^2/m^3
to get meters, you want something like kg / (kg/m^3 * m^2)

 

[Basic_Physics]

The problem statement gives the cross sectional area of the drum. So you need to determine what height of a cylindrical shaped amount of water has this weight. I would think that you need to use the density of sea water though for this calculation.