Ship draft (draught) word problem buoyancy

[late347]

Homework Statement

ship arrives from the Atlantic to the Baltic sea. How much does the draft of the ship deepen due to the saltyness of the water...

density of atlantic water = 1027kg/m^3
density of baltic water = 1005kg/m^3

at these depths the crosssectional area of the ship can be assumed to be 4000m^2
ship and cargo mass = 10000 metric tons
g= 9,81m/s^2

Homework Equations

N= ρ V g

G=mg

The Attempt at a Solution

I have no idea what ship draft is. I have only once been on a ship in my life. I have no idea why the the cross-sectional area would be relevant in this problem.

But I decided to calculate the volumes of water, which the ship displaces in atlantic,

And then displaces in the baltic.

I think those volumes will be different because water density changes.

N_atlantic = ρ_atlantic x V_atlantic x g

V_atlantic= 9737, 0983 m^3N_baltic = ρ_baltic x V_balticx g

V_baltic = 9950,2487 m^3I was later on thinking that maybe if I were to take all the water into a huge box

Then you could divide the displaced water volumes, by the area

Therefore you end up with height ( I think) Or something to this effect could be done.

If this is calculated so, then the difference between the heights of the displaced water is

(9950,2487m^3) / 4000m^2 = 2,4875 m = height baltic

9737,0983 m^3 / 4000m^2= 2,4342 m height atlantic.

calculate Δheight = 0,0533m

the deepening effect ought to be about 5,3 cm

please notify whether or not errors of understanding of judgement were made.

 

[SteamKing]

Homework Statement

ship arrives from the Atlantic to the Baltic sea. How much does the draft of the ship deepen due to the saltyness of the water...

density of atlantic water = 1027kg/m^3
density of baltic water = 1005kg/m^3

at these depths the crosssectional area of the ship can be assumed to be 4000m^2
ship and cargo mass = 10000 metric tons
g= 9,81m/s^2

Homework Equations

N= ρ V g

G=mg

The Attempt at a Solution

I have no idea what ship draft is. I have only once been on a ship in my life. I have no idea why the the cross-sectional area would be relevant in this problem.

You have a computer connected to the internet. "I have no idea" is no longer much of a reason for not looking stuff up. There are dictionaries on line which can help you find the meaning of unfamiliar terms.

BTW, the draft of a vessel is the distance it sinks in the water when it is floating.

But I decided to calculate the volumes of water, which the ship displaces in atlantic,

And then displaces in the baltic.

I think those volumes will be different because water density changes.

Good. Remember Archimedes' Principle.

N_atlantic = ρ_atlantic x V_atlantic x g

V_atlantic= 9737, 0983 m^3N_baltic = ρ_baltic x V_balticx g

V_baltic = 9950,2487 m^3I was later on thinking that maybe if I were to take all the water into a huge box

Then you could divide the displaced water volumes, by the area

Therefore you end up with height ( I think) Or something to this effect could be done.

If this is calculated so, then the difference between the heights of the displaced water is

(9950,2487m^3) / 4000m^2 = 2,4875 m = height baltic

9737,0983 m^3 / 4000m^2= 2,4342 m height atlantic.

calculate Δheight = 0,0533m

the deepening effect ought to be about 5,3 cm

please notify whether or not errors of understanding of judgement were made.

Sounds like some solid reasoning here.

 

[sidt36]

To begin with
Draft ; It is the distance between the waterline and the bottom of the ship(hull)
That should help you .And looking at your work it seems legit
Be sure to check your calculation

 

[Nidum]

Edit : Removed duplicate information