# Homework Help: Density of Water

Tags:
1. May 3, 2015

### Mongster

1. The problem statement, all variables and given/known data
A river car ferry boat has a uniform cross sectional area in the region of the water line of 720m^2. If sixteen cars of average mass 1100kg are driven on board, find the extra depth to which the boat will sink into the water.

2. Relevant equations
Density= m/V
Density= m/Area X Depth
3. The attempt at a solution

2. May 4, 2015

### stedwards

There seems to be missing information such as the length of the boat. Am I wrong? What if you assume the length is the variable 'L' meters?

Last edited: May 4, 2015
3. May 4, 2015

### phinds

You have enough info. Keep thinking about it.

EDIT: oh, I AM assuming that you can manage to look up the density of water on your own.

4. May 4, 2015

### CWatters

+1 to that.

The problem requires you to know (or look up) the density of water.

That's one of the things you have to calculate before you can answer the actual question.

5. May 6, 2015

### Mongster

Hey guy, I finally solved it! Yeap the density of water was all I needed!

Firstly I got the Density of Water, 1000kg /m^3

Secondly I calculated the volume of water displaced by the 16 cars of average mass 1100kg like this (16 X 1100)/1000 = 17.6m^3

With the volume of water displaced by the addition of cars and the uniform cross sectional area in the region of the water line of 720m^2, I found the increase in depth which the boat will sink like this 17.6/720 =0.0244m

I want to thank you all for chipping in and guiding me in this question. It wasn't that hard afterall! Cheers! :D

Last edited: May 6, 2015
6. May 6, 2015

### stedwards

No, sorry.

What you have calculated is the minimum length of a massless vessel, having a submerged cross sectional area of 720 square meters that will support 16 cars. It would be a very short vessel.

The problem statement is a result of confusion on the part of the author. (Buoyancy is one of the topics in science often misunderstood by educators.)

Last edited: May 6, 2015
7. May 6, 2015

### SteamKing

Staff Emeritus
I think you are confused here.

According to the OP, "A river car ferry boat has a uniform cross sectional area in the region of the water line of 720m2." Since no other information is given, this sentence suggests that the area of the waterplane of the ferry is 720 m2.

By assuming that the vessel does not change shape if it is immersed from its original draft by a small amount (i.e., the "uniform cross sectional area"), the change in draft, ΔT, multiplied by this waterplane area will give the additional volume of displacement required to support the 16 cars driving onto the ferry.

What the OP calculated in Post #5, 0.0244 m, was ΔT, the change in draft of the ferry with the 16 additional cars loaded aboard, not the length of the vessel.

Adding or removing cars from the ferry does not change the mass of the ferry taken by itself.

I agree that this problem could have been worded in a clearer fashion, and a sketch would have been helpful, too.

8. May 6, 2015

### CWatters

I agree with the OP's answer in #5 and Steamking in #7.

The problem states the "cross sectional area in the region of the water line" is 720sqm. So 720sqm is the horizontal cross sectional area not the vertical cross sectional area.

9. May 6, 2015

### stedwards

Harg. Land lubbers! Cross sectional area of a boat to the waterline means a vertical section.

10. May 6, 2015

### SteamKing

Staff Emeritus
Normally, you would take the cross section of a hull normal to the waterline.

However, in this case, the wording of the problem drops some vague hints about what the area stated in the OP represents. Also for this problem, there isn't nearly enough information provided about the ferry to work out the change in draft due to adding the cars, unless you assume that the "cross sectional area" refers to the "waterplane area".

Avast ye, matey! I'm a naval architect, so I know a little about botes.

11. May 7, 2015

### CWatters

Sure, but it doesn't say "to the waterline" it says "in the region of the waterline".

12. May 7, 2015

### phinds

Well, maybe it does but this is a physics forum. It never occurred to me in reading the problem that you could possibly interpret it as anything other that "parallel to the water, at the waterline" because for one thing, that's the only way it makes any sense in the context of the problem.

Specialized knowledge can sometimes distract you from what others mean.

13. May 7, 2015

### stedwards

Me brain tumor was acting up again.