# Change in Water Level after throwing over anchor

1. Feb 14, 2008

### TFM

[SOLVED] Change in Water Level after throwing over anchor

1. The problem statement, all variables and given/known data

An anchor is on a barge with bottom surface area: 8.5 m^2
The Anchor weighs 30kg and density 7860 kg/m^3
I know the water level will fall, but how do you calculate how much it falls?

2. Relevant equations

Density = mass/volume (?)
Pressure = Force/Area (?)
Pressure = Density*g*height (?)

3. The attempt at a solution

Not quite Sure,

TFM

2. Feb 14, 2008

### CaptainQuasar

Is this about the change in water level on the hull of the barge, maybe? Because if it's about the change in the level of a body of water itself, I would think you'd need to know something about the body of water the barge is in.

3. Feb 14, 2008

### dingpud

4. Feb 14, 2008

### mgb_phys

The density of the anchor is a red herring - you are just looking for how much volume of water 30kg displaces. Then use the area of the bottom of the boat to calculate the change in height.
You do have to know if it is fresh / salt water!

dingpud - Don't know why an aquaduct is in urban legends but the engineering feat isn't as good as this http://en.wikipedia.org/wiki/Falkirk_Wheel
Although a local council did announce that they were stengthening an aqueduct to allow it carry heavier barges!

5. Feb 14, 2008

### TFM

It is assumed to be a river. Here is the question:

'After the anchor is overboard and the barge has finally stopped bobbing up and down, has the barge risen or sunk down in the water?'

The Answer being the ship will sink down.

the next part:

'By what vertical distance?'

Initial Data:

'An iron anchor with mass 30.0kg and density 7860kg/m^3 lies on the deck of a small barge that has vertical sides and floats in a freshwater river. The area of the bottom of the barge is 8.50 m^2. The anchor is thrown overboard but is suspended above the bottom of the river by a rope; the mass and volume of the rope are small enough to ignore.'

TFM

6. Feb 14, 2008

### mgb_phys

That makes a difference - not the normal behaviour of an anchor!

In that case you do need the volume of the anchor - you need to work out the bouyant force on it.

7. Feb 15, 2008

### TFM

I have the volume of the anchor at 3.817*10^-3. What would be the best direction to go now?

TFM

8. Feb 15, 2008

### kamerling

I would say go up.

the amount of water displaced by the ship before you throw in the achor, is equal to the amount of water displaced by both the ship and the anchor after you throw in the anchor. Since the anchor now also displaces some water, the ship will now displace less water and thus float higher.

9. Feb 15, 2008

### TFM

I did put that down as the answer, but got them mixed up when i copied it over

The question now is, how do you calculate the change in water displaced?

TFM

10. Feb 15, 2008

### kamerling

If there are no other forces on the boat or the anchor then gravity and de buoyant force of the water, the weight of the water that is didplaced must be equal to the weight of the boat+the anchor. (make sure you understand why this is so).

If now the anchor displaces some water on its own, the amount of water that the boat displaces will decrease by the same amount

11. Feb 16, 2008

### TFM

What is the best equation to use to calculate the amount of water displaced?

TFM

12. Feb 16, 2008

### TFM

To continue on a nautical theme...

SOS
...---...

TFM

13. Feb 16, 2008

### Kurdt

Staff Emeritus
As other people have said the weight of the boat plus the anchor would displace a volume of water equal to that weight. When the anchor is in the water it only displaces water equal to its volume. Whats the difference in these two volumes?

14. Feb 16, 2008

### TFM

I have the volume of the anchor to be 0.003817 m^3

I have worked out the pressure casued by the anchor on the boat to be 34.588.

How can I convert this pressure into a volume?

TFM

15. Feb 16, 2008

### Kurdt

Staff Emeritus
You don't need the pressure, just the weight, which is the mass x acceleration due to gravity.

16. Feb 16, 2008

### TFM

The weight of the anchor is 294 Newtons.

TFM

17. Feb 16, 2008

### Kurdt

Staff Emeritus
What volume of water weighs 294 newtons?

18. Feb 16, 2008

### TFM

Using v=mass/density, I get 0.03,

but this doesn't involve the weight of water in Newtons...?

TFM

19. Feb 16, 2008

### Kurdt

Staff Emeritus
It doesn't matter anyway because the gravitational accelerations cancel.

So what is the change in the volume? Remember the volume displaced has decreased by 0.03 and increased by 0.00318.

20. Feb 16, 2008

### TFM

Taking the initial volume to be an arbituary 0, reducing by 0.03, then increasing by o.oo3817, the final volume is -0.026183206. do you now need to cube root this (taking out the minus as it is just signifying it is a dercrease)?

TFM