Change in Water Level after throwing over anchor

[TFM]

[SOLVED] Change in Water Level after throwing over anchor

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

Not quite Sure,

TFM

 

[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.

⚛​

 

[dingpud]

it would have to be a contained body of water...like a pond or something, otherwise you would think that any change would be negligible...

maybe a trick questions...

http://urbanlegends.about.com/library/bl_water_bridge.htm

 

[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!

 

[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

 

[mgb_phys]

The anchor is thrown overboard but is suspended above the bottom of the river by a rope;

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.

 

[TFM]

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

TFM

 

[kamerling]

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.

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.

 

[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

 

[kamerling]

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

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

 

[TFM]

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

TFM

 

[TFM]

To continue on a nautical theme...

SOS
...---...

Please?

TFM

 

[Kurdt]

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?

 

[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

 

[Kurdt]

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

 

[TFM]

The weight of the anchor is 294 Newtons.

TFM

 

[Kurdt]

What volume of water weighs 294 Newtons?

 

[TFM]

Using v=mass/density, I get 0.03,

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

TFM

 

[Kurdt]

Using v=mass/density, I get 0.03,

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

TFM

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.

 

[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

 

[Kurdt]

You were given the surface area of the bottom of the boat. You know the volume (don't worry about minus signs). What multiplied by surface area will give you that volume?

 

[TFM]

Volume/S.A. = 0.026183206/8.5 = 0.003080377

I have a feeling that this is the answer?

TFM

 

[Kurdt]

3 mm. Looks ok to me.

 

[TFM]

Apparntly its wrong. Mastering Physics doesn't accept it. The answer in the book (which uses slightly different variable, given below) is 5.57*10^-4, using

anchor mass: 35kg,
density: 7860
Area 8.00 m^2

Any ideas now?

TFM

 

[Kurdt]

It appears they're just taking into account the volume displaced by the anchor for some unknown reason.

 

[TFM]

Whats the best course of action from here?

TFM

 

[Kurdt]

Well apparently they're looking for just the displacement caused by the anchor. So if you work out the volume displaced by the anchor divided by the surface area of the boat.

Like most people on the thread I thought it involved the anchor being thrown out of the barge.

EDIT: the anchor is still attached to the barge which explains it.

 

[TFM]

Using the volume of the anchor 0.003817, divided by the area, 8.5m gives 0.000449, which is the answer

Thanks all,

TFM

 

[Kurdt]

Using the volume of the anchor 0.003817, divided by the area, 8.5m gives 0.000449, which is the answer

Thanks all,

TFM

Sorry about the confusion but a good exercise none the less.

 

[dingpud]

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!

I'm not sure why that was in there either. Maybe someone from ancient Rome didn't believe that what they use to use to move a small amount of water could move a ship...(you've got me)

The boat lift is freakin awesome. Could you imagine seeing one of those installed at the end of someones dock who has a small Jon boat or row boat? That would be cool...