Archimedes force: ship with(out) cargo

[Function]

Hello everyone

 

Here's a question in preparation for the med school approval exam:

 

A freighter, which has a constant horizontal sectional area of [math]4,500\text{ m}^2[/math], lowers [math]6[/math] more meters into the water when it's filled with cargo. The mass of the cargo is:

1. [math]27,000,000 \text{ kg}[/math]

2. [math]270,000,000\text{ kg}[/math]

3. [math]60,000\text{ kg}[/math]

4. [math]4,500,000\text{ kg}[/math]

 

To try to solve this, I called [math]l[/math] the length of the freighter, [math]h[/math] the height and [math]d[/math] the depth. Let [math]x[/math] be the height that is under water when the freighter is unloaded, [math]V_{o_1}[/math] the volume of the freighter that's immersed when unloaded, and [math]V_{o_2}[/math] when loaded. Let [math]m_1[/math] be the net mass of the freighter, and [math]m_2[/math] the mass of the freighter, when loaded. Let then [math]F_{A_1}[/math] be the magnitude of the Archimedes force on the ship when unloaded, [math]F_{z_1}[/math] the magnitude of the gravitational force on the ship when unloaded, [math]F_{A_2}[/math] the magnitude of the Archimedes force on the ship when loaded and [math]F_{z_2}[/math] the magnitude of the gravitational force on the ship when loaded.

 

 

 

[math]\begin{array}{rcccl}V_{o_1}&=&x\cdot l\cdot d&=&x\cdot \dfrac{4,500}{h}\cdot d \\ &&&&\\ V_{o_2}&=&(x+6)\cdot l\cdot d&=&(x+6)\cdot\dfrac{4,500}{h}\cdot d\end{array}[/math]

 

 

 

[math]\begin{array}{rrcl}&F_{A_1}&=&F_{z_1} \\ &&& \\ \Leftrightarrow &\rho_{H_2O}\cdot g\cdot V_{o_1}&=&m_1\cdot g \\ &&& \\ \Leftrightarrow & m_1&=&1,000\cdot x\cdot\dfrac{4,500}{h}\cdot d\end{array}[/math]

 

 

 

[math]\begin{array}{rrcl} &F_{A_2}&=&F_{z_2} \\ &&& \\ \Leftrightarrow & & \cdots &\\ &&& \\ \Leftrightarrow& m_2&=&1,000\cdot (x+6)\cdot\dfrac{4,500}{h}\cdot d\end{array}[/math]

 

 

 

[math]\begin{array}{rrcl}\Rightarrow& m_2-m_1&=&1,000\cdot (x+6)\cdot\dfrac{4,500}{h}\cdot d-1,000\cdot x\cdot\dfrac{4,500}{h}\cdot d \\ &&& \\ \Leftrightarrow &m_2-m_1&=& 27,000,000\cdot \dfrac{d}{h}\end{array}[/math]

 

 

 

Now I'd like to pick answer A, but since I don't know anything about [math]d[/math], nor [math]h[/math], I don't think I can just pick this one.

 

Could someone help me please?

 

Thanks.

 

Function

Edited May 11, 2014 by Function

[studiot]

Second thoughts.

Edited May 11, 2014 by studiot

[John Cuthber]

The 4500m² figure in the question isn't a length.

It is (confusingly) the product of the length and the depth.

So your initial equation isn't set up right.

[Function]

The 4500m² figure in the question isn't a length.

It is (confusingly) the product of the length and the depth.

So your initial equation isn't set up right.

 

But I thought: 4 500 = l * h; so x * l * d = V = x * 4 500/h * d?

With units: V_O1 = (xld)m³ = (x)m (4,500)m² (d)m / (h)m = (4,500 xd/h) m³

Edited May 11, 2014 by Function

[John Cuthber]

I am getting confused by the names you give to the dimensions of the ship

Let's make an arbitrary assumption (we can check it later if we need to).

The ship is 100 metres long and 45 metres wide. That gives an area of 4500 m²

It's a funny shape but...

 

What volume of water is displaces when it moves down by 6 metres?

 

How much does that volume of water weigh?

If the ship was twice as long , but only half as wide, (200 by 22.5) what would the displaced volume be?

[Function]

I am getting confused by the names you give to the dimensions of the ship

Let's make an arbitrary assumption (we can check it later if we need to).

The ship is 100 metres long and 45 metres wide. That gives an area of 4500 m²

It's a funny shape but...

 

What volume of water is displaces when it moves down by 6 metres?

 

How much does that volume of water weigh?

If the ship was twice as long , but only half as wide, (200 by 22.5) what would the displaced volume be?

 

Volume would be 27,000 m³, so m = 27,000,000 kg

 

Same for the other length and width. I think I get it.

Or it's a tricky question which hasn't got such a simple answer. (It's a bit too easy to be in the book I have...)

Probably my bad, but I think they could've explained it a bit better in the question what exactly is 4500 m²...

Edited May 11, 2014 by Function

[imatfaal]

It is not great wording - but I think you (and JC) have it down correct. For two reasons; firstly it is the only interpretation of the wording that gives a single answer, secondly a horizontal cross section is best interpreted as in the plane that does not have a vertical component ( ie the same plane as that of the deck).

 

FYI - these are not shipping terms. We would use the tonnes per centimetre (in this case 45 tonnes per cm) to express the information given in the question