Barge Buoyancy: Calculating Depth Change with Loaded Crate | Homework Help

[mogibb1]

Homework Statement

A rectangular barge, 5m long and 2m wide, floats in fresh water. Suppose that a 400-kg crate of auto parts is loaded onto the barge. Show that the barge floats 4cm deeper.

Homework Equations

Archimedes Principle
P=m/v

The Attempt at a Solution

I know that the area of the barge = 2(5)+2(2) = 14m
I know that the density of freshwater = 1000 kg/m³
Not real sure where to go from here.

 

[Borek]

Every time you see buoyancy you should think "volume".

 

[mogibb1]

Ok, so this is what I believe to be the answer:

400kg * 10 (for gravity) = 4000 / 1000 kg/m³

= 4cm

 

[Borek]

What you wrote doesn't make sense. Please put units everywhere.

 

[mogibb1]

To make more sense, here are the units:

400 kg * 10 m/s² = 4000 kg m/s²

4000 kg m/s² / 1000 g = 4 cm

I now I need to work more on the units/conversions, but I would like to know if my
thinking is correct or not on this. Thank you.

 

[Borek]

In case you have not noticed, it can't be right - you have seconds on the left, there is no miraculous way they can cancel out and left you with cm on the right. That's why I asked you add units, they often are a simple and sure way of telling you the answer must be wrong.

What mass of water must the barge displace to stay afloat after the crate has been added?

 

[mogibb1]

I appreciate your help. I really don't have much in the way of a formula (unless Archimedes principle is it) to go by and I'm trying to figure this out. The 400kg crate will displace 400kg of the water, isn't that right?

If I take the 400kg crate and divide it by 1000kg/m^3 water I get 0.4m^3 is this in the ballpark?

 

[Borek]

You are on the right track now.

 

[mogibb1]

Sorry to be so much trouble, but I'm not good with word problems and having to figure out what to plug into an equation. Thank you very much for your time and assitance.