# Capacity of a Barge

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1. Dec 11, 2014

### Mitchell Powell

1. The problem statement, all variables and given/known data
An open barge has the dimensions shown in the figure.
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/59/59574-46462a33e8dd3e05fa12094d99749d9c.jpg [Broken]
If the barge is made out of 5.2-cm-thick steel plate on each of its four sides and its bottom, what mass of coal can the barge carry in freshwater without sinking?
2. Relevant equations
p = m/V

3. The attempt at a solution
V = 22 * 40 * 12 m3 = 10,560 m3

pwater = 1,000 kg/m3

1,000 kg/m3 = m/10,560

m = 1,000*10,560 = 10,560,000 kg

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2. Dec 11, 2014

### SteamKing

Staff Emeritus
What about the steel plate from which the barge is constructed? Is this a new, weightless steel?

3. Dec 11, 2014

### Mitchell Powell

I wasn't sure what to make of that. The density of steel is not given, and answers online range from 7,750 kg/m3 to 8,050 kg/m3. Is it significant if I go with one vs. the other?

4. Dec 11, 2014

### SteamKing

Staff Emeritus
When in doubt, you can take an average density of steel, or use the heaviest steel to cover your bases.

As to whether the weight of steel is significant, you have enough information to determine the weight of steel in the barge.

5. Dec 11, 2014

### Mitchell Powell

I see that I can determine the volume of steel used,

H = 12 m
W = 22 m
L = 40 m
T = .052 m

V = 2(H * W * T) + 2(H * L * T) + (L * W * T) = 123.136 m3

I can't see how you would determine the weight of steel with the information given, without using the density (7,900 kg/m3 average).

6. Dec 11, 2014

### Mitchell Powell

I was actually able to work it out. Seems the steel density is insignificant.

mbarge = psteel * V = 7,900 * 123.136 = 972774.4 kg

Since we are adding weight to the barge by adding coal, the weight of the barge should be mbarge + mcoal

As long as the density of the barge is less than the density of water (1000 kg/3), the barge floats. So, setting the density of water equal to the changing density of the barge gives:

pwater = (mbarge + mcoal)/V

Solving for mcoal = (pwater * mbarge) - mbarge

The mass of the coal comes out to roughly 9.6 * 106

Thank you for your help! I greatly appreciate it!