Capacity of a Barge: 10.56 Million kg Coal

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Homework Help Overview

The problem involves determining the mass of coal that an open barge can carry in freshwater without sinking, given the dimensions of the barge and the thickness of the steel used in its construction. The subject area includes principles of buoyancy and density calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the volume of the barge and the implications of the steel's weight on its overall capacity. Questions arise regarding the density of steel and its significance in the calculations.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of average density for steel, and there is acknowledgment of the weight of the steel in relation to the coal's mass.

Contextual Notes

There is uncertainty regarding the density of steel, as participants note varying values found online. Additionally, the problem does not provide specific information on the density of the steel used, which influences the calculations.

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Homework Statement


An open barge has the dimensions shown in the figure.
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/59/59574-46462a33e8dd3e05fa12094d99749d9c.jpg
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?

Homework Equations


p = m/V

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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What about the steel plate from which the barge is constructed? Is this a new, weightless steel?
 
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?
 
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.
 
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).
 
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!
 

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