Capacity of a Barge: 10.56 Million kg Coal

  • Thread starter Thread starter Mitchell Powell
  • Start date Start date
  • Tags Tags
    Capacity Density
AI Thread Summary
The discussion centers on calculating the coal-carrying capacity of a barge made from steel, considering its dimensions and the weight of the steel itself. The volume of the barge is determined to be 10,560 m³, and the average density of steel is used to find the weight of the barge, which is approximately 972,774.4 kg. The key equation relates the weight of the barge and the coal to the density of water, leading to a calculated coal mass of roughly 9.6 million kg. The significance of the steel's weight is debated, with the conclusion that it has a minimal impact on the overall capacity. The final result confirms the barge can carry a substantial amount of coal without sinking.
Mitchell Powell
Messages
4
Reaction score
0

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
 

Attachments

  • YF-14-38.jpg
    YF-14-38.jpg
    7 KB · Views: 1,108
Last edited by a moderator:
Physics news on Phys.org
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!
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top