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...
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 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?
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...