A barge is in a rectangular lock on a freshwater river. The lock has length 58.0
ID: 1479149 • Letter: A
Question
A barge is in a rectangular lock on a freshwater river. The lock has length 58.0 m and width 24.0 m , and the steel doors on each end are closed. With the barge floating in the lock, a load of scrap metal weighing 2.15×106 N is put onto the barge. The metal has density 9000 kg/m3
When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise?
The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same?
If it rises or falls, by what vertical distance does it change? If it remains the same, enter 0.
Explanation / Answer
a.
the vertical distance is calculated as followsL
y = w/[water density]gA = [2.15×106] / [1000][9.8][58*24] = 0.1576 m
b.
the water level increase will fall.
the vertical distance is,
y' = w/[metal density]gA = w/[9*waterdensity]gA = y/9 =
hence, the required vertical distance is,
y' = y - y/9 = 8y/9 = 0.14 m
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.