- #1
ttb90
- 16
- 0
Homework Statement
A large barge of overall width 4m to be used at sea has a uniform empty weight per meter length of M=6327.19kg/m (length is into the page). This empty weight is uniformly distributed along the length of the barge. The center of gravity of the empty barge is at position G, 2m above the flat bottom of the vessel. The barge has a rectangular water-line area and the density of seawater is 1025 kg/m^3. Neglect the thickness of the hull.
Ballast of mass Mb kg per meter length is added to the barge in order to effect the stability. The weight of the ballast acts at point A (on the barge centerline at the bottom of the hull) and is uniformly distributed along the length of the barge. The addition of the ballast will shift the center of gravity from G to G (prime). Write out an expression for the height of the new center of gravity above the bottom of the hull in terms of Mb (distance AG(prime))
Homework Equations
I know that the key word in the question is 'acting on point A'. Adding Mb will shift G to G(prime) however G(prime) is lower than G. Because the weight was added below and not above.
I am aware that i need to use moments to find G (prime) but am not sure how to start it off..Really stuck
The Attempt at a Solution
I have found the draft value of h=1.5432m.
m of barge= 6327.19 kg
length (l) of barge =1m
width= 4m
ρ= 1025kg/m^3
For equilibrium, buoyancy force= 6327.19g
volume of displaced seawater V=wxhxl= 4xhx1=4h
If we equate the two equations above we get h= 1.5432m..
This value is calculated for G not sure if it stays the same when we have G (prime)??