QUOTE (noob @ Oct 09 2006, 03:53 PM)
It is hard to read this picture and the first one was un-readable!
====================
'''''''''''''''''''''''''''''|
'''''''A-----------------------G
'''''''''''''''''''''''''''''''''''''''''''''''''|
''''''''''''''''''''''H-------------------------B
'''''''''''''''''''''''|
'''''''''''D-----------C
I assume all horizontal strings are of equal length and attached to the centroid (center of mass) of every fish.
The sum of moments on the A-G rod needs to be 0.
The moment generated by A is Ma = Fa * 120 = 3.4*120=408 N*mm.
So the moment at G needs to be Mg = -408 N*mm in order to get Ma + Mg = 0
Just calculate the sum of moments at G to get the weigth of D.
Mg = -408 = Mg - Mb = (Md - Mc) - Mb = ( [Fd*70] - Mc ) - Mb = ...
Calculation of sum of moments is easy and gets you out of nearly any static equilibrium problem. Just make to use the right units. For instance, here notice I used mm everywhere. So the moments are in M*mm. Be carefull not to mix meters in there.
Believe it or not this was fun to do and allowed me to put my mind to something else than the maths I am doing right now.