Buoyant Force Problem: Equilibrium and Submerged Objects Explained

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Homework Statement


If two objects are suspended from strings on opposite side of a beam with a fulcrum halfway between and has reached equilibrium, if the entire system is submerged underwater will equilibrium be disturbed?
The volumes are identical. The mass's are not equal but it is in equilibrium because one is closer to the center of mass than the other.


The Attempt at a Solution


I said it won't be because each side will have an equal buoyant force due to identical volumes so the equilibrium will not be upset. However chegg.com says equilibrium will be upset.


originally
m1gL1 = m2gL2
m1L1 = m2L2
m1/m2 = L2/L1

Then it says something I'm having a hard time making sense of.
It says let the mass of reduction of each mass by c1 and c2 where c1≠c2
(m1 - c1)/(m2 - c2) ≠ m1/m2
therefore equilibrium will be disturbed.

I don't get it. c1 and c2 is the mass each loses? mass reduction wutt?? wouldn't the mass of reduction be the same since they displace and equal volume of water?
 
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PsychonautQQ said:
it is in equilibrium because one is closer to the center of mass than the other

I said it won't be because each side will have an equal buoyant force due to identical volumes so the equilibrium will not be upset.

Think.
 
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PsychonautQQ said:
It says let the mass of reduction of each mass by c1 and c2 where c1≠c2
(m1 - c1)/(m2 - c2) ≠ m1/m2
therefore equilibrium will be disturbed.
I don't like calling it mass reduction. Weight reduction perhaps, but I prefer to think of it as adding a buoyant force to each. Now, are you sure it says c1 ≠ c2? From the rest of the question, I would say you have c1 = c2 but m1 ≠ m2. Does it make sense to you then? ( It's also strange that at first you say the fulcrum is half way, but then later it clearly is not.)
 
Equilibrium doesn't just depend on force...:wink: