Elevator's Maximum Capacity

I think I understand the answer now. Think if it like a balance. If the motor were absent from the system, and the counterweight was 500kg, the elevator wouldn't move. Since the counterweight is 2500kg, there's 200kg more on that side, which would pull the elevator up and the counterweight down. That means the motor must be able to pull that 2000kg in order to move the elevator. Now if we add 2000kg to the elevator, the system is balanced, and without the motor, nothing would move. A tiny amount of force on either side would cause things to move up or down. Since we know the motor is at least capable of pulling that 200kg we figured out above, that means the elevator can be 2000kg more than the counterweight, or 4500kg. Therefore we can add at least 4000kg to the elevator for sure, possibly more, but that's all we know for sure.

Ye. Still not sure of that answer

I'm not an engineer, but that makes no sense at all. How does making the carriage weigh 2000 kg more than the cw mean that that the motor won't break just because it's "symmetric"?

Think of it this way, if we added 2000kg to the carriage they would be equal in weight and you would be able to pull the carriage up and down by hand with little effort. (I could be totally wrong as I'm an accountant not an engineer by any means but this is how I arrived at 4000kg as my answer, as that would, as Tim stated, make the wuarion equal.)

4000 kg can be added to elevator, but to carriage or counter weight? I think the correct answer is 2000 Kg only. Otherwise, the term 'Counter' weight is meaningless.

The problem explanation is actually confusing. I interpreted the motor to have a max capacity of 2500 kg, as in equal strength of the latter mentioned counterweight. In that case you can add 4500 kg ( - 2500 kg countered by the counterweight = 2000 kg plus carriage weight = motor strength).

We must assume that both the yield point and or the breaking strain of the cable is less than 4500 kg. Is the cable a single continuous one or are there two separate cables?

The question is wrong (or the solution?). It should have said minimum weight, not maximum. How can we know that the motor is not strong enough to move 100 tons without even a counterweight?

It's as if, seeing they can be suspended symmetrically while one side is 2000 kg heavier than the other (as pictured), we can assume the reciprocal must be true, because the motor is apparently functioning at symmetry as it is "pulling equal weight" from either direction, meaning in could be functioning with the 2000 kg discrepancy existing on the other side (i.e. this being the "work" or functional ability of the motor in and of itself). So because the question asks us how much more can be added to the lighter side of the diagram, without "breaking the motor" by having too much weight on one side in comparison to the other, we know we can add 2000 kg to the opposing side than in question (2500 kg + 2000 kg = 4500 kg) to get a total functioning weight of the motor at an increased, but still, reciprocal ratio of work exerted. Then the already present 500 kg from the side in question must be subtracted out of the total weight of 4500 kg to algebraicly arrive at the amount that could be "added" to the problem's stated givens, which leaves us at the result of 4000 kg. So the result ends in 2500 kg on the counterweight side and 4500 kg on the carriage side, with the 2000 kg discrepancy being performed by the motor as before, which we are "sure" it can handle having the same strength on both sides.