Question: Two pulleys of masses 12 kg and 8 kg are connected by a fine string hanging over a fixed pulley as shown. Over the 8 kg pulley is hung a fine string with masses 4 kg and M. Over the 12 kg pulley is hung another fine string with masses 3 kg and 6 kg. Calculate M so that the string over the fixed pulley remains stationary. Problem: The question places no demands on how long the setup must remain stationary. In time, it will snap out of equilibrium, i.e., when the heavier of the two masses in the two two-mass-string sub-system falls to one side, the 12 kg and 8 kg masses are all that will remain. And then that two will collapse to one side. So the question demands me to comment on the very brief period of time for which the system will be in equilibrium. This lack of explicit time-bound requirement on the solution discouraged me to proceed with the solution for a very long time. Can we all agree that the question could be clearer in its wordings? Or is the information sufficient for some of you? Would appreciate all thoughts. Best regards, wirefree P.S. I don't seek the solution to the question.
Yes i guess it is as you saying, but how "brief" that period can be, depends. For example if the length of the strings on each side of the sub system pulleys is too big that period may not be so brief afterall :). Of course one could say that in real life those strings cant be too big , but in theory we can make the strings as big in length as we want.
Actually, I don't think the wording is too sparse. This assumes that the question is aimed at people who are expected to be able to answer it. It is like the ideal circuit diagram, given in an electronics problem. The time involved in the answer is not really relevant, imo; you are asked to consider what will happen before any limit is reached, given the data, which is fair enough and will give an answer. As far as I can see, the only thing that is missing is the statement about the Moment of Inertia of the pulleys. I think a finite MI could affect the situation because some torque is required to get the pulleys to move and that must (?) affect the tensions in the strings. Giving the pulleys a mass but no MI seems strange. Perhaps someone will be able to put me right on that.
If the two hanging pulleys are treated as mass points then the stationary condition is met when the left weighs the same as the right; this occurs when M=9 kg. It doesn't matter where the weights on two pulleys end up, as long as they weigh the same. As sophiecentaur points out, there is insufficient information for balancing of torques.
I do not agree with M = 9. That assumed there are different tensions in the pulleys with unequal weights. But the tension must be the same on both sides of a pulley. There is less force needed to support the pulleys than you'd think because the heavier of the two weights is actually falling. It is not equilibrium, as the left hand pulley tells you.
If we dont assume a massless rope, the tension cant be the same everywhere cause there must be net force to accelerate the rope. If we assume massless rope i *think* there is a problem because the way i see it, the weights on each side of the rope in the sub pulley exert a force on the rope and the rope exerts a force on the pulley. Because the rope is in accelerating motion these two forces need not be the same.
The string is "fine" (=massless) in the OP. I have assumed the pulleys have no MI - but I will agree that the model is a bit selective. If you can't assume the above then there is no solution. Tension can be the same as long as the heavy mass is pulling the lighter mass up (and vice versa) the force is the difference in weights and the accelerated mass is their total mass.
Assuming massless rope and pulleys with no MI, i ve calculated the tension in the rope of the sub pulleys as [tex]T=\frac{2m_1m_2}{m_1+m_2}g[/tex]. Is this correct or not? Can we say that the phenomenal weight the sub pulley exerts to the main pulley is (sub pulley's weight)+2T?
I reckon that's the right starting point. Then you assume the top pulley remains stationary 'till one side or the other hits an end stop.