This is the diagram given in a question which I am solving now.Just to mention this isn't a homework question because I think I have actually got a doubt.Normally I use virtual work method to get some constraint relations , but I'm stuck here because that is a single stretch of string looped around,and given that all pulleys and the string are ideal the pulley Q cannot have tensions balanced on both sides equally (it being massless)but that should be necessary (as far as i think)
So am I wrong somewhere or is this system physically not possible?

  • $\begingroup$ I have done further analysis and updated my answer. It appears the problem does not assume "all pulleys and the string are ideal". This should put it to rest. $\endgroup$ – Bob D Nov 17 '19 at 15:58
  • $\begingroup$ This problem makes no sense. First of all, whats the question? Then, why is the heavier weight accelerating upward? Or is it? $\endgroup$ – R.W. Bird Nov 17 '19 at 19:56

Since $F_{Net}=ma$, if $m=0$ for pulley Q, then you can only have non zero acceleration of the pulley if $F_{Net}=0$ also.

If we can assume the string is ideal (massless and inextensible), and there is no friction between the string and the pulleys, the tension, T, throughout the string must be the same. (I said assume because the complete problem description did not specifically say so). That, however, would mean that $F_{Net}$ on pulley Q is not zero, since you have 2T acting downward and 1T acting upwards.

So unless I, too, am missing something, it seems there is an inconsistency and you are correct that the system does not appear to be physically possible. But I would wait to hear from others in case you and I have indeed missed something.


I reviewed the details of the problem. Given the upward acceleration of 3 m/s$^2$ on mass B you can calculate the tension in the string supporting mass B in the usual way. If you then calculate the tension in the string supporting mass A based on the answer to the problem which states the net force on mass A is 10.5 N downwards, you will get a different tension for the string supporting A.

Bottom line: the tension is not constant throughout the string. That means we cannot assume the string are ideal, that the pulleys are massless and/or that there is no friction between the string and pulleys.

One final comment. Their solution to the problem appears to involve constraints regarding the string, in order to determine the acceleration of mass A and from that the net force on A. I was not able to follow their analysis since they did not define the terms in their equation.

Hope this helps.

  • $\begingroup$ The actual question never asked for tension in the string , it only asked the velocity of a block (in don't remember which one).So the question boils down to constraint relation between the velocities of the three pulleys which can be easily derived by the 'point method' and thus never needing to write down tensions.In the constraint equation involving the 3 velocities , 2 were mentioned and hence 3rd was easy to find.The problem with me was that I don't prefer the point method because it's lengthy and virtual work method required the tensions. $\endgroup$ – Aditya Prakash Nov 17 '19 at 16:05
  • $\begingroup$ Its curious that two valid methods cannot be used interchangeably to solve the same question , is this discrepancy of any significance ? I think the virtual work method is a better one because its always true , even in real world.How does it fail here ? $\endgroup$ – Aditya Prakash Nov 17 '19 at 16:10
  • $\begingroup$ @ADITYAPRAKASH Actually the original question asked "Figure shows a system of four pulleys with two masses mA=3kg and mB=4kg. At an instant force acting on block A if block B is going up at an acceleration of 3m/s2 and pulley Q is going down at an acceleration of 1m/s2 is" So they were asking for the net force on A, which they get in their solution by determining the acceleration of mass A and multiplying it by the mass. I approached it from the tension perspective only to see if they were assuming all the ideal conditions you and I assumed. It turned out they weren't. $\endgroup$ – Bob D Nov 17 '19 at 16:16

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