# Two strings in pulleys supporting a mass [closed]

I am following a free online course on physics 101, where I found the problem posed below. The assumptions are that the system is static and that the two strings and three pulleys are ideal. The long string is tied to the floor on the left. The string tensions are labelled in different positions, and the task is to find expressions for all tensions $$T_i$$ in terms of the mass $$M$$ and the acceleration due to gravity $$g$$.

My immediate thoughts were: Firstly, since the tension is constant in a string, we have $$T_1=T_2=T_3=T_4$$. Secondly, since $$T_4$$ and $$T_5$$ jointly lift the mass $$M$$, we have $$T_4=T_5=\frac{Mg}{2}$$. Then I started wondering, and I have these two questions:

1. It looks like the long string is actually supporting the pulley to which the short string is attached. Does this affect the solution?
2. The problem makes no statement about where exactly the strings are attached to the mass $$M$$. Surely this will affect the solution as well? I am thinking that if one string is attached in the center of the mass, and the other at the corner, then they will not lift the mass equally. If this is true, how do you determine the tensions in the strings depending on where they are attached to the mass? I've attached an example with three strings (in this case, the mass might actually be slanted because of the asymmetry?). • I don't think I can rephrase my question in a way that doesn't ask about specific computations, since I am interested in how to derive the relation between $T_5$ and $T_4$. Is it custom here to delete closed questions, if they cannot be edited to be on-topic, or should I leave it? Jul 4, 2021 at 15:39

As @gandalf61 says, it does not look like forces $$T_4$$ and $$T_5$$ are equal. For simplicity assume that the $$T_2$$ and $$T_3$$ strings are vertical (and by the ideal assumption $$T_2=T_3=T_4$$) then $$T_5=2T_4$$. This (from the drawing) means a torque on $$M$$, and so it is not in equilibrium. Indeed it is clear from the drawing that by letting the $$T_4$$ bit of string to get longer at the cost of shortening the $$T_{2,3}$$ strings, the left bit of the $$M$$ will rise only half as far as the right bit will fall, so centre of mass of $$M$$ will be lowered. We are not in a local minumum of potential energy therefore, and the system will move.

The assumption of being static is false.

I assume that this is just an error on the part of the person who set the problem.

• "... so centre of mass of M will be lowered" - not necessarily - equilibrium is possible if attachment points of $T_4$ and $T_5$ are such that CoM is closer to $T_5$ than to $T_4$. An equivalent condition is that the torques of $T_4$ and $T_5$ about the CoM are equal and opposite. Jul 4, 2021 at 14:44
• Thanks. How do you come about that $T_5 = 2T_4$? Is it because the strings at $T_2$ and $T_3$ are sharing the load of carrying the string plus mass at $T_5$? (i.e. $T_2+T_3 = T_5$, and since $T_2=T_3$, we have $2T_2 = 2T_4 = T_5$. Jul 4, 2021 at 15:06
• @gandalf61. True if the attachment points are carefully chosen, but the drawing and the absence of such a statement suggests carelessness on behalf of the P101 course people. Of course we have not seen the original problem only the drawing. Jul 4, 2021 at 15:11
• @mikestone Agreed. There is important information missing from the problem as stated. Jul 4, 2021 at 15:15

Your assumption that $$T_4=T_5$$ is incorrect. Consider the forces on the middle pulley ...

• Finding a solution requires that you assume that $T_4$ and $T_5$ are at equal distances from the center of M, and that they exert equal torques. Jul 4, 2021 at 13:40
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review Jul 4, 2021 at 13:59
• @VincentThacker I intentionally avoided giving a fuller answer since "answers to homework questions with complete solutions may be deleted". Looks like Catch 22 is at work here. Jul 4, 2021 at 14:38
• @R.W.Bird $T_4$ and $T_5$ must exert equal and opposite torques about the CoM of $M$, but that does not mean that the attachment points are at equal distances from the CoM. Jul 4, 2021 at 14:49
• Without additional data this problem can only be solved if you assume symmetry as suggested by the sketch. If the distances are equal ,then the torques must be equal. Jul 4, 2021 at 18:33