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This was a question on a mechanics exam. Part i wants me to assume that the tension in both parts of the string is the same. Even though I got the correct answer( 18.9N) by assuming so, I don't understand how this assumption makes sense. Consider the bit of string under the ring. The net force on it must be zero ( light string ) so that implies the net vertical component is zero. This is not possible if I assume the tensions to be the same, or at least that's how I see it ( see image)

Basically in my mind: If the tensions are equal, the net force on the bit in contact with the ring can not be zero since the angles are different. But since the string is light, this does not add up. ( Or the system is at rest. Net force on that bit of rope must be zero, regardless of light/ not light)

But the question assumes the opposite. I would like to know what part of my argument here is flawed.

  • $\begingroup$ What am I missing ? Nothing. Now draw a free body/force diagram to prove that... $\endgroup$ – Gert Mar 27 '19 at 15:40
  • $\begingroup$ @Gert, please see the edited question. $\endgroup$ – Sal_99 Mar 27 '19 at 15:58
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    $\begingroup$ Even using using a free body diagram I am able to get only two equations(equilibrium in the horizontal and verical directions)for a question with 3 variables.Something is issing ... unless the answer is in terms of X. I think. $\endgroup$ – A.R.K Mar 27 '19 at 16:21
  • $\begingroup$ Tension in a massless string has to be same so T1=T2 $\endgroup$ – Tojrah Mar 27 '19 at 16:44
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    $\begingroup$ Regardless of whether this is on-topic or not (I'm inclined to think it is), the homework-and-exercises tag is appropriate, since the question is about a physics exercise. $\endgroup$ – Chris Mar 28 '19 at 20:31

First answering to your doubt of tension being same in both parts of string.
Young's modulus of a material is constant irrespective of shape length etc. (at least from the point of view of solving high school questions ) , you can study about it a bit on the internet if you don't know it already. $$Y={{F\over A}\over {\Delta L\over L}} $$where $Y$ is the Young's modulus of elasticity of your string.From here $$F={YA\over L}(\Delta L)$$As you can see from this equation, the force or the tension for a realistic string depends on its extension and the constants beside it.Moreover the Force equation depends inversely on the actual length of string too (see L in the denominator ), so unless you cut the string where the ring is and then attach both parts above and below the ring , the force equations are not going to change and the string as a continuous entity will have same tension everywhere , because $\Delta L$ here accounts for whole string.

The purpose of invoking elasticity in the picture is to explain the tension aspect in near ideal conditions , there is nothing like a mass less string but the above explanation is a close analysis.

Coming to the actual question (tension is now same throughout the string).
HORIZONTAL EQUILIBRIUM $$T(cos50+cos20)=X$$ VERTICAL EQUILIBRIUM $$T(sin50)=T(sin20)+(0.8)g$$

Upon solving these two , $T=18.66 N$ and $X=29.29 N$ approx.

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  • $\begingroup$ The issue with the tension idea is that since the string is light , net force on any bit must be zero ( net F= ma) . Let's suppose the tension is the same in this set up. Let's consider the bit of string under the ring. If both tensions pulling on this bit are the same then , due to different angles, the net vertical component on this bit isn't zero , and so net force isn't zero. I completely agree with your discussion on elastic properties but it is this bit I can not get across. True, this analysis is based on a massless string - a theoretical idea- but the question gives me permission $\endgroup$ – Sal_99 Mar 27 '19 at 18:46
  • $\begingroup$ @Sal_99 You missed a force , the weight of ring , the upward pulling tension balances weight of ring + downward pulling tension $\endgroup$ – Aditya Prakash Mar 27 '19 at 18:51
  • $\begingroup$ @Sal_99 You do want a non-zero net force on the part of the ring where the string threads through. If this was not the case then the weight of the ring and $X$ would not be able to be canceled out. $\endgroup$ – Aaron Stevens Mar 27 '19 at 19:17
  • $\begingroup$ @Sal_99 I'm trying to understand your confusion as well. Could it have something to do with external vs internal forces. Clearly the sum of the external forces (the fixed points A and B and whatever is applying the force X) has to be zero for this system (strings plus ring) to be in equilibrium. But tension is an internal force in the system, like the tension in a spring. If you cut section RA and tied the ends to a spring scale you would get the same scale reading if you cut section RB and put the scale there. $\endgroup$ – Bob D Mar 27 '19 at 19:18
  • $\begingroup$ @BobD Based on the comment "How is the weight of the ring acting on the bit of string?", it seems like the OP is under the impression that if we just focus (zoom in maybe?) on the part of the ring where the string is threaded through then you no longer have to consider the weight of the ring. This, of course, is not the case. $\endgroup$ – Aaron Stevens Mar 27 '19 at 19:20

First, since the ring doesn't involve any corners, sharp angles, etc. we can safely assume the tension is uniform throughout the massless string that is threaded through the ring.

Assuming that the vertical components of tension is equal to the vertical force the ring experiences due to each string, then $X$ must take a specific value consistent with Newton's laws in the horizontal direction. There is no issue in having a constant tension throughout the string.

Indeed, if you do the work for general angles (you can do the work yourself) you will find $$X=\left(\frac{\cos\theta_A+\cos\theta_B}{\sin\theta_A-\sin\theta_B}\right)mg$$ If we were to change the applied force $X$, then this would just result in different angles. Although notice that, for angles less between $0^\circ$ and $90^\circ$, we cannot have $\sin\theta_A<\sin\theta_B$ or else $X<0$. In other words, as we pull harder and harder ($X$ gets larger and larger) the angles will both approach $45^\circ$.

  • $\begingroup$ About point 1, what I don't I get is why. Yes a massless string means net force on any bit must be 0. If we set tension equal in both bits of strings above and below the ring, the net force on the bit under the ring isn't 0 since the angles are different. There are cases where tension in a massless string isn't the same where angles change, as my teacher points out. $\endgroup$ – Sal_99 Mar 27 '19 at 18:30
  • $\begingroup$ @Sal_99 What was the book answer for the tension? $\endgroup$ – Bob D Mar 27 '19 at 18:30
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    $\begingroup$ @Aaron Stevens , the ring is <i>smooth</i> , so no friction to support the ring vertically $\endgroup$ – Aditya Prakash Mar 27 '19 at 18:38
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    $\begingroup$ @Sal_99 Yes, that is correct. The net forces due to the string from A to B will have components upward and to the left, which are then canceled out by the weight of the ring and X when you consider the system as a whole. $\endgroup$ – Aaron Stevens Mar 27 '19 at 19:39
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    $\begingroup$ Yes. You've understood my confusion now. We accept that the net force on the bit of string ALONE isn't zero. Now this bit of string has no mass ( light) so it not possible for it to have a net force on it. And that is my concern $\endgroup$ – Sal_99 Mar 27 '19 at 19:41

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