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As per Newton's 3rd law

every action has an equal and opposite reaction, but both of these forces are on different bodies under consideration

Now when a single body (say rope in this case) is stretched by equal forces on the either ends of the rope, why is the tension the same as that of the applied force? (considering applied force doesn't exceed the maximum tension which that rope can hold).

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Your question has nothing to do with Newton's third law.

You are having trouble identifying forces. That's not unusual. It takes some time and practice. A good book would help. Evidently your book is not very good at all, because the sentence you quote is practically unintelligible. Please don't waste your energy trying to make sense of it.

Wikipedia has fairly good statements of Newton's laws, although I think the statement of the second law, while correct, could be even better.

First you have to define your system. You have implicitly chosen your system to be the rope. Then you have to find the forces on the system. In Newtonian mechanics, a force is either a non-contact force (gravity, electrostatic, magnetic) or a contact force (everything else: normal forces, tension, the push of your finger, friction...).

In your question, we are implicitly neglecting all of the non-contact forces. That makes things a little easier. We are looking for contact forces. These only occur when the system is in contact with something it its environment. Here, they occur where the forces are applied to the rope, say, by your hand pulling on it. So there are two forces on the rope, one on each end.

A force that is a "pull" is usually called a tension force. So you have two tension forces, and they happen to be applied in opposite directions. Now, the rope is at rest, is it not? It's acceleration is zero. Newton's second law tells us that the net force, the algebraic sum of the two forces, is zero. Thus, the forces must be of equal magnitude and opposite direction. This has nothing to do with Newton's third law, although it shares some of the words used in the statement of Newton's third law.

Note that if the rope were accelerating, then the two tension forces would be in opposite directions, but unequal magnitude.

So here's what we have: two forces on the rope. Each one is applying a tension force on the rope. Those two forces are equal in magnitude, and opposite in sense (direction).

The only introductory book that I've seen that spells all of this out clearly is out of print.

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  • $\begingroup$ okay so what you are saying is tension is the pulling force itself in magnitude, but what about its direction? $\endgroup$ – agha rehan abbas Sep 6 '14 at 18:24
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    $\begingroup$ Your left hand pulls to the left, your right hand pulls to the right. Don't think "tension in the rope". Think "tension on the rope". $\endgroup$ – garyp Sep 6 '14 at 19:09
  • $\begingroup$ As a system, the rope itself must be considered to be the entire rope. You can't ask questions about what happens inside the rope without dividing it up into subsystems. To talk about tension at a point in the length of the rope, say, at the midpoint, you have to define a new system: half the rope. Then, the tension at the boundary of that system (that is, the molecule just inside the boundary of the new system) is provided by the molecules of the other rope segment, the segement that is now outside of your system. $\endgroup$ – garyp Sep 6 '14 at 19:14
  • $\begingroup$ Just for my interest: which book is that? Is it any good as teaching material for introductory physics? $\endgroup$ – CuriousOne Sep 6 '14 at 20:03
  • $\begingroup$ The preliminary edition of Randall Knight's book. The preliminary edition was called "Physics: A Contemporary Approach". I may be unfair: I'm remembering his treatment of energy, which took a step backwards in the 1/e. The section on Forces may be ok. (My copy of 1/e has been lost.) He has a small paperback for instructors called "Five Easy Lessons" that contains condensed versions of his approach. However, I think it would serve as a great resource for students; it gives very succinct statements of physical principles without the distraction of examples, and sidebars, and colors. $\endgroup$ – garyp Sep 6 '14 at 21:29
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Think of the whole rope as consisting of two ropes, rope A and rope B, that are connected along an imaginary boundary. Rope A and rope B are connected by molecular bonds that twist and stretch a bit when you pull on the outer ends of rope A and rope B, increasing the potential energy of the molecular bonds along their connection. Because of the change in potential energy, there is a restoring force on the connected sides of rope A and rope B, which tends to draw the molecules along the connection back to their original configuration, where the potential energy is at a minimum. By Newton's third law, the force that pulls rope A toward rope B along the boundary is the same magnitude as the force that pulls rope B toward rope A along the boundary. And since the ropes aren't moving, the total force on rope A must be zero, so the force that's pulling rope A toward rope B along the boundary must be the same magnitude as the force with which you're pulling on the outer end of rope A.

Answer to your follow-up question:

Force is a vector quantity. The net force on rope A in my example is zero, but that doesn't mean that each of the two forces on rope A is zero, merely that the vector sum of the two forces on rope A is zero. The outward force on rope A due to your hand pulling on the outer end of rope A is the same magnitude, but in the opposite direction, as the inward force on rope A due to molecular bonds that pulls rope A toward rope B along the boundary. When you have two vectors that are equal in magnitude but point in opposite directions, their vector sum is zero, even though the sum of the magnitudes of the two vectors isn't zero.

Addendum:

Or are you talking about how the net force on the boundary between the two ropes is zero (the vector sum of the pair of forces to which Newton's third law can appropriately be applied)? The answer is similar: The force that pulls rope A toward rope B along the boundary is equal in magnitude but opposite in direction to the force that pulls rope B toward rope A, so the vector sum of those two forces is zero.

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  • $\begingroup$ yes, but my question was how tension developed when net force comes out to be zero $\endgroup$ – agha rehan abbas Sep 6 '14 at 18:25

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