I'm trying to understand tension. So here it goes:

I'll start from the beginning.

Let's assume I'm in space and can move around and apply forces.

Let's say a rope is attached to a body(which is in space).

1) Let's say the body is immovable. Then the force with which the rope is pulled will be the tension in the rope. Right?

2) Okay, now let's say the body has a mass M and I apply a force F. The body will accelerate towards me with a acceleration F/M. Obviously, the rope will slacken, so to keep the tension in the rope constant, I will have to accelerate away from the body, maintaining my distance to still apply the force. (I just mentioned that because no one really talks about how one can apply a constant force without having to 'keep up' with the body to keep applying the force. I just wanted to make sure that that truly is the case.)

Now let's say, I pull the rope and then let it slacken i.e. not try to maintain a constant force, then if I was able to measure the acceleration of the body and if I know it's mass, I will be able to find the force I applied and hence, the tension (=force applied) for that period of time on the rope. Is that correct?

3) Okay. Now, let's say I pull the rope with X newtons of force and another person holds the body and pulls it in the opposite direction with Y newtons of force, shouldn't be the tension in the rope now be (X+Y) newtons, even if the body accelerates in one direction? If the person instead pushes the body with Y newtons then shouldn't be the tension (X-Y) newtons in the rope now?

4) Let's say the ends of the rope are attached to two bodies and the rope is currently slackened. I then give both the bodies some initial velocity in the opposite direction to each other. Hence, the rope at some point will become taut. Is it possible to tell the force, the two bodies will experience when the rope becomes taut and hence find the tension the string will experience?

Please correct me if I'm wrong in any of these. I would really appreciate the help.

Thank you.

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    $\begingroup$ Hi Alraxite, and welcome to Physics Stack Exchange! This is a site for conceptual questions about physics, not general homework help. If you can edit your question to ask about the specific physics concept that is giving you trouble, I'll be happy to reopen it. See our FAQ and homework policy for more information. $\endgroup$
    – David Z
    Commented Sep 11, 2012 at 18:46
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    $\begingroup$ I apologize. But this is seriously not a homework question. I was studying tension and after doing many pulley-related problems I was pretty confident that I got a good grasp of tension. Until this question came to my mind. And for some reason I just couldn't do it. To me, it's more of a conceptual question really, as this clearly proves how my true understanding on tension really is. If you can give a simple explanation, I would really appreciate it! I just made the numbers up in the problem! $\endgroup$
    – Alraxite
    Commented Sep 11, 2012 at 18:59
  • $\begingroup$ If you pull on the string with 10N how is the tension any different from 10N. Unless you want to consider the inertial effects on the rope as it accelerates... $\endgroup$ Commented Sep 11, 2012 at 19:03
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    $\begingroup$ @Alraxite well, the issue is that you've just given us a physics problem without any information about what it is you don't understand about it. This question is not conceptual, it's exactly the kind of thing that we call a "do-my-homework question," regardless of whether it is actually a homework problem or not. A good start would be to explain what you tried, and then hopefully you can identify the one specific issue that is confusing you about this question. Ask about that. $\endgroup$
    – David Z
    Commented Sep 11, 2012 at 19:07
  • $\begingroup$ @DavidZaslavsky Okay. Here's what I don't understand: If a block is attached to a block on a friction-less plane and I pull on it with a force of 10 N, then the tension in the string would be 10 N, right? Well, if another person pulls on the block from the other side, then certainly the tension would increase! And in this case this other person is friction. So there is where my confusion results from. $\endgroup$
    – Alraxite
    Commented Sep 11, 2012 at 19:25

2 Answers 2


Strictly speaking, tension is not the same as force, although it is sometimes described as the magnitude of the 'pulling force' experienced by an element (such as a rope).

The important thing to remember when resolving forces in classical mechanics and to understand tension is to apply Newton's three laws of motion. They are:

1st Law: an object with no external force will not change velocity

2nd Law: Force = Mass x Acceleration

3rd law: Every applied force (action) has an equal and opposite force (reaction).

So for the one dimensional cases you've given, think of the 'tension' of the rope as the magnitude of any pulling force it would be experiencing, bearing in mind that this tension is not actually a force (it has no direction), whereas the force whose magnitude it has, would be appear to be pulling the rope in opposite directions (as per Newton's 3rd law).

$T\leftarrow\rightarrow T$

So, back to your questions:

1 - When you pull on a rope tied to an immovable object, applying a force $F$, it reacts with force $-F$ (Newton's 3rd law) and the 'tension' in the rope is the magnitude of this force $F$.

$F\leftarrow\rightarrow F$

2 - If you pull on a rope which is tied a mass $M$ (initially at rest and free to move) it will accelerate towards you (Newton's second law). If you keep keep pulling the rope, keeping it taut by applying a constant force $F$ for a time $t$ and then remove the force thereby slackening the rope (no tension), the final velocity of the mass will be $v=at$ (neglecting friction). You can determine the force applied by $F=Mv/t$.

3 - If you apply a force of $X$ Newtons pulling a rope tied to a mass $M$ which I am holding, the tension on the rope is $X$ as long as the mass isn't moving. If I increase my pulling force to $Y$, the resultant force, $F=Y-X$ will pull you along with the mass, towards me. Note that we subtract the forces because they are acting in opposite directions. The resultant force $F$ will accelerate both you and the mass towards me at a rate $a=F/(M+m)$, where $m$ is your mass (assuming the mass of the rope is negligible). The tension on the rope will be equal to the magnitude of resultant force on the rope, which is $T =\lvert X-ma\rvert = \lvert X-m\times \frac{F}{M+m} \rvert= \lvert X-\frac{(Y-X)m}{M+m}\rvert$. Note that if your mass, $m$ is negligible, the tension of the rope becomes $X$, whereas if the mass of the body $M$ is negligible, the tension of the rope becomes $Y$. If your mass is equal to the mass of the body $(m=M)$ then the tension on the rope is $(Y-X)/2 = F/2$.

If I apply a pushing force $Y$ directly to the body of mass $M$, while you pull on the rope tied to it by applying a force $X$, the resultant force on the mass will be $F=X+Y$ (in your direction). The two forces are added not subtracted (since they are applied in the same direction towards you). The body will therefore accelerate in your direction (Newton's second law) under the total force $a=F/M$ and the tension on the rope will be equal to the magnitude of the resultant force, $(F-Y)=X$. Note in this instance, your mass is irrelevant, because the rope does not transmit my pushing force $Y$ to you (a rope does not work under compression!).

4 - If two bodies of mass $M$ are tied together with a rope and are moving in opposite directions at a speed $v$, they will each have momentum with magnitude $Mv$ but in opposite directions. Since neither mass is experiencing a force, they will continue to move at at constant velocities in opposite directions (Newton's 1st law), until the rope between them becomes taut. At that point, they will quickly decelerate and travel back towards each other. The rate of deceleration and subsequent speed at which they will travel towards each other will depend upon the 'elasticity' of the rope as well as the amount of 'friction' in the rope. In the case of an 'inextensible' rope with no friction, the rope will have a non-zero 'impulse' tension only at the instant it is taut. The two bodies will then move towards each other with the same velocity as they were previously moving away from each other (due to conservation of momentum).

Until you realise that tension is not the same as force, you may experience a little tension yourself as you grapple with the concept!

As an aside, you may come across some textbooks on engineering mechanics or materials which describe tension as a type of pressure or stress (force per unit area) as in 'tensile stress' applied to a truss member. If we define the area as a vector whose magnitude is the cross sectional area of the material under stress and whose direction is normal (perpendicular) to the cross sectional area, then the resulting force is the product of stress and area. In the most general sense, since the tension may have a different effect in different directions (anisotropic), the resulting force is not necessarily in the same direction as the area. In a three-dimensional Euclidean space, the tension is a tensor of rank 2. This is a linear transformation (mapping) with $3^{2}$ co-ordinates, something like a (3x3) matrix, which when 'multiplied' by the "area vector" produces the resultant "force vector" (not necessarily in the same direction).

However, since your examples are all dealing with forces in 1 dimension only, we can treat tension as a scalar (that is, a tensor of rank 0) whose magnitude is that of the force exerted by the rope under tension.

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    $\begingroup$ Thank you for answering! I posted this a while back when I had no understanding about tension, but now I do understand it to a certain extent and your answer made me understand it even more. Though, I do want to ask some questions: In the 4th point, you said that they will move towards each other due to momentum conservation, but momentum conservation shouldn't guarantee that they will move at the same speed as before because initially there momentum was zero and afterwards it could still be zero even if they are at rest, right? $\endgroup$
    – Alraxite
    Commented Nov 8, 2012 at 12:36
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    $\begingroup$ That's true, conservation of momentum is not enough, we must also consider conservation of energy! Specifically, the kinetic energy of the masses has to go somewhere! Assuming no friction on the masses, it becomes 'potential' energy in the rope as the rope 'stretches' slightly. Assuming no 'friction' in the rope, the rope is then under tension, $T=m\bigtriangleup V / \bigtriangleup t$, which means each mass experience a 'reaction' force (inward) accelerating them towards each other with equal but opposite momentum. $\endgroup$
    – theo
    Commented Nov 8, 2012 at 20:21
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    $\begingroup$ I do have some doubts on the third point, though I will accept your answer since you cleared all the other points for me. Okay now my doubts: You are holding a block and I'm holding on a rope attached to that block. I pull with force X. Let's say you simply keep holding that block. Then you and the block will accelerate under my force X. So what will happen if you increase your force from 0 to X exactly? Will I accelerate? Will the tension change? I have asked similar questions on this topic before on this site but this part still confuses me. $\endgroup$
    – Alraxite
    Commented Nov 16, 2012 at 21:18
  • $\begingroup$ @theo , In point 3, you mentioned that tension is equal to the magnitude of resultant force on the rope which is |x-ma| I didn't quite get that because a rope that is accelerating at a has a net force ma acting on it right? Why should it be anything else? $\endgroup$ Commented Mar 22, 2016 at 12:54

Well, since the resisting force is the same as the force applied in pulling the rope, then then the body will be in equilibrium. I.e acceleration =0... In the case of increasing force from 0 to X, there will be acceleration until the resisting force reaches the magnitude XN used in pulling the rope, which halts the motion of the body

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    $\begingroup$ It seems you're only answering one portion of the question, could you amend your answer to include details on the other parts of the question? $\endgroup$
    – Kyle Kanos
    Commented Nov 17, 2015 at 18:39

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