# Infinite loop of forces? [duplicate]

I have a question regarding tension in a rope. Consider a situation where two masses $A$ and $B$ are tied by a massless string on a horizontal plane with no friction.

Suppose we pull mass $A$ by a force $F$ then the string exerts a backward force on the block; assume $T$. An equal amount of force is applied by the block on the string, which pulls the Block $B$ with force $T$. The Block $B$ exerts a force backward on the string with magnitude $T$.

So far the argument seems good, but why do we stop here? This backward force again pulls $A$ backward by the string. The string is again pulled forward by $A$ by Newton's Third Law, which again results in the same process...Thus there are forces acting in infinite directionsbetween the boxes..?

Where is the problem in the argument?

## marked as duplicate by sammy gerbil, Community♦Feb 13 '18 at 15:49

• Why B exerts a force backward? – ndrearu Feb 12 '18 at 15:28
• Due to Newton's third law...string applies force on block b and Block B applies force on string – user35508 Feb 12 '18 at 15:31
• Please don't make edits that do not change the content of the post (such as adding a second "?" where it is not needed). It bumps the question for activity; but that shouldn't happen unless there is an actual change to the question. – JMac Feb 12 '18 at 15:40
• Please do not edit question simply to bump them on the active queue. – dmckee Feb 12 '18 at 15:54
• Listen, every single person who posts here want their question answered. Yours is no different in that regard. I have an answer. I've even typed it up. But I'm going to click "discard" in the editor and move on because you've been needling the site as a whole and both people who've commented here in particular. – dmckee Feb 12 '18 at 15:59

1) A is pulled by force F

2) A pulls the the string with force T

3) String pulls A with force T

4) String pulls B with Force T

5) B pulls string with Force T

That's it. Newton's third law is in 2) & 3) and in 4) & 5), they cancel out so it does not continue any further.

• But when the string is pulled by B in 5) Doesn't it affect block A – user35508 Feb 12 '18 at 19:09
• 5) and 4) cancel out, so A doesn't feel that. – Daniel Feb 12 '18 at 19:20
• Newton's law is that every Force has a reaction, the reactive force, it stops there. You want the reactive force to also have a reactive force. – Daniel Feb 12 '18 at 19:28
• How do they cancel out when they act on different bodies...Are you perhaps considering Box B and string as the system to cancel internal forces? I am just asking if the reactive force has an effect on block A which would repeat the cycle – user35508 Feb 13 '18 at 5:01
• I see the flaw. When A pulls the string, the string doesnt pull back right away. the string transmits the force and will pull B. then there is newtons third law and b pulls the string. then there is the tension in the string that pulls A. – Daniel Feb 13 '18 at 5:31

I've always felt it helpful in example type problems to imagine strings as force redirection/transmission tools. When you pull on one end of a string with force, the force is transmitted to the other end of the string. Action at a distance, so to speak.

Now, let's put numbers to our scene as an example.

• A is a block with a mass of 5 kg.
• B is a block with a mass of 4 kg.
• S is a massless string connecting A and B.
• F is a force of 20 N.

F pushes on A, yielding an acceleration of 4$\frac{m}{s^2}$. Superfluous, but interesting nonetheless.

A is connected to SB (String and block.) With a little clever change of reference frame, it could be viewed as either A accelerating away from SB, or SB accelerating away from A. This is simply a reworking of Newton's 3rd law, but it's a concept I've always found useful.

If A pulls on SB with force T, then SB also pulls on A with force T. Since the force F is being applied to A (and causing the tension in S,) then $F=T$.

Now, on SB. In a classroom setting (e.g. ignoring thermodynamics, QED, friction, etc.) the tension on S is constant. A pulls on A with force F, S pulls on B with tension F. The reverse is true, in that B reacts with force T, which is transmitted through the string as tension T.

In effect, ASB can be regarded as a single block of mass 9kg. The string transmits force between A and B, without modulating it in any way.

• Does the string keep on transmitting forces between A and B? Moreover, Shouldn't it be $F-T$=$m_ Aa$ – user35508 Feb 13 '18 at 5:11
• Yeah, the string keeps on transmitting. F is the action force and T is the reaction force. Except, by a different reference frame t is the action force and f is the reaction force. Ergo, no feedback. – Jakob Lovern Feb 13 '18 at 15:13