Infinite loop of forces? [duplicate]

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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