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Astronauts on a tether Note that I have read this question but the answers do not make sense to me.

Suppose there are two astronauts in space. And they are holding each other's one hand (we could also say that a string is tied to both of them but I am avoiding massless string).

Now in first situation only one of them pulls the other.

In second situation both of them pull each other. Would the resulting acceleration be same in both cases? My calculations show that acceleration would be same. I don't understand what effect does second astronaut's pull produce?

Here's what I have done. Suppose they are tired by a string. Now A(on left) pulls rope with $ F_A$ and rope pulls her with $F_A$. Similarly B(on right) pulls rope with $ F_B$ and rope pulls her with $F_B$. Since string is mass less, the equation of string motion would be (right as positive) $$ F_B - F_A = M_s a$$ But since string is massless $M_s =0$, the above equation becomes, $$ F_B - F_A = 0$$ $$ F_B = F_A $$ So the acceleration of A would be, $$ a_a = \frac{F_A}{M_a}$$, And for B, $$ a_b = \frac{F_B}{M_b}= \frac{-F_A}{M_b}$$

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    $\begingroup$ Re, "I am avoiding massless string." Why? Simplified models (e.g., massless strings, frictionless planes, spherical cows) can be very helpful when you are trying to understand the underlying principles behind any real phenomenon. $\endgroup$ Jun 4, 2021 at 17:41
  • $\begingroup$ @Solomon I have added my calculations. Strings are confusing. But I think I should learn to use them. $\endgroup$
    – mum
    Jun 5, 2021 at 14:11

3 Answers 3

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Now in first situation only one of them pulls the other.

Per Newton's third law, if astronaut A pulls on astronaut B with a force $F$, astronaut B will experience an equal and opposite force of $F$. In effect, astronaut B will also be pulling on astronaut A with the same force $F$.

To determine the magnitude of the acceleration of each, given no external (e.g., gravitational) forces acting on the two astronauts, you apply Newtons second law to each astronaut individually, thus

$$a_{A}=\frac{F}{m_A}$$ $$a_{B}=\frac {F}{m_B}$$

In second situation both of them pull each other. Would the resulting acceleration be same in both cases ?

They are already "pulling on each other" in the first situation because of Newton's third law. The acceleration will only be greater if one or the other attempts to pull with a force greater than $F$ in the first situation, in which case both will experience the same greater force. In short, the force that each experiences will depend on who pulls harder

UPDATE (Astronauts pulling on a string):

With regard to your additional example incorporating a string, I believe the situation should not change, regardless of whether the string is massless, for the following reasons:

  1. The combination of A, B and the string are not subject to any external forces. That means the COM of the combination of the three will remain stationary or continue to move at constant speed in a straight line per Newton's first law.

  2. The COM of the combination of A, B and the string lies somewhere on the string between A and B. Since the COM cannot accelerate, and the COM is located on the string, the string cannot accelerate. If the string is extensible it may perhaps stretch, but that should not impact the motion of the COM of the combination.

Hope this helps.

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  • $\begingroup$ I have added my calculations . Would you please take a look. $\endgroup$
    – mum
    Jun 5, 2021 at 13:49
  • $\begingroup$ @mum I have updated my answer to respond to your conclusion. $\endgroup$
    – Bob D
    Jun 5, 2021 at 19:29
  • $\begingroup$ If each astronaut is pulling on the other with $F_1 + F_2$, why not just call it $F$? How can they be separated or differentiated? $\endgroup$
    – BowlOfRed
    Jun 5, 2021 at 22:25
  • $\begingroup$ @BowlOfRed Of course. $F=F_{A}+F_B$ $\endgroup$
    – Bob D
    Jun 5, 2021 at 23:09
  • $\begingroup$ I'm trying to understand what $F_A$ is. The person may be contributing more or less effort with their arm, but that doesn't mean it's a separate force. I see it as only a single force between the two. $\endgroup$
    – BowlOfRed
    Jun 5, 2021 at 23:12
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No they wouldn't be the same. Suppose only astronaut A pulls. Astronaut A is pulled with contractile force $f$ by himself, and astronaut B is pulled with contractile force $f$ by astronaut A (assuming she's holding on tight and resisting from breaking free). But now if both astronauts pull, each astronaut will get pulled with the original force $f$, plus an additional force $f$ coming from astronaut B.

Therefore if one astronaut pulls, the total mutual contractile force will be $f$, whereas when both pull there will be total mutual contractile force $2f$.

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    $\begingroup$ Astronaut A can only feel a force in the rope if Astronaut B is also pulling. There is no scenario where Astronaut B does not pull on the rope - if B isn't pulling, A is just reeling in a rope that's not attached to anything, and doesn't feel any force at all. $\endgroup$ Jun 4, 2021 at 15:36
  • $\begingroup$ @Nuclear. The string is tied to both of them. And in first case the B astronaut is not actively pulling she is like inanimate object. $\endgroup$
    – mum
    Jun 4, 2021 at 15:42
  • $\begingroup$ @NuclearHoagie I see your point, and I believe we are technically in agreement. But there is a difference between pulling and being pulled. In the case where only astronaut A pulls, astronaut B will indeed feel the force of the pull and resist it, but will not be pulling. In the case where B also pulls, she will feel a sort of double acceleration - due to the string pulling her, and due to her own pulling. $\endgroup$ Jun 4, 2021 at 15:46
  • $\begingroup$ @Arturo But each astronauts will experience only one force and that would be due to the rope tension. And tension would be uniform . So how can there be two forces. Should I show my calculations in question ? $\endgroup$
    – mum
    Jun 4, 2021 at 15:59
  • $\begingroup$ @mum Yes you're right. If one astronaut pulls with force $f$, the other must pull/resist with at least force $f$ if they are to remain attached to the rope. It is up to the astronaut to decide what is "pulling" and what is "resisting". However, the other astronaut is free to pull even harder, which would then require the original astronaut to pull just as hard in order to remain on the rope. $\endgroup$ Jun 4, 2021 at 16:11
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Would the resulting acceleration be same in both cases?

If the force that develops is the same, then yes the acceleration is the same.

I don't understand what effect does second astronaut's pull produce?

It doesn't have to produce anything different. But one obvious thing is that it would be easier to maintain the acceleration for a longer time (leading to an increased speed).

The simple way to think of this is if each pulls consecutively. Each one can get the same acceleration, but the acceleration is maintained for longer and there is more energy in the system at the end.

If they pull simultaneously, then its a bit more complicated. If we imagine both astronauts have the hand extended, then the COM is aligned with the hands. If one astronaut pulls, then that arm is shortened, and the COM moves toward the other astronaut. The contact point of the two astronauts has shifted toward the pulling person.

The result of the pull therefore is that in the inertial frame, the astronaut has exerted a constant force over less than one arm length. The astronaut has sped up by a certain amount.

But when the pull is simultaneous, the COM remains at the hand position. This time the astronaut pull is over a longer distance. In order to maintain the same force, the astronauts muscles must work longer and the total (inertial) speed at the end of the pull is greater than in the single-astronaut pull.

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