Rotational problem of arm - wrestling

In summary, the conversation discusses questions 6 and 7 about Newton's 3rd law and torque in arm wrestling. The summary explains that according to question 6, the force and torque applied on each arm is equal, but in question 7, the reasoning for question 6 seems to contradict the need for a bigger torque to win. It is also stated that Newton's 3rd law applies to the interface between arms in question 6, but not in question 7 where the two forces are on different interfaces. The summary ends with a mention of confusion about the correct answer for question 7 and a thank you to the participants for their help and explanations.
  • #1
songoku
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Homework Statement
Please see below
Relevant Equations
Torque = F x r
1689731084916.png


For number 6 I answer (a) and for number 7 I answer (d) but I can say I just take a guess.

In question 6, I tried to think about Newton 3rd law. The force exerted by each arm on the other is equal so assuming the distance from the point where the force acts to the pivot is also the same, the torque is the same.

But in question 7, my reasoning for question 6 seems absurd. To win, I need to exert bigger torque so the arms can rotate in the direction I want to but it seems to contradict the reasoning in question number 6 so I am confused.

Thanks
 
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  • #2
You should read #6 slowly : it says "you are winning", which implies that...

#7 I don't get : d) of course but a) looks reasonable as well : I don't think there's a way to nitpick English either way.

Both sets of answers are a bit dodgy.
 
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  • #3
hmmm27 said:
You should read #6 slowly : it says "you are winning", which implies that...
… Newton III is violated? I don't think so.
songoku said:
in question 7
The trouble with options b and c is that acceleration and velocity are vectors. What does it mean to say that one is greater than another? One vector can have a greater magnitude than another, but as long as the hands are clasped the magnitudes of the velocities, and of the accelerations, must match.

As written, I would reject all options for 7.
 
  • #4
hmmm27 said:
You should read #6 slowly : it says "you are winning", which implies that...
haruspex said:
… Newton III is violated? I don't think so.
Actually I missed the winning part. If I read that, I would choose (b) as the answer for question 6 but what about Newton's 3rd law?

haruspex said:
One vector can have a greater magnitude than another, but as long as the hands are clasped the magnitudes of the velocities, and of the accelerations, must match.
Sorry I don't really understand this part. What vector can have greater magnitude than another?

I am also confused about how to win in arm wrestling. What do I need to do? If I want to rotate both arms in the direction I want to, I have to exert more force so more torque than my friend? What about Newton's 3rd law?

Thanks
 
  • #5
@songoku , you can experiment this with your own hands and arms.
You can feel that moving or not, the force on each hand is the same.
The torque inducing the “winning” rotation is generated by a combination of muscles and bones of arm, shoulder and torso.

One set of muscles can contract in a stronger way than the counterpart on the opposite side, inducing the a net (differential) moment, a movement of masses, and the associated velocity and work and momentum.

The two hands are just links of a structure that is subjected to internal (third law) and external (first law) forces.

Please, see:
https://en.m.wikipedia.org/wiki/Newton's_laws_of_motion#Third

https://en.m.wikipedia.org/wiki/Newton's_laws_of_motion#First
 
  • #6
songoku said:
Actually I missed the winning part. If I read that, I would choose (b) as the answer for question 6 but what about Newton's 3rd law?
Question 6 is asking about the interface where the arms meet. About the torque being applied across that interface.

We see that because the question asks about the "torque you apply on your friend" and the "torque your friend applies on you". The interface between your friend and you is at your hands.

Newton's third law for torques applies here. The two torques across that interface must be equal and opposite.

Question 7 asks about the torque that you exert on the arm pair and the torque that your friend exerts on the arm pair. This time we are considering two separate interfaces to a common object. One interface at your elbow and the other at your friend's elbow. Newton's third does not apply to these two forces -- they are not between the same object pair. This is a job for Newton's second law for torques instead: ##\sum \tau = I \alpha##. You know what you need ##\alpha## to be. So what does that imply about ##\sum \tau##?

I do not understand why @haruspex objects to choice (d) for question 7. It seems manifestly correct. He is usually spot on, however, so perhaps I am missing something.
 
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  • #7
jbriggs444 said:
I do not understand why @haruspex objects to choice (d)
It's because I missed the "pair of arms" change in wording. Thanks for picking that up.
However, it still isn't quite satisfactory. The excess torque only needs to be transitory; thereafter it suffices to match the torques.
 
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  • #8
I am sorry for late reply

Thank you very much for the help and explanation hmmm27, haruspex, Lnewqban, jbriggs444
 
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