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We are to break this stick, by twisting our hands in opposite directions. What is the exact mechanism involved in the breaking of this stick?

I understand that we are applying two opposite couples, which results in sheer stress, which fractures the stick. Then why does it feel easier to break the stick, when we hold it away from the centre? Shouldn't the width of the palms be the only determining factor? Shouldn't the magnitude of the couples be independent of the distance from the centre?

Why must the stick even fracture at the centre? (I know it must by symmetry, but why by physics?)

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  • $\begingroup$ It is harder to do a pushup with the palms close rather than when they are shoulder width apart. Might be of interest. $\endgroup$
    – Sidarth
    May 28, 2021 at 10:07

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I understand that we are applying two opposite couples, which results in sheer stress, which fractures the stick.

"Opposite couples" does not make any sense. You are not applying a couple to the stick: a couple is a pair of forces equal in magnitude and acting in opposite directions. You cannot break a stick with a couple; you will only cause it to rotate.

That aside, the stick breaks because you are effectively trying to rotate one half of the stick in one direction and the other half in the opposite direction about the middle of the stick, which is the center of rotation. This leads to enormous stresses at the center, causing it to break. It is easier to break the stick by applying the forces farther out toward its ends than near the center, since the former leads to a larger torque, allowing you to do a large amount of (angular) work while expending relatively little force.

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  • $\begingroup$ Isn't each hand individually applying a couple? If we release one of our hands, then the stick would rotate about the centre of the second hand. That's what I meant by couple. $\endgroup$
    – Aspirant
    May 28, 2021 at 7:42
  • $\begingroup$ Maybe "couple" was the wrong word to use; But as per the diagram, each thumb seems to act as an individual hinge point, and twisting about this hinge point would account for the applied torque; the distance from the centre wouldn't matter in this case either... $\endgroup$
    – Aspirant
    May 28, 2021 at 8:46

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