Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question says it all. Ripping a piece of tape along the width is tough, stretches the tape, messes it up, etc, but if you put the tiniest nick at the top, it rips without any problems. Why is that?

share|cite|improve this question
up vote 23 down vote accepted

This is a problem in the theory of cracks, but let me try to give an intuitive discussion on the level of linear elasticity.

Imagine a rectangular sheet of material with two opposite ends (which I'll call East and West) being pulled apart slowly. The stress in the material due to the tension from the boundary conditions will be symmetric across the North South axis by symmetry, and is intuitively "spread out" on the rectangle (though the distribution will not be uniform, of course).

Though I won't do it, this stress distribution can be calculated in linear elasticity theory, as part of the theory of plates.

On the other hand, if we cut a notch in say the North edge of the rectangle, the symmetry of the material is broken and we should expect the stress distribution to be asymmetrical as well. A calculation similar to above (but harder due to the funnier geometry) in linear elasticity will probably show that the stress will become focused near the notch. The wikipedia article above seems to claim that the stress near a very sharp notch will actually come out infinite! (That is, linear elasticity theory breaks down)

Obviously the actual tearing process goes outside of linear elasticity and into plasticity theory somewhat, but I think what happens can be described as follows. The high concentration of stress near the notch means that when the material gives, it will begin to tear at the "corner" of the notch. Then this tearing causes the notch to get larger, and weakens more nearby material at that corner, which then becomes torn, etc. This is a bit like a positive feedback loop.

I can't quite relate it to the question you're asking yet, but I remember it's possible to do some cute dimensional analysis to show that cracks below a certain size in a material tend to shrink, and those above the threshold size tend to grow.

Let me just add that the shapes of tears in thin sheets has been a recently popular topic in the field of "extreme mechanics" in soft matter physics. See for instance this recent article by Audoly, Reis and Roman and citing references.

share|cite|improve this answer

Dead simple! Materials tear when the stress in them gets above a certain level. The stress at the tip of even a small but sharp crack is huge! So it doesn't take much extra force to cause the small crack to tear through the material. How do you tear a sheet of paper? Put a small crack in it then pull!

However, I have to expand on this to clear some misconceptions in @j.c. Firstly, if a crack is introduced in the North edge and the loading is East-West, the stress field is still symmetric, about the crack. Secondly, cracking doesn't require plasticity: linear-elastic fracture mechanics (LEFM) describes cracking in materials like glass that undergo virtually no plastic deformation. Cracks below a certain size don't tend to shrink - they just stay as they are. Nor does the presence of the crack weaken the nearby material - it raises the local stress. At a certain combination of stress $\sigma$ and crack length $a$ , the crack will grow. The 'certain combination' is the critical stress intensity factor $Kc= \sigma*\sqrt(\pi*a)$. Calculating this doesn't involve dimensional analysis, it requires that the strain energy released by crack advance is greater that the surface energy needed to create new crack flank. If there is substantial crack-tip plasticity, e.g as with most metals, LEFM doesn't work and we have to use what's called elastic-plastic fracture mechanics (EPFM).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.