Adhesion is strong at fast peeling rates and weak at slow peeling rate (paper, video; Newtonian adhesive, non-Newtonian adhesive?), i.e. it is rate-sensitive. Why?

enter image description here(source: Meitl et al.: thanks to rate-sensitive adhesion energy, objects were transferred from one substrate to another -- peel-off at 10 cm/s has higher adhesion than peel-off at 1 cm/s)

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    $\begingroup$ Maybe because adhesion works like Kinetic friction. Until a particular limit, "The more force you apply, the more repulsive it becomes." $\endgroup$ Apr 17, 2016 at 13:23

4 Answers 4


The crucial word here is rate. When pulling one gives a dp/dt, momentum to the pulled tape, which is the force that is applied. This can be given as delta(p)/delta(t). For the same transfer of momentum delta(p) to the unstuck tape, a smaller delta(t) (faster) will result in a greater force. This can be exacerbated because one gives a bigger momentum to the unstuck part when in a hurry to do it fast.

  • $\begingroup$ So the work of adhesion is independent of the rate? $\endgroup$
    – Sparkler
    Apr 11, 2016 at 6:04
  • $\begingroup$ if you read the adhesion article in wikipedia there are many elements entering. This is a simplified explanation, assuming the same energy needed in a smaller time interval to unstick the surfaces $\endgroup$
    – anna v
    Apr 11, 2016 at 6:12

Adhesion (wiki) can be due to several phenomena. In some, the adhesive is a very viscous liquid which will slowly deform (or "flow") with a small applied force.

  • $\begingroup$ @rmhleo In the video there is adhesion between the goop and the chest hairs and adhesion between the goop and the bandage. These adhesions are not broken, it is the viscosity of the goop that accounts for the larger force with fast peeling. $\endgroup$ Apr 14, 2016 at 4:28

Adhesion is not stronger at faster peeling rate. If we formally associate adhesion as "the amount of energy needed to separate two parts", then we can describe it physically by the potential joining those two parts together.

In classical mechanics, where potentials are independent of time, like in gravitation or electrostatics, you can still explain why it "feels harder" to separate two parts the faster you do it: it "feels harder" because the force needed is bigger.

But why do we need bigger force to separate two parts attached by a potential which is independent of the external force applied or the time?

Because the effect expected (parts separated to a non-interaction distance) involves using an amount of energy equal to the potential energy bounding the parts. And although this amount of energy is independent of time, the force needed does depend on the time desired.

In mathematical expression, the work performed by the external force:

$$W_e = \int_0^d F_e(s)ds$$

is a time independent magnitude which does not care about the force involved nor its time dependence. But if we put $ds = v(t)dt$ we get:

$$W_e = \int_0^{t_d} F_e(t) v(t)dt$$

it can be seen how, the shorter is the time desired to achieve full separation $t_d$, the higher values of the integrand is needed. This integrand is a magnitude of power, or energy per unit of time, thus the faster the separation, the higher the power needed.

However, the power is the product of the force and velocity here $F_e(t) v(t)$, so is not clear yet why the force has to be higher. After all, a higher $v(t)$ would also contribute to a higher power so $F_e(t)$ does not have to be necessarily higher for faster separation.

Well actually the speed of separation depends on the balance of the external force to the force related to the bounding potential $F_b$:

$$v(t) = \int_0^{t_d} \frac{F_e - F_b}{m_a}dt$$

which is always smaller than the maximum velocity $v(t) < v_m(t)$ that can be achieved by $F_e(t)$, which is:

$$v_m(t) = \int_0^{t_d} \frac{F_e}{m_a}dt$$

and therefore, whatever the force you apply $F_e(t)$


it will not be as effective in increasing speed, as if there were no bounding potential (nor the related force $F_b(t)$).

Thus although $v(t)$ will be higher the faster is the removal, it will not suffice to account for the necessary power, and an increase in $F_e(t)$ will always occur such that the total work performed is constant however small the time employed.

  • $\begingroup$ This approach is incorrect, however, when the force depends on the shear rate, as in the cited Nature paper. $\endgroup$
    – Sparkler
    Apr 18, 2016 at 2:44

The reason is the same as with non-newtonian fluids.

As you can see there, the work done is not independent to the rate.

If you do it with lower velocity, you can get through with less work. There is a lot of Video's from these "non-newtoniand Fluids" in youtube.

The reason why this is so, lies on the understaning of Turbulence; Turbulent water is less hard than laminar/still water.

This is not mainstram physics, but Turbulence is a cut/crack in fluid; there is surface's inside the fluid. And to produce these surface's you need surface energy. When your penetration rate is too high, you are producing these surfaces/cuts on bigger volume of the fluid. (conical distribution) If you penetrate with less velocity, you only make a single cut, and thus you need less energy, as less surfaces/cuts is produced.

You can think this even with glass. Normal Glass vs. Safety Glass; you need more force to break the safety glass, cause it needs the force to split it completely when the normal glass just needs the force to cut it in few pieces.

Answer; With the lower rate there is minimum amount of atomic bonds separated, with the higher rate there is much more atomic bonds that are being separated, and thus the adhesion appears much stronger.

  • $\begingroup$ If "turbulent water is less hard than laminar/still water" then I would expect faster peeling to be easier. What am I missing here? $\endgroup$
    – Sparkler
    Apr 17, 2016 at 19:58
  • $\begingroup$ @sparkler Causing (and maintaining) turbulence needs (continuous) energy input. Adding energy is here expected "less easier". Still water without "cuts" is the minimum energy situation which nature seeks. The surfaces are "welded" togehther and the surface energies are released as heat. ie. Vibration reduces adhesion; sciencedirect.com/science/article/pii/S0022489898000147 But this vibration is also added energy. $\endgroup$
    – Jokela
    Apr 17, 2016 at 20:09
  • $\begingroup$ I like the non-Newtonian explanation; it's difficult for me to understand how the rest of your answer is related to it, in the context of sticky tape for example. $\endgroup$
    – Sparkler
    Apr 17, 2016 at 20:13
  • $\begingroup$ @Sparkler Don't worry about your difficulties. This "rest" is the solution to "unsolved" Turbulence-problem, which is not understood by anyone else either. If you get the idea from "non-newtonian Fluid" in makro-level, you don't really need to go to molecular level and down to the single atoms. As It just add's complexity. But here's some video to help with the "rest" youtube.com/watch?v=XSQeVT2fCJo compare it to this; youtube.com/watch?v=TxqpqeWO3WY $\endgroup$
    – Jokela
    Apr 17, 2016 at 20:23
  • $\begingroup$ I don't see at all how the last two videos are related... :/ $\endgroup$
    – Sparkler
    Apr 17, 2016 at 20:30

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