Is this how it works?

The fingers generate moment.
The load generates moment.
Finger moment > load moment, so the load is cut.

How do I use an equation to identify the most vulnerable point of the product?

More questions on this topic:

What is the typical accelaration of scissors? $F=ma$, so that I can assume $F$? Other method of giving a logical assumption of F is also acceptable.

  • 1
    $\begingroup$ Very much related: meta.engineering.stackexchange.com/questions/121/… $\endgroup$
    – user16
    Aug 12 '15 at 17:22
  • $\begingroup$ Have you tried to work out an equation yourself? How far have you gone? $\endgroup$
    – hazzey
    Aug 13 '15 at 1:20
  • $\begingroup$ @hazzey about a few hours trying to figure out $\endgroup$ Aug 15 '15 at 10:10
  • $\begingroup$ If all you're looking for is a free body diagram, you can find plenty of images on Google. Your edit did not make the title better or the question more ueful, and it had worse grammar than the previous title, so I rolled it back. $\endgroup$
    – Air
    Aug 15 '15 at 21:06
  • $\begingroup$ I'm putting this question on hold as I have a number of concerns with it. While homework questions are currently on-topic for the site, there is an expectation of a clear problem statement as well as an attempt to solve the problem. This question fails to list any equations that have been devised, and doesn't demonstrate attempting to solve the problem with those equations. Future visitors would not be helped by this question as they won't be able to understand where the problem was at in the analysis. $\endgroup$
    – user16
    Aug 16 '15 at 11:44

Your general idea is correct. Your fingers apply a force which can be translated to a moment at the rotation axis. The reaction forces applied by the object to be cut must generate an equal moment at the axis. If the necessary reaction forces are greater than the object's shear resistance forces, then the object is sheared through.

For this reason, it is best to place the object as close to the axis as possible, so that the forces required to balance the moment are maximized. Too close, however (at which point the angle between the blades becomes greater than 90 degrees), and you run into the problem that the forces start becoming primarily horizontal (pushing the object out of the scissors) as opposed to vertical (cutting through the object).

This is because the blade's force is applied perpendicular to the blade, so the greater the angle between the blades, the greater the horizontal component of the force becomes. This component is useless since it does not serve to cut through the object, but instead simply tries to force the object out of the scissors. Only the vertical component actually shears the object. This is because the horizontal components of both blades point in the same direction, while the vertical components point in opposite directions.

  • $\begingroup$ Close to the axis means the pin? How does it make the forces become horizontal ? $\endgroup$ Aug 12 '15 at 17:13
  • 2
    $\begingroup$ Yes, I mean the pin. The pin doesn't make the forces become horizontal. What happens is that, as the object comes closer to the pin, the blades must increase the angle between them. Since scissors always apply their force on the object in a direction perpendicular to the blade, an increased angle between the blades implies in an increased horizontal component. This is why you can often feel objects being pushed out of the scissors instead of being cut. $\endgroup$
    – Wasabi
    Aug 12 '15 at 17:28
  • $\begingroup$ can you translate the moment produced by my hand to the forces on the blade? how do i use write the equilibrium of the scissors? $\endgroup$ Aug 15 '15 at 10:03
  • $\begingroup$ @problematic, the moment at the pin due to the forces of your hand is equal to $M_h = F_h d_h$ and the moment at the pin due to the blade forces is $M_b = F_b d_b$. Since $M_h = M_b$, $F_b$ can be trivially found. $\endgroup$
    – Wasabi
    Aug 15 '15 at 11:09

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