I have a question about the direction of forces when a scissors is closed.

I open a scissors so that there is, let's say, a 30 degree angle formed. Assume that then
a given amount of force is applied to closing the scissors using the grips. And, assume
that the blades on the scissors measure 12" long from where they meet -- which i think is
not critical, but just to get an image.

If I open the scissors to 30 degrees, put a square peg with a 1"x1" profile between the
blades (with one side of the peg flat against one of the blades), and then try to close
the scissors, in general, layman's terms there is a force that tries to cut the peg and a
force that tries to push the peg outwards towards the tips of the blades. (I suppose that
is the same force.) If instead of the 1" square peg I use a 2" square peg -- same 30
degree angle to start -- the "cutting" force against the peg goes down I think, but what
about the force pushing the peg towards the tips of the blades?

And, if the force outwards changes, is there a relationship of that change to the downward
force? That is important because, for a given material, I am trying to figure out changes
in friction between the flat face of the peg that is against one of the blades of the
scissors and that blade. So, for example, if the downward force dropped when going from
the 1" to 2" peg and the outward force stayed the same, then the friction would seem to go
down.

My experience tells me that as the angle of the scissors' blades gets smaller the outward
force drops for a peg of a given size, but I cannot figure out what happens with a given
angle for two different sized pegs.

Thanks for any help with this. And, if any more facts need to be provided, just ask.

CB wrote:

I have a question about the direction of forces when a scissors is closed.

I open a scissors so that there is, let's say, a 30 degree angle formed. Assume that then
a given amount of force is applied to closing the scissors using the grips. And, assume
that the blades on the scissors measure 12" long from where they meet -- which i think is
not critical, but just to get an image.

If I open the scissors to 30 degrees, put a square peg with a 1"x1" profile between the
blades (with one side of the peg flat against one of the blades), and then try to close
the scissors, in general, layman's terms there is a force that tries to cut the peg and a
force that tries to push the peg outwards towards the tips of the blades. (I suppose that
is the same force.) If instead of the 1" square peg I use a 2" square peg -- same 30
degree angle to start -- the "cutting" force against the peg goes down I think, but what
about the force pushing the peg towards the tips of the blades?

And, if the force outwards changes, is there a relationship of that change to the downward
force? That is important because, for a given material, I am trying to figure out changes
in friction between the flat face of the peg that is against one of the blades of the
scissors and that blade. So, for example, if the downward force dropped when going from
the 1" to 2" peg and the outward force stayed the same, then the friction would seem to go
down.

My experience tells me that as the angle of the scissors' blades gets smaller the outward
force drops for a peg of a given size, but I cannot figure out what happens with a given
angle for two different sized pegs.

Thanks for any help with this. And, if any more facts need to be provided, just ask.

All forces drop in proportion to the distance from the pivot (assuming
the hand force is constant).

The total force that the blade applies in a direction tangent ot the
swing of the blade can be broken arbitrary components that add
vectorially to this tangential force. For instance, one component
directed toward the opposite blade at the same distance from the pivot
as the point in question, and one that is directed radially outward
from the pivot. The division of the original force into these two
components is dramatically dependent on the angle between the two
blades.

--
John Popelish