# Torque contacts

Does anyone know the answer? Or is it rocket science or unsolvable?

I experienced yesterday that when the slotted screw driver head was exact fit to the slotted screw, it was easier to turn the screw. I'd like to understand the physics of it.

Is the bottom easier to turn because there is even contact, while the top is harder to turn because the force is applied to a small area which damaged the screw by shredding the contact region?

But supposed the screw is infinitely rigid and can't deform. Would they give you same effort or force to turn the screws in the top case?
(If No, so even without the small region of metal deforming under the force in the top case, fewer contacts would make it harder to turn it? Why?)

• If perfectly rigid, there is always going to be three points of contact (one on each side of the slot and one on the bottom). So the contact mechanics only change when you consider the parts as elastic, where contacts can distribute themselves along lines or areas instead of singular points. Commented Jan 27, 2020 at 13:55

## 2 Answers

While the screw head is turning, too keep it turning at a constant angular speed we must make sure that the total torque is 0.

In the top case the normal reaction that we get from the groove walls in the screw head will be applied only at the outer most part of the screw driver, while in the bottom case, the normal reaction will be spread over a larger part of the screw driver.

Clearly, we can say that the moment of the force in the upper case will be more than that in the lower case. {moment = (Force)(distance of application)}

The moment of force mentioned above is equivalent to the torque that must be applied by us to the turn the screw head at a constant speed.

This can help explain why we might require more effort to turn the screw in the top case.

The torque required to drive a screw depends on the screw tip and threads, and the nature of the material being screwed into. A normal force on the screw may help it penetrate. A poor fit between the driver and the slot will require an extra normal force to keep the driver in the slot.