FRICTION ON A CURVED SURFACE

F=µN requires the "ideal case" condition of perfectly smooth surfaces with no physical interlocking of surface irregularities and absolutely rigid surfaces with no deformation. Since we don't want to be so nit-picky and geeky about, yes it is just F=µN with a sufficiently fudged coefficient value.

Depends on the clearance. Generous clearance will result in a line contact at the bottom of the groove. For this case the normal force is vertical and the force required to slide the rod is uN (also = uMg). Reducing clearance results in a wider contact patch and some of the normal force becomes non-vertical so the force required to slide the rod will be higher. At some idealised clearance, the pressure will be uniform across the entire semi cylinder and can be calculated by integrating the vertical component and setting the result equal to Mg. Further reduction in clearance results in "wedging", a rapid increase in the force required to slide the rod and shortly thereafter a situation where the rod no longer touches the bottom of the groove under self-weight.

Engineering is the art of creating things you need, from things you can get.

Hi MRSSPOCK

Friction is independent of contact area untill very heavy loads and subesquently heavy pressures are encountered therefore I would agree that F= mu*N is applicable in your case as a reasonable estimate.

desertfox

Thanks for each reply, but I'm sorry, I'm still not really clear what is true.

The integration answer seems to make most sense theoretically, though.

Maybe if I spell out a simplified realistic example, it should help clarify why I'm having difficulty grasping this issue.

(Just ignore the round bar scenario for a minute, and ignore the finer details found in the real world, like contamination and physical interlocking at the interfaces etc.).

Suppose I have a length of square steel bar, 70.71mm x 70.71mm, and 2000mm long.



Taking steel density as 7860kg / m³, the bar in question has a mass of 78.6kg.



I now lay this bar down onto a steel flat floor where there is a THEORETICAL coefficient of friction of 0.1 between the two.



The force N where the bar meets the floor will be 78.6 x 9.81 = about 770N.



So using the equation F=µN, I calculate a force of 77N is needed to push the bar along the floor.



Okay, now suppose another bar, 100mm x 100mm x 1000mm long.



It's mass is identical.



This bar lying on the same steel floor will obviously THEORETICALLY also require 77N to push it along.



The former example has 70.71mm x 2000mm surface area in contact with the steel floor, i.e. 141420mm²,



while the second example has 100mm x 1000mm surface area in contact, i.e. 100000mm².



So, we see that surface area is irrelevant with regards to the equation F=µN



Okay so here is the question. (Just forget the 100mm² square bar now, it was just to emphasise the irrelevance of surface area).





Imagine I take my 70.71mm x 70.71mm x 2000mm bar and rotate it 45° along it's 2000mm length, so that it no longer is sitting flat on the floor, but sitting on one of the 2000mm long edges.



I now lift this bar, and place it into a long steel Vee block which is welded to the steel floor.



My question is, will it take a greater force to push this bar along now, than while it was resting directly and flatly on the floor?



My line of reasoning is as follows.



The 770N resolves into two forces acting on the the two Vee block faces, i.e. two forces of 554N each.



So, therefore am I right to suppose that each of these 554N forces will need 55.4N applied to push this steel bar along while resting on the Vee block,



i.e. 55.4 x 2 = 110.8N



Does that seem to make sense to you?



It does to me, but I can't help but think I have left some factor out.



My gut instinct says the force required to push it along the flat floor, or along while in the Vee block, ought to be the same, but my line or reasoning tells me different.

Since the round bar scenario is really just an infinitely faceted Vee block scenario, that would suggest it will indeed take a greater load.

Is there no recognised school of thought regarding these issues?

I suppose if push comes to shove, (pardon the pun), I will simply carry out the Vee block experiment for myself to see what happens in reality.

Thanks

Hi MRSSPOCK

Yes if the two sides are resting on the sides of the vee block then theoretically you need more force although I calculate the forces resolved from the 770N at 45 degrees to be 54.4 and not 55.4 as in your post.

desertfox

"I calculate the forces resolved from the 770N at 45 degrees to be 54.4 and not 55.4 as in your post."

It's 54.4 on my jotter...honest.

The vertical components of force must remain equal to the weight of the bar. However, in the v-block the horizontal components must increase in response to support the bar. Therefore the resulting normal force at the contact surface must increase. The force to slide the bar is dependant on the normal force which is increasing with the angle. The normal force is no longer equal to the weight of the bar. The sum of vertical forces remains equal to the weight of the bar but not the surface normal forces.
Interesting question.

Ted

Hi MRSSPOCK

If its 54.4N on your jotter honest, then I agree with that figure but you put it at 55.4N in your earlier post:-

MRSSPOCK said:

So, therefore am I right to suppose that each of these 554N forces will need 55.4N applied to push this steel bar along while resting on the Vee block, i.e. 55.4 x 2 = 110.8N"

desertfox

Click to expand...

area vs friction - Best not start thinking about wide tires, or quickest/fastest quarter mile dragstrip times.

If the V-block angle is small enough it will be self locking.


====================
If this bar/groove with "minimal clearance" is real, and does not account for non-cylindricity or accuracy of profile then a physically locked or even un-assemble-able condition can result.

I just took a length of Ø25mm stainless bar, about 600mm long and set it on my vee block.

I then pushed it with my digital kitchen scales, about ten times, and it required somewhere between 550g-650g to slide it along, on each occassion.

I then leaned the vee block over by about 85°, so that the face that the round bar was resting on was almost horizontal. It now required between 375g-400g to slide along.

And just for the desertfox..

Hi MRSSPOCK

So your test shows that the bar resting on two faces of the vee block requires more force as we have both calculated and agreed earlier.
By the way I didn't need your jotter loading up on here,all you needed to say was that you made a typing mistake in your earlier post.

desertfox

My gut instinct says the force required to push it along the flat floor, or along while in the Vee block, ought to be the same, but my line or reasoning tells me different.

Since the round bar scenario is really just an infinitely faceted Vee block scenario, that would suggest it will indeed take a greater load.

Is there no recognised school of thought regarding these issues?

I suppose if push comes to shove, (pardon the pun), I will simply carry out the Vee block experiment for myself to see what happens in reality.
Your V block example is dead right, the increased normal force required to produce the same vertical component results in increased friction. Likewise with the circular groove - except for the special case of large clearance and line contact, there will be horizontal components and friction will be higher. So the answer (as I said previously) depends very much on the clearance. Except for the special case where the pressure is uniform over the entire surface (and the line-contact case) the problem is very complex. I think the uniform pressure case gives the same result as a 90 deg V block (Think L/R symmetry - each quarter circle resolves to a 45 deg resultant.)

Engineering is the art of creating things you need, from things you can get.

If the shaft rides on the bottom of the circular groove,there is no wedging action at all, so it is equivalent to a flat surface in behavior. If the curvature of the groove is slightly less than the shaft there will be a severe wedging effect, depending on how deep the groove is.

Thanks for everyone's comments. I think this more or less wraps this one up for now.