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A basic steel ball contact calculation
2

A basic steel ball contact calculation

A basic steel ball contact calculation

(OP)
Referring to the attached sheet one. I am designing a simple mechanism that consists of a rotating arm that will lock in position using a steel ball mated in a machined conical cup and a spring force will hold the ball in place.

http://files.engineering.com/getfile.aspx?folder=c619cce3-2d0a-44a3-b8fc-3a1ad04edf06&file=sheet_1.jpg

I have completed a basic calculation which uses the standard friction equation with a weight on an incline. This is detailed in Sheet two. A very basic approximation of a mass in contact with a single inclined surface.

http://files.engineering.com/getfile.aspx?folder=61f9e191-4d3e-47af-94fe-62fb6baac697&file=Sheet_2.jpg

I calculated that it would take a horizontal force of 2.97 Newtons to move the mass.

So If I have a horizontal spring force of lets say 5 Newtons (holding the ball in the conical cup) my mechanism will surely stay in the horizontal position as illustrated in sheet one.


But realistically I am using a Steel ball in contact in a machined conical cup (steel). Im sure there is a better method to calculate and reassure me that the steel ball will not slip out of the conical cup and rotate further (intuitively it looks like it will stay in place). I am not too concerned about the numbers I have used in the calculation for sheet two. The method is what concerns me.

I am using the steel ball/ conical cup to locate the mechanism at a position accurately and it's a convenient mechanism for the working conditions.

Could you recommend a better method/ give some insight on this issue.
 

RE: A basic steel ball contact calculation

I'm a bit confused with your method so I would follow the links that unclesyd used. Several reasons:

The inclined angle method is usually used for estimating the coefficient of friction. The values are known, such as the inclined angle, weight of the object, and the amount of force used to move the object.

The amount of force used to move the object you may have measured as you moved it in the incline, but this is a static friction based on a flat surface, not a rolling ball. So I wonder how you measured this.

 

Charlie
www.facsco.com

RE: A basic steel ball contact calculation

(OP)
Thank you UncleSyd your prompt reply is much appreciated. Hi FACS yes I was using a wrong method, I didnt know what else to try.

RE: A basic steel ball contact calculation

This is a classical statics problem of rolling resistance, a drum hits a dent along a horizontal floor and the reader is asked to compute that horizontal force necessary to initiate motion.  In your case you have the spring which adds to "the weight" of the spherical ball, but the same problem.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

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