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Calculating the pressure force of a lever roller printing press

Hello Everyone. I need some advice in calculating the pressure of a roller printing machine. The machine is operated by a geared rotating lever which moves the impression roller across a print surface. I had thought it was a case of using the equation for rolling friction to work out the normal force and then just to use the normal force in the pressure formula but its not making sense to my mind. The way I have been doing it is by calculating the torque on the roller axle to work out a tangential force acting on the roller which I have been assuming is equal to the rolling resistance force, then using a coefficient of rolling friction to get the normal force from the roller onto the surface but I am sure something is not right. I know the deformation of the roller is important but again I was not sure exactly how it comes into play. I am assuming the roller is made of ebonite rubber with a young's modulus of 500MPa and an input force on the handle of 100N. I also assumed an area of 5mmx127mm because the 127mm is the roller length and I just used the 5mm to give something to work with, I realise thats not correct but I just did it to get an estimated pressure area. Here is how i saw it.


F_n(normal force)=F_r(friction force)⁄CRF(coefficient of friction)

F_n=100⁄0.01

F_n=10000N


P=F_n⁄Area

P=10000⁄0.00127

P=7.8 MN⁄m^2

I have attached a couple of sketches of the problem. Ignore the 200mm length as I changed it afterwards to 127mm. Any help with this would be much appreciated. Cheers.

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