Simple coefficient of friction question

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SUMMARY

The discussion focuses on calculating the coefficient of friction (CoF) for a plastic material in contact with polished steel under specific conditions. The no-load torque on the motor is 3.19 Nm, while the loaded torque is 3.49 Nm. The calculated CoF is 0.0428, derived from the resistive force of 1.7928 lbs divided by the rotational force of 20.92 lbsf, further divided by 2. This calculation is validated by the context of the materials involved and the conditions of use.

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  • Understanding of torque calculations in mechanical systems
  • Familiarity with the concept of coefficient of friction
  • Knowledge of force and rotational dynamics
  • Basic grasp of material properties, specifically plastic and steel interactions
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jmart157
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Here's my givens... i need to find the coefficient of friction on the material.

No load torque on motor - 3.19 Nm
Loaded torque on motor - 3.49 Nm
Surface area in contact with motor shaft - .1885"
Mass of loaded material on shaft - 17.16 lbs = 76.33 N
Rads/sec - 31.42Looking to find the coefficient of friction. I need formulas and assistance please, not answers :)

Also, if I need to get/provide more information, I believe I can.

Jon
 
Last edited:
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alright, this came too easy. please confirm...

torque = Radius x force

which means rotational force = torque/radius

loaded rotational force = 41.83/2" = 20.92 lbsf
no load rotational force = 38.25/2" = 19.123 lbsf

resistive force from load = 20.92 lbsf - 19.123 lbs f = 1.7928 lbs

coefficient friction is (resistive force divided by rotational force) divided by 2

CoF = (1.7928/20.92)/2 = 0.0428

This seems realistic as this is plastic wearing against polished steel, in hot water. What do you guys think?
 

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