- #1
RichyRich85
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Hi everyone
This is my first post on the forum. I work in engineering and use this forum to answer questions from time to time.
I am looking for confirmation of my understanding of static rotation torque (not something I deal with in day to day work).
Imagine 2 identical discs or washers that can rotate. The i/d is 0.095m and o/d is 0.105m. The coefficient of friction is 0.4. They are held together by an invisible force of 10kN. I believe the torque required to start rotation is (coefficient of friction x radius x force) 0.4x0.05×10000=200Nm
Next scenario: A third identical disc is inserted between the other two, and the two outer discs are unable to rotate. The same invisible 10kN force is applied on both of the outer discs. Would the torque required to rotate the middle disc be the same 200Nm, or would it be the sum of the 2 contact faces, I.e. 400Nm. I believe the latter, but just wanted to check.
Thanks in advance.
Richard
This is my first post on the forum. I work in engineering and use this forum to answer questions from time to time.
I am looking for confirmation of my understanding of static rotation torque (not something I deal with in day to day work).
Imagine 2 identical discs or washers that can rotate. The i/d is 0.095m and o/d is 0.105m. The coefficient of friction is 0.4. They are held together by an invisible force of 10kN. I believe the torque required to start rotation is (coefficient of friction x radius x force) 0.4x0.05×10000=200Nm
Next scenario: A third identical disc is inserted between the other two, and the two outer discs are unable to rotate. The same invisible 10kN force is applied on both of the outer discs. Would the torque required to rotate the middle disc be the same 200Nm, or would it be the sum of the 2 contact faces, I.e. 400Nm. I believe the latter, but just wanted to check.
Thanks in advance.
Richard