Friction coefficient on a rotating cylinder [mechanics]

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SUMMARY

The discussion centers on calculating the friction coefficient on a rotating cylinder, specifically addressing problem #4 from a homework assignment, which has a provided solution of 0.432. Participants emphasize the importance of accurately drawing free body diagrams, particularly the placement of normal forces (Fn) and gravitational forces (Fg). It is established that reaction forces should be drawn through the point of contact rather than the center of gravity, which is a common misconception. Clarification on these concepts is crucial for arriving at the correct solution.

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  • Understanding of Newton's laws of motion (F=ma)
  • Familiarity with free body diagrams and force analysis
  • Knowledge of frictional forces and coefficients
  • Basic mechanics of rotating bodies
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  • Learn about the calculation of friction coefficients in rotating systems
  • Explore the role of normal and gravitational forces in mechanics
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albatross84
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Homework Statement


Here is a picture of the problem I am trying to solve in a book. The problem I am having trouble with is #4 (the solution is provided in the brackets below the question -> 0.432)
http://img15.imageshack.us/img15/241/questiona.jpg

Homework Equations


F=ma
Ff=uma
etc...

The Attempt at a Solution


I have tried accounting for all the possible cases (including all free body diagrams I could think of) - Fg, and two Fn's and Ff's on both sides of contact (as well as individually). I have drawn all of my Fn's and Fg's through the center of gravity, which I am not entirely sure about. All of my attempts have so far given me a different number from the suggested solution.
In particular, I was hoping someone could help me out with the present forces and how they are drawn on the free body diagram (center of gravity?). Thanks.

 
Last edited by a moderator:
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Welcome to PF!

Hi albatross84! Welcome to PF! :smile:
albatross84 said:
… I have drawn all of my Fn's and Fg's through the center of gravity, which I am not entirely sure about …

Yes, you're right not to be sure :wink:

reaction forces always go through the point of contact (so in this case they'll also go though C) …

only the weight goes through the centre of mass :smile:
 

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