Friction coefficient on a rotating cylinder [mechanics]

AI Thread Summary
The discussion focuses on solving a mechanics problem involving the friction coefficient of a rotating cylinder. The original poster struggles with the free body diagrams, particularly in accurately representing the forces acting on the cylinder, including normal forces and gravitational force. A key point made is that reaction forces should be drawn through the point of contact rather than the center of gravity, while only the weight acts through the center of mass. The poster seeks clarification on the correct representation of these forces to arrive at the provided solution of 0.432. Understanding the proper application of free body diagrams is essential for solving this problem accurately.
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.

 
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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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