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Rotation and Friction Problem

  1. May 29, 2005 #1
    This problem is from Vectorial Mechanics for engineers:Statics of Beer and Johnston

    it's about friction

    it says: suppose that a cylinder of weight W and radius R has the same coefficient of static friction in A and B. Determine the magnitude of the maximum momentus M that can be applied to the cylinder so it doesn't rotate.

    The cylinder is in a corner and it touches only a wall (A) and the floor (B) (this is in case the attachment does not display properly)

    I think i figured out what happens when the cylinder doesn't touch point (A) or when the coefficient of static friction (U*) in A is zero, but I really have no clue of what happens when we consider point A.

    my guess is that when U*A is zero the answer is:

    M = WRU*B

    the complete answer is:

    M = WRU*(1+U*)/(1+(U*^2))

    i'd really appreciate if you could point me in the right direction.

    thanks in advance. :shy: :shy:
     

    Attached Files:

  2. jcsd
  3. May 29, 2005 #2

    OlderDan

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    The applied moment is going to be opposed by moments from the frictional forces at points A and B. The frictional force at B will be a horizontal force applied to the cylinder that must be opposed by the contact force at point A. Hence the normal force at A must be equal in magnitute to the frictional force at point B. The frictional force at point A will be vertical, opposing the weight and reducing the normal force at point B. Can you take it from there?
     
  4. May 30, 2005 #3
    Uhmm, just one question

    thanx very much for the info

    just one question... what about the forces that act in A and B as a result of the Momentum applied (M = Fd, so M=Fr and F=M/r) Would F be equal in A and B (and of course F in A would be opposing to friction vertically, and in B, F would oppose the horizontal friction force, so the normal force in A wouldn't be equivalent to the friction force in B), and should they exist, would F in B be affected my the friction in A

    should i take them into account?

    please do not despair, i really need to get this problem done, but i'm really clueless about what to do or why.
     
  5. May 30, 2005 #4
    ok that thing about clueless :shy: sorry, i was out of line, just gettin frustrated....coz i'm realising that it's true!!!

    what i'm doing now is:

    FA* =Friction force in A
    FB* =Friction force in B

    M = -R(FA*+FB*)
    = -R(U*NA + U*NB)
    SINCE NA = FB* = U*NB
    = -R(U*U*NB + U*NB)
    = -R U* NB (1+U*)

    AND NB = -W+FA*

    = (-W+FA*)(-R U*)(1+U*)
    = WRU*(1+U*) - FA*(RU*)(1+U*)

    and.... i'm stuck again :cry:

    is my approach wrong?,

    thank you in advance for any help that you can give me
     
  6. May 30, 2005 #5

    OlderDan

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    It's a bit tough to follow the algebra when you don't use the Tex format, but it looks to me like you are really close. In your previous post, the issue of the Fd is a matter of interpretation of what M means. I took it to mean a moment applied to the axis of the cylinder in such a way that there was no force applied to they cylinder other than the forces at A, B, and the weight. It appears from your first equation that is what you are assuming now.

    How do you make this step???
    What I ended up with is

    [tex]M = \frac{\mu + \mu^2}{1 + \mu^2}RW[/tex]
     
    Last edited: May 30, 2005
  7. May 30, 2005 #6
    wow, thanx, could you post a link where i can learn how to use the Tex characters (i take it that they are LaTex characters, I've heard of the software but never used it)

    oh by the way, could you tell me something that could take me foward, because i don't know what to do with the two last terms (left to right) of the equations, given that this equation is on the right track

    = WRU*(1+U*) - FA*(RU*)(1+U*)

    Edit: well, sorry if i really made a basic mistake but i balanced the forces in the y axis:

    Friction force in A + Normal force in B = Weight

    oops there goes the minus sign, sorry

    thanks!
     
    Last edited: May 30, 2005
  8. May 30, 2005 #7

    OlderDan

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    Look back at my previous post. I edited it. I think I see where you got off track. Here is a link for the Latex stuff

    https://www.physicsforums.com/showthread.php?t=8997

    OK I misinterpreted your flow. You just made a substitution. Now replace the FA* and see what that does for you.
     
    Last edited: May 30, 2005
  9. May 30, 2005 #8
    ok so now i have:

    U* = Coefficient of static friction
    FA* = Frictional force in A

    -W R U* - W R U*^2 + FA* r U* + FA* r U*^2 = M

    is this equation the one from which i can continue working or i'm missing something?

    thanks in advance
     
    Last edited: May 30, 2005
  10. May 30, 2005 #9

    OlderDan

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    You need to substitute for FA*, and you might need to do something with your sign convention. I'm not sure. Here is mine, with all quantities positive.

    [tex] M = R\left( {f_B + f_A } \right) [/tex]

    [tex] M = R\mu \left( {N_B + N_A } \right) [/tex]

    [tex] N_A = \mu N_B [/tex]

    [tex] M = R\mu \left( {N_B + \mu N_B } \right) [/tex]

    [tex] M = RN_B \left( {\mu + \mu ^2 } \right) [/tex]

    [tex] N_B = W - \mu N_A [/tex]

    [tex] N_B = W - \mu ^2 N_B [/tex]

    [tex] N_B \left( {1 + \mu ^2 } \right) = W [/tex]

    [tex] N_B = \frac{W}{{\left( {1 + \mu ^2 } \right)}} [/tex]

    [tex] M = RN_B \left( {\mu + \mu ^2 } \right) = \frac{{\left( {\mu + \mu ^2 } \right)}}{{\left( {1 + \mu ^2 } \right)}}RW [/tex]

    I have to close out for tonight.
     
  11. May 30, 2005 #10
    i really do not want this to sound exagerate, but i think God sent you.

    Thank you. On the other hand: I was so blind :mad: I really need to go over the basics and start making things more simple, thanx again
     
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