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Coefficient of friction

  1. Nov 12, 2005 #1
    If the coefficient of static friction between a table and a rope is [tex] \mu_{s} [/tex], what fraction of the rope can hang over the edge of a table without the rope sliding?

    Ok, so I declared two variables, P and 1-P . From here, all I know is that mass and weight are not of any concern in this problem. Could someone please offer some help in solving this problem? I know the answer is [tex] \frac{\mu_{s}}{1+\mu_{s}} [/tex]

    Thanks
     
  2. jcsd
  3. Nov 12, 2005 #2
    Are you sure the answer you have is right? i get something slightly different.

    In any case, I think you should start by equating the two forces acting on your rope, [itex] F_g[/itex] and [itex]F_f[/itex]. You know that
    [tex] F_g=mg[/tex], where m is the mass that's hanging, and that
    [tex] F_f=\mu_s(M-m)[/tex], where M is the total mass

    With these equations in hand, you can now find the critical percentage, M/m.

    Hope it's useful, but once again, this leads to a different answer from that which you've got.
     
  4. Nov 12, 2005 #3

    daniel_i_l

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    Gold Member

    Lets say that p is hanging of the table and 1-p is on the table. Think how much force (mg) p is pulling down with and how much friction is resisting due to the 1-p on the table. Then equate the two. Oops! once again I post a second after someone else!
     
  5. Nov 13, 2005 #4
    Did you really mean [tex]F_f=\mu_s(M-m)[/tex]? The part in brackets has to be a force for the equation to be homogenous, so I think you are missing a 'g' in this equation.

    I agree with the answer that you are looking for

    Regards,
    Sam
     
  6. Nov 13, 2005 #5
    Oops, yes, there's a g missing. So yeah, the answer is perfectly right.
    Sorry, my bad.
     
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