What is the fraction of a rope that can hang over a table without sliding?

[courtrigrad]

If the coefficient of static friction between a table and a rope is $$\mu_{s}$$, 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 $$\frac{\mu_{s}}{1+\mu_{s}}$$

Thanks

 

[SN1987a]

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, $F_g$ and $F_f$. You know that
$$F_g=mg$$, where m is the mass that's hanging, and that
$$F_f=\mu_s(M-m)$$, 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.

 

[daniel_i_l]

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!

 

[BerryBoy]

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, $F_g$ and $F_f$. You know that
$$F_g=mg$$, where m is the mass that's hanging, and that
$$F_f=\mu_s(M-m)$$, 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.

Did you really mean $$F_f=\mu_s(M-m)$$? 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

 

[SN1987a]

Did you really mean $$F_f=\mu_s(M-m)$$? 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

Oops, yes, there's a g missing. So yeah, the answer is perfectly right.
Sorry, my bad.