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

AI Thread Summary
The discussion centers on determining the fraction of a rope that can hang over a table without sliding, given the coefficient of static friction, μ_s. Participants analyze the forces acting on the rope, specifically the gravitational force (F_g) and the frictional force (F_f), leading to the equation F_g = mg and F_f = μ_s(M - m)g. There is some confusion regarding the correct formulation of the frictional force, with one participant pointing out a missing 'g' in the equation. Ultimately, the correct answer is confirmed to be μ_s / (1 + μ_s). The conversation highlights the importance of accurately equating forces to solve the problem.
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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
 
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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.
 
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!
 
SN1987a said:
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
 
BerryBoy said:
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.
 
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