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

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Homework Help Overview

The problem involves determining the fraction of a rope that can hang over the edge of a table without sliding, given the coefficient of static friction between the table and the rope.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the rope, including gravitational force and frictional force, and suggest equating these forces to find the critical fraction. There are questions about the correctness of the original answer and variations in the derived equations.

Discussion Status

There is an ongoing exploration of different interpretations of the problem, with some participants questioning the initial answer and providing alternative equations. Guidance is offered regarding the relationship between the forces involved, but no consensus has been reached on the final answer.

Contextual Notes

Participants note the importance of ensuring that equations are dimensionally consistent, particularly regarding the inclusion of gravitational force in the friction equation.

courtrigrad
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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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