Coefficient of friction

Tags: coefficient, friction
 P: 1,236 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
 P: 37 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.
 PF Gold P: 867 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!
P: 183
Coefficient of friction

 Quote by 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.
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
P: 37
 Quote by BerryBoy 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.