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Coefficient of friction

by courtrigrad
Tags: coefficient, friction
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courtrigrad
#1
Nov12-05, 11:33 PM
P: 1,236
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
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SN1987a
#2
Nov12-05, 11:41 PM
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, [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.
daniel_i_l
#3
Nov12-05, 11:42 PM
PF Gold
daniel_i_l's Avatar
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!

BerryBoy
#4
Nov13-05, 03:06 AM
P: 183
Coefficient of friction

Quote 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, [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.
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
SN1987a
#5
Nov13-05, 12:49 PM
P: 37
Quote Quote by BerryBoy
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
Oops, yes, there's a g missing. So yeah, the answer is perfectly right.
Sorry, my bad.


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