Analyzing Friction in a Hanging Mass System

[darkSun]

Homework Statement

A mass of 5 kg has a hook attached to it, and it is placed on an initially slack string that is suspended from two poles.

Due to friction between the rope and the hook, the mass does not slide to the middle of the two poles; it hangs closer to one side so that the string makes an angle of 50 degrees with the horizontal on that side, and 30 degrees with the other side.

Question is: What is the force of friction between the hook and the string?

Is this description clear enough?

Homework Equations

Is this correct? Friction between the string and the hook = (mu*Tension)

The Attempt at a Solution

The first thing I'm wondering is whether the tensions in the string on either sides of the mass will be the same. They are both the same string, but... I'm not sure.

Also, when separating tension into components, ie

T1sin 50 + T2 sin 30 = 5g
and
T1cos 50 = T2cos 30

T1 is the tension in the shorter side of the string, the side closer to the pole and inclined at a higher angle. T2 is the tension in the other side.
Would friction be included in these equations? I think so, and I think it would be along the same direction as T1. (so replace T1 by T1 + Friction)

But then I get two equations in three unknows, namely T1, T2, and Friction. I think I am going about this the wrong way...

 

[darkSun]

I didn't mention that the hook is cylindrical, but apparently that has a great deal to do with the problem.

Basically I have to find an expression for friction on a curved surface... I think that expression looks like e^(mu) or something similar (friends told me). Let's see if I can do that now!

 

[PhanthomJay]

Homework Statement

A mass of 5 kg has a hook attached to it, and it is placed on an initially slack string that is suspended from two poles.

Due to friction between the rope and the hook, the mass does not slide to the middle of the two poles; it hangs closer to one side so that the string makes an angle of 50 degrees with the horizontal on that side, and 30 degrees with the other side.

Question is: What is the force of friction between the hook and the string?

Is this description clear enough?

Homework Equations

Is this correct? Friction between the string and the hook = (mu*Tension)

, no friction is a function of the normal bearing force between the hook and string. The lower tension, T2, must lie within the range of the higher tension,T1 (no friction) and $$T_1e^{u_sk}$$, but that equation is not needed in this problem.

The Attempt at a Solution

The first thing I'm wondering is whether the tensions in the string on either sides of the mass will be the same. They are both the same string, but... I'm not sure.

they are not the same; friction accounts for the difference. If they were the same, the hook would slide to the low point of the wire, and there would be no friction.

Also, when separating tension into components, ie

T1sin 50 + T2 sin 30 = 5g
and
T1cos 50 = T2cos 30

T1 is the tension in the shorter side of the string, the side closer to the pole and inclined at a higher angle. T2 is the tension in the other side.

yes

Would friction be included in these equations? I think so, and I think it would be along the same direction as T1. (so replace T1 by T1 + Friction)

the equations do not include friction which is internal to the system. T1 is T1 everywhere on the shorter side, and T2 is T2 everywhere on the longer side, regardless of whether the string is at the hook or away from it. Rather, the resulting tension differential is due to friction.

But then I get two equations in three unknows, namely T1, T2, and Friction. I think I am going about this the wrong way...

Calculate T1 and T2. They would be equal if there were no friction, and the hook would slide to the low point, but since there is friction which prevents that, the friction force must be________?

 

[darkSun]

I see, so you would just solve for the tensions like in a problem without friction, and find the difference. That would be the friction.

I'll try to think about why the friction is not included in

T1sin 50 + T2 sin 30 = 5g
and
T1cos 50 = T2cos 30

..Oh, I think I just got it. These equations are used when the mass is at rest regardless of what friction or anything else is acting, to find the tensions.

And also, could you give me a hint as to where that T1e^mu came from? That looks very interesting... and useful for the next part of the problem, which asks for the minimum coefficient of static friction that will allow the hook to sit there.

Thank you very much PhanthomJay!

 

[PhanthomJay]

I'm not much into the calculus, but for the derivation of $$T_2 = T_1e^{\mu\beta}$$, see
http://ocw.mit.edu/NR/rdonlyres/Physics/8-01TFall-2004/84DD1138-93A4-47E0-8F54-CB45EBE8351D/0/exp05b.pdf

 

[darkSun]

Interesting link, I appreciate it PhanthomJay.