# Homework Help: Mechanics of smooth rings and string

1. Jun 6, 2010

### thereddevils

1. The problem statement, all variables and given/known data

A smooth ring with a mass m is threaded through a light inextensible string .The ends of the string are tied to two fixed points A and B on a horizontal ceiling so that the ring is suspended and can slide freely on the string.A hotizontal force acts on the ring in a vertical plane through the string .THe ring is in equilibrium with the parts of the string AP and BP making angles of 60 degrees and 30 degrees with the vertical respectively. Find the magnitude of the horizontal force.

2. Relevant equations

3. The attempt at a solution

Call tension of string AP , T1 and of BP , T2

T1(0.5)+T2((root 3)/2)=10 m

T1=20m-T2root(3) ---1

Horizontal components are equal ,

T2 sin 30 +F = T1 sin 60 ---2

i ended up with F=-2T2+10root(3)m

?

2. Jun 6, 2010

### PhanthomJay

Re: mechanics

You still have an unknown T2 in your equation. Since there is no friction between the ring and string, what can you say about the relationship between T1 and T2?

3. Jun 6, 2010

### thereddevils

Re: mechanics

tension is 0 ?

4. Jun 6, 2010

### PhanthomJay

Re: mechanics

No, there must be tension in the string. Think of the ring as a frictionless massless pulley with a weight hanging from it. What do you know about the magnitude of tensions on either side of a massless, frictionless pulley?

5. Jun 6, 2010

### thereddevils

Re: mechanics

thanks , the tensions would be the same in this case .

I am quite confused as to when would the tensions in the string be the same and when is it different. Could you explain a little on this ?

6. Jun 6, 2010

### PhanthomJay

Re: mechanics

They are the same if there is no friction or clamping force between the weight and string. If you released the horizontal force F, the ring would slide to the center. Otherwise, to keep the ring where it it is when releasing the force F, you'd have to clamp the ring to the string or tie a knot, and the tension forces would not be the same.

7. Jun 6, 2010

### thereddevils

Re: mechanics

thanks !

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