Mechanics of smooth rings and string

[thereddevils]

Homework Statement

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.

Homework Equations

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

?

 

[PhanthomJay]

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?

 

[thereddevils]

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?

tension is 0 ?

 

[PhanthomJay]

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?

 

[thereddevils]

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?

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 ?

 

[PhanthomJay]

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.

 

[thereddevils]

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.

thanks !