Finding Tension in a Ring on a Smooth Hoop

[omegasquared]

Homework Statement

A ring of mass m slides on a smooth circular hoop with radius r in the vertical plane. The ring is connected to the top of the hoop by a spring with natural length r and spring constant k.

By resolving in one direction only show that in static equilibrium the angle the spring makes to the vertical is θ where:

cos\theta = \frac{1}{2}\cdot \frac{1}{1-\frac{mg}{kr}}

Homework Equations

T=kx

The Attempt at a Solution

To find the equation for tension:

L=2rcos\theta\\<br /> T = kr(2cos\theta - 1)

Resolving vertically I got:

Tcos\theta = Rcos\theta + mg

and horizontally:

Rsin\theta = Tsin\theta

The above can't be correct and also I've resolved in two directions where the question has only asked for one. The forces I have drawn on my diagram are tension and reaction parallel but in opposite directions and weight straight down.

 

[TSny]

Hello. Welcome to PF!

A diagram of the setup would be helpful. Based on some of your work, I think I understand the system.

Are you sure the reaction force (normal force), R, makes the angle θ to the vertical?

Since you are asked to resolve forces in one direction only, try to resolve the forces in an appropriate direction that is neither horizontal nor vertical.

 

[omegasquared]

I've attached the provided diagram of the system.

I assume the action force would make the same angle θ as it's been pulled in that direction by the tension.

Resolving parallel to the reaction & tension would yield a result still involving R though.

 

[TSny]

Thanks for the diagram. That makes the setup clear.

I don't think your assumption about the direction of the reaction force is correct. You are apparently supposed to assume no friction between the ring and the hoop. So, the reaction force is a "normal force".

You won't be able to make one resolution parallel to the reaction and tension since those two forces are not in the same direction.

 

[omegasquared]

Thanks for the diagram. That makes the setup clear.

I don't think your assumption that about the direction of the reaction force is correct. You are apparently supposed to assume no friction between the ring and the hoop. So, the reaction force is a "normal force".

You won't be able to make one resolution parallel to the reaction and tension since those two forces are not in the same direction.

If it's a normal force, what is it normal to? The surface/edge of the hoop?

 

[TSny]

Yes, it's normal to the surface of the hoop.

 

[omegasquared]

Yes, it's normal to the surface of the hoop.

Like in the modified diagram attached?

 

[TSny]

No. In your diagram, R is not perpendicular to the surface of the ring.

 

[omegasquared]

No. In your diagram, R is not perpendicular to the surface of the ring.

Sorry, noticed that almost straight away. Thanks for your help there.

I'm still unsure how it is possible to resolve to get a value for cosθ which doesn't involve R though.

 

[TSny]

To get a relation that does not involve R, resolve along a direction for which R has zero component.

 

[omegasquared]

To get a relation that does not involve R, resolve along a direction for which R has zero component.

Thanks, that pointed me in the right direction! Much appreciated!