# Ring attached to elastic and inextensible strings

## Homework Statement:

Two fixed points A and B are in the same vertical line with A at a distance 2L metres above B. Small smooth ring P of mass m kg is threaded on a light elastic string of modulus of elasticity 3 mg N and natural length L metres, and both ends of the string are fixed at A. The ring P is also attached to one end of another inextensible string of length L metres with the other end fixed at B.
a. Find the tension in the string PB when the system is in equilibrium with A, P and B in the same vertical line
b. The inextensible string is being cut while the system is in equilibrium. Find in terms of g and L the speed of P just before it hits A

## Relevant Equations:

F = k.x

k = λ / L

λ = modulus of elasticity
k = force constant
x = change in length
L = original length
I imagine the system where P is in the middle between A and B, and P is also in the middle of light elastic string.
Between P and B, there is tension force acting downwards on P.
Between A and P, there is tension force acting upwards and because P is in the middle of the elastic string, there will be two tension forces acting upwards on P (each from half part of elastic string and the extension of each part is L/2)

a. Resultant force acting on P = 0
Tension upwards = weight + tension downwards
k.x + k.x = mg + tension downwards
( λ / L) (1/2 L) + ( λ / L) (1/2 L) = mg + tension downwards
λ = mg + tension downwards
3 mg = mg + tension downwards
tension downwards = 2mg

Is this correct?

Thanks

Related Introductory Physics Homework Help News on Phys.org
kuruman
Homework Helper
Gold Member
It looks correct, but next time please post a diagram, preferably a free body diagram.

songoku

The left picture is how I imagine the system will be. The right picture is the free body diagram. P is the ring and the two F's is the upwards tension resulting from the ring threaded to the middle of elastic string.

For (b), I tried to use conservation of energy but failed:
Taking the initial position of P as reference point (height = 0), total energy at this point will be the same as total energy at point A after the inextensible string being cut.

elastic potential energy due to two components of elastic string = kinetic energy at A + gravitational potential energy at A (the elastic string is slack before the object reaches A so there is no elastic potential energy at A)

1/2 k x2 + 1/2 k x2 = 1/2 m v2 + mgh

1/2 (λ / L) (1/2 L)2 + 1/2 (λ / L) (1/2 L)2 = 1/2 m v2 + mgL

λL/4 = 1/2 m v2 + mgL

3mgL / 4 = 1/2 m v2 + mgL

1/2 m v2 = - 1/4 mgL

I got the kinetic energy is negative. Where is my mistake?

Thanks

Last edited:
kuruman
Homework Helper
Gold Member
3mgL / 4 = 1/2 m v2 + mgL
Compare the left hand side with the right hand side. The left hand side can never be equal to the right hand side because (3/4)mgL is always less than mgL even if the kinetic energy at A is zero. What does this suggest to you?

songoku
Compare the left hand side with the right hand side. The left hand side can never be equal to the right hand side because (3/4)mgL is always less than mgL even if the kinetic energy at A is zero. What does this suggest to you?
The maximum height P can travel is 3/4 L so it will never reach A. But how to answer question (b)? The speed of P just before it hits A is zero? Does not really make sense to me

Thanks

haruspex
Homework Helper
Gold Member
1/2 k x2 + 1/2 k x2
You originally wrote k=λ/L, which is correct if k is the string constant of the whole string. But in the energy expression above you need k to be the constant for only half the string.
It will be less confusing to think in terms of the energy stored in the string length L stretched to length 2L.

Last edited:
songoku
You originally wrote k=λL, which is correct if k is the string constant of the whole string. But in the energy expression above you need k to be the constant for only half the string.
It will be less confusing to think in terms of the energy stored in the string length L stretched to length 2L.
You mean the working should be like this?
1/2 k x2 = 1/2 m v2 + mgh
1/2 (λ/L) (2L)2 = 1/2 m v2 + mgL
1/2 (3mg / L) (4L2) = 1/2 m v2 + mgL
6mgL = 1/2 m v2 + mgL
1/2 m v2 = 5 mgL
v = √(10 gL)

Thanks

haruspex
Homework Helper
Gold Member
1/2 (λ/L) (2L)2
Not quite. The extension of the whole string is L, not 2L.

songoku
Not quite. The extension of the whole string is L, not 2L.
Oh yes, my bad. I will change the extension to L

Thank you very much for the help kuruman and haruspex