# 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

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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