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songoku
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- 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
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