Elasticity of a string

Question

An elastic string $AB$ consists of two porions $AC$ and $CB$ each of natural length $l_{0}$, their moduli of elasticity being $\lambda$ and $\lambda'$ respectively. Show that when the whole string is stretched from its endpoints $A$ and $B$, it behaves like a uniform elastic string of natural length $2l_{0}$ whose modulus of elasticity is $\lambda''$ with $$\frac{2}{\lambda''} = \frac{1}{\lambda} + \frac{1}{\lambda'}.$$

From Hooke's law we know that $T_{AC} = \lambda \frac{l_{AC}}{l_{0}}$ and $T_{CB} = \lambda' \frac{l_{BC}}{l_{0}}$.

At this point can we also say that: $T_{AB} = \lambda'' \frac{l_{AB}}{2l_{0}} = \lambda'' \frac{l_{AC} + l_{BC}}{2l_{0}}$?

Now I'm not sure what to do, I have attempted substituting $l_{AC}$ and $l_{BC}$ which gives $$T_{AB} = \frac{\lambda''}{2} \left(\frac{T_{AC}}{\lambda} + \frac{T_{BC}}{\lambda'}\right)$$

From this point I have no idea what to do.

score 2 · Accepted Answer · answered Feb 17 '14 at 15:14

2

You are there already. Mechanical equilibrium dictates $T_{ac} = T_{bc} = T_{ab}$, which allows the immediate conclusion.

answered Feb 17 '14 at 15:14

Buco

76

How do we know we are in the mechanical equilibrium state? – user2850514 Feb 17 '14 at 15:19
"Show that when the whole string is stretched from its endpoints A and B , it behaves like a uniform elastic string", the problem formulation suggests so by itself: no mention of body forces, accelerations, etc. – Buco Feb 17 '14 at 15:22
Ah ok that makes sense, thanks. – user2850514 Feb 17 '14 at 15:25

Elasticity of a string

1 Answers1