Question

1. A 1.60 m long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.25 cm when tightened?

Answer 1

The relation between the tension and stretch is given by the Hooke's law:

$$\sigma = Y\varepsilon$$

where $\sigma = \dfrac{F}{A}$ is the stress (tension per unit area), $\varepsilon = \dfrac{\Delta L}{L}$ is the relative stretch ($\Delta L = 0.25\times10^{-2}m$ and $L = 1.6m$ ) and $Y = 200\times10^9Pa$ is the Young's modulus for steel.The cross-sectional area of the string is $A = \pi d^2/4$

Combining it all together, obtain:

$$\dfrac{F}{A} = Y\dfrac{\Delta L}{L}$$

Expressing the tension, obtain:

$$F = \dfrac{AY\Delta L}{L} = \dfrac{\pi d^2Y\Delta L}{4L}= \dfrac{\pi 0.002^2\cdot 200\cdot 10^9 \cdot 0.0025}{1.6} \approx 3.93\times 10^3 N$$

Answer. 3.93*10^3 N.