Problem

A 1.50 cm length of piano wire with a diameter of 0.25 cm is stretched by attaching a 10 kg mass to one end. How far is the wire stretched?

Data

• Length of wire, $L = 1.50 \, \mathrm{cm} = 1.50 \times 10^{-2} \, \mathrm{m}$
• Diameter of wire, $D = 0.25 \, \mathrm{cm} = 0.25 \times 10^{-2} \, \mathrm{m}$
• Radius of wire, $r = \frac{D}{2} = 0.125 \times 10^{-2} \, \mathrm{m}$
• Mass attached, $m = 10 \, \mathrm{kg}$
• Young’s modulus for steel, $Y = 2.0 \times 10^{11} \, \mathrm{N/m}^2$

To find:

• Change in length, $\Delta L$

Prerequisite Concepts

• Young’s modulus is defined as:

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

Rearranging for $\Delta L$:

$$\Delta L = \frac{F \cdot L}{Y \cdot A}$$

• Force due to the weight of the mass:

$$F = W = m \cdot g$$

• Cross-sectional area of the wire (circular):

$$A = \pi r^2$$

Solution

1. Calculate the force applied:

$$F = m \cdot g = 10 \, \mathrm{kg} \cdot 9.8 \, \mathrm{m/s^2} = 98 \, \mathrm{N}$$

2. Calculate the cross-sectional area:

$$A = \pi r^2 = 3.14 \cdot (0.125 \times 10^{-2})^2$$ $$A = 3.14 \cdot 1.5625 \times 10^{-6} \, \mathrm{m^2} = 4.91 \times 10^{-6} \, \mathrm{m^2}$$

3. Substitute values into the formula for $\Delta L$:

$$\Delta L = \frac{F \cdot L}{Y \cdot A}$$ $$\Delta L = \frac{98 \cdot 1.50 \times 10^{-2}}{2.0 \times 10^{11} \cdot 4.91 \times 10^{-6}}$$ $$\Delta L = \frac{1.47 \times 10^{-1}}{9.82 \times 10^5}$$ $$\Delta L = 1.498 \times 10^{-4} \, \mathrm{m}$$

Answer

The wire is stretched by $\Delta L = 1.498 \times 10^{-4} \, \mathrm{m}$ or $0.1498 \, \mathrm{mm}$.