Answer to Question #236071 in Physics for Abraham

Question

A 50kg mass is suspended from one end of a wire of length 4m and diameter of 3mm whose other end is fixed. What will be the elongation of the wire? Take the young modulus of the material to be 7 × 10^10 Nm^-2

Answer 1

According to the Hooke's law, the elongation is given as follows:

$$E\dfrac{\Delta l}{l} = \sigma\\ \Delta l =\dfrac{ \sigma}{lE}$$

where $l$ $=4m$ is the length of the wire, $E = 7\times 10^{10}Nm^{-2}$ is the young modulus of the material, and $\sigma$ is the stress in the wire. By definition, this stress is given as follows:

$$\sigma = \dfrac{F}{S}$$

where $F = mg$ is the weight of the mass $(m=50kg, g = 9.8N/kg)$, and $S = \pi d^2/4$ is the cross-sectional area of the wire ($d = 3mm = 3\times 10^{-3}m$ is the diameter of the wire).

Putting it all together, obtain:

$$\Delta l = \dfrac{4mgl}{\pi d^2E}\\ \Delta l = \dfrac{4\cdot 50kg\cdot9.8N/kg\cdot 4m}{\pi \cdot (3\times 10^{-3}m)^2\cdot 7\times 10^{10}N/m^2} \approx 3.96\times 10^{5}m$$

Answer. $3.96\times 10^{5}m$.