Answer to Question #272625 in Physics for sajidur

Question

A copper rod of length 𝟒. 𝟓 𝒎 and cross-sectional area 𝟐. 𝟑𝟎 𝒄𝒎𝟐 is stretched by a heavy load of 𝟓𝟓𝟎 𝒌𝒈 attached to the rod’s lower end. What is the tensile stress in the rod and the final length after the elongation of the rod under the stress? Young’s modulus for copper is 𝒀 = 𝟏. 𝟐 𝒙 𝟏𝟎𝟏𝟏 𝑵𝒎−𝟐 .

Answer 1

1) We can find the tensile stress in the rod as follows:

$$\sigma=\dfrac{F}{A},$$ $$\sigma=\dfrac{550\ kg\times9.8\ \dfrac{m}{s^2}}{2.3\ cm^2\times(\dfrac{1\ m}{100\ cm})^2}=2.34\times10^7\ \dfrac{N}{m^2}.$$

2) Let's first find the strain of the rod:

$$E=\dfrac{\sigma}{\epsilon},$$ $$\epsilon=\dfrac{\sigma}{E}=\dfrac{2.34\times10^7\ \dfrac{N}{m^2}}{1.2\times10^{11}\ \dfrac{N}{m^2}}=1.95\times10^{-4}.$$

Finally, we can find the final length after the elongation of the rod under the stress as follows:

$$\epsilon=\dfrac{\Delta l}{l_0},$$ $$\Delta l=\epsilon l_0=1.95\times10^{-4}\times4.5\ m=8.77\times10^{-4}\ m,$$ $$l=l_0+\Delta l=4.5\ m+8.77\times10^{-4}\ m=4.500877\ m.$$