Question

Two rods are identical, except that one is brass (Y = 9.0 × 1010 N/m2) and one is tungsten (Y = 3.6 × 1011 N/m2). A force causes the brass rod to stretch by 3.0 × 10-6 m. What is the amount of stretch ΔLTungsten for the tungsten rod when the same force is applied to it?

Answer 1

To solve this problem it is necessary to apply the concepts related to Young's Module, and find the radius that gives the ratio between the two given materials. Young's module can be defined as,

`Y=(FL)/(A \Delta L)`

Where,

F= Force

L = Initial Length

A = Cross-sectional Area

`\Delta L =` Change in Length

Re-arrange the equation to find the change in Length we have,

`\Delta L = (FL)/(AY)`

If both the Force, as the Area and the initial length are considered constant, we can realize directly that the change in length is inversely proportional to Young's Module, therefore

`\Delta L \propto (1)/(Y)`

Applying this concept to that of the two materials (Brass and Tungsten),

`(\Delta L_T)/(\Delta L_B) = (Y_B)/(Y_T)`

`(\Delta L_T)/(\Delta L_B) = (9*10^(10))/(3.6*10^(11))`

`(\Delta L_T)/(\Delta L_B) = 0.25`

If the force caused
`3 * 10^ {- 6}m` to be stretched, the tungsten will stretch 0.25 of that ratio

`L_T = 3*10^(-6)*0.25`

`L_T = 7.5*10^(-7)m`

Therefore the amount of stretch of Tungsten is 7.5*10^{-7}m