Question

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of the rod be under this load?

Answer 1

0.29 m

Step-by-step explanation:

9 mm = 0.009 m in diameter

Cross-sectional area
`A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36* 10^(-5) m^2`

Let the tensile modulus of Nickel
`E = 170 * 10^9Pa`.

The elongation of the rod can be calculated using the following formula:

`\Delta L = (F L)/(A E) = (6283*50)/(6.36* 10^(-5) * 170 * 10^9) = (314150)/(1081200) = 0.29 m`