Question

A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causing it to stretch by 1.10 mm . You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A stretching elevator cable. Part A What is the stress in this wire? Express your answer in pascals.

Answer 1

The stress is calculated as
`1.003* 10^(9)\ Pa`

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm =
`0.560* 10^(- 3)\ m`

Mass of the weight attached, m = 25.2 kg

Elongation in the wire,
`\Delta l = 1.10\ mm = 1.10* 10^(- 3)\ m`

Now,

The stress in the wire is given by:

`Stress,\ \sigma = (Force,\ F)/(Area,\ A)` (1)

Now,

Force is due to the weight of the attached weight:

F = mg =
`25.2* 9.8 = 246.96\ N`

Cross sectional Area, A =
`\pi ((d)/(2))^(2) = \pi ((0.560* 10^(- 3))/(2))^(2) = 2.46* 10^(- 7)\ m^(2)`

Using these values in eqn (1):

`\sigma = (246.96)/(2.46* 10^(- 7)) = 1.003* 10^(9)\ Pa`