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

Answer:

The stress is calculated as $1.003\times 10^{9}\ Pa$

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 = $0.560\times 10^{- 3}\ m$

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, $\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m$

Now,

The stress in the wire is given by:

$Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}$          (1)

Now,

Force is due to the weight of the attached weight:

F = mg = $25.2\times 9.8 = 246.96\ N$

Cross  sectional Area, A = $\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}$

Using these values in eqn (1):

$\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa$