Question

The maximum stress that can be applied to the material of a wire employed to suspend an elevator is $\dfrac{3}{\pi } \times {10^8}N/{m^2}$. If the mass of the elevator is 900kg and it moves up with an acceleration of $2.2m/{s^2}$then calculate the minimum radius of the wire is:

Answer 1

Hint: Any elastic material can hold its elasticity property up to a certain limit, known as elastic limit. Beyond the elastic limit, the material breaks up (deformed). So, the force per unit area applied to the elastic body is defined as the stress.

Formula used:
\[stress = \dfrac{{force}}{{area}} = \dfrac{F}{A}\]
Here, $F$ is force and $A$ is the area.

Complete step by step answer:
The stress on an elastic body is defined as force exerted per unit cross section. It is synonymous to pressure and appears as pressure in the mathematical definition of Bulk modulus. Hooke’s law states a direct proportionality of stress and strain. In a stress and strain graph after a point, the graph becomes non-linear. This point is described as the elastic limit and also the point of maximum stress that can be applied to an elastic body without deformation. So, we have,
\[stress = \dfrac{{force}}{{area}} = \dfrac{F}{A}\]
For the stress to be maximum, the cross-sectional area has to be minimum. For the cross-sectional area to be minimum, the radius has to be minimum. And the force is defined by Newton’s Second law as mass times acceleration. So, on substituting the sub expressions in the main equation, we have,
\[
stres{s_{\max }} = \dfrac{{force}}{{are{a_{\min }}}} \\
{r_{\min }} = \dfrac{{(ma)}}{{2\pi {{(stress)}_{\max }}}} \\
\]
On substituting the data from the question, we have,
\[
{r_{\min }} = \dfrac{{900 \times 2.2}}{{2\pi \left( {\dfrac{3}{\pi } \times {{10}^8}} \right)}}m \\
{r_{\min }} = 3.3 \times {10^{ - 6}}m \\
or \\
{r_{\min }} = 3.3\mu m \\
\]
Therefore, the minimum area of cross-section for the given maximum stress is 3.3 micro-meter.

Note: The examiner may ask the same question in terms of strain. In place of stress, the value of maximum strain may be given. Then we have to use Hooke’s law which states that stress is directly proportional to strain.so,
\[
Strain \times Y = Stress \\
Strai{n_{\max }} \times Y = \dfrac{F}{{{A_{\min }}}} \\
{r_{\min }} = \dfrac{F}{{2\pi \times Y \times strai{n_{\max }}}} \\
\]