Question

A $1000\,kg$ lift is tied with metallic wire of maximum safe stress of $1.4 \times {10^8}\,N{m^{ - 2}}$, Calculate the minimum diameter of wire required to move the lift with maximum acceleration of $1.2\,m{s^{ - 2}}$?

Answer 1

The diameter of the wire is determined by using the stress formula in the wire. First the tension of the wire is determined by using the given information in the question. By using the tension and the area formula in the stress equation, the diameter of the wire can be determined.

Useful formula
The tension in the wire or force of the wire is given by,
$T = mg$
Where, $T$ is the tension of the metallic wire, $m$ is the mass of the lift and $g$ is the sum of the acceleration due to gravity and the acceleration of the lift.
The stress of the metallic wire is given by,
$\sigma = \dfrac{T}{A}$
Where, $\sigma$ is the stress of the metallic wire, $T$ is the tension of the metallic wire and $A$ is the area of the metallic wire.

Complete step by step solution
Given that,
The mass of the lift is, $m = 1000\,kg$,
The maximum safe stress is, $\sigma = 1.4 \times {10^8}\,N{m^{ - 2}}$,
The maximum acceleration of the lift is, $a = 1.2\,m{s^{ - 2}}$.
Now,
The tension in the wire or force of the wire is given by,
$T = m\left( {g + a} \right)\,...................\left( 1 \right)$
By substituting the mass of the lift, acceleration due to gravity and the acceleration of the lift in the above equation (1), then
$T = 1000\left( {9.8 + 1.2} \right)$
By adding the terms in the above equation, then
$T = 1000\left( {11} \right)$
By multiplying the terms in the above equation, then
$T = 11000\,N$
Now,
The stress of the metallic wire is given by,
$\sigma = \dfrac{T}{A}\,....................\left( 2 \right)$
By substituting the maximum stress and the tension of the wire in the above equation (2), then
$1.4 \times {10^8} = \dfrac{{11000}}{A}$
Now the area formula is written in the above equation, then
$1.4 \times {10^8} = \dfrac{{11000}}{{\pi {r^2}}}$
By keeping the radius in one side and the other terms in the other side, then
${r^2} = \dfrac{{11000}}{{\pi \times 1.4 \times {{10}^8}}}$
By substituting the value of $\pi$ in the above equation, then
${r^2} = \dfrac{{11000}}{{\left( {\dfrac{{22}}{7}} \right) \times 1.4 \times {{10}^8}}}$
By rearranging the terms in the above equation, then
${r^2} = \dfrac{{11000 \times 7}}{{22 \times 1.4 \times {{10}^8}}}$
By multiplying the terms in the above equation, then
${r^2} = \dfrac{{77000}}{{30.8 \times {{10}^8}}}$
By dividing the terms in the above equation, then
${r^2} = \dfrac{{2500}}{{{{10}^8}}}$
Then above equation is also written as,
${r^2} = \dfrac{{{{50}^2}}}{{{{\left( {{{10}^4}} \right)}^2}}}$
By taking square root on both sides, then
$r = \dfrac{{50}}{{{{10}^4}}}$
By rearranging the terms in the above equation, then
$r = 50 \times {10^{ - 4}}\,m$
The diameter is,
$d = 2r$
Now substituting the value of $r$ in the above equation, then
$d = 2 \times 50 \times {10^{ - 4}}\,m$
By multiplying the terms in the above equation, then
$d = 100 \times {10^{ - 4}}\,m$
On further simplification in the above equation, then
$d = 0.01\,m$

Note: The minimum diameter is required to move the lift with the maximum acceleration of the $1.2\,m{s^{ - 2}}$ is $d = 0.01\,m$.
As the diameter of the metallic wire increases, the area of the metallic wire increases, so that the stress of the metallic wire decreases.