Question

A steel cable with a radius of $1.5cm$ supports a chairlift at a ski area. If the maximum stress is not to exceed ${10^8}N{m^{ - 2}}$, what is the maximum load the cable can support?

Answer 1

Hint: By using the formula of maximum stress and formula of area of cross-section we can calculate the value of maximum force.
The formula for maximum stress is given by, ${\text{Maximum stress}} = \dfrac{{{\text{maximum force}}}}{{{\text{maximum area}}}}$.
Also the area of the cross section is given by ${\text{A}} = \pi {r^2}$, where ${\text{A}}$ is the area of the cross section and $r$ is the radius.

Complete step by step solution:
Maximum load that the cable can support can be calculated by the formula given by,$
  {\text{Maximum stress}} = \dfrac{{{\text{maximum force}}}}{{{\text{maximum area}}}} \\
  {\text{maximum force}} = {\text{maximum stress}} \cdot {\text{maximum area}} \\
 $
Step 1:
As given that radius of cable is$1.5cm$. Let us convert the unit of radius from $cm$ to$m$. As $1m = 100cm$ therefore the radius of the cable is,
$
  r = 1.5cm \\
  r = \dfrac{{1.5}}{{100}}m \\
  r = 1.5 \times {10^{ - 2}}m \\
 $
Step 2:
Therefore the area of cross section would be, calculated by
${\text{A}} = \pi {r^2}$
Replace the value of $r = 1.5 \times {10^{ - 2}}m$ in above equation,
$
  {\text{A}} = \pi {r^2} \\
  {\text{A}} = \pi {\left( {1.5 \times {{10}^{ - 2}}} \right)^2} \\
  {\text{A}} = \pi \times {\left( {{\text{0}}{\text{.015}}} \right)^2} \\
  {\text{A}} = {\text{7}}{\text{.069}} \times {\text{1}}{{\text{0}}^{ - 4}}{m^2} \\
 $
Step 3:
The maximum stress on the cable is not to exceed${10^8}N{m^{ - 2}}$ and the area of cross section is ${\text{A}} = {\text{0}}{\text{.047}}{m^2}$, therefore
In the equation,
${\text{maximum force}} = {\text{maximum stress}} \cdot {\text{maximum area}}$
Replace the value of ${\text{maximum stress}}$ as ${10^8}N{m^{ - 2}}$ and ${\text{maximum area}}$ as ${\text{A}} = {\text{0}}{\text{.047}}{m^2}$
$
  {\text{maximum force}} = {\text{maximum stress}} \times {\text{maximum area}} \\
  {\text{maximum force}} = {10^8} \times {\text{7}}{\text{.069}} \times {\text{1}}{{\text{0}}^{ - 4}} \\
  {\text{maximum force}} = {\text{7}}{\text{.069}} \times {\text{1}}{{\text{0}}^4}N \\
 $
So, the maximum force the cable with radius $r = 1.5cm$ and maximum stress capacity of ${10^8}N{m^{ - 2}}$ experiences is equals to ${\text{7}}{\text{.069}} \times {\text{1}}{{\text{0}}^4}N$.

Note:
While calculating maximum force the students should remember the unit of maximum stress and the area of cross-section should be similar. The value of maximum stress is taken as ${10^8}N{m^{ - 2}}$ because in the problem it is given that cable cannot take the value of stress above than the given stress.