Question

If the elastic limit of steel is 5.0\times10^8 Pa,determine the minimum diameter a steel wire can have if it is tosupport a 60 kg circus performerwithout its elastic limit being exceeded.

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asked 2021-04-23
If the elastic limit of steel is \(\displaystyle{5.0}\times{10}^{{8}}\) Pa,determine the minimum diameter a steel wire can have if it is tosupport a 60 kg circus performerwithout its elastic limit being exceeded.

Answers (1)

2021-04-25
elastic limit is the maximum strees that can be applied to thesubstance before it gets permenantly deformed
E = maximum stress
as we know
stress \(\displaystyle={\frac{{{F}}}{{{A}}}}\)
or \(\displaystyle{E}={\frac{{{F}}}{{{A}}}}\)
for stress to be maximum area should be minimum hence here A is the minimum area of cross-section thesubstance and F is the maximum force applied on the subtance now in the given problem
\(\displaystyle{E}={5}\times{10}^{{8}}\) Pa=elstic limit of steelwire
mass of the circus performer m = 60 kg
downward force F acting on the steel wire = weight of the man= m.g =588N
now let A be the minimumarea of cross-section ofthe steel wire
then using
\(\displaystyle{E}={\frac{{{F}}}{{{A}}}}\)
or \(\displaystyle{A}={\frac{{{F}}}{{{E}}}}={117.6}\times{10}^{{-{8}}}{m}^{{2}}\)
now area \(\displaystyle{A}=?{r}^{{2}}={117.6}\times{10}^{{-{8}}}\)
\(\displaystyle{r}={6.12}\times{10}^{{-{4}}}{m}\)
where r is the radius of the wire
hence the diameter of the wire \(\displaystyle{d}={2}.{r}={12.24}\times{10}^{{-{4}}}{m}={1.224}{m}{m}\)
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