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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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.

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 $$\displaystyle{d}={2}.{r}={12.24}\times{10}^{{-{4}}}{m}={1.224}{m}{m}$$