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To penetrate armor, a projectile's point concentrated force in a small area, creating a stress large enough that the armor fails. A species of jellyfish launches a pointed needle that can penetrate the hard shell of a crustacean. The rapid deceleration on impact creates a 32 μN force on the tip, which has a very small 15 nm radius.How does this compare to the ultimate stress of steel?

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We know that the equiation of Stress is,

[tex]\sigma = \frac{F}{A}[/tex]

Where

[tex]F= Force[/tex]

[tex]A= Area[/tex]

Here the Force is basically,

[tex]F=32\mu N= 32*10^{-6}N[/tex]

And we know as well, that

[tex]A= \pi r^2 =  \pi 15*10^{-9}m[/tex]

So,

[tex]\sigma = \frac{32.10^{-6}}{\pi (15*10^{-9})^2}[/tex]

[tex]\sigma = 4.53*10^{10}N/m^2[/tex]

For this question, we know that the ultimate stress of steel is 1020Mpa

[tex]\sigma_{steel}=1020Mpa=1020*10^6Pa[/tex]

So the ratio,

[tex]R=\frac{\sigma}{\sigma_{steel}}=\frac{4.53*10^{10}}{0.084*10^{10}}[/tex]

[tex]R= 44.38[/tex]

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