Question

A Type I muscle fibers have an average diameter of 98 µm, and can exert a compressive load of 532 µN[1]. If they are modeled as a perfect cylinder, what is the stress in an individual fiber?

Answer 1

Answer:

stress is 70529.30 N/m²

Explanation:

given data

diameter = 98 µm = 98 × $10^{-6}$ m

compressive load = 532 µN = 532 × $10^{-6}$ N

to find out

stress in an individual fiber

solution

we know that stress formula that is

stress = $\frac{load}{area}$     ................1

put here value we get  stress

stress = $\frac{load}{area}$

stress = $\frac{532*10^{-6}}{\frac{\pi }{4} (98*10^{-6})^2}$

stress = 70529.30 N/m²

so stress is 70529.30 N/m²