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

stress is 70529.30 N/m²

Step-by-step 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 =
`(load)/(area)` ................1

put here value we get stress

stress =
`(load)/(area)`

stress =
`(532*10^(-6))/((\pi )/(4) (98*10^(-6))^2)`

stress = 70529.30 N/m²

so stress is 70529.30 N/m²