Physics
posted by Brittnieee. on .
When standing erect a person's weight is supported chiefly by the larger of the two leg bones. Assuming this bone to be a hollow cylinder of 2.5 cm internal diameter and 3.5 cm external diameter, what is the compressive stress on the bone (in each leg) in the case of a person whose mass (excluding legs) is 70kg?
So far I have
inner radius = 0.0125 m^2
outer radius = 0.0175 m^2
m = 70kg
F= 686 N
t= outer  inner radius = 5x10^3m^2
not sure if i use this equation:
t= 2piGr^3tsigma/l

Compressive stress is F/A
F=mg=70•9.8=686 N
A=πR^2πr^2=π(R^2r^2)=
=3.14(12.256.25)•10^4=
=18.84•10^4 (m^2)
sigma = F/A=364118 (N/m^2)