Homework Help: Physics

Posted by Brittnieee. on Sunday, March 4, 2012 at 2:06pm.

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

Physics - Elena, Sunday, March 4, 2012 at 3:06pm
Compressive stress is F/A
F=mg=70•9.8=686 N
A=πR^2-πr^2=π(R^2-r^2)=
=3.14(12.25-6.25)•10^-4=
=18.84•10^-4 (m^2)
sigma = F/A=364118 (N/m^2)

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