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
To calculate the compressive stress on the bone, you can use the following equation:
σ = F / A
Where:
- σ is the compressive stress
- F is the force acting on the bone (which equals the person's weight supported by each leg, since you mentioned the weight is supported by the larger leg bone)
- A is the cross-sectional area of the bone
Since the bone is assumed to be a hollow cylinder, we can calculate its cross-sectional area using the formula:
A = π(R^2 - r^2)
Where:
- A is the cross-sectional area
- R is the outer radius of the bone
- r is the inner radius of the bone
Based on the given information:
- R = 0.0175 m
- r = 0.0125 m
- F = weight of the person = 70 kg * 9.8 m/s^2 (gravitational acceleration) = 686 N
Now, let's substitute these values into the equations to find the compressive stress (σ):
A = π((0.0175)^2 - (0.0125)^2)
≈ π(0.00030625 - 0.00015625)
≈ π(0.00015)
≈ 0.000471 m^2
σ = 686 N / 0.000471 m^2
≈ 1,456,702.128 N/m^2 (or Pascal, which is the unit for stress)
Therefore, the compressive stress on the leg bone is approximately 1,456,702.128 N/m^2 (or Pascal) in each leg.
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)