How Much Force Can a Human Femur Withstand Before Breaking?

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SUMMARY

The maximum force that a human femur can withstand before breaking is calculated to be 73,631 N, based on a maximum stress of 1.5x10^8 N/m² and an effective diameter of 2.5 cm. The calculation utilizes the formula Y=F/A, where Y represents the maximum stress. Additionally, when this force is applied compressively to a 25 cm long femur, the bone shortens by 0.0025 m. It is crucial to distinguish between maximum stress and Young's modulus in these calculations.

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UrbanXrisis
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Young's modulus for bone is 1.5x10^10 N/m^2 and that bone will fracture if more than 1.5X10^8 N/m^2 is exerted.

What is the max force that can be exerted on the femur if the effective diameter is 2.5cm?

Y=F/A
1.50x10^8N/m^2 =\frac{F}{\pi (0.0125m^2)^2}
F=73631N

is that correct?

if this force is applied compressively, by how much does the 25cm long bone shorten?

\Delta L=\frac{FL}{AY}
\Delta L=\frac{73631N*0.25m}{\pi (0.0125m^2)^2 * 1.5x10^{10} N/m^2}
\Delta L=0.0025m

I think I did some of this wrong but I'm not sure how to approch these problems. Any guidance?
 
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UrbanXrisis said:
Y=F/A
1.50x10^8N/m^2 =\frac{F}{\pi (0.0125m^2)^2}
F=73631N

is that correct?
Yes. But realize that 1.5x10^8 N/m^2 is not Y (which stands for Young's modulus) but is the maximum stress the bone can support. (Stress = F/A). Also: round off to a sensible number of significant figures.

if this force is applied compressively, by how much does the 25cm long bone shorten?

\Delta L=\frac{FL}{AY}
\Delta L=\frac{73631N*0.25m}{\pi (0.0125m^2)^2 * 1.5x10^{10} N/m^2}
\Delta L=0.0025m
Looks good. But realize you could have saved a bit of arithmetic by starting with the maximum stress (F/A) instead of the force.
 
so what was the point in giving me 1.5X10^8 N/m^2? Was that to throw me off?
 
I don't understand your question. You were given two numbers: max stress and Young's modulus. You used them both.
 

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