Physics in biology and medicine (spring/compression) PLEASE HELP, THANK YOU

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SUMMARY

The discussion centers on calculating the height from which a person can fall before risking broken fingers when catching themselves. The formula used is h=(area)(length)(stress^2 of broken finger)/(2)(young's modulus)(mass)(gravity). Given a total body mass of 70 kg, a cross-sectional area of 3.0 cm² per finger, and a Young's modulus of bone at 14 x 10^10 dyn/cm², the calculated height is approximately 2 cm or 0.064 ft. The assumption that all energy is absorbed by the finger bones is deemed acceptable for this calculation.

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Physics in biology and medicine (spring/compression)... PLEASE HELP, THANK YOU

Homework Statement


A person falls from some height before catching themselves by their fingers (one hand) on a lodge. If each of the four fingers is treated as a 3 cm long bone with a cross-sectional area of 3.0cm^2 per finger, a fall from what height would result in broken fingers? Assume that they have a total body mass of 70 kg.

Homework Equations


h=(area)(length)(stress^2 of broken finger)/ (2)(young's modulus)(mass)(gravity)
Cross-sectional area = 3.0cm^2 per finger
total body mass = 70kg
stress of broken fingers/rupture strength = 100 x 10^7 dyn/cm^2
length of each of the 4 fingers = 3cm
young's modulus of bone = 14 x 10^10 dyn/cm^2
gravity = 9.8 m/s^2

The Attempt at a Solution


change mass (kg) to mass (g) = 70kg x (1000g)/(1kg) = 70000g/4 fingers = 17500g
change stress (x 10^7) to stress (x 10^9) = move decimal 2 places = 1 x 10^9 dyn/cm^2
change gravity (m) to gravity (cm) = 9.8m/s^2 x (1000cm/s^2)/(1m/s^2) = 980cm/s^2
now I plug in the problem:
h= (3.0cm^2)(3cm)(1x10^9dyn/cm^2)^2/(2)(14x10^10dyn/cm^2)(17500g)(980cm/s^2)
h=2cm or 0.064ft

Please help me by letting me know if I am headed in the right direction or not... Thanks in advance for your help
 
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Looks fine to me. (That is, as long as the assumption that all the energy gets stored in the finger bones is acceptable for solving this problem.)
 


Vela... Thank you so much for the reply
 

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