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3 years ago

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The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresp

onds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?

1 answer:

IRISSAK [1]3 years ago

4 0

Answer:

a

$k = 11600000 N/m$

b

$\Delta L = 3.2323 *10^{-5} \ m$

c

$F = 3750.28 \ N$

Explanation:

From the question we are told that

The Young modulus is $E = 1.4 *10^{10} \ N/m^2$

The length is $L = 0.35 \ m$

The area is $2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2$

Generally the force acting on the tibia is mathematically represented as

$F = \frac{E * A * \Delta L }{L}$ derived from young modulus equation

Now this force can also be mathematically represented as

$F = k * \Delta L$

So

$k = \frac{E * A }{L}$

substituting values

$k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}$

$k = 11600000 N/m$

Since the tibia support half the weight then the force experienced by the tibia is

$F_k = \frac{750 }{2} = 375 \ N$

From the above equation the extension (compression) is mathematically represented as

$\Delta L = \frac{ F_k * L }{ A * E }$

substituting values

$\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }$

$\Delta L = 3.2323 *10^{-5} \ m$

From the above equation the maximum force is

$F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}$

$F = 3750.28 \ N$

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