While slowly (i.e., assume static equilibrium) walking, a 51 kg woman places all

Question

While slowly (i.e., assume static equilibrium) walking, a 51 kg woman places all of her weight, W, on one foot, as shown. Determine the magnitude of the vertical compressive force F_B that the tibia exerts on the talus. An equal and opposite compressive force therefore acts on the tibia. If the diameter of the tibia is 0.23m. determine the resultant stress in the tibia. If the elastic modulus of bone is 12 GPa, determine the resultant strain in the tibia. If the woman's tibia is initially 0.6 m long, how much does it shorten by? Does your answer for part (d) seem reasonable? Why or why not?

Explanation / Answer

Here,

W = 51 * g

W = 51 * 9.8 N

W = 500 N

a)

for the magnitude of vertical compressive force Fg

balancing moment of forces about A

W * (15 + 100) - FB * 15 = 0

FB = 500 * 115/15

FB = 3833 N

the magnitude of vertical compressive force Fg is 3833 N

b)

Resultant Stress =FB/area

Resultant Stress = 3833.33/(pi * (0.23/2)^2)

Resultant Stress = 92281 N/m^2

c)

as elastic modulus = stress/strain

strain = 92281/(12 *10^9)

strain = 7.69 *10^-6

d)

let the length shotened is x

strain = x/Lo

7.69 *10^-6 = x/.60

x = 4.614 *10^-6 m

the bone will shorten by 4.614 *10^-6 m

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