Question

Assume that Young's Modulus for bone is 1.50 *10^10 N/m2 and that the bone will fracture if more than 1.50 *10^8 N/m2 is exerted. (b) If a force of this magnitude is applied compressively, by how much does the 27.0 cm long bone shorten? in "mm"

Answer 1

Clarification: I'm guessing that the statement "the bone will fracture if more than 1.50x10^8 is inserted" really means that the bone will fracture if more than 1.50x10^8 N/m^2 pressure is exerted (applied).

a)What is the maximum force that can be exerted on the femur bone in the leg if it has a minimum effective diameter of 2.70 cm?

First, calculate the cross-sectional area A of the 2.70cm-diameter bone:
(1) A = ? * r^2
= ? * (1.35x10^-2)^2
= 3.1416 * 1.82 x 10^-4 = 5.73 x 10^-4 m^2

Then, the fracture force equation is given by:

(2) Ff = PressureFracture * A
= 1.50x10^8 [N/m^2] * 5.73 x 10^-4 [m^2] = 8.59*10^4 N <<==Answer


b) If a force of this magnitude is applied compressively, by how much does the 26.0 cm long bone shorten?

Use the equation for the definition of Young's modulus (reference 1):
(3) ?L = F * L / (A * E),

where
F = the (compressive) force = 8.59*10^4 N from (2)
L = the length = 0.26m
A = the crossectional area = 5.73 x 10^-4 m^2 from (1)
E = Young's modulus for bone = 1.50x10^10 N/m^2

Substituting the values above into (3):

(4) ?L = (8.59*10^4) * (0.26) / [(5.73 x 10^-4) * (1.50x10^10 )]

= (2.23*10^4) / (8.59 * 10^6)

= 0.260 * 10-2 m = 0.26 cm,
the amount the 26cm bone shortens, or by 1% <<==Answer to (b)

Answer 2

Given, Y=1.50*10¹⁰ N/m²

Stress=1.50*10⁸ N/m²

Strain=Stress/Y=1.50*10⁸/1.50*10¹⁰

?L/L=1/100

?L=27.0/100=0.27cm

Answer 3

we know that , stress / strain = Young's modulus = E

P/(e/l)=E

P=stress=1.5*10^8 ,l=27.0cm = 0.27m

e = compression

e= Pl/E = 0.27 cm shortened by