Answer: 4.74 mm
Explanation:
We can solve this problem with the following equation:
[tex]Y=\frac{stress}{strain}[/tex] (1)
Where:
[tex]Y=1.8(10)^{10} Pa[/tex] is the Young modulus for femur
[tex]stress=\frac{F}{A}=1.58(10)^{8} Pa[/tex] is the stress (force [tex]F[/tex] applied per unit of transversal area [tex]A[/tex]) on the femur
[tex]strain=\frac{\Delta l}{l_{o}}[/tex]
Being:
[tex]\Delta l[/tex] the compression the femur can withstand before breaking
[tex]l_{o}=0.54 m[/tex] is the length of the femur without compression
Writing the data in equation (1):
[tex]Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l_{o}}}[/tex] (2)
[tex]1.8(10)^{10} Pa=\frac{1.58(10)^{8} Pa}{\frac{\Delta l}{0.54 m}}[/tex] (3)
Isolating [tex]\Delta l[/tex]:
[tex]\Delta l=\frac{(1.58(10)^{8} Pa)(0.54 m)}{1.8(10)^{10} Pa}[/tex] (4)
[tex]\Delta l=0.00474 m[/tex] (5) This is the compression in meters
Converting this result to millimeters:
[tex]\Delta l=0.00474 m \frac{1000 mm}{1 m}=4.74 mm[/tex]
