Answer:
[tex]1.9*10^{-5}m[/tex]
Explanation:
The elongation in bones is calculated from Euler’s equation of
ΔL=[tex]\frac{FL_{o}}{A \gamma }[/tex] where F is force, A is cross section area, ΔL is elongation of length, [tex]L_{o}[/tex] is initial length
To find cross sectional area of the bone, [tex]A=\pi r^{2}[/tex] and the radius of the leg is given as 0.018m
[tex]A= \pi * (0.0180)^{2}=1.018*10^{-3}[/tex]
Since the upward force on lower performer is 3 times her weight,
Total force, [tex]F_{total}= 3mg[/tex] where m is mass of performer provided as 60.0kg
[tex]F_{total}= 3*60*9.8= 1764 N[/tex]
The above force is balanced by two legs hence for each leg,
[tex]F_{leg}= \frac {F_{total}}{2}= \frac {1764}{2}=882N[/tex]
From the formula for elongation initially provided as
ΔL=[tex]\frac{F_{leg}L_{o}}{A \gamma }[/tex]
Substituting [tex]L_{o}[/tex] as 0.350m, [tex]F_{leg}[/tex] as 882N [tex] \gamma = 16*10^{9}[/tex]
ΔL=[tex]\frac{F_{leg}L_{o}}{A \gamma }[/tex]
ΔL= [tex]\ frac {882N * 0.35}{16*10^{9}*1.018*10^{-3}=1.90*10^{-5}m[/tex]
Therefore, elongation is [tex]1.9*10^{-5}m[/tex]
