In a human pyramid in a circus,the entire weight of the balanced group is supported by the legs of a performer who is lying on his back. The combined mass of all the persons performing the act,and the tables,plaques,etc.,involved is $280 \; kg$. The mass of the performer lying on his back at the bottom of the pyramid is $60 \; kg$. Each thighbone (femur) of this performer has a length of $50 \; cm$ and an effective radius of $2.0 \; cm$. Determine the amount by which each thighbone gets compressed under the extra load. (Take Young's modulus for bone $Y = 9.4 \times 10^9 \; N/m^2$)

(N/A) Total mass of all the performers,tables,plaques,etc. $= 280 \; kg$.
Mass of the performer at the bottom $= 60 \; kg$.
Mass supported by the legs of the performer at the bottom of the pyramid $= 280 \; kg - 60 \; kg = 220 \; kg$.
Weight of this supported mass $F = 220 \; kg \times 9.8 \; m/s^2 = 2156 \; N$.
Weight supported by each thighbone of the performer $= \frac{1}{2} \times 2156 \; N = 1078 \; N$.
Young's modulus for bone $Y = 9.4 \times 10^9 \; N/m^2$.
Length of each thighbone $L = 50 \; cm = 0.5 \; m$.
Radius of thighbone $r = 2.0 \; cm = 2.0 \times 10^{-2} \; m$.
Cross-sectional area of the thighbone $A = \pi r^2 = \pi \times (2.0 \times 10^{-2})^2 \; m^2 \approx 1.26 \times 10^{-3} \; m^2$.
The compression in each thighbone $(\Delta L)$ is given by $\Delta L = \frac{F \times L}{Y \times A}$.
Substituting the values: $\Delta L = \frac{1078 \times 0.5}{9.4 \times 10^9 \times 1.26 \times 10^{-3}} \; m$.
$\Delta L \approx 4.55 \times 10^{-5} \; m = 4.55 \times 10^{-3} \; cm$.