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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.
Solution
Answer Total mass of all the performers, tables,
$\text { plaques etc. } \quad=280 kg$
Mass of the performer $=60 kg$ Mass supported by the legs of the performer at the bottom of the pyramid
$=280-60=220 kg$
Weight of this supported mass
$=220 kg wt .=220 \times 9.8 N =2156 N$
Weight supported by each thighbone of the performer $=1 / 2(2156) N =1078 N$
the Young's modulus for bone is given by
$Y=9.4 \times 10^{9} N m ^{-2}$
Length of each thighbone $L=0.5 m$ the radius of thighbone $=2.0 cm$ Thus the cross-sectional area of the thighbone
$A=\pi \times\left(2 \times 10^{-2}\right)^{2} m ^{2}=1.26 \times 10^{-3} m ^{2}$
the compression in each thighbone $(\Delta L)$ can be computed as
$\Delta L =[(F \times L) /(Y \times A)]$
$=\left[(1078 \times 0.5) /\left(9.4 \times 10^{9} \times 1.26 \times 10^{-3}\right)\right]$
$=4.55 \times 10^{5} m \text { or } 4.55 \times 10^{-3} cm .$
This is a very small change! The fractional decrease in the thighbone is
$\Delta L / L=0.000091$ or $0.0091 \%$
