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 (as shown in Fig. 9.5). The combined mass of all the persons performing the act, and the tables, plaques etc. involved is 280 kg and of each person 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.

Explanation

Total mass of all the performers, tables, plaques etc. = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid `=280-60=220kg`
Weight of this supported mass = 2156 N
Weight of this supported mass = 2156 N
Weight supported by each thigh bone of the performer`=(1)/(2)(2156N)=1078N`
From table, the Young's modulus for bone is given by, `Y=9.4xx10^(9)Nm^(-2)`
Length of each thigh bone L = 0.5 m
The radius of each thigh bone L = 2.0 cm
Thus the cross-sectional area of the thigh bone
`A=pi(2xx10^(-2))^(2)m^(2)=1.26xx10^(-3)m^(2)`
The compression in each thigh bone `(DeltaL)` can be computed as
`DeltaL=((FxxL))/((YxxA))=((1078xx0.5))/((9.4xx10^(9)xx1.26xx10^(-3))`
`Delta=4.55xx10^(-5)m or 4.55xx10^(-3)cm`
This is a very small change! The fractional decrease in the thigh bone is
`(DeltaL)/(L)=0.000091 or 0.0091%`