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 table, 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 ehich each thighbone gets compressed under the extra load.

In a human pyramid in a circus, the entire weight of the balanced grou

1.	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 table, 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 ehich each thighbone gets compressed under the extra load.
Answer» Solution :Total mass of all performers, tables PLAQUES etc. =280kg 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."=220xx9.8" N = 2156 N"`. Weight supported by each thighbone of th performer `=1//2(2156)N=1078N`. From Table 9.1, the Young.s MODULUS for bone is GIVEN by `Y""=9.4xx10^(9)"N m"^(-2)`. Length of each thighbone `L=0.5` m the radius of thighbone `=2.0cm` Thus the cross - sectional area of the thighbone `A=pixx(2xx10^(-2))^(2)m^(2)=1.26xx10^(-3)m^(2)`. Using Eq. (9.8), the compression in each thighbone `(DeltaL)` can be computed as `DeltaL=[(FxxL)//(YxxA)]` `=[(1078xx0.5)//(9.4xx10^(9)xx1.26xx10^(-3))]` `=4.55xx10^(5)" m or "4.55xx10^(-3)" cm"`. This is very small change! The fractional decrease in the thighbone is `DeltaL//L=0.000091` or `0.0091%`