in human pyramid in a circus the entire weight of the badlands group is supported by the legs of former who is lying on his back the combined mass of all persons performing this act and tables flakes extra involved is to it kilogram the mass of the performer lying on his black at the bottom of the pyramid is 60 kg bone is this performer has length of 50 cm and an effective radius of 2.2 CM determine the amount by which each type on gets compressed under extra load
Answers
The fractional decrease in the thighbone is ΔL/L = 0.000091 or 0.0091%.
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 = 220 kg
Weight of this supported mass
= 220 kg wt. = 220 × 9.8 N = 2156 N.
Weight supported by each thighbone of the performer = ½ (2156) N = 1078 N.
The Young’s modulus for bone is given by
Y = 9.4×10^9 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=π×(2×10^−2)^2 m^2 = 1.26×10^−3 m^2
Using Eq. (9.8), the compression in each thighbone (ΔL) can be computed as
\ΔL=[(F×L)/(Y×A)]
= [(1078×0.5)/(9.4×10^9×1.26×10^−3)]
= 4.55×10^−5 m
= 4.55×10^−3 cm
This is a very small change! The fractional decrease in the thighbone is
ΔL/L = 0.000091 or 0.0091%.
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