Ankush Patil
Last Activity: 4 Years ago
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.
Young’s modulus for bone is given by Y = 9.4 × 109 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 m2 = 1.26 × 10-3 m2.
The compression in each thighbone (∆L) can be computed as ∆L = [(F × L)/(Y × A)]
= [(1078 × 0.5)/(9.4 × 109 × 1.26 × 10-3)]
= 4.55 × 10-5 m or 4.55 × 10-3 cm.
This is a very small change! The fractional decrease in the thighbone is ∆L/L = 0.000091 or 0.0091%.