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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. (g = 9.8 ms ^(-2))Young's Modulus for bone Y=9.4xx 10 ^(9) Nm^(-2). |
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Answer» Solution :Tota mass of all the PERFORMERS tables, plaques etc. Is M = 280 kg Mass of performer at the bottom of the PYRAMID `m = 60 kg ` Mass of SUPPORTED by the legs of the performer at the bottom of the pyramid `m _(1) = M - m` `therefore m _(1) =280 -60 = 220kg` Weight of this supported mass, `W = m _(1) G = 220 xx 9.8 =2156N` Weight suppported by each thingbone of the proformer (weight) `= W/2 = (2156)/(2) =1078 N(because` Two thingbone) Length of each thingbone `L = 0.5m` The radius of thingbone `r=2.0 m =2 xx 10 ^(-2) m` Young.s modulus, `Y = (F //A)/(Delta L //L)` `therefore Delta L = (F)/(pi r ^(2)) xx (L)/(Y)` where, `Delta L =` compression in each thingbone. `= (1078)/(3.14 xx (2 xx 10 ^(-2))^(2)) xx (0.5)/(9.4 xx 10 ^(9))=4.565 xx 10 ^(-5) m= 4.56 xx 10 ^(-5)m` The fractional decrease in the thingbone, `(Delta L)/(L) = (4.56xx 10 ^(-5))/(0.5) = 9.12 xx 10 ^(-5) ` `therefore (Delta L )/(L) xx 100 % 9.12 xx 10 ^(-3) %` |
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