InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performero 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 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. |
|
Answer» 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 `xx` 9.8 N = 2156 N. Weight supported by each thighbone of the performer = ½ (2156) N = 1078 N. 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 = 2.0 cm 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 `Delta L=[(FxxL)//(YxxA)]` `" "=[(1078xx0.5)//(9.4xx10^(9)xx1.26xx10^(-3))]` `" "=4.55xx10^(-5)"m" or 4.55xx10^(-3)"cm"` This is a very small change! The fractional decrease in the thighbone is `DeltaL//L` = 0.000091 or 0.0091%. |
|