Question

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 280kg . 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 50cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.​

Answer 1

✨$\huge\mathcal\color{purple} AnSwEr::$

➣Given:-

•Tᴏᴛᴀʟ Mᴀss Oғ Aʟʟ Tʜᴇ Tᴇʀғᴏᴍᴇʀs, Tᴀʙʟᴇs, Pʟᴀǫᴜᴇs ᴇᴛᴄ. = 280 kg

•Mass of the performer = 60 kg

➣Find out:-

•Mass supported by the legs of the performer at the bottom of the pyramid

$= 280 - 60 =$ 220 kg

•Weight of this supported mass

$= 220 \: kg \: wt. = 220 \times 9.8N =$ 2156 N

•Weight supported by each thighbone of the performer $= \frac{1}{2}(2156) \: N = 1078N.$

$\bold{ Y = 9.4 \times {10}^{9} \: N \: \: {m}^{2}}$

•Lenght of each thighbone L = 0.5 m

•The radius of thighbone = 2.0 cm

Thus the cross - sectional area of the thighbone

$\bold{A = \pi \times (2 \times {10}^{ - 2} ) ^{2} \: {m}^{2} = 1.26 \times {10}^{ - 3} {m}^{2}}$

∆L $\bold{=[ \: (F \times L) \setminus \: (Y \times A)}$ ]

$\bold{= (1078 \times 0.5 ) \setminus(9.4 \times {10}^{9} \times 1.26 \times {10}^{ - 3})}$

$\bold{= 4.55 \times {10}^{ - 5} \: m \: or \: 4.55 \times {10}^{ - 3} cm}$

•This is a very small change! The fractional decrease in the thighbone is

∆L = 0.000091 or 0.0091%.

${\boxed{\huge{\red{\mathcal{Be \ Brainly...!}}}}}$