Question

During a circus act, one performer swings upside down hanging from a trapeze while holding another performer, also upside-down, by the calves. The legs (femurs) of the lower performer undergo deformation due to the upward force from the upper performer.
a. If the upward force on the lower performer's legs is 3mg (three times her weight), how much do the bones (the femurs in her upper legs stretch in meters?
You may assume each bone is equivalent to a uniform rod 35 cm long and 1.5 cm in radius and the legs are massless, serving only to transmit the upward force on the torso. Her mass is 56 kg and Young's modulus for tension on bones is 1.6 x 10^10 N/m^2.

Answer 1

Answer:

ΔL = 5.09x10^-5 m

Explanation:

data provided by the exercise:

m = 56 kg

L = 35 cm = 0.35 m

r = 1.5 cm = 0.015 m

Y = 1.6x10^10 N/m^2

F = 3 m*g

A = pi*r^2

The Young´s modulus is equal to:

Y = (F/A)/(ΔL/L) = (F*L)/(A*ΔL)

Clearing ΔL, we have:

ΔL = (F*L)/(A*Y) = (3*56*9.8*0.35)/(pi*(0.015^2)*1.6x10^10) = 5.09x10^-5 m