During a circus act, one performer swings upside down hanging from a trapeze holding another, also upside-down, performer by the legs. If the upward force on the lower performer is three times her weight, how much do the bones (the femurs) in her upper legs stretch? You may assume each is equivalent to the uniform rod 35.0 cm long and 1.8 cm in radius with 60 kg weight.

The amount of stretch in the performer's leg is 0.019 mm.

Explanation:

Given:

- Mass of the performer m = 60 kg

- The radius of each leg r = 1.8 cm

- The length of each leg L = 35.0 cm

- The modulus of elasticity of bones E = 16 GPa

- Upward Force P = 3 times the weight.

Find:

how much do the bones (the femurs) in her upper legs stretch?

Solution:

- We will model the legs of the person hanging upside down grabbed from legs as two rods under tension with a force P. Where,

P = 3*W = 3*m*g

- The tensile force F in each leg (rod) is half the total force applied b/c its distributed equally between two legs:

F = P / 2 = 1.5*m*g

- Next we express the cross-sectional area of each leg (rod), assumed to be circular.

A = pi*r^2

- The elongation of bones assumed to be rods can be approximated by Euler's formula as follows:

x = F*L / A*E

Where, x: Elongation (extension)

x = 1.5*m*g*L / E*pi*r^2

- Plug in the given values and compute x:

x = 1.5*60*9.81*0.35*10^3 / (10^916*pi*0.018^2)

x = 0.019 mm

Answer: The amount of stretch in the performer's leg is 0.019 mm.