Question

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 (in m) do the bones (the femurs) in her upper legs stretch? You may assume each is equivalent to a uniform rod 31.0 cm long and 1.78 cm in radius. Her mass is 55.0 kg.

Answer 1

Answer: 1.65 x 10^-5 m

Explanation:

Using the equation

Delta L = (1/Y x F/A ) Lo

Y = 1.6 x 10^10 N/m^2

Lo = 0.31 m

A= pi x r^2 = pi x (0.0178m)^2 = 9.9 x 10^ - 4 m^2

Ftot = 3 x w = 3 x 55 x 9.8 = 1617N

Fleg = Ftot/2 = 1617/2 = 808.5N

Delta L = 1/1.6 x 10^10 x 808.5/9.9 x 10^-4 x 0.31

= 0.65 x 10^-10 x 81.7 x 10^4 x 0.31 = 1.65 x 10^-5 m

The fire leg is stretched by 1.65 x 10^-5 m