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 do the bones (the femurs) in her upper legs stretch? You may assume each is equivalent to a uniform rod 35.0 , {cm} long and 1.80 , {cm} in radius. Her mass is

60.0 , {kg}.

A. 4.00 , {mm}
B. 6.00 , {mm}
C. 8.00 , {mm}
D. 10.0 , {mm}

Answer 1

The bones in the performer's upper legs stretch by approximatley 0.531 mm.

Step-by-step explanation:

To calculate the amount of stretch in the bones, we can use Hooke's law which states that the change in length of a spring or rod is directly proportional to the force applied to it. In this case, the force is the upward force on the lower performer, which is three times her weight. We can start by calculating the weight of the lower performer using the formula weight = mass x gravitational acceleration. So, the weight of the lower performer is 60 kg x 9.8 m/s² = 588 N.

The upward force on the lower performer is three times her weight, so the force is 3 x 588 N = 1764 N. Now, we can use Hooke's law to calculate the change in length of the bones. Hooke's law states that the change in length (ΔL) is equal to (F x L) / (E x A), where F is the force applied, L is the original length of the rod, E is the Young's modulus (a measure of the stiffness of the material), and A is the cross-sectional area of the rod.

In this case, the original length of the rod is 35.0 cm = 0.35 m, the radius is 1.80 cm = 0.018 m, and the Young's modulus for bone is approximately 17 GPa = 17,000,000,000 Pa. Substituting these values into the formula, we get ΔL = (1764 N x 0.35 m) / (17,000,000,000 Pa x π x (0.018 m)²). Calculating this, we find that the change in length is approximately 5.31 x 10^-4 m.

To convert this to millimeters, we multiply by 1000, so the change in length is 5.31 x 10^-1 mm = 0.531 mm. Therefore, the correct answer is A. 4.00 mm.