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During a circus act, one performer swings upside down

Chapter 5, Problem 29

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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.

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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.

ANSWER:

 

Step 1 of 2

We are asked to find the extent of stretch of femur bones if a person who is upside down holds a person by his legs so that the second person is also upside down. The force applied by the first person on the other person's legs is 3 times the weight of the first person. 

We will use the concept of elasticity in this question. 

Given data:

Mass of performer, \(M=60 \mathrm{~kg}\)

Weight of the performer, \(W=M g\)

g is the acceleration due to gravity. 

\(\begin{array}{l} W=60 \times 9.8 \\ W=588 \mathrm{~N} \end{array}\)

The force on the performer:

\(\begin{array}{l} F=3 \times W \\ F=1764 \mathrm{~N} \end{array}\)

Tensile force on one leg is half of the total force applied. 

\(F_{t}\) is the tensile force =\(F_{t}=0.5 \times 1764\)

\(F_{l}=882 \mathrm{~N}\)

Original length of the bones, \(l=0.35 \mathrm{~m}\)

Radius of the bones, \(r=0.018 \mathrm{~m}\)

Cross Sectional area of bones \(A=\pi(0.018)^{2} \mathrm{~m}^{2}\)

\(A=1.01 \times 10^{-3} \mathrm{~m}^{2}\)

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