What is the Young's modulus of a nylon rope used for climbing?

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SUMMARY

The Young's modulus of a nylon rope used for climbing was calculated based on a 50 m long rope with a diameter of 1.0 cm and a 90 kg climber. The elongation of the rope under load was measured at 1.6 m. Using the formula Y = F*Lo / (A*delta L), the calculated Young's modulus was found to be 1.87 x 10^7 Pa, which is significantly lower than the expected value of 3.5 x 10^8 Pa. The discrepancy arose from an incorrect calculation of the cross-sectional area of the rope.

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  • Understanding of Young's modulus and its significance in material science.
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  • Knowledge of geometric calculations for circular cross-sections.
  • Proficiency in using the formula Y = F*Lo / (A*delta L) for elasticity calculations.
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  • Review the calculation of the cross-sectional area for cylindrical objects, specifically using A = πr².
  • Study the implications of Young's modulus in different materials, focusing on nylon and its applications in climbing gear.
  • Learn about the effects of load and elongation on the mechanical properties of materials.
  • Explore the differences between static and dynamic loads in climbing scenarios.
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Homework Statement


For safety in climbing, a mountaineer uses a nylon rope that is 50 m long and 1.0 cm in diameter. When supported, a 90 kg climber, sthe rope elongates 1.6m. Find it's Young's modules?


Homework Equations



F/A= Y * delta L/ Lo -> Y= F*Lo/ A*delta L

The Attempt at a Solution


I have Lo= 50m, delta L = (50+1.6)= 51.6m
m=90kg, g=9.8m/s^2
So, F= 882N, B

But I don't have Area, A? Even when I do this -> diameter,d =1 cm= .01m , r= .005m

So, Area,A= pi r^2 = .0157.

When I plus in the numbers into the equations, Y= F*Lo/ A*delta L

the answer comes out to be 1.87*e^7 pa WHILE the real answer is 3.5*e^8 pa. What am I doing wrong?
 
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Pi * r^2 is much smaller than 0.0157. Also delta L is just 1.6m.
 

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