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

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In summary, a 50m long nylon rope with a diameter of 1.0 cm is used by a 90 kg climber for safety in climbing. When supported, the rope elongates 1.6m. Using the equation Y=F*Lo/A*delta L, the Young's modulus of the rope is calculated to be 1.87*e^7 pa, which is significantly lower than the actual value of 3.5*e^8 pa. This may be due to incorrect calculations, as the area of the rope was calculated using the formula pi*r^2, which yielded a much smaller value than the actual area of the rope.
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sowmit
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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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  • #2
Pi * r^2 is much smaller than 0.0157. Also delta L is just 1.6m.
 

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